heat_tr_manual.pdf

A Manual for the

HEATTRANSFER

LABORATORY

William S. JannaDepartment of Mechanical Engineering

University of Memphis

2

©2000 William S. Janna

All Rights Reserved.No part of this manual may be reproduced, stored in a retrieval

system, or transcribed in any form or by any means—electronic, magnetic,mechanical, photocopying, recording, or otherwise—

without the prior written consent of William S. Janna

3

TABLE OF CONTENTS

Item Page

Cleanliness and Safety........................................................................................................... 4

Code of Student Conduct ......................................................................................................... 5

Report Writing ...................................................................................................................... 6

Experiment 1 Thermocouples and Instrumentation .................................................. 11

Experiment 2 One Dimensional Axial Heat Conduction .......................................... 13

Experiment 3 Measurement of Thermal Conductivity of a Metal ............................. 15

Experiment 4 Effect of Area on One Dimensional Conduction................................... 17

Experiment 5 Measurement of Thermal Conductivity of an Insulator ....................... 18

Experiment 6 Radial One Dimensional Conduction ................................................ 21

Experiment 7 Heat Transfer from a Fin .................................................................. 22

Experiment 8 Graphical Solutions to Unsteady Heat Transfer Problems.................. 25

Experiment 9 Determination of Conduction Shape Factor ....................................... 28

Experiment 10 Transient Conduction with Convection .............................................. 32

Experiment 11 Natural Convection from a Vertical Plate......................................... 36

Experiment 12 Natural Convection Heat Transfer: Flat, Finned, and Pinned Plates... 38

Experiment 13 Forced Convection Flat Plate to Air .................................................. 40

Experiment 14 Forced Convection in Pipe Flow......................................................... 42

Experiment 15 Analysis of a Double Pipe Heat Exchanger........................................ 44

Experiment 16 Analysis of a Plate and Frame Heat Exchanger ................................. 48

Experiment 17 Analysis of a Shell and Tube Condenser ................................................

Experiment 18 Analysis of a Cross Flow Heat Exchanger..............................................

Experiment 19 Radiation Heat Transfer .................................................................. 54

Experiment 20 Emissivity of Black and Gray Surfaces.............................................. 58

Experiment 21 Radiation View Factor..................................................................... 61

Experiment 22 The Vapor Compression Refrigeration Cycle ..................................... 66

Appendix ........................................................................................................ 69

Statistical Treatment of Experimental Data.............................................................................

Error Analysis..........................................................................................................................

4

CLEANLINESS AND SAFETY

CleanlinessThere are “housekeeping” rules that the user

of the laboratory should be aware of and abideby. Equipment in the lab is delicate and eachpiece is used extensively for 2 or 3 weeks persemester. During the remaining time, eachapparatus just sits there, literally collecting dust.University housekeeping staff are not required toclean and maintain the equipment. Instead, thereare college technicians who will work on theequipment when it needs repair, and when theyare notified that a piece of equipment needsattention. It is important, however, that theequipment stay clean, so that dust will notaccumulate too badly.

The Heat Transfer Laboratory containsequipment that uses water or air as the workingfluid. In some cases, performing an experimentwill inevitably allow water to get on theequipment and/or the floor. If no one cleaned uptheir working area after performing anexperiment, the lab would not be a comfortable orsafe place to work in. No student appreciateswalking up to and working with a piece ofequipment that another student or group ofstudents has left in a mess.

Consequently, students are required to cleanup their area at the conclusion of the performanceof an experiment. Cleanup will include removalof spilled water (or any liquid), and wiping thetable top on which the equipment is mounted (ifappropriate). The lab should always be as cleanor cleaner than it was when you entered. Cleaningthe lab is your responsibility as a user of theequipment. This is an act of courtesy that studentswho follow you will appreciate, and that youwill appreciate when you work with theequipment.

SafetyThe layout of the equipment and storage

cabinets in the Heat Transfer Lab involvesresolving a variety of conflicting problems. Theseinclude traffic flow, emergency facilities,environmental safeguards, exit door locations,

etc. The goal is to implement safety requirementswithout impeding egress, but still allowingadequate work space and necessary informalcommunication opportunities.

Distance between adjacent pieces ofequipment is determined by locations of watersupply valves, floor drains, electrical outlets,and by the need to allow enough space around theapparatus of interest. Immediate access to theSafety Cabinet is also considered. Emergencyfacilities such as showers, eye wash fountains,spill kits, fire blankets and the like are not foundin the lab. We do not work with hazardousmaterials and such safety facilities are notnecessary. However, waste materials aregenerated and they should be disposed ofproperly.

Not infrequently, specimens under study areheated by the use of a hot plate. The student isadvised to use caution when conductingexperiments that involve heated surfaces,because usually there is no visual indication thata specimen is hot.

Safety Procedures. There is unmistakenlyonly one, clearly marked exit in this laboratory.It has a single door and leads directly to thehallway on the third floor of the EngineeringBuilding. In case of fire, exit the lab to thehallway. After closing the door, take the stairsdown to first floor, and leave the building.

There is a safety cabinet attached to thewall of lab adjacent to the door. In case ofpersonal injury, the appropriate item should betaken from the supply cabinet and used in therecommended fashion. If the injury is seriousenough to require professional medical attention,the student(s) should contact the MechanicalEngineering Department in EN 312, Extension2173.

Every effort has been made to create apositive, clean, safety conscious atmosphere.Students are encouraged to handle equipmentsafely and to be aware of, and avoid beingvictims of, hazardous situations.

5

THE CODE OF STUDENT CONDUCTTaken from the University of Memphis

1998–1999 Student Handbook

Institution Policy StatementThe University of Memphis students are citizensof the state, local, and national governments, andof the academic community. They are, therefore,expected to conduct themselves as law abidingmembers of each community at all times.Admission to the University carries with itspecial privileges and imposes specialresponsibilities apart from those rights andduties enjoyed by non-students. In recognition ofthis special relationship that exists between theinstitution and the academic community which itseeks to serve, the Tennessee Board of Regentshas, as a matter of public record, instructed “thepresidents of the universities and colleges underits jurisdiction to take such action as may benecessary to maintain campus conditions…and topreserve the integrity of the institution and itseducational environment.”

The following regulations (known as the Codeof Student Conduct) have been developed by acommittee made up of faculty, students, and staffutilizing input from all facets of the UniversityCommunity in order to provide a secure andstimulating atmosphere in which individual andacademic pursuits may flourish. Students are,however, subject to all national, state and locallaws and ordinances. If a student’s violation ofsuch laws or ordinances also adversely affects theUniversity’s pursuit of its educational objectives,the University may enforce its own regulationsregardless of any proceeding instituted by otherauthorities. Additionally, violations of anysection of the Code may subject a student todisciplinary measures by the University whetheror not such conduct is simultaneously violative ofstate, local or national laws.

The term “academic misconduct” includes, butis not limited to, all acts of cheating andplagiarism.

The term “cheating” includes, but is not limitedto:a. use of any unauthorized assistance in taking

quizzes, tests, or examinations;

b. dependence upon the aid of sources beyondthose authorized by the instructor in writingpapers, preparing reports, solving problems,or carrying out other assignments;

c. the acquisition, without permission, of testsor other academic material before such

material is revealed or distributed by theinstructor;

d. the misrepresentation of papers, reports,assignments or other materials as the productof a student’s sole independent effort, for thepurpose of affecting the student’s grade,credit, or status in the University;

e. failing to abide by the instructions of theproctor concerning test-taking procedures;examples include, but are not limited to,talking, laughing, failure to take a seatassignment, failing to adhere to starting andstopping times, or other disruptive activity;

f . influencing, or attempting to influence, anyUniversity official, faculty member,graduate student or employee possessingacademic grading and/or evaluationauthority or responsibility for maintenance ofacademic records, through the use of bribery,threats, or any other means or coercion inorder to affect a student’s grade orevaluation;

g. any forgery, alteration, unauthorizedpossession, or misuse of University documentspertaining to academic records, including, butnot limited to, late or retroactive change ofcourse application forms (otherwise known as“drop slips”) and late or retroactivewithdrawal application forms. Alteration ormisuse of University documents pertaining toacademic records by means of computerresources or other equipment is also includedwithin this definition of “cheating.”

The term “plagiarism” includes, but is not limitedto, the use, by paraphrase or direct quotation, ofthe published or unpublished work of anotherperson without full or clear acknowledgment. Italso includes the unacknowledged use ofmaterials prepared by another person or agencyengaged in the selling of term papers or otheracademic materials.

Course PolicyAcademic misconduct (acts of cheating and ofplagiarism) will not be tolerated. The StudentHandbook is quite specific regarding the course ofaction to be taken by an instructor in cases whereacademic misconduct may be an issue.

6

REPORT WRITING

All reports in the Heat Transfer Laboratoryrequire a formal laboratory report unlessspecified otherwise. The report should be writtenin such a way that anyone can duplicate theexperiment and obtain the same results as theoriginator. The reports should be simple andclearly written. Reports are due one week afterthe experiment was performed, unless specifiedotherwise.

The report should communicate several ideasto the reader. First the report should be neatlydone. The experimenter is in effect trying toconvince the reader that the experiment was

performed in a straightforward manner withgreat care and with full attention to detail. Apoorly written report might instead lead thereader to think that just as little care went intoperforming the experiment. Second, the reportshould be well organized. The reader should beable to easily follow each step discussed in thetext. Third, the report should contain accurateresults. This will require checking and recheckingthe calculations until accuracy can be guaranteed.Fourth, the report should be free of spelling andgrammatical errors. The following format is to beused for formal Laboratory Reports.

Sample Title Page

Experiment Number

TITLE OF THE EXPERIMENT

Name of the Author

Name of Partner #1Name of Partner #2

etc.

Date the Experiment was Performed

Due date for this Report

The University of MemphisDepartment of Mechanical Engineering

ABSTRACTThis report was designed to contain the instructions on how to write a report, and to serve as an

example of the format and style expected in all reports. It was based on the style and format ofengineering reports used in the writing of professional engineering publications. The Title Page andAbstract are the first two components of the report.

The Abstract summarizes the information in the report. It provides a brief summary of the objectiveof the experiment, the procedures, the results, conclusions and recommendations. It should not referenceany tables, figures or appendices. A short abstract may appear on the title page as in this example. Alonger abstract would appear on the sheet following the Title Page.

The Abstract allows the reader to determine whether to read the report. It is written in the pasttense, except for the recommendations, which may be written in the present or future tense.

This report was proofread by

(signature of proofreader)

7

INTRODUCTIONThis section tells the reader what the experimentis about. It begins with a description of theproblem that is being investigated. It includesthe background and refers to related experimentsand publications. When the work of others isquoted, it is done by appropriately referencingtheir work. For example, the idea for writing areport in this way was obtained from Professor JoeCallinan at Loyola Marymount University. Hewrote a paper about it which was published in1992, and it could be referenced in one of severalways, as the examples that eventually followwill show. The report will contain a list ofreferences near the end, and the reader could thenrefer to the report written by Callinan in 1992.The references are listed alphabetically by theauthor’s last name. The format and style isdescribed in the reference section of this example.

If the reference has two authors, include the lastnames of each author in the citation. If thereference has three or more authors, use the firstauthor’s last name and the abbreviation “et al.”

Examples:Callinan (1992) provides an example of reportwriting, and has found that it improves thewriting skills of the students.

A detailed report writing method was found toimprove the writing skills of the students(Callinan, 1992).

In 1992, Callinan showed that a detailed reportwriting method improved the writing skills ofthe students.

Çengel and Boles (1998) provide a derivation ofthe energy equation.

Paré et al (1997) investigated the graphicalsolution method for a number of descriptivegeometry problems.

Subheadings. The beginning of this sectionshowed how a major heading should appear in areport: all capital letters, boldface type, and left

justified. It may be necessary to use subheadings,and the format for these is shown at thebeginning of this paragraph. Note also, that thebeginning of a paragraph is not indented, butinstead is preceded by a blank line.

Write for the Reader. Consider that the report iswritten for a technically competent person who isunfamiliar with the specific subject matter, butwill be after he/she reads the report. Alsoconsider that the reader is not as closelyassociated with the test as you are. Checkgrammar and spelling. Check continuity of page,figure, and table numbers. Have an associate whodid not perform the experiment with you, but whohas technical competency, proofread your report.

Report Preparation. Reports must be composed ona word processor. Use white paper and black text.Use only one side of a page. All margins should be1 in. Do not right justify the text. Each sectiondoes not need to begin on a new page.

Each page is to be numbered with an Arabicnumeral centered at the bottom of the page. Donot number the title page. Begin numbering withpage 2.

Figures should be numbered sequentially usingArabic numbers. Each figure is to have adescriptive title. Figures should be drawn using acomputer and a drawing program, or use thefigures available with the lab manual. Figuresare to be located near the place in the text wherethey are first referred to. Figures should becentered left-to-right either on the page (singlecolumn) or within the column (two or morecolumns). The figure number and title shouldappear centered just below the figure itself.

Tables should contain as much information aspossible. They are to be enclosed in a border. Theycan be placed in the text or at the end of thesection where they are first referred to. Tablesare to be numbered consecutively with Arabicnumerals. An acceptable table format is asfollows:

TABLE 1. Reduced data for heat transferredpast a flat plate.

TrialVelocityV in m/s

Heattransferred

q in WTemperature T

in °C1 0.5 13.6 74.82 1.0 16.2 75.53 1.5 17.5 75.34 2.0 18.9 74.5

8

Note carefully the following features regardingthis table:• The first column is “trial” or “run.”• Each column heading is of a parameter,

followed by the symbol and the unit.• Each column heading is centered within the

column.• The table is centered left to right within the

page or column.• A border has been placed around the table,

and around each cell.• The font size is smaller than that used for the

text in the report.• “TABLE 1” is in all capital letters, and the

actual title is in italics.

Graphs. In many instances, it is necessary tocompose a plot in order to graphically present theresults. Graphs must be drawn neatly following aspecific format. Figure R.1 shows an acceptable

graph prepared using a computer. There are manycomputer programs that have graphingcapabilities. An acceptably drawn graph hasseveral features of note. These features aresummarized next to Figure R.1.

Graphs especially should have descriptivetitles. A graph of temperature versus time, forexample, should not have a title of:

FIGURE 1. Temperature versus time.

The reader can see by looking at the graph thatthis is so. A better title would be:

FIGURE 1. Temperature variation with time for abrass sphere cooling in air.

Note that “FIGURE 1” is in all capital letters,and the actual title is in italics.

Features of note

• Border is drawn about the entire graph.• Axis labels defined with symbols and

units.• Grid drawn using major axis divisions.• Each line is identified using a legend.• Data points are identified with a

symbol: “ ´” on the Qac line to denotedata points obtained by experiment.

• The line representing the theoreticalresults has no data points represented.

• Nothing is drawn freehand.• Title is descriptive, rather than

something like Q vs ∆h.

0

0.05

0.1

0.15

0.2

0 0.2 0.4 0.6 0.8 1

Qth

Qac

Q

∆ hhead loss in m

flow

rat

e

in m

3 /s

FIGURE R.1. Theoretical and actual volume flow ratethrough a venturi meter as a function of head loss.

Writing Style. Use simple words with exactmeanings. Use technical terms to express a precisetechnical meaning. Do not use a large and unusualterm to add false importance to the report or toyourself. Do not use slang words or expressions.

Use simple sentences that have a subject and apredicate. Add adjectives as required. Avoidextra long sentences. Never use “I” regardless ofwhat you were taught in any previous courses.“We” is acceptable. Insert only one space betweena period and the beginning of the next sentence.

Do not bind the report in a folder. Staple thepages together in the upper left-hand corner.

The introduction section should conclude with abrief statement of what the objective of theexperiment is.

The Introduction is written in the past or presenttense.

The report can be written using 1 column or twocolumns per page.

9

THEORY AND ANALYSISThis section explains the theory associated withthis experiment. The theory should be discussedin much greater detail in this section than in theintroduction. It should contain an explanation ofthe theoretical model. For example, if anexperiment was performed with a pendulum, theninclude a brief derivation of the mathematicalmodel of a pendulum. Put the significant portionsof the derivation in this section. Cite referencesusing the guidelines previously described. Includesimple sketches or diagrams to help the readervisualize the physical phenomenon beingstudied.

If there is little or no theory involved in thisexperiment, include the theory with theIntroduction section.

All equations in the report should be indentedand numbered consecutively with Arabicnumerals. Each symbol in the equations should benamed and its dimensional unit given. Anexample:

“Newton’s Second Law of Motion (Resnick andHalliday, 1966) is:

F = ma (1)

where F is the unbalanced external force in N, mis the mass of the block in kg, and a is theacceleration in m/s2.”

There are several inconspicuous but extremelyimportant details associated with this example,specifically in the way equations and units arewritten. Note that the letters used in theequation are in italics. Every reference to force,for instance, is in italics. The units used for eachvariable are in normal type (e.g., non italics).Numerical subscripts and superscripts are innormal type as well. However, subscripts andsuperscripts that are variables are italicized.

When a number is written with a unit, a spaceshould separate the two. For example, 5 N, or17.3 kPa. Numbers are written in normal type.

These features are very important in reportwriting. It is these features that will make awell done technical report appear professional inevery way.

The theory section in its entirety is written in thepast or present tense.

PROCEDUREThis section describes the equipment used in theexperiment and the test procedures. Theequipment setup should be shown in a figure. Thetest equipment and instrumentation used shouldbe listed with model and/or serial numbers, andthe expected instrument precision. Figures ofspecific components should be provided ifnecessary to help the reader to better understandthe test procedure.

Briefly describe the steps of the experimentalprocedure in the order in which they wereconducted. Include sufficient detail in this sectionsuch that the reader could repeat the experiment.

The procedure is written in the past tense.

RESULTS AND DISCUSSIONThe section should discuss the results. Summarizeyour outcome in the topic sentence, then supportthat summary with the results. Use graphs andtables to concisely present the results. Do notdraw conclusions in this section; only list anddiscuss results. This is also the section where acomparison of results with referenced valuesshould be presented.

A sample calculation should also be provided.Start with raw data obtained while performingthe experiment, and show the calculationsinvolved in finding one of the numbers in thissection.

The Results and Discussion section should bewritten in the past tense.

CONCLUSIONSThis section is a clear and concise qualitative andquantitative summary of the experiment andresults. It includes conclusions, observations,trends, and recommendations. Recommendationsare especially valuable if the experiment failedor was impaired. Do not refer to tables or figuresin this section. Coordinate the material in thissection with the Introduction section. If there wasa clear objective in this experiment, statewhether the objective was reached. Makerecommendations regarding the experiment.

Do not use sentences such as “We learned a lot inthis experiment.” Remember that yourperspective is that of an engineer writing atechnical report to others who are technicallyminded. It is not that of a student writing to a labreport grader.

10

The conclusions should be written in the past orpresent tense, except for the recommendationswhich are in the future tense.

APPENDICESThe Appendix section contains its own title page,with a list of what the reader will find inside.

References. This portion of the appendix listsreferences used in the preparation of the report.You must cite the source publication for the workof all others which you include. This gives themdue credit for their work, and shows the researcheffort you put into your report. Do not list the labmanual as a reference. An example of analphabetical Reference list follows:

Bannister, L. (1991). University Style Manual,Fourth edition, Los Angeles: LoyolaMarymount University, 36–39 and 58–61.

Kovarik, M. (1989). “Optimal Heat Exchangers,”Journal of Heat Transfer, 111, 287–293.

Main, B. W. and A. C. Ward. (1992). “What doDesign Engineers Really Know AboutSafety?” Mechanical Engineer, 114, 8, 44–51.

Resnick, R. and D. Halliday. (1966). Physics.New York: John Wiley.

Original Data Sheet. The data sheet completedwhen the experiment was conducted is includedhere.

Calibration Curves. If provided by the instructoror the manufacturer of the lab equipment,calibration curves for each meter used should beincluded in this section.

SHORT FORM REPORT FORMATOnce in a while the experiment requires not a

formal report but an informal report. An informalreport includes the Title Page, ExperimentObjective, Procedure, Results, and Conclusions.Other portions may be added at the discretion ofthe instructor or the writer.

SUMMARY

TITLE PAGE Experiment Number, Title of the Experiment, Name of the Author, Name ofPartners, Date the Experiment was Performed, Due date for this Report, TheUniversity of Memphis, Department of Mechanical Engineering, proofreader’ssignature.

ABSTRACT Brief summary of the objective of the experiment, the procedures, the results,conclusions and recommendations. Present or future tense.

INTRODUCTION Description of the problem, references, objective. Past tense.

THEORY ANDANALYSIS

Theory associated with this experiment, equation derivation. Past or presenttense.

PROCEDURE describes the equipment used, equipment setup, model and/or serial numbers,experimental procedure. Past tense.

RESULTS ANDDISCUSSION

Summarize your outcome, graphs and tables, sample calculation. Past tense.

CONCLUSIONS Conclusions, observations, trends, and recommendations. Past or present tense;recommendations in future tense.

APPENDICES Title page, references, original data sheet, calibration curves.

11

EXPERIMENT 1

THERMOCOUPLES & INSTRUMENTATION

Thermocouples are devices used to measuretemperature electrically. They consist of twodissimilar metals (wires, primarily) joinedtogether and appropriately connected to areadout device. The readout device can be amillivolt meter or an electronic thermometer. Useof a thermocouple in making temperaturemeasurements is the subject of this experiment.

Figure 1.1 shows two dissimilar metals joined(melted) together to form a junction. The otherend of each wire is connected to a readout deviceat the posts. When two dissimilar metals arejoined together, an emf will exists between thetwo posts. The emf is a function of the junctiontemperature. This phenomenon is called theSeebeck effect. If the thermocouple is connected toan external circuit and a current passes throughthe wires, then the emf generated will be alteredslightly. This is known as the Peltier effect. If atemperature gradient exists along either or bothwires, the emf again will be altered slightly.This is called the Thomson effect.

Of primary interest in the measurement oftemperature is the Seebeck effect because theSeebeck emf is a function of the junctiontemperature. So if the Seebeck emf is measuredcarefully as a function of junction temperatureusing a certain combination of metals, then thiscombination can be used repeatedly in otherapplications for the reliable and precisemeasurement of temperature.

When the Seebeck emf is measured, adifficulty arises in measuring the potentialitself. For example in Figure 1.1, if the posts arepart of a millivolt meter, they are likely to bemade of copper or brass. Metal A joined to copperforms another thermocouple with its own Seebeckemf. Metal B joined to copper forms still anotherthermocouple with its own Seebeck emf as well.These two may affect the emf generated at thejunction and provision must be made to take intoaccount the additional potential. Moreover, thetemperature of the binding posts can affect theemf of the junction. These effects must beaccounted for and the setup of Figure 1.1 couldlead to some problems. On the other hand, therenow exists readout devices that electronicallycompensate for the undesirable emf’s andthermocouples are connected to them just as inFigure 1.1. Such devices are convenient to use andare generally very accurate. The readout devicein this instance is usually digital and the output

is in °C or °F.

Metal A

Metal Bjunction

voltage or temperaturereadout device

posts

FIGURE 1.1. Two dissimilar metals connected toform a thermocouple.

The traditional and accepted method of usingthermocouples and eliminating the undesirableeffects involves using a second junction. If thetemperature of the second junction is known andcarefully controlled, the undesirable effects arecompensated for and the results are repeatable.The known temperature is referred to as areference temperature. An ice bath consisting ofcrushed ice and distilled water at atmosphericpressure is commonly used as a referencetemperature in a configuration that is illustratedin Figure 1.2. If the binding posts are at the sametemperature, then the emf’s generated there willnot affect the emf of the circuit. Also thetemperature of interest is measured in this circuitrelative to the temperature of the ice bath, 0°C or32°F. It should be noted, however, that thereadout device using the setup of Figure 1.2 willbe given in millivolts. Standard tables have beenprepared on this basis in order to convert mV to °Cor °F. Obviously, the digital/electronic, directreadout devices are far more convenient to usethan a millivolt meter.

There are a number of types of thermocouplesmanufactured to close tolerances to give accurate,reliable and repeatable results. The types areusually designated with a letter which refers tothe two dissimilar metals used. The standardsare established and recorded by ANSI. The typesare:

Iron-Constantan (ANSI Symbol J)Copper-Constantan (ANSI Symbol T)Chromel-Alumel (ANSI Symbol K)Chromel-Constantan (ANSI Symbol E)Platinum-Rhodium Alloys (ANSI Symbols S,

R and B)Tungsten-Rhenium Alloys (G and C, no ANSI

symbols to date)

12

Metal A

Metal Bjunction

voltagereadout device

posts

reference junction

flask

cork

thermometer

ice bath(water/crushed

ice mix)

Metal A

FIGURE 1.2. Traditional method for measuringtemperature with a thermocouple and a

millivolt meter.

Each type has certain advantages and uses overrequired temperature ranges.

Experiments1. Using a millivolt meter as a readout device,

connect two thermocouples to the posts as inFigure 1.2 and measure room temperature withrespect to an ice bath. Use a chart asappropriate to convert emf to temperatureunits. Reverse the leads and determine how theoutput is affected. Alter the temperature of oneof the connections, if possible, and determinehow the output is affected.

2. Using a digital thermometer as a readoutdevice, connect a thermocouple to the posts as inFigure 1.1 and measure room temperature.Reverse the leads and determine how theoutput is affected. Alter the temperature of oneof the connections, if possible, and determinehow the output is affected.

3. Using a recorder in which output emf is printedas a function of time on chart paper, connect athermocouple to the posts appropriately andobserve the results.

Thermocouple Color Code

ThermocoupleANSI

DesignationPositive Wire Negative Wire Outside

TemperatureRange °F

Iron-Constantan J White Red Brown 32 to 1382

Copper-Constantan T Blue Red Brown 32 to 662

Chromel-Alumel K Yellow Red Brown 32 to 2282

13

EXPERIMENT 2

ONE DIMENSIONAL AXIAL HEAT CONDUCTION

Energy in the form of heat travels from aregion of high temperature to a region of lowtemperature. When one dimensional flow exists,the heat transfer rate can be described byFourier’s Law of Heat Conduction. Fourier’s Lawis an experimentally observed law whichpredicts a linear temperature distribution in onedimension for a constant heat flux. Itsexperimental verification is the subject of thisexperiment.

Figure 2.1 is a sketch of the apparatus used inthis experiment. It consists of three separablesections. The center section is removable. The leftend section contains a brass rod, and an electricalheater. The heat input to the heater can becontrolled and measured. The right end section isalso made of brass, and contains a hollowed outcavity with water tubes attached. Thus heatflows through from the heater through the leftend section, then through the center section, andfinally through the right end section to thewater.

The entire apparatus is insulated so that onedimensional heat conduction is wellapproximated. The end sections containinstrumentation for measuring temperature. Therods in the end sections have a diameter of 25 mmwhile the distance between adjacent temperaturemeasurements is 10 mm. The center section is 30mm long.

Temperature versus length readings can beobtained with this apparatus. Severalexperiments can be performed depending on whatis used in the center section.

ProcedureInstall the center section that is instrumented

for temperature measurement. The heater controlis turned on to some value on the rheostat. Wateris circulated through the cavity. Once steadystate is reached, record temperature versusdistance. The steady state data can be used toverify the accuracy of Fourier’s Law, and tocalculate the thermal conductivity of thematerial.

AnalysisFourier’s Law of Heat Conduction is most

easily verified (or tested) in the one dimensionalconfiguration of this experiment. In equationform, Fourier’s Law is

q = – kA ∂T∂x

(2.1)

where q is the heat flowing through the rod ofdimensions F·L/T (BTU/hr or W), k is thethermal conductivity of the material ofdimensions F·L/(T·L·t) [BTU/(hr·ft·°R) orW/(m·K)], A is the cross sectional area ofdimensions L2 (ft2 or m2 ), and ∂T/∂x is thetemperature gradient of dimensions t/L (°R/ft orK/m). Because temperature decreases withincreasing distance, the gradient ∂T/∂x isnegative, and – ∂T/∂x is actually positive.

Temperature is measured at discrete pointsalong the rod in this experiment. It is thereforeappropriate to rewrite Equation 2.1 in a differentform:

brass rod brass rod

t

temperaturemeasurement

water coolingchamber

center sectionelectrical heaterwater inwater out

A

A

insulation

25 mmdiameter

section A-A

FIGURE 2.1. A schematic of the apparatus used to verify Fourier’s Law of Conduction.

14

q = – kA ∆T∆x

(2.2)

where ∆T is the temperature difference betweenany two thermocouples (adjacent or not) and ∆x isthe distance between the two thermocouples ofinterest. If in fact Fourier’s Law is accurate, then∆T/∆x is a constant for any number of pairs ofthermocouples within the same material.

Referring to Equation 2.2, we assume thatthe heat transferred q is found from the meterreading, the cross sectional area A is calculatedknowing the diameter, and the gradient ∆T/∆x iscalculated from the temperature versus distancedata. With all these parameters known, thethermal conductivity k is easily calculated usingEquation 2.2.

ResultsThermal Conductivity. Plot temperature

versus distance along the rod. Determine thethermal conductivity of the metal in the left end,the center, and the right end sections. (The metalis brass in all cases.) If there are deviations,identify and describe the parameters that causethe discrepancy.

Contact Resistance. Figure 2.2 is a sketch oftemperature versus distance for two materialstouching each other at a location labeled“interface.” Data points are graphed, and a “bestline” is drawn through them for both materials.The lines are extrapolated to the interface. Datafor the material on the left indicates that theinterface temperature is Ti., while data for theother material indicates an interfacetemperature of Tir. The difference in these twotemperatures is due to what is called the“contact resistance” to heat transfer.

T

z

data point extrapolatedprofiles

Til

Tir

interface

FIGURE 2.2. A sketch of temperature versusdistance.

The contact resistanceRtc is found from the onedimensional conduction equation:

qA

= Til - Tir

Rtc(2.3)

The apparatus used in this experiment has 3materials in series, and so there are twointerfaces. Determine the associatedtemperatures for both interfaces, and calculatethe contact resistances. Compare to values foundin a textbook.

QuestionsDoes thermal conductivity vary with

temperature? If so, should ∆T/∆x vary instead ofbeing constant? If the rod itself is tapered, thenwhat should be constant in addition to or insteadof ∆T/∆x?

Are the values for both contact resistancesequal? Should they be?

15

EXPERIMENT 3

MEASUREMENT OF THERMAL CONDUCTIVITY OF A METAL

In order to make accurate predictions of heattransfer rates through materials, it is necessaryto first know the value of the thermalconductivity of the material itself. Thermalconductivity can be measured using standardmethods, devices and techniques. In thisexperiment, we will measure thermalconductivity of a metal, and in addition,calculate an overall heat transfer coefficient forthree metals in series.




ProcedureInstall the center section that is not

instrumented for temperature measurement. Theheater control is turned on to some value on therheostat. Water is circulated through the cavity.Once steady state is reached, record temperatureversus distance. The steady state data can be usedto verify the accuracy of Fourier’s Law, and tocalculate the thermal conductivity of thematerial.



q = – kA ∂T∂x

(3.1)



brass rod brass rod

t




A

A

insulation

25 mmdiameter

section A-A

FIGURE 3.1. A schematic of the apparatus used to verify Fourier’s Law of Conduction.

16

q = – kA ∆T∆x

(3.2)

where ∆T is the temperature difference betweenany two thermocouples (adjacent or not) and ∆x isthe distance between the two thermocouples ofinterest.

For one dimensional heat flow, we can writethe following:

qA

= kLT1 – TIL

xL = k

TIL – TIR

x = kR

TIR – T6

xR(3.3)

where T1 is the temperature at the warmest pointon the rod at the left end, TIL is the interfacetemperature between the left end rod and thecenter rod, TIR is the interface temperaturebetween the center rod and the right end rod, T6 isthe temperature at the coolest point of the rod onthe right, and the x’s correspond to theappropriate distances. Thermal conductivity kvalues correspond to the appropriate materials.The first and third materials are brass. Thecenter section is stainless steel.

By manipulation of Equation 3.2 and 3.3, it ispossible to express the heat transferred along therods in terms of a heat transfer coefficient, U, as

qA

= U(T1 – T6) (3.4)

where

1U =

xL

kL +

xk

+ xR

kR(3.5)

The preceding equations do not include the effectof contact resistance.

ResultsPlot temperature versus distance along the

rods. Determine from your plot the interfacetemperatures. Using published thermalconductivity values for brass, and Equation 3.3,find the thermal conductivity of stainless steeland compare your results to published values. D oinclude the effects of contact resistance, thevalues of which should be obtained from atextbook. Also calculate the overall heat transfercoefficient.

17

EXPERIMENT 4

EFFECT OF AREA ON ONE DIMENSIONAL CONDUCTION

In some geometries, one dimensional heatconduction occurs through a material that has anarea that is not constant. In such cases, it is stillnecessary to be able to make accurate predictionsof heat transfer rates through the material. Wewill investigate the effect of variable area on onedimensional conduction in this experiment.Thermal conductivity of the material will beknown.




ProcedureInstall the center section that a smaller cross

sectional diameter than the end sections. Thiscenter section is not instrumented for temperaturemeasurement. The heater control is turned on tosome value on the rheostat. Water is circulatedthrough the cavity. Once steady state is reached,record temperature versus distance.

AnalysisFourier’s Law of Heat Conduction is easily

verified (or tested) in the one dimensionalconfiguration of this experiment. In equationform, Fourier’s Law is

q = – kA ∂T∂x

(4.1)



brass rod brass rod

t




A

A

insulation

25 mmdiameter

section A-A

FIGURE 4.1. A schematic of the apparatus used to measure the effect of area on one dimensionalconduction.

18

q = – kA ∆T∆x

(4.2)


For one dimensional heat flow, we can writethe following:

qA

= kLT1 – TIL

xL = k

TIL – TIR

x = kR

TIR – T6

xR(4.3)

where T1 is the temperature at the warmest pointon the rod at the left end, TIL is the interfacetemperature between the left end rod and thecenter rod, TIR is the interface temperaturebetween the center rod and the right end rod, T6 isthe temperature at the coolest point of the rod onthe right, and the x’s correspond to theappropriate distances. Thermal conductivity kvalues correspond to the appropriate materials.

The first and third materials are brass. Thecenter section is stainless steel, with dimensionsof 13 mm diameter by 30 mm long.

By manipulation of Equation 4.2 and 4.3, it ispossible to write an equation in terms of area andtemperature gradient:

AL

∆T

∆xL

= A

∆T

∆x = AR

∆T

∆xR

(4.4)

Note that the preceding equations do not includethe effects of contact resistance.

ResultsPlot temperature versus distance along the rods.Determine from your plot the interfacetemperatures. Using published thermalconductivity values for brass and stainless steel(if you need them), verify whether Equation 4.4is correct. Include the effects of thermal contactresistance in your calculations.

19

EXPERIMENT 5

MEASUREMENT OF THERMAL CONDUCTIVITY OF AN INSULATOR

In order to make accurate predictions of heattransfer rates through materials, it is necessaryto first know the value of the thermalconductivity of the material itself. Thermalconductivity can be measured using standardmethods, devices and techniques. In thisexperiment, we will measure thermalconductivity of an insulator.

Figure 5.1 is a sketch of the apparatus used inthis experiment. It consists of two separablesections, between which is placed the insulator ofinterest. The left end section contains a brass rod,and an electrical heater. The heat input to theheater can be controlled and measured. The rightend section is also made of brass, and contains ahollowed out cavity with water tubes attached.Thus heat flows through from the heater throughthe left end section, then through the centersection, and finally through the right end sectionto the water.

The entire apparatus is insulated so that onedimensional heat conduction is wellapproximated. The end sections containinstrumentation for measuring temperature. Therods in the end sections have a diameter of 25 mmwhile the distance between adjacent temperaturemeasurements is 10 mm.


ProcedureInstall an insulating material between the

two end sections. The heater control is turned on tosome value on the rheostat (ensure that thetemperature at the hottest point does not exceed100°C). Water is circulated through the cavity.Once steady state is reached, record temperatureversus distance. The steady state data can be usedto calculate the thermal conductivity of thematerial.



q = – kA ∂T∂x

(5.1)



brass rod brass rod



insulatingmaterial

electrical heaterwater inwater out

A

A

insulation

25 mmdiameter

section A-A

t

FIGURE 5.1. A schematic of the apparatus used to measure thermal conductivity.

20

q = – kA ∆T∆x

(5.2)


We can apply Equation 5.2 across theinsulating material to obtain:

k = qA

∆x

∆T(5.3)

Thus, ∆x is the thickness of the insulatingmaterial while it is clamped in position, and ∆Tis the difference between the interfacetemperatures; i.e., the temperature differenceacross the insulating material.

Thermal conductivity should be expressed inthe appropriate units.


rods. Determine from your plot the interfacetemperatures. Use them with Equation 5.3 todetermine thermal conductivity. If publishedvalues of the thermal conductivity are availablecompare your results to them. Include the effectsof thermal contact resistance.

21

EXPERIMENT 6

RADIAL ONE DIMENSIONAL CONDUCTION

Conduction in the radial direction is made tooccur in a number of geometries. In such cases, it isimportant to be able to make accurate predictionsof the heat transfer rate. In this experiment, wewill investigate the temperature profile andheat transfer rate for a radial system.

Figure 6.1 is a sketch of the apparatus used inthis experiment. It consists of a 55 mm diameterdisk 3 mm in thickness, which has a hole in itscenter that is 4 mm in diameter. The disk has anelectrical heater attached to its center, andcontains a circumferential water channel near itsouter edge. With the heater on and cooling waterflowing through the channel, heat flowsradially through the disk. The disk and heaterare well insulated to prevent heat losses in otherdirections. Provision is made to measuretemperature at selected, evenly spaced locationsalong the disk.


disk

insulation

electricalheater

circumferentialwater channel

55 mm 4 mm

FIGURE 6.1. A schematic of the apparatus used toinvestigate radial, one dimensional conduction.

ProcedureTurn the heater and the cooling water on and

allow sufficient time to elapse for steady state tobe reached. Take readings of temperature at allradial locations. Measure the distance betweenadjacent thermocouples.

AnalysisFourier’s Law of Heat Conduction is verified

(or tested) in the radial, one dimensionalconfiguration of this experiment. In equationform, Fourier’s Law is

q = – kA ∂T∂r

(6.1)

where q is the heat flowing through the disk, ofdimensions F·L/T (BTU/hr or W), k is thethermal conductivity of the material ofdimensions F·L/(T·L·t) [BTU/(hr·ft·°R) orW/(m·K)], A is the cross sectional area normal tothe heat flow, of dimensions L2 (ft2 or m2 ), and∂T/∂x is the temperature gradient of dimensionst/L (°R/ft or K/m). Because temperaturedecreases with increasing distance from the heatsource, the gradient ∂T/∂r is negative, and(–∂T/∂r) is actually positive.

The cross-sectional area is given by

A = 2πrL

where r is the radial coordinate (R1 < r < R2), andL is the disk thickness. Substituting into Equation6.1 and separating variables for integration gives

⌡⌠

T1

T2

dT = ⌡⌠

R1

R2

– 1

2πkL

qrdrr

where the subcripts “1” and “2” are at any twodifferent points in the disk. Integrating gives

T2 – T1 = – qr

2πkL ln

R2

R1

or T1 – T2 = qr

2πkL ln

R2

R1(6.2)


disk. Use the data with Equation 6.2 to calculatethe rate of radial heat conduction. Compare theresults with the heat input read from thewattmeter. Are you expecting to get a straightline graph?

22

EXPERIMENT 7

HEAT TRANSFER FROM A FIN

Conduction is the primary mode of heattransfer through solids. When a heated solid isexposed to a lower temperature fluid, heat istransferred from the solid to the fluid byconvection. The amount of heat transferred at thesurface is calculated by means of a convectioncoefficient h. Measuring the convectioncoefficient for a fin is the subject of thisexperiment.

Figure 7.1 is a sketch of a fin, also known asan extended surface. Fins are used to increase thesurface area of a solid. When the surface areaAs is increased, the rate of heat transfer q isincreased because q is directly proportional to As.

In the apparatus of this experiment, thereare three fins attached to a chamber into whichsteam is admitted. The steam will heat the endof each fin. Heat will be conducted axially alongeach fin and will be transferred by naturalconvection to the air. Thermocouples areembedded at intervals along each fin so thattemperature is known at selected points. It is atthese points where the convection coefficient willbe determined.

Abbreviated TheoryFins are usually characterized with a ratio of

parameters denoted as m which is defined as

m = √hcP k A

where hc is the convection coefficient [W/(m2·K)or BTU/(hr·ft2·R)], P is the perimeter of the fin[m or ft], k is the thermal conductivity of finmaterial [W/(m·K) or BTU/(hr·ft·R)], and A isthe cross sectional area of the fin [m2 or ft2]. It canbe shown that m has units of 1/m or 1/ft.

By making suitable assumptions regardingthe fins of this experiment, a differentialequation can be derived. Boundary conditionsmust then be written. In all cases, thetemperature at the steam end (to correspond to awall temperature) is denoted as Tw. At the otherend of each fin, we could have one of threevariations:

1. The fin tip is uninsulated.2. The fin extends to infinity where the fin

temperature equals the air temperature.3. The fin tip is insulated.

steam chest

steam inlet

outlet

fin

T

z

L

z

FIGURE 7.1. Schematic of one of the three fins of the apparatus, and a corresponding steady state temperature profile.

23

The simplest formulation corresponds to case 2and will be the one used here. When thedifferential equation is solved subject to theboundary conditions, the following temperatureprofile results

T – T∞Tw – T∞

= exp (– mz) (7.1)

whereT is the temperature at any axial location zT∞ is the temperature of the ambient airTw is the temperature of the wall (the first

thermocouple)z is any axial location at which there is a

thermocouple.Equation 7.1 is dimensionless.

The efficiency of a fin is defined as

ηe = actual heat transferred from wall with fin attachedheat transferred if entire fin is at wall temperature

It can be shown that the fin efficiency is

ηe = tanh (mL)

mL(7.2)

where L is the length of the fin. Anotherparameter of importance in the analysis of a finis the fin effectiveness, defined as

ηf = heat flux from wall after adding fin

heat flux from wall before adding fin

or ηf = √ kP hcA

tanh (mL) (7.3)

The preceding equations apply to Case 2. Similarexpressions have been developed for Cases 1 and3.

Experiment and AnalysisSteam is admitted into the end chamber and

each fin receives heat. Monitor temperaturereadings at 20 minute intervals to determinewhen steady state is reached. The temperaturereadings at steady state are those of significance;the rest may be discarded. Ambient temperatureand pressure, and fin dimensions should all bemeasured and recorded. For the fin assigned,

1. Construct a graph of temperature T versus z.

2. Calculate values of T – T∞Tw – T∞

, m and using the

equations that are appropriate for the finused in the experiment. Do not the Case 2equations unless they indeed do apply.

3. Use values of m to determine the convectioncoefficient hc for each location wheretemperature T is measured. Graph convectioncoefficient (vertical axis) as a function oftemperature.

4. Calculate the efficiency and theeffectiveness of the fin.

5. With respect to the worded definitions ofefficiency and effectiveness, comment on therelative efficiency and effectiveness of eachfin. Determine which is the most efficientand the most effective and explain whatfactors make it so.

6. For the report, state the differential equationfor fins and write the boundary conditions asthey would apply to the fins of thisexperiment.

QuestionsWhat values are expected for the convection

coefficient hc? Is the convection coefficient aconstant as is so often assumed? How does theconvection coefficient vary with temperature?

ExampleData were obtained on the system of Figure

7.1 for a copper rod that is 1 cm in diameter and 1m long. At a distance z of 30 cm, the temperaturewas measured to be 50°C. The wall and ambienttemperatures were measured to be 95°C and 20°C,respectively. Calculate the convectioncoefficient, the effectiveness and the efficiency.

Assumptions1. Heat is transferred along the rod by

conduction only.2. One dimensional conduction exists along the

rod.3. Heat is transferred only by convection to the

surrounding air; radiation to the surroundingsis neglected.

4. Properties of the copper are constant.5. The system is at steady state.

Fin Geometry Factors and PropertiesD = 0.01 mA = πD 2/4 = π(0.01)2/4 = 7.85 x 10-5

P = πD = π(0.01) = 3.14 x 10-2

z = 0.3 mL = 1 mkCu = 399 W/(m·K) at 20°C

24

Calculations for Case 2

T – T∞Tw – T∞

= 50 – 2095 – 20 = 0.4 = exp (– mz)

or – m(0.3 m) = ln (0.4) = – 0.916

Therefore,

m = 3.05 = √hcP k A

Solving for the convection coefficient gives

hc = k AP

(3.05)2 = 399(7.85 x 10-5)

3.14 x 10-2 (3.05)2

hc = 9.29 W/(m2·K)

The fin efficiency is

ηe = tanh (mL)

mL =

tanh (3.05(1))3.05(1)

ηe = 0.326

The fin effectiveness is

ηf = √ kP hcA

tanh (mL)

= √399(3.14 x 10-2) 9.29(7.85 x 10-5) tanh (3.05(1))

ηf = 130

The interpretation of these results is left to thereader.

25

EXPERIMENT 8

GRAPHICAL SOLUTIONS TO UNSTEADYHEAT TRANSFER PROBLEMS

Prior to the existence of computers andcalculators, graphical methods were usedextensively to solve engineering problems. In somecases, a graphical solution was the only way tosolve the problem. Graphical methods in heattransfer require little mathematics and canafford acceptable accuracy. They permit thedesigner the luxury of having an overall view ofhow temperature varies with time through anobject for unsteady problems. In addition,mistakes can easily be found and corrected. TheSchmidt Plot was one of the very first graphicalsolution methods devised. A modernizedgraphical solution method known as the Saul’evMethod will be used in this experiment.

The one dimensional unsteady conductionequation in heat transfer can be solvedgraphically in a step-by-step manner. Theprocedure involves:

1. Describing the problem with thedifferential equation and appropriate boundaryconditions.

2. Rewriting the differential equation in analgebraic form.

3. Solving the problem by graphicallyfollowing the algebraic equation.The method is easier to illustrate than to explainand so an example problem will be solved.

Example ProblemA brick wall 5 cm thick is part of the chimney

section of an incinerator. When in operation, thetemperature of the exhaust gases inside thechimney are 460°C, and the outside wall surfaceof the chimney is not to exceed 100°C. Initially,the entire chimney is at 20°C. Determine the timerequired to meet this condition if the brickproperties are:

Cp = 840 J/(kg·K)sp. gr. = 1.8

k = 0.5 W/(m·K)

and the convection coefficient hc = 50 W/(m2·K).

Solution: We must first make a number ofassumptions in order to solve this problem. Theseinclude:

1. The exhaust gas temperature of 460°C isuniform and constant.

2. The flow of heat is one dimensional.

3. The convection coefficient is constant.4. Material properties are constant.5. Radiative effects are negligible.

The differential equation that describes the flowof heat through the chimney is

∂T ∂t

= α ∂ 2T∂x 2 α =

kρCp

(8.1)

The boundary condition at the inside face is

x = 0 qA

= – k ∂T ∂x

= hc(T∞ – Tw)

The boundary condition at the outside face wasnot given in the problem statement so we assume a“worst case” condition which is that the outsideface is insulated:

x = L qA

= – k ∂T ∂x

= 0

The condition on time is that

t < 0 Tw = 20°C

t > 0 T∞ = 460°C

We are seeking the time required for the rightface under these conditions to be heated from 20°to 100°C.

To make the problem suitable for graphicalsolution, the differential equation must beconverted to an algebraic form or finitedifferenced. Of the several methods available,we use the Saul’ev formulation. Equation 1 infinite difference form is

Tmp+1– Tm

p

∆ t = α

Tm+1p – Tm

p – Tmp+1 + Tm-1

p+1

(∆x)2

(8.2)

in which the m subscript refers to an x position,and the p superscript refers to time. Thus an msubscript indicates the position of current interest,m + 1 indicates a position that is one ∆x furtherthan m away from the origin, and m - 1 is one ∆xcloser to the origin. Also, p indicates the presenttime, and p + 1 indicates a time that is one ∆t into

26

the future. Equation 7.2 can be rearranged to give

T p+1m =

α ∆t(∆x)2 (T p

m+1 – T pm – T p+1

m + T p+1m-1 – T p

m )

(8.3)

Many finite difference equations like the oneabove depend heavily on the value of the FourierNumber α∆t/(∆x)2 for their stability.Fortunately Equation 8.3 is unconditionally stableso we can select virtually any reasonable valueand obtain good results. Equation 3 simplifiesgreatly if we use α∆t/(∆x)2 = 1, yielding

T p+1m =

Tm+1p + Tm-1

p+1

2 (8.4)

Equation 8.4 states that we use adjacenttemperatures at m - 1 and m + 1 to find thetemperature at m. The temperature at m + 1 is atthe present time and the temperature at m - 1 isat the future time. So when a new temperature atm - 1 is found graphically (or numerically), it isused immediately to find the temperature at m.

Next, we must rewrite the boundaryconditions in finite difference notation. It must beremembered that in the graphical method, wewill solve the problem by constructing a graph oftemperature T versus the space variable x forvarious times. Therefore we rewrite the firstboundary condition by solving it for ∂T/∂x, theslope at the boundary. The first boundarycondition thus becomes:

x = 0

∂T

∂x

m

p = –

hc

k (T p

∞ – T pw ) = –

(T∞p – Tw

p) k / h c

(8.5)

This equation states that at the convectiveboundary, the slope ∂T/∂x will equal the ratio offree stream minus wall temperatures to k/hc. Thesecond boundary condition becomes

x = L ∂T ∂x

= 0 (8.6)

So at the insulated boundary, the slope is 0.Equations 8.4, 8.5 and 8.6 are sufficient to solvethe problem graphically. Figure 8.1 shows two

drawings of the system. The first drawing showsthe brick wall arbitrarily divided into five 1 cmwide intervals. (It could be divided into 2, 10 or100 intervals.) The boundary of each interval islabeled, beginning at 0 on the left and increasingto 5. The temperature axis drawn to scale isshown also. To the left of the m = 0 line is astarting point for all temperature profiles. Theexhaust gas temperature of 460°C is located adistance of k/hc = 0.5/50 = 0.01 m = 1 cm from theleft face. At time zero, the temperature of theexhaust gases is 460°C and the wall temperatureis 20°C.

The second drawing shows the time t = 0profile and four subsequent profiles. We begin byaligning the 460°C point with the temperature atm = 1 and draw a straight line until the m = 0 lineis intersected. Next we align this newtemperature at m = 0 with the temperature at m =2 and draw a straight line until the m = 1 line isintersected. This process is continued to completethe t = 1 profile. To complete the t = 2 profile, webegin with the 460°C value again. This process isthe graphical solution method described by theabove equations.

By inspection, we see that after 4 timeintervals (or t = 4∆t), the right face has exceeded100°C. To obtain better accuracy, we would haveto use more than five intervals on x. We now usethe Fourier number selected to determine thesolution to the problem:

α ∆t(∆x)2 = 1

By substitution

k

ρCp

∆ t(∆x)2 =

0.51 800(840)

∆ t(0.01)2 = 1

Solving gives

∆t = 302.4 s = 5 min

The time required for the right face to reach100°C is 4∆t or

4∆t = 20 min

For the report, derive the finite differenceddifferential equation. The instructor will providean unsteady problem to solve.

27

01 2 3 4 5 = m

= 0t

x

T

01 2 3 4 5 = m

t

x

T

43

21 0 =

1 cm

100

200

300

400

0

100

200

300

400

0

FIGURE 8.1. The Saul’ev Plot of the ExampleProblem.

Problems

1. A large slab of cast iron 12 in. thick isinitially at 400°F. Suddenly its left face isreduced to 40°F while its right face is kept at400°F. Using the graphical method,determine the temperature at the midplaneafter 15 min have elapsed.

2. A large of slab of copper 10 cm wide isinitially at a uniform temperature of 25°C.Suddenly, the temperature of its left andright faces are increased to 200°C. How muchtime elapses before the midplanetemperature reaches 140°C.

3. It is desired to construct a fireproof wall for asafe. The wall consists of two 12 gage piecesof sheet metal with insulation sandwichedbetween them. The criterion for protection isthat if the temperature of the outside pieceof sheet metal is suddenly raised to 1000°F,then no less than 2 hr should elapse beforethe temperature of the inside piece of sheetmetal rises to 200°F from an initial anduniform temperature of 75°F. Determine therequired thickness of insulation if it isasbestos.

4. A stainless steel wall serves as part of asteam conveying line. It is 120 mm thick andat an initial temperature of 18°C. Its left faceis exposed to a high speed stream of steam ata temperature of 250°C, with a convectioncoefficient of 500 W/(m2·K). The right face isexposed to a stream of water at 15°C with aconvection coefficient of 250 W/(m2·K).Determine the left and right surfacetemperatures when the midplanetemperature reaches 75°C.

5. A stainless steel (type 304) wall is 8 in. thickand is part of the chimney section of afurnace. When the furnace is in operation, thetemperature of the inside wall surface is600°F, while the outside wall temperature is95°F. the profile within the wall is at steadystate. The chimney section is to be cleaned, sothe furnace is shut down and cooling fans areused to move 75°F air upward through thechimney. The right face stays at 95°F duringthe cooling process. Determine the time ittakes for the stainless steel slab to cool to thepoint where the temperature at any pointwithin the wall is less than 200°F. Use aconvection coefficient of 85 BTU/(hr·ft2·°R).

28

EXPERIMENT 9

DETERMINATION OF CONDUCTION SHAPE FACTORUSING THE ANALOG FIELD PLOTTER

In many two dimensional geometries forwhich a heat transfer analysis is to beperformed, it is sometimes necessary to use atechnique for solution other than an analyticalone. One such method involves the use of theconduction shape factor, which is generated froma field plot for the geometry. An analog fieldplotter is an electronic aid used to help generate afield plot, which is the subject of thisexperiment.

TheoryFor heat transfer in a two dimensional

material having constant properties, theapplicable differential equation for temperatureis

∂ 2T∂x 2 +

∂ 2T∂y 2 = 0 (9.1)

subject to appropriate boundary conditions. For asimple geometry with manageable boundaryconditions, the above equation can be solved by aseparation of variables method. For a complexgeometry, the above equation might not besolveable and so an alternative method isnecessary to obtain a solution.

In cases where Equation 9.1 cannot be solved,it is rationalized that a temperature distributionwithin the material is not really needed. If atemperature distribution is found, we would use itprimarily to find the heat transfer rate anyway.So if there is a method we could use from thebeginning to find the heat transfer rate, we wouldbe content with it being the solution to theproblem.

For steady two dimensional electricalconduction within a conductive material havingconstant properties, the applicable differentialequation for the electric potential is

∂ 2E∂x 2 +

∂ 2E∂y 2 = 0 (9.2)

When Equations 9.1 and 9.2 are solved subject toanalogous boundary conditions, the solutionsobtained will have the same form. Lines ofconstant voltage within the material willcorrespond to lines of constant temperature.

A device known as an analog field plotter can

be used to generate lines of equal potential in atwo dimensional geometry of virtually anyshape. The field plotter consists of a drawingboard onto which electrical conducting paper isplaced, a low voltage-difference DC source andassociated wiring. Figure 9.1 is a schematic of theoperation of an analog field plotter. Theobjective in solving a heat transfer problem in atwo dimensional geometry now becomes one ofgenerating equipotential lines in the samegeometry.

Insulatedboundary orline of symmetry

Null detector

Stylus orprobe

Wired andpainted boundary(isothermal)

E or T2 2

E or T1 1

E - E = T - T1 22 1

2 E = E - E

FIGURE 9.1. Schematic of the operation of ananalog field plotter in generating points.

ProcedureThe procedure involved in using a field

plotter first requires construction of the twodimensional geometry with a sheet of paper thatconducts electricity (several makes areavailable). The conductive paper is held down ona drawing board with thumbtacks at the corners.Metal wire is attached to thumbtacksappropriately so that the wire is in contact withthe paper along the necessary boundaries. Silverconducting paint is then brushed on over the wiresand the paper. Generally, isothermal boundariesin the geometry are wired and painted; insulatedboundaries are left alone. As shown in Figure 9.1,a voltage potential is imposed on the regionacross which a temperature difference exists. A

29

null detector is then set at a desired value ofpotential difference ∆E = E - E2, and pointscorresponding to the set difference are located inthe region by a probe. Points having an equalpotential difference ∆E are then connected toobtain lines of constant voltage. These lines areanalogous to lines of equal temperature, calledisotherms.

Once isotherms are obtained, the fieldplotter is then used to obtain heat flow lines, alsoknown as adiabats. The same geometry is laidout, except in this case the insulated boundariesare wired and painted while the isothermalboundaries are left alone. (See Figure 9.2.)

(a ) (b)

FIGURE 9.2. Wired and painted boundaries foruse with analog field plotter in obtaining (a)isotherms and (b) adiabats for the geometryof Figure 7.1.

ResultsWhen the adiabats and isotherms are known,

they are combined into one drawing which is usedto help generate a field plot. The field plotconsists of isotherms and adiabats drawn infreehand in the geometry of interest. The actualsketching is aided by paying strict attention tosome key concepts, namely:

• Heat flow lines and constant temperaturelines are perpendicular to each other.

• Heat flow lines are parallel or coincident toinsulated surfaces or lines of symmetry.

• Isotherms intersect an insulated surface orline of symmetry at a right angle.

Ensuring that the above rules are incorporatedinto the sketch is a matter of trial and error.However, once the field plot (or flux plot or flownet) is complete and the above rules are satisfied,the heat transferred from one boundary toanother can be approximated. The accuracy of theresults depends on how fine a mesh is used and onthe skill of the analyst. Note that two analystscan use differing meshes and still obtain results

that are in good agreement. The results obtainedwith the field plotter merely give an indicationof the direction of the isotherms and heat flowlines within the system.

The flux plot contains what are known ascurvilinear squares. Counting the squares allowsfor finding what is called a conduction shapefactor. The flux plot contains heat flow laneswhich are bounded by adjacent adiabats. Thenumber of squares within each heat flow lane andthe number of heat flow lanes are to be counted.The conduction shape factor S is then found with

S = MLN

(9.3)

where M is the number of heat flow lanes, L is thedepth of the system into the page and N is thenumber of curvilinear squares per lane. The heattransferred is then easily found with

q = kS(T1 – T2) (9.4)

The above discussion is now illustrated by anexample.

ExampleConsider that we have a two dimensional

geometry, illustrated isometrically in Figure 9.3,for which we are to find the heat transfer rate.Heat will be transferred from the surface attemperature T1 to the surface at T2 in an amountgiven by Equation 9.4. It is assumed that thetemperatures and the thermal conductivity areknown. All we need in order to calculate the heattransferred is the conduction shape factor for thegeometry. This particular geometry was selectedbecause its shape factor is known from analyticalmethods to be

S = 2πL

ln (0.54W/R) (9.5)

In studying the cross section, lines of symmetrywere identified and as a result, only one fourth ofthe cross section is used for analysis. Twoidentical pieces corresponding to the geometryare cut out of conducting paper. The field plotteris used to generate lines of constant temperatureand adiabats. The results are shown in Figure 9.4.Superimposed results are shown in Figure 9.5.

Now using Figure 9.5 as a guide (a backingsheet), the sketch of Figure 9.6 was preparedfreehand. An effort was made to ensure that thecurvilinear squares were approximately “square”.

30

WL

W

radius R

FIGURE 9.3. Schematic of the system of theExample Problem.

The resulting flux plot of Figure 9.6 shows thatthere are 10 heat flow lanes in one fourth of thegeometry. Therefore, M = 10(4) = 40. Also, thereare 7 curvilinear squares within each heat flowlane and so N = 7. The conduction shape factor perunit length is then found to be

SL

= MN

= 407

or

SL

= 5.71

Note that integral values were selected for M andN. It is acceptable if the opinion of the analyst isthat N = 6.75. Fractions of a square can be used.By substituting W = 8.4 and R = 1.3 into Equation5, we obtain

SL

= 2π

ln (0.54W/R) = 2π

ln (0.54(8.4)/1.3)

or

SL

= 5.02

The % error between 5.02 and 5.71 is 14%.

ExperimentFor the cross section assigned, use the analog

field plotter to obtain isotherms and adiabats.Superimpose the graphs to obtain a rough plot.Use the rough plot to generate a sketched andrefined flux plot. The final plot is to be donefreehand. Determine the number of heat flowlanes and the number of curvilinear squares perlane, and calculate the conduction shape factor.Each group member is to compose a different plotand the results should be compared to each otherand if available to an analytical result. Commenton the accuracy of the method.

31

FIGURE 9.4a. Lines of constant temperature forthe geometry of Figure 9.3, generated withan analog field plotter.

FIGURE 9.4b. Adiabats for the geometry of Figure9.3 generated with an analog field plotter.

FIGURE 9.5. Adiabats and isothermssuperimposed.

FIGURE 9.6. Sketched field plot for thegeometryof Figure 9.3.

32

EXPERIMENT 10

PREDICTING TEMPERATURE AT THE CENTER OF A CYLINDERUSING TRANSIENT CONDUCTION WITH CONVECTION

When a heated (or cooled) substance isimmersed in a bath having a differenttemperature, heat transfer takes place. Heat isconducted throughout the substance, andconvection occurs at the surface. There aredifferent ways to model this unsteady problem,depending on the relative magnitudes of theconvection coefficient and the thermalconductivity of the material. In this experiment,unsteady temperature versus time is measured fora material, and the convection coefficient iscalculated. The results are then used withHeisler charts in order to predict the expectedtemperature, and the results are compared.

Transient Heating of a CylinderFigure 10.1 is a sketch of the apparatus used

in this experiment, which consists of dualconcentric open top tanks filled with water. Thetank apparatus also contains an immersion heaterand a circulating pump. The heater is turned on ato a pre-determined setting. The circulating pumpis turned on also, and the water temperature ismonitored. The water bath is ready when thewater temperature reaches steady state(approximately 3 hours).

WaterReturn

ImmersionHeater

InnerCylinder

InsulatedTank

TestCylinder

WaterFlow

SupportSupport

Motor

Pump

MotorShaft

Valve

FIGURE 10.1. Schematic of the apparatus used inthis experiment.

The test cylinder (with known properties),initially at some temperature Ti, is immersed inthe water bath. The cylinder temperature ismeasured (using thermocouples and a digitalthermometer) every 10 seconds. The experimentconcludes when the cylinder has reached a

steady state temperature.The temperature within the cylinder is

assumed to be uniform (to be checked later), andso we assume that the lumped analysis applies.In the lumped analysis method, temperature canbe predicted with

T – T∞Ti – T∞

= exp

–

hAstρV—c

(10.1)

where T is the measured temperature, T∞ is thetemperature of the surroundings (the water bath),h is the convection coefficient, As is the surfacearea of the cylinder, V— is the volume of thecylinder, and c is its specific heat. All terms inthis equation are known except for convectioncoefficient h which can therefore be calculatedfor every instant that temperature is known.Values for the convection coefficient aredetermined for all times, and can be averagedtogether if necessary.

Once the convection coefficient is determined,the Biot number is calculated. The Biot number isdefined as

Bi = hLc

k

where the characteristic length Lc is the ratio ofvolume to surface area (= V—/As). If the Biotnumber is less than 0.1, the lumped analysisequation applies.

CalculationsThe objective in this experiment is to use the

Heisler chart for an infinite cylinder and that fora semi-infinite plate to predict the temperatureobtained by experiment. We could assume thecylinder has an infinite length to diameter ratio,but this would introduce some error into thecalculations.

At any recorded time, use the data tocalculate the value of dimensionless temperature,(Tc – T∞)/(Ti – T∞), in which Tc is the temperatureat the cylinder center. With the lumped analysisassumption, we assume that the centertemperature Tc is equal to the temperature Tanywhere within the cylinder. Use Equation 10.1to find the value of the convection coefficient. We

33

will use the calculated value of the convectioncoefficient in subsequent calculations, even if thelumped analysis does not apply.

An exact solution for the problem of unsteadyheating (or cooling) of an infinite cylinder hasbeen obtained. Results are graphed in what isknown as a Heisler chart for the cylinder. Adimensionless group called the Fourier numberappears on the horizontal axis, whiledimensionless temperature is on the vertical axis.The Fourier number for the cylinder is

Fo = αtR2

A number of curves on the axes of the Heislerchart correspond to the inverse of the Biotnumber, 1/Bi. Note that the Biot number for thisgraph (Bi = hR/k) is different from the Biotnumber used in the lumped analysis method.

If we use only the Heisler chart for theinfinite cylinder in order to predict temperature,we will encounter an error, because the cylinder ofthis experiment is finite in length. In order tomake the necessary correction, we refer to aHeisler chart for a semi-infinite plate. Thischart is, again, a graph of dimensionlesstemperature versus the Fourier number, with theinverse Biot number appearing on various lines ofthe graph. The Fourier number for the semi-infinite plate is given by

Fo = αtL2

while the Biot number is Bi = hL/k, where L inthese equations is half the plate thickness.

The infinite cylinder problem is one

dimensional and unsteady. The semi-infiniteplate problem is also one dimensional andunsteady. The combination of these solutions canbe used to model the two dimensional, unsteadycylinder of this experiment. It can be showntheoretically that the product of the twosolutions gives us what we are seeking. A visualrepresentation of this is provided in Figure 10.2.

A number of cylinders are available for thisexperiment, some are noncircular in cross section,and some are made of nonmetals. Also availableare various spheres. Conduct the experiment forthe object(s) assigned.

Sample CalculationData obtained on a steel cylinder (D = 2 in., L

= 8 in.) yielded the following data:

Ti = 79°F T∞ = 149°F

at t = 100 s, Tc = 138°F

What temperature does the Heisler chartspredict? What is the % error?

Solution: We apply Equation 10.1, which is

T – T∞Ti – T∞

= exp

–

hAstρV—c

Calculations are:

T – T∞Ti – T∞

= 138 - 14979 - 149 = 0.157

As = 0.3927 ft2 V— = 0.01454 ft3

finite cylinder = infinite cylinder x semi-infinite plate

R R

2L 2L

FIGURE 10.2. Finite cylinder solution as product of infinite cylinder and semi-infinite plate solutions.

34

and

Lc = 0.014540.3927 = 0.037 ft

For steel, k = 24.8 BTU/(hr·ft·°R), c = 0.113BTU/(lbm·°R), and ρ = 487 lbm/ft3. With theseproperties, and the preceding calculations, wesubstitute into Equation 10.1 and solve to find

h = 135.8 BTU/(hr·ft2·°R)

The Biot number for the lumped analysiscalculation is

Bi = hLc

k =

135.8(0.037)24.8 = 0.203

Thus the lumped analysis does not strictly apply,but we assume that we can still use this result forthe convection coefficient.

In order to use the Heisler charts, we mustcalculate two Fourier numbers and two inverseBiot numbers. Thus for the cylinder,

1/Bi = k

h R =

24.8135.8(0.0833) = 2.2

Fo = kρc

t

R2 = 1.8

Using these values, we obtain from the Heislerchart for a cylinder, a dimensionless temperatureof

Tc – T∞Ti – T∞

≈ 0.21 (infinite cylinder)

For the semi-infinite plate, remembering thatthe length we need is half the actual length ofthe cylinder, we find

1/Bi = k

h L =

24.8135.8(4/12) = 0.55

Fo = kρc

t

L2 = 0.113

Using these values, we obtain from the Heislerchart for a semi-infinite plate, a dimensionlesstemperature of


≈ 0.85 (semi-infinite plate)

So, for the finite cylinder of this example, we get


≈ 0.21(0.85) = 0.179 (finite cylinder)

The calculated temperature then is

Tc = 149 + 0.179(79 - 149)

Tc = 137°F

The % error is

% error = 100 x 138 - 137

138 = 1.1%

This error can be attributed to several sources.

ExperimentFor this experiment, obtain temperature versustime for an object submerged in a water bath.Determine a value for the convection coefficientfor all instants in time. Select 2 data points anduse the Heisler charts to determine how wellthey predict the temperature of the cylinder.

Questions1. Is the convection coefficient a constant? Is it

within an “acceptable” range for theconditions of the experiment?

2. For the 2 data points you selected, what isthe error if we use only the infinite cylindersolution without correction from the semi-infinite plate solution?

2. What is the difference between the Biotnumber for the lumped capacitance methodand the Biot number used in the Heislercharts?

3. Reading the Heisler charts with a great dealof accuracy is very difficult. What is yourestimate of the error involved in using thegraphs?

4. Radiation was not accounted for in theanalysis associated with this experiment.Why not?

35

0.01

0.1

1

(Tc -

T∞)/

(Ti -

T∞)

Fourier Number αt / R2

0.20.4

= 1/Bi = k/hR

0.02

0.04

0.06

0.08

0.2

0.4

0.6

0.8

0

0 1 2 3 4 6 8 10 12 14 16 18 20 22 2624 3028

2

3

108

Infinite Cylinder

0.6 0.81.0

1.2 1.4

1.6

2.5

4

6

1214

16

20

25

18

0.01

0.1

1

(Tc -

T∞)/

(Ti -

T∞)

Fourier Number αt /L2

0.2

0.4

0.6

0.8

1.0

1/Bi = k/hL 0.02

0.04

0.06

0.08

0.2

0.4

0.6

0.8

= 0

0 1 2 3 4 6 8 10 12 14 16 18 20 22 2624 3028

1.52.0

3.0

4.0

6.0

10.0

8.0

Semi-infinite Plate

36

EXPERIMENT 11

NATURAL CONVECTION FROM A VERTICAL PLATE

Consider a heated vertical plate that isallowed to cool in air. Heat will be transferred tothe surrounding air by convection and to otherparts of the surroundings by radiation. If theentire plate can be considered at any time to havea uniform temperature, then the simplifiedlumped analysis can be applied with little error.The objective of such an analysis would be todetermine a natural convection coefficient whichcan then be used to make predictions in similarproblems. Measurement of the natural convectioncoefficient for heat transfer from a vertical plateis the subject of this experiment.

AnalysisA heated plate loses energy to the

environment. Under certain conditions, the plateis uniform in temperature and in this experimenthas an attached thermocouple to get temperaturereadings. Because the problem is unsteady, anoutput of temperature versus time will beobtained. In modeling this problem, we assumethat :

1. The plate temperature is uniform, whichmust be verified by calculating the Biot number.

2. The air and the surroundings are at thesame temperature, T∞.

An energy balance performed for the plate iswritten as

heat lost

by naturalconvection

+

heat lost

byradiation

=

decrease in

internal energyof plate

In equation form, we have

– hcAs(T – T∞) – εσAs(T 4 – T∞4) = – mc

–

dTd t

(11.1)

whereh c = convection coefficientAs = total surface area of plate (or

object) including edgesT = absolute temperature of plate at

any timeT∞ = absolute temperature of

surroundings, assumed equal to airtemperature

ε = emissivity of plate surfaceσ = Stefan-Boltzmann Constant

m = mass of platec = specific heat of plate materialt = time

Several simplifications can be made. First, werewrite the radiation loss in terms of a radiationcoefficient, similar to the convection coefficient;thus

qrad = εσAs(T 4 – T∞4) = hrAs(T - T∞) (11.2)

Also, the mass can be written as the product ofdensity and volume. Equation 9.1 now becomes

– ρV—c dTd t

= (hc + hr)As(T - T∞)

We define an overall coefficient h = hc + hr.Substituting into the preceding equation,separating variables and integrating gives

T – T∞Ti – T∞

= exp

–

hAstρV—c

(11.3)

The objective of this experiment is to obtaindata and calculate the convection coefficient hc.In analyzing the equation, we note that thesurface area is obtained by measuring thephysical dimensions of the plate. Thetemperature of the surroundings is measured witha thermometer or thermocouple. The emissivityof the plate surface, the specific heat, density(for calculating mass), and the Stefan BoltzmannConstant are readily found in a table ofproperties. The plate temperature as a function oftime is measured while the initially heatedplate is cooling in air. It can be measured with arecording device or with a digital thermometerwhich is read at selected time intervals.

ProcedurePlates that have been coated with a black

surface covering are available. For the oneassigned, attach the thermocouple leads to areadout device and heat the plate to over 200°F.Suspend the plate in a vertical configuration onthe stand provided. Temperature vs timereadings should be taken at regular intervalsuntil the plate reaches room temperature.Alternatively, if a recorder is used, a

37

temperature vs time graph will be produceddirectly.

Calculations and Data RepresentationFor every data point, apply Equation 11.3 to

find the overall coefficient h. Then use Equation11.2 and solve for hr. Calculate the convectioncoefficient hc, which is the difference between hand hr.

With the convection coefficient known, agraph of convection coefficient versus time couldbe constructed. Such a graph would have littlemeaning because it would apply only to theconditions of this particular experiment. Adifferent graph would be required for everyplate, for every plate surface emissivity, forevery ambient temperature, etc. It is therefore toour advantage to produce a dimensionlessrepresentation of the results. A dimensionlessgraph could apply to any configuration, not just tothe one of this experiment.

The dimensionless ratios of significance forthis experiment are the Nusselt number Nu, theRayleigh number Ra and the Prandtl number Pr:

Nu = hcLk

Ra = gβ(T – T∞)L 3

α ν

Pr = να

whereL = plate length in vertical directionk = thermal conductivity of airg = gravitational accelerationβ = 1/T∞ = coefficient of thermal

expansionα = thermal diffusivity of air = k/ρcν = kinematic viscosity of air

Using the data obtained in this experiment,calculate Nusselt number Nu, Rayleigh number

Ra and Prandtl number Pr. Graph Nusselt number(vertical axis) versus Rayleigh number using thedata. Generate an equation to fit the data; thatis, determine the constants C and n in thefollowing equation:

Nu = C Ran (11.4)

Other GeometriesThis experiment can be conducted using other

objects either heated or cooled and placed in adifferent temperature environment. Some possiblecombinations are:

1. A heated, horizontal cylinder losing heat tothe environment. Apply the Nusselt-Rayeigh equation to the data and determinethe constants.

2. A heated, vertical cylinder losing heat to theenvironment. Apply the Nusselt-Rayeighequation to the data and determine theconstants.

The preceding equations apply to an object coolingin air, but conceptually they apply equally wellto a heated object cooling by natural convection inany fluid. In addition, they could apply to acooled object being heated by natural convection.As per directions given by the instructor, performthe experiment that is assigned, whether itrequires a flat plate or a cylinder. The method ofanalysis is the same as that for the verticalplate.

QuestionsIt was assumed that the temperature of the

object is uniform at any time. What is thisassumption called and how is its use justified?

Several assumptions were made in writingEquation 11.3. What were they? Include them inthe derivation of Equation 11.3.

38

EXPERIMENT 12

NATURAL CONVECTION HEAT TRANSFER:FLAT PLATE, FINNED PLATE, AND PIN FINS

Heat transfer by natural convection occurs inmany situations, and so it is important tounderstand it and be able to successfully model it.Studying natural convection for variousgeometries is the subject of this experiment.

ApparatusConsider a heated flat plate oriented

vertically and transferring heat by only one of itssurfaces to the surrounding air. Air near the platebecomes warm and its density decreases. Buoyantforces within the air act to move this less denseair upward, and it is replaced with cooler air.The motion of the air is due to the presence of theheated plate, and so we call this a naturalconvection problem.

Figure 12.1 is a sketch of the apparatus usedin this experiment. It consists of a rectangularduct that is held in the vertical direction. Flowstraighteners are at the bottom of the duct, andan electrically operated fan is located at the top.Air flow is upward through the system. Alongthe front of the duct is a viewing window made ofclear plastic. Behind the viewing window is an

air flowdirection

duct

viewingwindow

supportstand

baseplate

opening into which aheated model can beplaced, as indicated inFigure 12.2.

In this experiment,the heated model is aflat plate. Wheninstalled, the flat plateis heated by an internalresistance heater, andthe energy to the heateris controlled by arheostat. The back andsides surrounding theheated flat plate arewell insulated so thatall energy from therheostat is transferred tothe air by the front faceof the plate. Othermodels that could beused are a finned plate,or a plate that has pinfins attached.


Finned and Flat PlateModels

viewingwindow

heatedmodel

baseplate

flowstraightenerswithin

fan andmotorwithin

FIGURE 12.2. Heated models and duct.

The apparatus is instrumented withthermocouples at selected locations. There is athermoccouple located upstream of the viewingwindow, and one placed within the heated modelthat gives a reading of the surface temperature.There is also provision for measuring the airvelocity within the duct.

ExperimentInsert the flat plate model into the duct and

turn the rheostat to a wattage of 20 W. (Notethat if the plate temperature exceeds 100°C, asafety switch will turn the heater off.) Allowthe system to reach steady state. Take readingsof ambient temperature T∞, and plate surfacetemperature Tw. Repeat this procedure forwattages of 40, 60, and 80 W. Measure thephysical dimensions of the flat plate.

ResultsConstruct a graph of power (vertical axis) as

a function of temperature difference, Tw - T∞. Theconvection equation for heat transfer from a flatplate to surrounding air is

q = hcA(Tw - T∞)

where A is the area of the plate in contact withthe air. Using this equation and the data

39

obtained, determine an average value for theconvection coefficient, hc.

With the convection coefficient known, adimensionless representation of the results can bemade. A dimensionless graph could apply to anyflat plate, not just to the one of this experiment.

The dimensionless ratios of significance forthis experiment are the Nusselt number Nu, theRayleigh number Ra and the Prandtl number Pr:

Nu = hcLk

Ra = gβ(T – T∞)L 3

α ν

Pr = να

whereL = plate length in vertical directionk = thermal conductivity of airg = gravitational accelerationβ = 1/T∞ = coefficient of thermal

expansionα = thermal diffusivity of air = k/ρcν = kinematic viscosity of air

Using the data obtained in this experiment,calculate Nusselt number Nu, Rayleigh numberRa and Prandtl number Pr. Graph Nusselt number(vertical axis) versus Rayleigh number using the

data.There are a number of equations that could be

used to relate the above parameters in thisexperiment. Here, however, we will use the datato test the validity of the following equation:

Nu = 0.68 + 0.67 Ra1/4

1 +

0.492

Pr

9/16 4/9(12.1)

Given values of Rayleigh and Prandtl numbers,the Nusselt number can be calculated withEquation 12.1. Using the calculated Nusseltnumber, determine the convection coefficient, andcompare the results to those obtained byexperiment. Calculate the percent error.

Other GeometriesThis experiment can be repeated with a flat

plate that has longitudinal fins attached, asshown in Figure 12.2, or with a flat plate thathas pin fins attached. Perform this experimentfor the heated model assigned. (Note: For theseshapes, a comparison of results with an equationlike 12.1 is not possible, because such equationsare unavailable for plate fins and pin fins.)

40

EXPERIMENT 13

FORCED CONVECTION HEAT TRANSFERFROM A FLAT PLATE TO AIR

Heat transfer by forced convection is acommon problem, and so it is important to be ableto successfully model it. Modeling forcedconvection for a flat plate transferring heat to airis the subject of this experiment.

ApparatusConsider a heated flat plate oriented

vertically and transferring heat by only one of itssurfaces to the surrounding air, which is beingforced past the plate by a fan. Air near the platereceives energy and is moved away.

Figure 13.1 is a sketch of the apparatus usedin this experiment. It consists of a rectangularduct that is held in the vertical direction. Flowstraighteners are at the bottom of the duct, andan electrically operated fan is located at the top.

air flowdirection

duct

viewingwindow

supportstand

baseplate

Air flow is upwardthrough the system.Along the front of theduct is a viewingwindow made of clearplastic. Behind theviewing window isanopening into which aheated flat plate modelcan be placed, asindicated in Figure 13.2.

In this experiment,the flat plate is heatedby an internal resistanceheater, and the energyto the heater iscontrolled by a rheostat.The back and sides ofthe flat plate housingare well insulated sothat all energy from therheostat is transferred tothe air by the front ofthe plate.


The apparatus is instrumented withthermocouples at selected locations. There is athermoccouple located upstream of the viewingwindow, and one placed inside the flat platehousing that gives a reading of the surface

Flat Plate Model

viewingwindow

heatedmodel

baseplate

flowstraightenerswithin

fan andmotorwithin

FIGURE 13.2. Heated model and duct.

temperature. There is also provision formeasuring the air velocity within the duct.

ExperimentInsert the flat plate model into the duct and

turn the rheostat to a preselected wattage, say 20W. (Note that if the plate temperature exceeds100°C, a safety switch will turn the heater off.)Turn the fan on and use the controller to get an airvelocity of 0.5 m/s. Allow the system to reachsteady state. Take readings of ambienttemperature T∞, and plate surface temperatureTw. Repeat this procedure for air velocities of 1.0,and 1.5 m/s. Measure the physical dimensions ofthe flat plate.

Next, change the wattage of the heater toanother value, and repeat the experiment for airvelocities of 0.5, 1.0, and 1.5 m/s. Repeat theexperiment again for a third heater wattage.

ResultsConstruct a graph of velocity (vertical axis)

as a function of temperature difference, Tw - T∞,and label each line according to the wattageselected. The convection equation for heattransfer from a flat plate to surrounding air is

q = hcA(Tw - T∞)

where A is the area of the plate in contact with

41

the air. Using this equation and the dataobtained, determine an average value for theconvection coefficient, hc for each wattage.

With the convection coefficient known, adimensionless representation of the results can bemade. A dimensionless graph could apply to anyflat plate, not just to the one of this experiment.

The dimensionless ratios of significance forthis experiment are the Nusselt number Nu, theReynolds number Re and the Prandtl number Pr:

Nu = hcLk

Ra = ρVLµgc

Pr = να

whereL = plate length in vertical directionk = thermal conductivity of airρ = air density

V = air velocityµ = absolute viscosity of airα = thermal diffusivity of air = k/ρcν = kinematic viscosity of air

Using the data obtained in this experiment,calculate Nusselt number Nu, Reynolds number Reand Prandtl number Pr. Graph Nusselt number(vertical axis) versus Reynolds number using thedata.

There are a number of equations that could beused to relate the above parameters in thisexperiment. Consult a heat transfer textbook andselect an equation that fits the conditions of thisexperiment. Given values of Reynolds andPrandtl numbers, the Nusselt number can becalculated with the equation that you haveselected. For the same Reynolds and Prandtlnumbers, the Nusselt number was also determinedwith the data. Using these two values of Nusseltnumber, calculate the percent error.

Other GeometriesThis experiment can be repeated with a flat

plate that has longitudinal fins attached, orwith a flat plate that has pin fins attached.Perform this experiment for the heated modelassigned. (Note: For these other shapes, acomparison of results with a textbook equationmight not possible, because such equations areunavailable for fins and pin fins.)

42

EXPERIMENT 14

FORCED CONVECTION IN PIPE FLOW

Heat transfer to a fluid flowing in a pipe isan important problem to consider, because thereare many applications in which this method isused to warm (or cool) a fluid. A double pipe heatexchanger, for example, is a device that usesforced convection in pipe flow to effect a heatexchange. So it is helpful to understand thisrocess.

Figure 14.1 is a schematic of the apparatusused in this experiment. Water flows through avariable area meter, and then through either oftwo tubes. One tube is chrome plated, and theother is blackened. Ball valves are used to directthe water through the tube of interest. The twotubes are housed in a “box,” insulated on top,sides, and back. The box has a transparent glassfront. After flowing through a tube in the box, thewater then exits to a drain. For this experiment,water should be made to flow through theblackened tube.

Steam is admitted to the box, and as itcondenses on the surfaces inside, it heats thewater in the tube.

ExperimentWith both the steam and the water made to

flow through the system, record all temperatures,and the flow rate of the water through theblackened tube. Repeat this procedure for threedifferent water flow rates, all at a constant flowrate of steam. Next, re-direct the water flowthrough the chromed tube at the same threewater flow rates, and record T1 and T3. For allcases, make sure that the temperature increase isat least 10°F for water flowing through eithertube.

AnalysisFor forced convection heat transfer through a

tube, there are two general solutions that apply.The first of these is the constant wall fluxproblem, and the second is the constant walltemperature problem. These two models areconsidered extremes, and all practical problemsof interest will be described by a model that fallsbetween them. In this experiment, we will assumethat the constant wall temperature case exists(although not strictly accurate) and makecalculations accordingly.

General Equations. Consider the flow of waterthrough a tube that is being heated from anexternal source. The energy absorbed by the wateris given by

q = m· Cp(Tin - Tout) (14.1)

For a constant wall temperature Tw, we can alsowrite an equation for the heat transfer rate interms of an average convection coefficient as

q = h–

As [(Tw - Tout) – (Tw – Tin)]ln[(Tw – Tout)/(Tw – Tin)] (14.2)

where the surface area As is πDL for a circularduct, and the temperature term is commonlyknown as the log-mean temperature difference.An equation that relates temperatures to averagevelocity is

Tw – Tout

Tw – Tin = exp

– 2αL h

–

VkfR(14.3)

water in water out

flowmeter

steam in

condensateout

ball valve

boxthermocoupleT1 T2 T3

T4

T5

T6T8T7

T1 = inlet temperature chrome tubeT2 = steam temperatureT3 = outlet temperature chrome tubeT4 = surface temperature black tubeT5 = water temperature black tubeT6 = outlet temperature black tubeT7 = inlet temperature black tubeT8 = surface temperature black pipe

FIGURE 14.1. Schematic of the apparatus used to study forced convection in pipe flow

43

where L is the heated length of tube, V is theaverage water velocity, and R is the rube radius.Equation 14.3 is obtained by combining Equations14.1 and 14.2.

The objective in these calculations is tocalculate the Nusselt number for flow in the tube,and graph it as a function of the inverse Graetznumber, and ultimately compare the results tothose that have been published. Substitute thedata obtained in the experiment into Equation14.3 and determine the convection coefficient. Ifthe wall temperature is not constant (as permeasurements obtained), then use the arithmeticaverage of the wall temperatures as Tw. Forwater properties, determine the average of Tinand Tout and interpolate. The average velocity Vis found by dividing cross sectional area into flowrate. With the convection coefficient known,calculate the Nusselt number for the dataobtained:

Nu = h–

Dk f

The inverse Graetz number is defined as

1Gz =

LD Re Pr

where Re (= VD/ν) is the Reynolds number, andPr (= ν/α) is the Prandtl number. Calculate theinverse Graetz number for the data.

Construct a semi-log graph of Nusselt numberversus inverse Graetz number. Compare results tothose found in textbooks noting that fullydeveloped conditions probably do not exist in thisexperiment.

Empirical Equations. Based on experimentalwork, a number of empirical equations have beenwritten to model forced convection heat transferin a duct, for both laminar and turbulent flow.Some of the many equations developed areprovided in the following list. Use theappropriate equation (one or more) to calculatethe Nusselt number, and compare the result tothat which was calculated from the data.Comment on the results.

Seider-Tate Equation for laminar flowConstant wall temperature

Nu = 1.86

D Re Pr

L

1/3

µ

µw

0.14

where µw is the viscosity of the fluid evaluatedat the wall temperature, and the remainingproperties are all evaluated at the average ofinlet and outlet temperatures. Other conditions,

0.48 < Pr < 16 7000.004 4 < (µ/µw) < 9.75Re ≤ 2100

Hausen Equation for laminar flowConstant wall temperature

Nu = 3.66 + 0.0668(D/L) Re Pr

1 + 0.4[(D/L) Re Pr]2/3

Seider-Tate Equation for turbulent flowConstant wall flux or constant wall temperature

Nu = 0.027 Re4/5 Pr1/3 (µ/µw)0.14

0.7 ≤ Pr ≤ 16 700Re ≥ 10 000L/D ≥ 60

Dittus-Boelter Equation for turbulent flowConstant wall flux or constant wall temperature

Nu = 0.023 Re4/5 Prn

n = 0.4 if Tw > Tfluidn = 0.3 if Tw < Tfluid0.7 ≤ Pr ≤ 160Re ≥ 10 000L/D ≥ 60

44

EXPERIMENT 15

ANALYSIS OF A DOUBLE PIPE HEAT EXCHANGER

A heat exchanger is a device used to transferheat from one fluid to another. There are manydifferent types of heat exchangers includingshell-and-tube, cross flow, and double pipe. Theanalysis of a double pipe heat exchanger is thesubject of this experiment.

The Double Pipe Heat ExchangerA double pipe heat exchanger consists of two

concentric, different diameter tubes with fluidflowing in each as indicated in Figures 15.1 and15.2. If the two fluids travel in oppositedirections as illustrated in Figure 15.1, theexchanger is a counter flow type. If the fluidstravel in the same direction as shown in Figure15.2, parallel flow exists. The same apparatus isused for either flow configuration. The objectivein using a heat exchanger is to be able to transferas much heat as possible for as small a cost asnecessary. In sizing or selecting a particularexchanger, all that will be known are:

• The tube sizes and areas (surface and crosssectional areas)

• The physical properties of the fluids• The inlet temperatures of the fluids

In order to predict the amount of heat that willbe exchanged, it is necessary to know the outlettemperatures of both fluids. Predicting outlettemperatures involves an in depth analysis ifthose parameters above are all that are known.Presenting an entire analysis here is too lengthy.

Consequently in this experiment, the equationsare given but not derived. Following now is ananalysis for double pipe heat exchangers inwhich the outlet temperatures are to be found.Calculations are presented in a suggested formatand order for a specific example problem.

Analysis of Double Pipe Heat Exchangerswith Outlet Temperatures Unknown

Problem. Water at a temperature of 195°F and amass flow rate of 5000 lbm/hr is to be used to heatethylene glycol. The ethylene glycol is availableat 85°F with a mass flow rate of 12,000 lbm/hr. Adouble pipe heat exchanger consisting of a 11/4

standard type M copper tubing inside of 2standard type M copper tubing. The exchanger is 6ft long. Determine the outlet temperature of bothfluids using counterflow and again using parallelflow.

Discussion. Water loses energy only to theethylene glycol and as heat is transferred, fluidproperties change with temperature changes.Outlet temperatures are unknown so in order toevaluate properties, we use either the inlettemperatures or the average of both inlettemperatures. The fluid with the higher flowrate should be placed in the passage (annular ortubular) having the greater cross sectional area.In this way, pressure losses are minimized.

tube fluid

outlet

annulusfluid inlet

annulusfluid outlet

tube fluid

inlet t1

t2

T1 T2

L

t2

t1

T1

T2

length or distance

tem

per

atu

re

FIGURE 15.1. A double pipe heat exchanger setup in counterflow and the correspondingtemperature profile.

L

tube fluid

outlet

annulusfluid inlet

annulusfluid outlet

tube fluid

inlet t1

t2

T1

T2

t1 t2

T1

T2

length or distance

tem

pera

ture

FIGURE 15.2. A double pipe heat exchanger setup in parallel flow and the correspondingtemperature profile.

45

Assumptions1. Steady state conditions exist.2. Fluid properties remain constant and are evaluated at 140°F [= (195 + 85)/2].

Nomenclature 1. T refers to the temperature of the warmer fluid.2. t refers to the temperature of the cooler fluid.3. h subscript refers to the warmer fluid, or hydraulic diameter4. c subscript refers to the cooler fluid.5. a subscript refers to the annular flow area or dimension.6. p subscript refers to the tubular flow area or dimension.7. 1 subscript refers to an inlet condition.8. 2 subscript refers to an outlet condition.9. e subscript refers to equivalent diameter.

Fluid Properties

Water m· h = 5000 lbm/hr T1 = 195°F140°F ρ = 0.985(62.4) lbm/ft3 Cp = 0.9994 BTU/lbm·°R

k f = 0.376 BTU/hr·ft·°R α = 6.02 x 10-3 ft2/hrν = 0.514 x 10-5 ft2/s Pr = 3.02

Ethylene m· c = 12,000 lbm/hr t1 = 85°FGlycol ρ = 1.087(62.4) lbm/ft3 Cp = 0.612 BTU/lbm·°R140°F k f = 0.150 BTU/hr·ft·°R α = 3.61 x 10-3 ft2/hr

ν = 5.11 x 10-5 ft2/s Pr = 51

Tubing Sizes (See Figure 8.3 for a definition of these diameters.)2 std M IDa = 0.1674 ft

11/4 std M IDp = 0.1076 ft ODp = 1.375/12 = 0.1146 ft

Flow Areas Ap = πIDp2/4 = 0.00909 ft2

Aa = π(IDa2 - ODp

2)/4 = 0.0117 ft2

Fluid Velocities [Because Aa > Ap, we route the ethylene glycol (the fluid with the higher flow rate)through the annulus.]

Water Vp = m· h/ρAp = 2.48 ft/s

Ethylene Va = m· c/ρAa = 4.20 ft/sGlycol

Annulus Equivalent DiametersFriction Dh = IDa – ODp = 0.0528 ft

Ht Transfer De = (IDa2 – ODp

2)/ODp = 0.1299 ft

Reynolds NumbersWater Rep = VpIDp/ν = 5.2 x 104

Ethylene Rea = VaDe/ν = 1.07 x 104

Glycol

Nusselt Numbers (The recommended expressions are listed at the end.)Water Nup = 0.023(Rep)4/5 Pr0.3 = 190

Ethylene Nua = 0.023(Rea)4/5 Pr0.4 = 185Glycol

46

IDa/2

IDp/2ODp/2

R12 R23 R34

T t

r

T

L

FIGURE 15.3. Definition sketch of the diameters associated witha double pipe heat exchanger cross section.

Convection CoefficientsWater hp = Nup kf/ODp = 623 BTU/hr·ft2·°R

Ethylene ha = Nua kf/De = 214 BTU/hr·ft2·°RGlycol

Exchanger Coefficient1

Uo =

1h p

+ 1h a

Uo = 159 BTU/hr·ft2·°R

Outlet Temperature Calculations (Exchanger length L = 6 ft)

R = m· cCpc

m· hCph

= 1.47 Ao = πODpL = 2.16 ft2

Counterflow T2 = T1(R – 1) – Rt1[1 – exp(UoAo(R – 1)/m· cCpc)]

R exp[UoAo(R 1)/m· cCpc] – 1

t2 = t1 + T1 – T2

R

Parallel Flow T2 = {R + exp[UoAo(R + 1)/m· cCpc]}T1 + Rt1{exp[UoAo(R + 1)/m· cCpc] – 1}

(R + 1)exp[UoAo(R + 1)/m· cCpc]

t2 = t1 + T1 – T2

R

Water T2 = 188°F CounterflowEthylene

Glycol t2 = 90°F Counterflow

Water T2 = 188°F Parallel FlowEthylene

Glycol t2 = 90°F Parallel Flow

Heat Balance (as a check on the results)

Water qh = m· hCph(T1 – T2) = 3.5 x 104 BTU/hr

Ethylene qc = m· cCpc(t2 – t1) = 3.67 x 104 BTU/hr (discrepancy due to roundoff error)Glycol

47

The results show that no difference existsbetween parallel flow and counterflow for thisexample. This is not always the case, however.Counterflow is usually the preferred flowconfiguration.

Experimental MethodFigure 15.4 is a schematic of the apparatus

used in this experiment. Two double pipe heatexchangers are arranged in a U-turn setup (calleda two pass exchanger) mounted vertically on aboard. Thermocouples are located at variousplaces in the tubes and in the annular flowpassages. Of interest here, however, are the inlettemperatures of both streams. Variable areameters are located in each flow line. Water is thefluid in the tube and in the annulus. The warmerfluid stream is heated with an electric heater justbefore the inlet to the exchangers. Valves arepresent in the system (not shown in the figure) inorder to set up counterflow or parallel flow. Thevalve arrangement gives control over the fluidflow direction in the annuli.

tube fluidoutlet

annulusfluid inlet

annulusfluid outlet

tube fluidinlet

t1

t2T1

ti Ti

T2


For one predetermined flow rate of thewarmer and one of the cooler fluid, establish asteady state operation of the heat exchangers incounterflow. Take all temperature readings andflow rate readings. Next, reverse the flowdirection of one of the fluids and establish steadystate again. Make sure that the flow rates inparallel flow are identical to those used inparallel flow so that a comparison can be made.

Construct graphs of temperature versus axialdistance for both double pipe heat exchangers

when each operates in counterflow and inparallel flow (2 graphs). Prepare an analysislike that given above for both configurations,using as total length the length of bothexchangers. Compare the outlet temperaturepredicted by the analysis to that measured on thedevice. Identify sources of error.

Nusselt Number EquationsSeider-Tate Equation for laminar flow:

Nu = hDk f

= 1.86

D Re Pr

L

1/3

Re = VDν < 2 200

D = IDp if cross section is tubularD = De if cross section is annular

0.48 < Pr = να < 16 700

µ changes moderately with temperatureProperties evaluated at the average fluidtemperature [ = (inlet + outlet)/2]

Modified Dittus-Boelter Equation forturbulent flow:

Nu = hDk f

= 0.023(Re)4/5 Prn

n = 0.4 if fluid is being heatedn = 0.3 if fluid is being cooled

Re = VDν ≥ 10 000

D = IDp if cross section is tubularD = De if cross section is annular

0.7 ≤ Pr = να ≥ 160

L/D ≥ 60

Properties evaluated at the average fluidtemperature [ = (inlet + outlet)/2]

48

EXPERIMENT 16

PLATE AND FRAME HEAT EXCHANGER

A heat exchanger is a device used totransfer heat from one fluid to another. Thereare many different types of heat exchangersincluding shell-and-tube, cross flow, and doublepipe. The analysis of a shell and tube heatexchanger, and of a plate and frame heatexchanger, is the subject of this experiment.

Figure 16.1 shows the apparatus used inthis experiment. The device contains a watersupply tank, a pump-motor combination, tworotameters, two heat exchangers, and the

associated tubing with fittings.There are three fluid streams flowing

through the apparatus. Steam produced by asteam generator is piped to the shell of a 1-2shell and tube heat exchanger. Steam condenseson the outside surface of the tubes, and isdischarged into a drain. Another fluid streamis tap water, which is passed through theplate and frame heat exchanger, and then isdischarged to a drain.

A

B

CD

E

F

motor

pumpsupply tank

rotameter

tap water in

tap water out(to drain)

steam in

condensate out

plate and frameheat exchanger

shell and tubeheat exchanger

return line

pump inlet line

FIGURE 16.1. Schematic of the heat exchanger apparatus.(Notes: pump inlet line is 3/4 nominal schedule 80 pipe;pump outlet line is 1/2 standard copper tubing.)

Tap WaterWater (recirculating)From steam generatorThermocouple

49

The third fluid stream is water stored in thesupply tank, which is pumped throughout thesystem. It is first routed through the shell andtube heat exchanger, where the water is heatedby the condensing steam. The heated water isthen routed through the plate and frame heatexchanger where it transfers heat to the tapwater.

There are two experiments that can beperformed with this apparatus. One experimentis the analysis of the plate and frame heatexchanger in which published equations are usedin an effort to predict the outlet temperatures ofboth fluid streams under suitable operatingconditions. The second is similar to the first, butperformed for the shell and tube heat exchanger.Both experiments can be performedsimultaneously. All fluid flows are establishedfor predetermined conditions, and temperaturesand flow rates are recorded. Data are then usedwith the mathematical models to predict outlettemperatures.

Procedure• Turn the steam generator on to its maximum

setting. The valve at B should be fully open.The valve at D should be closed and thevalve at C should be open.

• Turn the pump on and adjust the valve at A sothat the flow rate reading on the manometerdoes not exceed the maximum reading. Thevalve at F should be closed.

• Turn the floor valve on to allow tap water tocirculate through the plate and frame heatexchanger. The flow rate of tap water shouldnot exceed the maximum on the rotameter.The valve at E should be fully open. The flowrate of tap water is to be controlled by thefloor valve.

• Allow the system to reach steady state.Monitor the progress toward steady state byobserving the temperature of anythermocouple.

• When steady state is achieved, record allfluid temperatures, and both flow ratereadings.

The Plate and Frame Heat ExchangerA plate and frame heat exchanger consists of

several metal sheets with corrugated surfacesthat are clamped together. Figure 16.2 shows afrontal view of a plate having a herringbonepattern; other patterns are also in use (such as onereferred to as a “washboard”). In the herringbonepattern, the angle made between adjacent ribsand the vertical is called the chevron angle, θ.Plates can be made with small chevron angles

(low θ plates), or large angles (high θ plates).Performance of the heat exchanger is a function ofthe chevron angle. High θ plates provide highheat transfer rates with high pressures losses.The converse is also true.

frontal view

profile view ofseveral plates

warmerfluid

coolerfluid

herringbonepattern

gasketmaterial

t s

b

L

FIGURE 16.2. Frontal view of one plate, andprofile views of several plates showing acountercurrent flow configuration.(Countercurrent flow is represented.)

Plates have rubber gaskets glued to them inthe pattern shown in the figure, or in somesimilar way. Also shown in Figure 16.2 is aprofileview that indicates how the two fluids flowabout adjacent plates. Figure 16.3 shows how theplates are arranged and how two fluids arerouted as they pass through the heat exchangeritself. As indicated in these two figures, eachsheet separates the cold and warm fluids whichcan be made to flow in either a countercurrent(Figure 16.2) or a parallel flow pattern (Figure16.3).

Plate exchangers are well suited for liquid-liquid flows under turbulent conditions. Theseexchangers can also be used as condensing unitsoperating at moderate flow rates and pressures(to 60 psia). Large volume flow rates in acondenser application, however, are betterhandled by shell and tube heat exchangers. Thegreatest advantages of a plate and frameexchanger is that it can be modified readily, andis small in size. If an existing exchanger can nottransfer the required amount of heat, forexample, one or more plates can be added easilyon an as-needed basis.

50

FIGURE 16.3. Flow through aplate & frame heatexchanger. (Parallelflow is represented.)

coolerliquidinlet

seals arranged differently onadjacent plates to direct the flow

coolerliquidoutlet

warmerliquidoutlet

warmerliquidinlet

end plateis sealed

Plate Construction and MaterialsThe surface of each plate is important

because it is the heat transfer area. Corrugatingthe plates imparts a certain degree of stiffnessand provides contact points between adjacentplates when they are clamped together.Furthermore, a dimpled or corrugated surface oneach plate causes turbulent mixing to occur thatwill enhance the heat transferred. Platethicknesses can be as small as 0.6 mm (0.024 in.).

GasketsThere are at least two gaskets that separate

the fluids. If there is a gasket failure, theleaking fluid is discharged to the atmosphereand the two fluids seldom mix. Gasket materialsinclude natural rubber styrene, resin cured nitrile,and silicone and butyl rubbers. Neoprene andcompressed asbestos are also used.

FramesThe sheet metal plates are clamped together

with nuts and long bolts in a frame that containspipe connections for both fluids. Frames areusually free standing, and because of the waythey are constructed, the entire heat exchangercan be taken apart in a very short time. Framesare usually made of carbon steel, painted orcoated to protect against corrosion. Connections

are usually made of the same material as theplates in order to prevent electrochemicalproblems.

Analysis of Plate and Frame Heat Exchangerswith Outlet Temperatures Unknown

The analysis of plate and frame heatexchangers follows the same format as that forthe double pipe heat exchanger. Following is asuggested order of calculations for this heatexchanger, including recommended equations.Specifications for the heat exchanger of thisexperiment appear within the calculation sheets.

Assumptions1. Steady state conditions exist.2. Fluid properties remain constant and are

evaluated at the average of inlet and outlettemperatures for both fluid streams.

Nomenclature1. T refers to the temperature of the warmer

fluid.2. t refers to the temperature of the cooler fluid.3. w subscript refers to the warmer fluid.4. h subscript refers to hydraulic diameter.5. c subscript refers to the cooler fluid.6. 1 subscript refers to an inlet condition.7. 2 subscript refers to an outlet condition.

51

Tap WaterWater (recirculating)Thermocouple

tap water out(to drain)

plate and frameheat exchanger

return line

warm water inlet

cool water inlet

T1

T2t2

t1

FIGURE 16.4. Schematic of flow parallel flow through a plate & frame heat exchanger.

A. Fluid Properties·mw = T1 =ρ = Cp =k f = α =ν = Pr =

·mc = t1 =ρ = Cp =k f = α =ν = Pr =

B. Plate Dimensions and Propertiesb = plate width = 0.1 mL = plate height = 0.35 ms = plate spacing = 0.004 8 mt = plate thickness = 0.001 mAo = plate surface area = bL =A = flow area = sb =Dh = hydraulic diameter of flow passage = 2s =Ns = number of plates = 19k = thermal conductivity of plate = 14.3 W/(m·K)plate material = stainless steel

C. Fluid Velocities

Odd number of plates

V = ·m/ρA

(Ns + 1)/2 for both fluids

Vw =

Vc =

D. Reynolds NumbersRew = VwDh/ν =

Rec = Vc Dh/ν =

52

E. Nusselt Numbers

Modified Sieder-Tate Equation for laminar flow:

Nu = hDh

k f = 1.86

DhRe Pr

L

1/3

Re < 100 0.48 < Pr = ν/α < 16 700

Modified Dittus-Boelter Equation for turbulent flow:

Nu = hDh

k f = 0.374 Re0.668 Pr1/3

Re > 100; Pr = ν/α > 0;

Conditions:µ changes moderately with temperatureProperties evaluated at the average fluid

temperature [= (inlet + outlet)/2]

Nuw =

Nuc =

F. Convection Coefficientshi = Nuwkf/Dh =

ho = Nuckf/Dh =

G. Exchanger Coefficient1

Uo =

1h i

+ tk

+ 1h o

Uo =

H. Capacitances

(m· Cp)w =

(m· Cp)c =

(m· Cp)min =

I. Number of Transfer Units, NTU, and Correction Factor

N = UoAoNs

(m· Cp)min

= F =1 - 0.016 6 N =

J. Outlet Temperature Calculations

R = ·mcCpc

·mwCpw

= Ecounter = exp[UoAoNsF(R - 1)/m· cCpc] =

T2 = T1(R - 1) - Rt1(1 - Ecounter)

REcounter - 1 =

53

t2 = (T1 - T2)/R + t1 =

K. Log Mean Temperature Difference

Counterflow LMTD = (T1 - t2) - (T2 - t1)

ln [(T1 - t2)/(T2 - t1)] =

L. Heat Balance for Fluids and for the Exchanger

qw = ·mwCpw (T1 - T2) = qc = ·mcCpc (t2 - t1) =

q = UoAoNsF LMTD =

M. Pressure Drop Calculations

Darcy-WeisbachReynolds number range Friction factor

1–10 f = 280Re

10–100 f = 100Re0.589

> 100 f = 12Re0.183

fw = fc =

∆pw = fwLDh

ρωVw

2

2gc + 1.3

ρwVp2

2gc =

∆pc = fcLDh

ρcVc

2

2gc + 1.3

ρcVp2

2gc =

N. Comparison of Calculated to Measured Outlet Temperatures

Measured Calculated % ErrorT2 =t2 =

Experimental MethodA plate and frame heat exchanger is

mounted on a support system containing a tankof water with immersion heaters and a pump.The pump moves heated water through theheat exchanger. (There are controls for theheaters.) Cold tap water is also routed throughthe exchanger. Thermocouples are located atthe inlet and outlet of each fluid stream.Variable area meters are located in each flowline also. Valves are present in the system inorder to control the flow rates

For one predetermined flow rate of thewarmer and one of the cooler fluid, establish a

steady state operation of the heat exchangersin countercurrent flow. Take all temperaturereadings and flow rate readings. Next, reversethe flow direction of one of the fluids andestablish steady state again. Make sure thatthe flow rates in parallel flow are identical tothose used in counter flow so that a comparisoncan be made.

Prepare an analysis like that given abovefor both configurations. Compare the outlettemperature predicted by the analysis to thatmeasured on the device. Identify sources oferror.

54

EXPERIMENT 19

RADIATION HEAT TRANSFER

Conduction and convection are heat transfermechanisms involving a material mediumthrough which energy travels. However, energycan also be transferred through a region in whicha perfect vacuum exists. This mechanism or modeof energy transfer is commonly calledelectromagnetic radiation. All radiation ispropagated at the speed of light in a vacuum.

The many types of electromagnetic radiationinclude X rays, gamma rays, the visible spectrum,radio waves, thermal radiation, andmicrowaves. Our concern is soley with thethermal component. Net heat transfer byradiation is a result of a temperature difference,but it should be remembered that radiation

energy propagates from a body (a source) with orwithout the presence of a second body (orreceiver).

ApparatusThe apparatus used in this experiment is

manufactured specifically for making radiationmeasurements, and is sketched in Figure 19.1. Itconsists of several pieces. One is a track and scaleonto which various items can be attached.Another is an instrument controller, whichcontains a rheostat to control heat input to theheat source, and digital readout instruments. Inthis experiment, we will use only the heat source,and not the light source.

instrument consoleand controller

heatertransformer heat source light source

track and scale

FIGURE 19.1. A sketch of the apparatus used in this experiment.

RadiometerFigure 19.2 shows the same apparatus with a

radiometer in place. A radiometer is used tomeasure radiant energy emitted from a heatedsurface. A device such as this can made of anarray of thermocouples connected in series to givea high voltage output, compared to the output ofonly one thermocouple. (The thermocouples couldalso be connected in parallel to give a highercurrent output.) A radiometer is any of a broadclass of temperature measuring devices, includingthermocouples (as used here), thermistors, andresistance pyrometers.

heat source

track and scale

radiometer

L

FIGURE 19.2. Sketch of the apparatus with aradiometer.

Several words of caution regarding theradiometer are appropriate. It is a sensitiveinstrument, and a cover is supplied for its openend to keep out dust and dirt. Objects should not beplaced in the open end so that the finish and theshield will not become scratched or damaged.

Because of its sensitivity, the performance ofthe radiometer can be significantly affected bylocal heat sources or air currents about its openend. If the radiometer is exposed for long periodsof time to a radiant heat source, a zero shift in itsreadout will occur. So between readings, thepolished disc is to be placed over the open end.The radiometer reads directly in W/m2.

Using the Sensor. The technique for using theradiometer or sensor to obtain readings from theheat source is to allow the heater to reach asteady temperature with the radiometer inposition, and the polished disc in place. Whensteady state has been reached, the polished discis removed without touching the sensor ifpossible. The reading will increase rapidly,

55

eventually reaching a steady value. The elapsedtime could take up to two minutes. Once a readinghas been taken, replace the polished disc overthe open end to prevent excessive heating of thesensor, and subsequent erroneous readings.

The radiometer is mounted on a post that inturn is attached to a slide mechanism, whichallows the radiometer to be moved close to oraway from the heat source. The slide mechanismis scribed with a line that is used with the trackscale to measure position. The sensor is 2 cm closerto the heat source than the scribed line indicates.Thus all length readings taken from the scalemust have 2 cm subtracted from them.

Incident and Emitted RadiationThe digital readout indicates the intensity of

radiation received by the radiometer in W/m2.This value is different from that which wasemitted by the heat source at which theradiometer was aimed. Therefore, readings mustbe modified to account for this difference.

We will use the concept of configuration orview factor to make corrections. The heat sourceitself is a flat plate, and the radiometer sensor ismuch smaller in area. The actual setup betweenthe heat source and sensor is sketched in Figure19.3, in which the sensor is denoted as dA1.

The view factor is defined as the fraction ofenergy that leaves the heat source that isintercepted by the sensor. So a reading at thesensor (dA1 in Figure 19.3) must be modified inorder to determine the heat that is emitted bythe source. We do not have results for the systemof Figure 19.3, but we can obtain them from a viewfactor geometery that we do have. Figure 19.4shows a view factor between a plane A2 andasmall area dA1. The small area is aligned with

dA1

FIGURE 19.3. Actual setup between heat sourceand sensor.

one of the corners of the plane. We can apply thisresult 4 times to the system of Figure 19.3 toobtain the view factor for the apparatus of thisexperiment. This number will be needed for allexperiments in which the radiometer is used. Wehave by definition

qincident = qemitted • AFd12

so that the energy emitted at the plate surface is

qemitted

A =

radiometer

reading • 1

Fd12(19.1)

where A is the area of the emitter.

ExperimentsThere are two parts to this experiment:

verification of the inverse square law; and,verification of the Stefan-Boltzmann law.

dA1

A2

a

c

b

Defining x = a/c, y = b/c, we have

Fd12 = 1

2π

x

√1 + x2 tan-1

y

√1 + x2 +

y

√1 + y2 tan-1

x

√1 + y2

FIGURE 19.4. View factor between a differential area and aplane.

56

Inverse Square LawThe transfer of radiant energy is a function of

the distance between source and receiver. Theenergy received is inversely proportional to thesquare of the distance between the two. This isknown as the Inverse Square Law, which will beverified in this experiment.

This experiment is conducted with the heatsource (a flat plate) and the radiometer. Figure19.2 shows the setup. The objective is to showthat the intensity of radiation on a surface isinversely proportional to the square of thedistance between the source and the receiver:

q” ∝ 1L2 (19.2)

where q” is the energy per unit area.The procedure is as follows:

• Turn the power selector switch to some value.• Set the radiometer anywhere on the scale.• Allow approximately 15 minutes for steady

state to be reached.• Remove the cover from the sensor and allow

the reading to stabilize.• Record the incident radiation and the

distance between the source and the sensor.• Repeat the procedure for other distances, all

taken at the same power setting.

Results. Note that the length measurementsmust be corrected for this part of the experiment,but the reading from the radiometer must nothave the view factor applied. Construct a graphof intensity of radiation q” versus distance L. Isthis graph linear? Construct also a log-log graphof q” versus D. Is this graph linear, and if so whatis its slope? Has Equation 19.2 been verified?

Using the same data, calculations can bemade to determine the validity of the viewfactor calculations. The length measurementsmust be corrected, and the reading from theradiometer must have the view factor equationapplied. With these calculations, determine thevalue of the emitted radiation for every distanceused. Should the emitted radiation be the samefor all distances?

Stefan-Boltzmann LawIn order to model radiation heat transfer, it is

useful to define a surface or substance that emitsradiation ideally. An ideal radiator is known asa black body, and it will emit energy at a ratethat is proportional to the fourth power of itsabsolute temperature:

qA

∝ T4

With a proportionality constant, the precedingequation becomes

q = σAT4 (19.3)

where σ is called the Stefan-Boltzmann constant,and has the value of

σ = 5.67 x 10-8 W/(m2·K4)σ = 0.1714 x 10-8 BTU/(hr·ft2·°R4)

Temperature in Equation 19.3 must be expressed inabsolute units.

In the case of most real materials, surfaces donot emit electromagnetic radiation ideally; thatis, they are not ideal radiators as black bodiesare. To account for the “gray” nature of realsurfaces, we introduce a dimensionless factor ε,called the emissivity. Emissivity values forvarious surfaces varies with finish (e.g.,polished versus painted). Modifying Equation19.3 to account for gray-body behavior yields

q = σAεT4

The emissivity is 1 for an ideal radiator, and lessthan 1 for a gray surface. This equation is knownas the Stefan-Boltzmann Law.

When a real body exchanges heat byradiation only with a black body, the net heatexchanged is proportional to the difference in T4.For such a system, the net heat exchanged is

q = σA1ε1(T14 – T2

4) (19.4)

where the “1” subscript refers to the source, and“2” refers to the receiver. The objective of thisexperiment is to verify Equation 19.4; i.e., thatheat transferred by radiation varies with adifference in the fourth power of temperature.

The setup for this part of the experiment isshown in Figure 19.5. A black plate is located onthe track/scale at a distance L1 from the source.The distance from the radiometer to the blackplate is L2. The heater radiates in all directions,and some of this energy is absorbed by the blackplate. The black plate radiates in all directions,and the radiometer intercepts a fraction of thisenergy.

• Let L1 = 5 cm, and L2 ≥ 4L1.• Record temperature of the black plate and

radiometer reading for ambient conditions.

57

• Set the power control to its maximum value.• When conditions become steady, record black

plate temperature and radiometer reading.• Move the power control to another setting and

repeat the procedure to obtain 4 more sets ofreadings.

• Install the polished disc over the sensorbetween readings.

heat source

track and scale

radiometer

L1 L2

blackplate

FIGURE 19.5. Setup for verification of theStefan-Boltzmann Law.

Results. For the conditions of this experiment,Equation 19.4 applies to the black plate, where ε1= 1.0, and T2 is the ambient temperature. Equation19.4 is solved for intensity to give

qA

= σ(T14 – T2

4) (19.5)

Use the temperatures obtained with Equation19.5 to calculate values of q/A. The results givevalues of the energy emitted by the black plate.

Make appropriate corrections to the readingsfrom the radiometer, and compare the results tothose obtained with Equation 19.5. Calculate %errors and account for all error sources.

Referring to Equation 19.4, and substitutingthe Stefan-Boltzmann constant in SI units we get

q = 5.67 x 10-8A1ε1(T14 – T2

4)

It is sometimes convenient to group the powerterm with the temperatures as in

qA1

= 5.67[(T1/100)4 – (T2/100)4] (19.6)

For the results, produce a graph of [(T1/100)4

– (T2/100)4] as a function of corrected radiometerreading. (Temperature must be expressed inabsolute units.) The area of the source and itsemissivity are constants. So in essence, we aregraphing (T1

4 – T24) versus q. Should the graph be

linear? Would a semilog or a log-log graphproduce a linear result?

58

EXPERIMENT 20

EMISSIVITY OF BLACK AND GRAY SURFACES


The many types of electromagnetic radiationinclude X rays, gamma rays, the visible spectrum,radio waves, thermal radiation, andmicrowaves. Our concern is soley with thethermal component. Net heat transfer byradiation is a result of a temperature difference,but it should be remembered that radiation

energy propagates from a body (a source) with orwithout the presence of a second body (orreceiver).

ApparatusThe apparatus used in this experiment is

manufactured specifically for making radiationmeasurements, and is sketched in Figure 20.1. Itconsists of several pieces. One is a track and scaleonto which various items can be attached.Another is an instrument controller, whichcontains a rheostat to control heat input to theheat source, and digital readout instruments. Inthis experiment, we will use only the heat source,and not the light source.

instrument consoleand controller

heatertransformer heat source light source

track and scale

FIGURE 20.1. A sketch of the apparatus used in this experiment.

RadiometerFigure 20.2 shows the same apparatus with a

radiometer in place. A radiometer is used tomeasure radiant energy emitted from a heatedsurface. A device such as this can made of anarray of thermocouples connected in series to givea high voltage output, compared to the output ofonly one thermocouple. (The thermocouples couldalso be connected in parallel to give a highercurrent output.) A radiometer is any of a broadclass of temperature measuring devices, includingthermocouples (as used here), thermistors,resistance pyrometers, and radiation pyrometers.

heat source

track and scale

radiometer

L

FIGURE 20.2. Sketch of the apparatus with aradiometer.

Several words of caution regarding theradiometer are appropriate. It is a sensitiveinstrument, and a cover is supplied for its openend to keep out dust and dirt. Objects should not beplaced in the open end so that the finish and theshield will not become scratched or damaged.

Because of its sensitivity, the performance ofthe radiometer can be significantly affected bylocal heat sources or air currents about its openend. If the radiometer is exposed for long periodsof time to a radiant heat source, a zero shift in itsreadout will occur. So between readings, thepolished disc is to be placed over the open end.The radiometer reads directly in W/m2.

Using the Sensor. The technique for using theradiometer or sensor to obtain readings from theheat source is to allow the heater to reach asteady temperature with the radiometer inposition, and the polished disc in place. Whensteady state has been reached, the polished discis removed without touching the sensor ifpossible. The reading will increase rapidly,

59

eventually reaching a steady value. The elapsedtime could take up to two minutes. Once a readinghas been taken, replace the polished disc overthe open end to prevent excessive heating of thesensor, and subsequent erroneous readings.

The radiometer is mounted on a post that inturn is attached to a slide mechanism, whichallows the radiometer to be moved close to oraway from the heat source. The slide mechanismis scribed with a line that is used with the trackscale to measure position. The sensor is 2 cm closerto the heat source than the scribed line indicates.Thus all length readings taken from the scalemust have 2 cm subtracted from them.

Incident and Emitted RadiationThe digital readout indicates the intensity of

radiation received by the radiometer in W/m2.This value is different from that which wasemitted by the heat source at which theradiometer was aimed. Therefore, readings mustbe modified to account for this difference.

We will use the concept of configuration orview factor to make corrections. The heat sourceitself is a flat plate, and the radiometer sensor ismuch smaller in area. The actual setup betweenthe heat source and sensor is sketched in Figure20.3.

The view factor is defined as the fraction ofenergy that leaves the heat source that isintercepted by the sensor. So a reading at thesensor (dA1 in Figure 20.3) must be modified inorder to determine the heat that is radiated bythe source. We do not have results for the systemof Figure 20.3, but we can obtain them from a viewfactor geometery that we do have. Figure 20.4shows a view factor between a plane A2 andasmall area dA1. The small area is aligned withone of the corners of the plane. We can apply this

dA1

FIGURE 20.3. Actual setup between heat sourceand sensor.

result 4 times to the system of Figure 20.3 toobtain the view factor for the apparatus of thisexperiment. This number will be needed for allexperiments in which the radiometer is used. Wehave by definition

qincident = qemitted • AFd12

so that the energy emitted at the plate surface is

qemitted

A =

radiometer

reading • 1

Fd12(20.1)

where A is the area of the emitter.

BackgroundRadiant energy is characterized by

electromagnetic waves that travel at the speedof light. According to wave theory, radiation canbe thought of as many waves, all oscillating atdifferent wavelengths and frequencies. Theradiation properties considered in this

dA1

A2

a

c

b

Defining x = a/c, y = b/c, we have

Fd12 = 1

2π

x

√1 + x2 tan-1

y

√1 + x2 +

y

√1 + y2 tan-1

x

√1 + y2

FIGURE 20.4. View factor between a differential area and aplane.

60

experiment are generally functions of theproperties of the surface of a material, or itsopacity. We measure emissivity in thisexperiment for three different surfaces.

Radiative PropertiesThe surface of a substance highly influences

its radiation characteristics, and the amount ofradiant energy the surface emits, absorbs,reflects, and transmits. Polished gold, forexample, has an emissivity of (measured normalto the surface) of 0.025. An unpolished goldsurface, on the other hand, has a normalemissivity of 0.47. It is important to realize thatsuch differences exist, and that surface treatmentis a significant factor in how the materialbehaves in radiative heat transfer studies.

Emissivity is a property that describes howradiant energy interacts with the surface of amaterial. Other properties that are importantare reflectivity, absorptivity, andtransmissivity. As implied by these names,reflectivity is the fraction of incident radiantenergy reflected by the surface; absorptivity isthe fraction of incident radiant energy absorbedby the surface; and, transmissivity is the fractionof incident radiant energy transmitted through alayer of the material.

The radiative properties discussed in thepreceding paragraphs are in general functions ofwavelength. Properties that describe surfacebehavior as a function of wavelength are calledmonochromatic or spectral properties. Inaddition, radiative properties can be a function ofdirection. Such properties are referred to asdirectional properties. In the analysis ofradiation heat transfer, accounting for the exactbehavior of a surface or material can be complexenough to make a solution elusive. A simplifiedapproach must therefore be formulated. Thisinvolves the use of radiative properties that areaverages over all wavelengths and all directions.These properties are called total orhemispherical properties. Use of total propertiesis accurate enough in a majority of cases for mostengineering problems.

Measurement of EmissivityFor a system in which radiation is being

accounted for, the net heat exchanged is given by

q = σA1ε1(T14 – T2

4) (20.2)

where the “1” subscript refers to the source, and

“2” refers to the receiver. The objective of thisexperiment is to use this equation to measure theemissivity of various surfaces.

Figure 20.5 is a sketch of the apparatus used.Shown is the track and scale with a black platemounted a distance L1 from the heat source, andthe radiometer located a distance L2 from theblack plate. Energy is radiated from the heatsource, a portion of which is interecepted by theblack plate. The black plate is warmed, and itradiates energy in all directions. The radiometerintercepts a portion of this energy.

heat source

track and scale

radiometer

L1 L2

blackplate

FIGURE 20.5. Setup for measuring emissivity of ablack plate.

Procedure:• Locate the plate and radiometer at distances

of L1 = 5 cm and L2 ≥ 2L1.• Obtain plate temperature and radiometer

reading at ambient conditions.• Move the heater power control to its

maximum setting.• When conditions become steady, record

surface temperature and radiometer reading.• Set the heater power control to a new setting,

and obtain 3 more sets of readings.• Install the silver disc over the radiometer

between readings.• Repeat the entire procedure for two other

plates. Equation 20.2 can be rearranged to solve forthe emissivity:

ε1 = q / A

σ(T14 – T2

4) (20.3)

in which the numerator is the reading obtainedfrom the radiometer, T1 is the plate surfacetemperature, and T2 is the ambient temperature.Solve for emissivity for the 4 readings of theblack plate; average them to obtain an averageemissivity. Repeat the calculations for the otherplates. Note that the radiometer was placednormal to the plates in all cases, and so it is notuncommon to refer to the results as normalemissivities.

61

EXPERIMENT 21

RADIATION HEAT TRANSFER: THE VIEW FACTOR


The many types of electromagnetic radiationinclude X rays, gamma rays, the visible spectrum,radio waves, thermal radiation, andmicrowaves. Our concern is soley with thethermal component. Net heat transfer byradiation is a result of a temperature difference,but it should be remembered that radiationenergy propagates from a body (a source) with orwithout the presence of a second body (orreceiver).

Stefan-Boltzmann LawIn order to model radiation heat transfer, it is

useful to define a surface or substance that emitsradiation ideally. An ideal radiator is known asa black body, and it will emit energy at a ratethat is proportional to the fourth power of itsabsolute temperature:

qA

∝ T4

With a proportionality constant, the precedingequation becomes

q = σAT4

where σ is called the Stefan-Boltzmann constant,and has the value of

σ = 5.67 x 10-8 W/(m2·K4)σ = 0.1714 x 10-8 BTU/(hr·ft2·°R4)

Temperature in the equation for heat transfermust be expressed in absolute units.

In the case of most real materials, surfaces donot emit electromagnetic radiation ideally; thatis, they are not ideal radiators as black bodiesare. To account for the “gray” nature of realsurfaces, we introduce a dimensionless factor ε,called the emissivity. Emissivity values forvarious surfaces vary with finish (e.g., polishedversus painted). Modifying the heat transferequation to account for gray-body behavior yields

q = σAεT4

The emissivity is 1 for an ideal radiator, and lessthan 1 for a gray surface. This equation is knownas the Stefan-Boltzmann Law.

When a real body exchanges heat byradiation only with a black body, the net heatexchanged is proportional to the difference in T4.For such a system, the net heat exchanged is

q = σA1ε1(T14 – T2

4)

where the “1” subscript refers to the source, and“2” refers to the receiver.

View FactorThe view factor (also called the

configuration factor, the shape factor, or thegeometry factor) is a geometry term for a systemin which two surfaces exchange energy byradiation. Radiation waves travel in straightlines, and if one surface cannot “see” another,then there is no direct radiation from the firstsurface to the second. Thus position (or thegeometry) between source and receiver isimportant in determining the radiative energytransfer rate. The view factor expresses thegeometry of position that exists between sourceand receiver. For a net heat exchange in whichwe must account for geometry, the heattransferred becomes

q = σA1ε1F1-2(T14 – T2

4)

where F1-2 represents the geometry effectsinherent in the problem; i.e., how surface 1 viewssurface 2. In essence, F1-2 is the fraction of energyleaving surface 1 that goes to surface 2.

In this experiment, we will determine theview factor between a differential area and aplane using what is known as the Nusselt SphereMethod.

TheoryConsider that we are trying to obtain the

configuration factor between a differential areadA1 and the finite area A2, as shown in Figure21.1. Initially, we identify a differential areadA2 on A2, and draw lines normal to bothdifferential areas. We also draw a line from dA1to dA2, and this line has a length S. The angles β1

62

and β2 are measured from S to the lines that arenormal to dA1 and dA2, respectively. Next, weconstruct lines from dA1 to all points on theboundary of dA2. These lines form a solid angledenoted as dω1, which is given by

dA1

dA2

dAs

A2

As

Ab

Normal to dA2

r

S

dAs cos 1

1

1

2

1d

FIGURE 21.1. Configuration factor between adifferential and a finite area.

dω1 = dA2 cos β2

S2 (21.1)

The view factor between dA1 and dA2 is found as

FdA1-dA2 = cos β1 cos β2 dA1 dA2

πS2 (21.2)

Integrating over the area A2 results in theconfiguration factor that we are trying toevaluate:

FdA1-A2 = ∫A2

∫

cos β1 cos β2 dA2

πS2 (21.3a)

or in terms of solid angle, with Equation 21.1, weget

FdA1-A2 = 1π ∫

A2

∫

cos β1 dω1 (21.3b)

where the integration limits on A2 extend overthe portion that can be viewed by dA1. Nowsuppose that we construct a hemisphere over thearea dA1, again referring to Figure 21.1. Theprojection of dA2 onto the surface of thehemisphere is dAs which is given by

dAs = r2dω1

from the definition of solid angle. Combiningwith Equation 21.3b gives

FdA1-A2 = 1

πr2 ∫As

∫

cos β1 dAs (21.4)

where the integral is to be evaluated over As.The term dAs cos β1 is the projection of the areadAs onto the base of the hemisphere, and if thisterm is integrated over As, we obtain the area Ab.As shown in Figure 21.1, the area Ab is theprojection of As onto the base of the hemisphere.Equation 21.4 therefore becomes

FdA1-A2 = Ab

πr2 (21.5)

This equation states that the configuration factorbetween dA1 and A2 equals the ratio of theprojection of A2 onto the base of the hemisphere(= Ab) to the area of the base of the hemisphere:

FdA1-A2 = projection of As onto the base of the hemisphere

area of a circle with radius of the hemisphere

A plan view of Figure 21.1 would reveal the twoareas. Measurements or calculations could then beperformed to find the view factor.

Graphical Solution to aSpecific Problem

The sphere method has a graphicalequivalent, as described here. The sphere methodcan also be used with a photographic technique,which is described later. Both methods are usedin this experiment. In some geometries, there aretheoretical solutions which have been developedand catalogued. One such case is described here.It has been investigated with the sphere method,and the experimental results are to be comparedto those obtained by equation.

Figure 21.2 shows the plan and profile viewsof a differential area located at the origin, and aplane, selected to illustrate the method. Wewill use results obtained from the sphere methodto obtain a graphical solution for theconfiguration factor between the differentialarea dA and the plane.

The graphical solution method first involvesdrawing a hemisphere of radius r in both views.The hemisphere is constructed with its center atthe area dA, and the radius of the hemisphere isselected arbitrarily. The hemisphere canencompass the entire plane, or pass through it or

63

exclude it. Next lines are constructed from theorigin to every point on the edges of the plane.The lines connecting the origin with the edges ofthe plane intersect the hemisphere and form anarea that we have identified in Figure 21.1 as As.

PlanProfile

dA

plane

FIGURE 21.2. Plan and profile views of adifferential area and a plane.

As mentioned previously, after drawing ahemisphere, we then draw lines from the originto a number of points on the edges of the plane.The lines will intersect the hemisphere on theedge of the projection plane. In order to find apoint of interesection, the true length of the linemust be constructed in the profile view.

The true length of the line is found here by arotation of the line as illustrated in Figure 21.3.We draw line 0-1 in both views. Line 0-1 is

0

1

2

2

3

3

1´

1´

0

1

Plan

Profile

dA

plane

FIGURE 21.3. Plan and profile views of adifferential area, a plane, and a hemisphere.

rotated in the plan view to the point labeled 1´.The line from 0 to 1´ is drawn to locate point 2 inthe profile view. Point 2 is projected back to theplan view and rotated back to line 0-1 to locatepoint 3. Thus 3 is the point where the line 0-1intersects the hemisphere.

This process of finding intersecting points isrepeated for many points on the periphery of theplane until something like Figure 4 results.Shown is the plan view of the completeddrawing, which includes the areas dA, the plane,and projection of the plane onto the base of thehemisphere. The view factor is the ratio of theprojected area to that of the circle whose radiusequals that of the hemisphere.

dA

plane

projection of plane ontothe plane of dA

FIGURE 21.4. Plan view of the completeddrawing.

Figure 5 again shows the solution but with agrid imposed. The projected area can bedetermined either by counting squares or by somenumerical technique.

FIGURE 21.5. Plan view of the completeddrawing with grid imposed for determiningarea .

64

Photographic MethodThe photographic method is based on the

concepts discussed in the theory section.Regarding the problem solved graphically in thepreceding section, we can use a reflectinghemisphere which is placed over the area dA,and view directly above dA. The reflection of theplane in the surface of the hemisphere willappear like that shown in Figure 21.4. So aphotograph taken in this way can be used toobtain the areas, and the configuration factor canthen be calculated.

Figure 21.6 is a sketch of the apparatus usedhere. Shown is the relative mounting of allpieces. The plane area is mounted on the frame,and a camera is used to take a photograph in theviewing direction.

Viewingdirection

Plane area

FIGURE 21.6. Sketch of the assembled apparatus.

Various plane areas are available, made of16 gage sheet metal. These planes include twodisks, a square, two octagons, and an equilateraltriangle. All plane areas have been painted aglossy white so they would reflect effectively.

Correction FactorAs mentioned earlier, the hemisphere used is

not quite a true hemisphere; it has beentruncated, as shown in Figure 21.7. The actualknown dimensions, referring to the figure, areradius ra and height d2. In order to makeappropriate calculations, it is necessary todetermine the true radius r. For the dimensions ofFigure 21.7, we can write:

r d1

r

ra

d2

truncatedhemisphere

FIGURE 21.7. Sketch of the truncatedhemisphere used in this experiment.

r2 = ra2 + d1

2

in which only ra is known. We can also write

d1 + d2 = r

The preceding two equations contain twounknowns (d1 and r). Solving simultaneously, weget

r = ra

2 + d22

2d2(21.6)

and d1 = r - d2. For this experiment, measure raand d2. Then calculate r and d1.

Comparison of Results for a DiskA mathematical solution has been obtained

for the view factor when the plane area is a disk,or when it is a square (or rectangle).

When the plane area is a disk oriented in acertain way, an exact expression can be used tofind the view factor. Figure 21.8 shows a drawingof two surfaces. One is a differntial area dA1 andthe other is a disk A2. It is desired to find theview factor FdA1-A2 The exact solution to thisproblem is given by:

FdA1-A2 = A2

1 + B2 + A2

√(1 + B2 + A2)2 – 4B2 – 1

(21.7)

where A = a/c and B = b/c. A graph of thisequation is shown also in Figure 21.8.

ExperimentTwo plane areas are to be used for this

experiment. One must be a disk and the other is tobe any of the remaining noncircular planes (yourchoice). Mount the planes on the frame at theappropriate position. Take a photograph in the

65

viewing direction, and transfer the result to acomputer that can read photographs. Determinethe areas in any convenient way from thephotographs for both planes. Calculate the viewfactors.

For the disk, use the information in thepreceding section to calculate the view factor by

equation. Compare the result to that obtained bythe photograph. Calculate % error.

For the noncircular plane, use the graphicalmethod to determine the view factor. Comparethe result to that obtained from the photograph.Calculate % error.

0

0.1

0.2

0.3

0.4

0.5

0.01 0.1 1

F

a/c

dA

1-A2

0.02 0.04 0.06 0.2 0.4 0.6 2 4

1.0

b/c = 0.975

0.9

0.95

0.8

0.7

0.6

0.40.5

0.2

dA1

A2

c

b

a

FIGURE 21.8. A graph of Equation 21.7 for the view factor between a differential area and a disk.

66

EXPERIMENT 22

THE VAPOR COMPRESSION REFRIGERATION CYCLE

The vapor compression refrigeration cycle isused in what is commonly known as an airconditioner. Such devices are thought ofprimarily as something needed to cool a room.The air conditioner circulates a fluid called arefrigerant through heat exchangers. Therefrigerant absorbs heat at a low temperature atone of the exchangers and discharges heat at ahigh temperature at the other exchanger.Evaluation of the vapor compressionrefrigeration cycle is the subject of thisexperiment.

The Air Conditioner ApparatusFigure 22.1 is a schematic of the apparatus

used in this experiment. How the apparatusworks in ‘pumping heat’ is easily understood ifwe follow a finite mass of refrigerant around thecircuit. The compressor is a component that takesin low temperature, low pressure refrigerantvapor and compresses it. The compressedrefrigerant vapor is discharged as a highpressure, high temperature vapor. One wayvalves in the flow lines of the compressor controlthe flow direction.

Upon leaving the compressor, thesuperheated vapor enters a finned tube heatexchanger. In the apparatus of this experiment,the heat exchanger is a cross flow type. Air flowsin a duct across which are located small diametertubes that convey the refrigerant. The smalldiameter tubes have fins in the form of thin sheetaluminum attached. The superheated vapor ishigher in temperature than the air flowing acrossthe finned tubes and so heat will be transferredfrom the refrigerant to the air. During this heattransfer process, the refrigerant will condense andso the heat exchanger is commonly called acondenser. Refrigerant leaving the condenser hasa very low quality.

After leaving the condenser, the part liquid–part vapor refrigerant enters the top of areceiving tank. Within this tank, we have vaporrising to the top and liquid falling to the bottom;the tank thus acts as a separator. Liquid exitsthis tank at its bottom and then enters athrottling device. There are two such devices onthis apparatus. One is called a thermostaticexpansion valve and the other is a capillary tube(very small inside diameter tube). The object of athrottling device is to cause a significant loss in

pressure of the refrigerant. (There are actuallytwo thermostatic expansion valves. The second isused in reversing the flow direction of therefrigerant, which will not be done for thisexperiment.)

When the refrigerant leaves the throttle, ithas a low pressure and temperature. Therefrigerant now enters another finned tube, crossflow heat exchanger. The refrigerant is coolerthan the air passing through the exchanger andso heat is transferred to the air from the liquidrefrigerant. During this process, the refrigerantvaporizes and so this heat exchanger is commonlycalled an evaporator.

Refrigerant that has vaporized in theevaporator now enters the top of a reservoircalled an accumulator tank. This tank also acts asa vapor–liquid separator. Vapor leaves this tankthrough another opening at the top of the tankand then returns to the compressor.

Air that passes through the condenser absorbsenergy from the refrigerant and is thereforeheated in the process. Air that passes throughthe evaporator transfers energy to the refrigerantand is cooled in the process. The refrigerant thuspumps heat from the evaporator and dischargesit at the condenser. In addition, the refrigerantgained energy from the compressor which isdischarged also at the condenser.

Experimental ProcedureThe unit used in the laboratory contains

several pressure and temperature gages at variouslocations throughout the piping system. A numberof valves are also in the circuit for flow control.Operation of the unit will be shown by theinstructor. For the experiment, start and operatethe unit as per directions given. Identify thecompressor, condenser, receiving tank, throttlingdevice, evaporator and the accumulator tank.Begin at the compressor and follow the closedloop made by the refrigerant. Observe the processof condensation and vaporization in the heatexchangers.

Fans are used to move air past the cross flowheat exchanger tubes. Set the fans at a desiredsetting and with the compressor operating, allowthe system to reach steady state. Use either thecapillary tube or the expansion valve. Oncesteady state is reached, take temperature andpressure readings at all the gage locations.

67

1

2

3

4

compressor

condenser

evaporator

capillarytube

expansionvalve

receivertank

accumulatortank

filter/dryersightglass

valve

FIGURE 22.1. A schematic of the air conditioner.

Remember that readings from the pressure gagesare in psig while the reduction of data mightrequire psia.

AnalysisA pressure–enthalpy diagram will be

provided for the refrigerant. Identify and labelall points in the cycle and sketch the cycle on aseparate graph. For the calculations involved,take readings of properties directly from thepressure–enthalpy diagram. By applying theenergy equation to a compressor, we get thefollowing equation assuming that the process isadiabatic

(compressor work F·L/M)

4w1 = h4 – h1 (22.1)

where the subscripts refer to gage locations aslabeled in Figure 9.1. For the condenser we have

(heat transferred at condenser F·L/M)

1q2 = h1 – h2 (22.2)

For the throttle, we write

(throttle F·L/M)

h2 = h3 (22.3)

For the evaporator, we get

(heat transferred at evaporator F·L/M)

3q4 = h3 – h4 (22.4)

Sample CalculationsData taken on an air conditioner yielded the

pressures and temperatures provided in Table22.1. A sketch of the cycle was made on apressure–enthalpy diagram which is given inFigure 22.2. Readings from a (more detailed)pressure–enthalpy diagram and from a table ofproperties of Freon-12 yielded the enthalpiesshown in column 4 of Table 22.1. Also shown inTable 11.1 are the entropy and specific volumevalues for each point in the cycle. Using thosedata gives the following results:

Compressor Work:

4w1 = h4 – h1 = 84 – 90 = – 6 BTU/lbm

Heat Transferred at Condenser:

1q2 = h1 – h2 = 90 – 31 = 59 BTU/lbm

Throttle:

h2 = h3 = 31 BTU/lbm

Heat Transferred at Evaporator:

3q4 = h3 – h4 = 31 – 84 = – 53 BTU/lbm

cycle operating between the evaporator andcondenser temperatures.

68

For the data obtained on the air conditioningapparatus, perform an analysis like that shownin the sample calculations above. Sketch p–h, p–v and T–s diagrams for the refrigerant. Sketchthe T–s diagram for the corresponding Carnotcycle.

QuestionsWhat is the relationship between 4w1, 1q2

and 3q4? Why? On the T–s diagram for the

refrigerant, sketch also the Carnot cycle. Atwhat points and processes does the actual cyclediffer from the Carnot cycle? What is theefficiency of the Carnot cycle for thetemperatures of the cycle for which your datawere obtained? What is the actual efficiency ofthe cycle?

TABLE 22.1. Refrigerant properties for each point in the cycle of the example.

Pressure Temperature enthalpy entropy specific volumePoint psia °F BTU/lbm BTU/lbm·°R ft3/lbm

1 136 116 90 0.169 0.252 136 96 31 0.078 0.0143 56 42 31 0.08 0.0134 48 64 84 0.175 0.85

12

34

pressurepsia

enthalpyBTU/lbm

satu

rate

d liq

uid

satu

rate

d va

por

FIGURE 22.2. Pressure–enthalpy diagram for the data of the example.

69

Appendix

70

ANALYSIS OF DOUBLE PIPE HEAT EXCHANGERSSUGGESTED ORDER OF CALCULATIONS

Problem Complete problem statementDiscussion Potential heat losses; other sources of difficulties

Assumptions 1. Steady state conditions exist.2. Fluid properties remain constant and are evaluated at atemperature of

Nomenclature 1. T refers to the temperature of the warmer fluid.2. t refers to the temperature of the cooler fluid.3. w subscript refers to the warmer fluid.4. h subscript refers to hydraulic diameter5. c subscript refers to the cooler fluid.6. a subscript refers to the annular flow area or dimension.7. p subscript refers to the tubular flow area or dimension.8. 1 subscript refers to an inlet condition.9. 2 subscript refers to an outlet condition.10. e subscript refers to equivalent diameter.

A. Fluid Properties

m· w = T1 =ρ = Cp =k f = α =ν = Pr =

m· c = t1 = ρ = Cp =k f = α =ν = Pr =

B. Tubing SizesIDa =IDp = ODp =

C. Flow Areas Ap = πIDp2/4 =

Aa = π(IDa2 - ODp

2)/4 =

D. Fluid Velocities [Route the fluid with the higher flow rate through the flowcross section with the greater area.]

Vp = m· /ρA = Gp = m· /A =

Va = m· /ρA = Ga = m· /A =

71

ODp

IDp

IDa

E. Annulus Equivalent DiametersFriction Dh = IDa - ODp =

Ht Trans De = (IDa2 - ODp

2)/ODp =

F. Reynolds NumbersRep = VpIDp/ν =

Rea = VaDe/ν =

G. Nusselt Numbers

Modified Seider-Tate equation for laminar flow:

Nu = hDk f

= 1.86

D Re Pr

L

1/3

Re = VDν < 2 200 D = IDp if cross section is tubular

D = De if cross section is annular

0.48 < Pr = να < 16 700

µ changes moderately with temperatureProperties evaluated at the average fluid

temperature [ = (inlet + outlet)/2]

72

Modified Dittus-Boelter Equation for turbulent flow:

Nu = hDk f

= 0.023(Re)4/5 Prn

n = 0.4 if fluid is being heatedn = 0.3 if fluid is being cooled

Re = VDν ≥ 10 000 D = IDp if cross section is tubular

D = De if cross section is annular

0.7 ≤ Pr = να ≥ 160

L/D ≥ 60

Properties evaluated at the average fluid temperature [ = (inlet + outlet)/2]

Nup =

Nua =

H. Convection Coefficientshp = Nup kf/ODp =

ha = Nua kf/De =

I. Exchanger Coefficient1

Uo =

1h p

+ 1h a

Uo =

J. Outlet Temperature Calculations (Exchanger length L )

R = m· cCpc

m· hCph

= Ao = πODpL =

Counterflow Ecounter = exp[UoAo(R - 1)/m· cCpc] =

T2 = T1(R - 1) - Rt1(1 - Ecounter)

REcounter - 1

t2 = t1 + T1 - T2

R

73

Parallel Flow Epara = exp[UoAo(R + 1)/m· cCpc] =

T2 = (R + Epara )T1 + Rt1(Epara - 1)

(R + 1)Epara

t2 = t1 + T1 - T2

R

T2 =

t2 =

K. Log Mean Temperature Difference

Counterflow LMTD = (T1 - t2) - (T2 - t1)

ln [(T1 - t2) /(T2 - t1)] =

Parallel Flow LMTD = (T1 - t1) - (T2 - t2)

ln [(T1 - t1) /(T2 - t2)] =

L. Heat Balance

qw = m· wCpw(T1 - T2) =

qc = m· cCpc(t2 - t1) =

q = UoAoLMTD = (clean)

M. Fouling Factors and Design Coefficient

Rdi = Rdo =

1U

= 1

Uo + Rdi + Rdo U =

N. Heat Transfer Area and Tube Length (unless already known)

Ao = q

U (LMTD) =

L = Ao

π (ODp) =

74

O. Friction Factors

Rep = VpIDp/ν =

εIDp

= fp =

Rea = VaDh/ν =

εDh

= fa =

Laminar Flow Equations

Laminar flow in a tube fp = 64

RepRep =

VpIDp

ν ≤ 2 200

Laminar flow in an annulus κ = ODp

IDaRea =

VaDh

ν ≥ 10 000

1fa

= Rea

64

1 + κ 2

1 - κ + 1 + κ ln(κ)

Turbulent Flow Equations D = IDp if cross section is tubularD = Dh if cross section is annular

Chen Equation

1

√ f = - 2.0 log

ε

3.7065D -

5.0452Re log

1

2.8257( )εD

1.1098 +

5.8506Re0.8981

Churchill Equation

f = 8

8

Re12

+ 1

(B + C)1.5

1/12

where B =

2.457ln

1(7/Re)0.9 + (0.27ε/D)

1 6

C =

37 530

Re

1 6

75

P. Pressure Drop Calculations

∆pp = fpLIDp

ρpVp

2

2gc =

∆pa =

faL

Dh + 1

ρaVa2

2gc =

Q. Summary of Information Requested in Problem Statement

heat_tr_manual.pdf

Documents

radiation heat transfer

frame heat exchanger

dimensional conduction

double pipe heat exchanger

cross flow heat exchanger

forced convection flat

transient conduction

pipe flow