multidisciplinary engineering senior design project 05424 high temperature pizza oven 2005 critical...

Multidisciplinary Engineering Senior Design

Project 05424High Temperature Pizza Oven

2005 Critical Design ReviewMay 13, 2005

Project Sponsor:Abraham Fansey / VP Office of Finance and

AdministrationTeam Members:

Izudin Cemer – Electrical EngineeringAdam George – Mechanical Engineering

Nathan Mellenthien – Mechanical EngineeringDerek Stallard – Mechanical Engineering

Team Mentor:Dr. Satish Kandlikar

Kate Gleason College of EngineeringRochester Institute of Technology

Mission Statement Design and build a high

temperature pizza oven to replicate the unique results of a coal oven High temperature Crispy crust Fast cook time

For use at R.I.T.’s future pizzeria or other universities

Design Process

Define problem Data collection/Research Concept development/Brainstorming Feasibility assessment Performance objectives &

specifications Analysis & synthesis Prototype detailed design

Key Requirements & Critical Parameters Achieve comparable results to a coal

oven No coal High internal temperatures Mixture of traditional baking methods and

current technology Evenly cooked pizza User friendly Capable of high production Safe Oven

Major Design Challenges Time

20 weeks Money

Budget Suppliers

Reliability Manpower

Performance Specifications Cooking time: no longer than five

minutes per pizza Stone deck must reach a minimum

temperature of 650°F Internal air temperature must reach a

minimum temperature of 850°F Deck must be rotating and have a variable

speed Oven insulation: outside surface is no higher

than 120°F Minimum production capacity: 40 pizzas/hour

Analysis of Design Thermal Analysis Mechanical

Analysis Electrical Analysis

Thermal Analysis

Thermal Model Pizza Heat

Transfer Methods

Heat loss Heat

generation

Thermal Model Elements

Flue

Flame

Dome

Stone Deck

IR Burner

Door

Pizza

Pizza Heat Transfer Model

Pizza

Radiation from dome Convectio

n from air

Conduction from stone

Conduction Detail Assume

1-D conduction Standard pressure Constant Area and Thickness Avg. temp of pizza=330.7 K

Values λ(k)=3.43 W/mK (Experiment) A=.07297 m2 (D=.3048m) W=.00635m T1=616.5 K T2=330.7 K

Q=11264.9 J/s

21 TTAw

Q

Radiation Detail Assume

1-D radiation Standard pressure Constant Area and Thickness Avg. temp of pizza=330.7 K

Values ε=.75 W/m2K A=.07297 m2 (D=.3048m) σ=5.67x10-8 W/m2K4

T1=697.3 K T2=330.7 K

Q=710.7 J/s

42

41 TTQ

Convection Detail Assume

1-D convection Steady State Standard pressure Constant area and thickness Free Convection Avg. temp of pizza=135.5°F

Values α=3.43 W/mK A=.07297 m2 (D=.3048m) W=.00635m T=727.6 K TW=330.7 K

Q=144.8 J/s

wTTAQ

Pizza Heat Transfer Summary Conduction

11264.9 J/s 92.9% of total heat

transfer Radiation

710.7 J/s 5.9% of total heat

transfer Convection

144.8 J/s 1.2% of total heat

transfer

21 TTAw

Q

42

41 TTQ

wTTAQ

Heat Loss Elements

Flue

Flame

Dome

Stone Deck

IR Burner

Door

Pizza

Heat Loss Through Door

Heat Loss Through Dome

Heat Loss Through Flue

Heat Loss to Pizza

Heat Loss Summary Heat loss to pizzas*: 50,222 J/s Heat loss through walls: 176.32 J/s Heat loss through door:

Open: 942.5 J/s Closed: 26.81 J/s

Heat loss through flue: 44.42 J/s Total Heat loss Range during operation:

50,469 J/s to 51,385 J/s

*Oven is operating at capacity of 100 pizzas/hour

Heat Generation

Mass rate of propane required

Preheat conditions

Mass Rate of Propane Propane rate required=mp

mp=Qneeded/HHV HHV=50,350 kJ/kg (Incropera & Dewitt)

Qneeded=Qwalls+Qpizza Equations developed via curve fitting in Excel

Closed Door mp=236.94·(pizzas/hour)-0.9868

3.607 kg/hr / 2.515 hr tank life* Open Door

mp=190.08·(pizzas/hour)-0.9419

3.672 kg/hr / 2.470 hr tank life*

*100 pizza per hour load/20 lb tank

Preheat Conditions Mass of propane required

m=300 kg Cp=900 J/kgK ΔT=434.4 K Mp=2.33 kg

Preheat time (Based on 3.65 kg/hr mass rate) : 38 min

Tank Drain (20 lb tank) : 25.7%

HHV

TmcM p

p

Design of Prototype

Design of Prototype Total weight of dome = 495 lbs

Design of Prototype

Oven base support Constructed of

3”x3”x3/8” angle iron

Total height = 30”

Design of Prototype

Design of Prototype

Mechanical Analysis

Using COSMOS finite element analysis Top load of 600 lbs

Concrete dome Lower load of 50 lbs

Deck, deck support, shaft, etc.

StrainMax of 3.807e-005

Mechanical Design Concerns Thermal Expansion

Enough clearance during thermal expansion of deck shaft

Oven being top heavy Extended base footprint

Cracking of concrete dome Un-reinforced concrete

Thermocouple and Microcontroller Based Temperature Monitoring

Use of thermocouples and microcontroller to measure, and display temperature

Send data through RS232 to a PC

Electrical Overview Introduction to microcontrollers and thermocouples

• Purpose of the microcontroller in the design• How thermocouples work• Implementation circuitry

Representing thermocouple temperature voltage relationship

• Use of linear approximation Cold junction compensation

• Hardware• Software• Typical application circuitry used in the design

Introduction to Microcontrollers

General purpose microprocessors that control external devices

The execute use program loaded in its memory

Under the control of this program data is received as an input, manipulated, and then sent to an external output device

Temperature Sensors Classical temp. sensors are thermocouples, RTDs and

thermistors New generation of sensors are integrator circuit sensors

and radiation thermometry devices Choice of sensor depends on the accuracy, temperature

range, speed of response, and cost

Advantages/Disadvantages of a Thermocouple

Advantages• Wide operating temp. range• Low cost

Disadvantages• Non-linear• Low sensitivity• Reference junction compensation required• Subject to electrical noise

Thermocouple

Thermocouples Thermoelectric voltage is produced and an

electric current flows in a closed circuit of two dissimilar metals it the two junction are held at different temperatures

The current depends on the type of metal and temp. difference between hot and cold junction (not an absolute temp.)

Thermocouples Continued Voltage measuring device measures the temp To know the absolute temp. we need to keep

the reference temp. stable and known Temperature of the reference junction is not

known and not stable We used cold junction compensation method

to take care of this problem

Cold Junction Compensation Done through hardware using IC’s LT1025 It has a built in temp. sensor that detects the

temp. of the reference junction Produces voltage proportional to voltage

produced by thermocouple with hot junction at ambient temp. and cold junction at 0 °C

This voltage is added to thermocouple voltage and net effect is as if the reference junction is at 0 °C

Compensating Circuitry

Linear Approximation Method of representing thermocouple

temp. voltage relationship V=sT + b

V-thermocouple voltage S is the slope T is the temperature ‘b’ is an offset (b=0)

Equation then becomes V= sT where s is now Seeback coeffcient

In order to obtain a 10 mV output from an amplifier we will need a gain of G=10mv/51.71uV=193

Electrical Conclusion

Electrical Block Diagram

Desired Outcomes Cooking Time < 5

min. Dome Temp. = x °C Deck Temp. = x °C Budget < $3000.00 Rotating Deck Exterior Temp< 49

°C

Actual OutcomesCooking Time=Dome Temp=Deck Temp=Budget=Rotating DeckExterior Temp=

Results

Questions?

Backup Slides and References

Trial 1

Toven=232.2 °C Mi=.805 kg Mf=.715 kg T=11 minutes Heat=2034 kJ

Trial 2

Toven=260 °C Mi=.655 kg Mf=.585 kg T=10 minutes Heat=1582 kJ

Trial 3

Toven=287.8 °C Mi=.800 kg Mf=.720 kg T=8 minutes Heat=1808 kJ

(back)

Determination of Thermal Conductivity Lack of availability of

specific k value for pizza Standard oven, pizza stone,

and measuring devices required

Set area and thickness dQ=(mi-mf)*L Values

L=2260 kJ/kg A=.07297 m2 (D=.3048m) dt=240 s mi=.300 kg mf=.290 kg Ti=23.2°C Tf=65.5°C

Solving for k yields k=3.43 W/mK

(back)

x

TkA

dt

dQ

Experimentation Value of k=3.43

W/mK Heat Required=(mi-

mf)*L=1808 kJ Total Heat Supplied

= Heat Rate * Cooking Time

Cooking Time = 149s (2 min, 29 sec)

Heat Loss to Pizza

Aim: 100 pizzas per hour Each pizza takes 1808 kJ to bake

Experimentally Determined Average heat lost to pizzas=

180,800 kJ/hr=50,222 J/s 171,365 BTU/hr=47.6 BTU/s

Back

Heat Transfer Through Door (Open)

Assume 1-D radiation Standard pressure Steady State

Values ε=.75 W/m2K (concrete) A=.096774 m2 (door) σ=5.67x10-8 W/m2K4

T1=697.3 K T2=293.2 K

Q= 942.5 J/s

42

41 TTQ

Back

Heat Transfer Through Door (Closed) AISI 304 Stainless Steel

k=16.6 W/mK .003175 m thick (1/8”)

on both sides Insulation (Durablanket S

Ceramic Fiber Blanket) k=.087 W/mK .1016 m thick (4”)

between Stainless Steel plates

Using Program Q=26.81 J/s TSurface=168.1 °F

Back

Heat Transfer Through Wall Reflective Concrete

k=.80 W/mK .1016 m thick (4”)

Insulation (Durablanket S Ceramic Fiber Blanket)

k=.087 W/mK .2032 m thick (8”)

Air k=28.5*10^3 W/mK .0254m thick (1”)

AISI 304 Stainless Steel k=16.6 W/mK .003175 m thick (1/8”)

Using Program Q=176.32 J/s TSurface=123.0 °F

Back

Photos from experiment

Photos from experiment (Continued)

Heat Loss Through Conduction (Closed door and Wall)

Compound Wall Unsure of insulation thickness

desired Wanted to be able to try different

values Plugging numbers into equations

would be time consuming and inefficient

Code for Wall CalculationsPrivate Sub CommandButton1_Click()s = steelthickness.Valuet = insulationthickness.Valueks = ksteel.Valueki = kinsul.Valueq = (727.6 - 293.2) / ((1 / (5 * 0.09677)) + (s / (ks * 0.09677)) +

(t / (ki * 0.09677)) + (s / (ks * 0.09677)) + (1 / (5 * 0.09677)))tinsul = 727.6 - (q * ((1 / (5 * 0.09677)) + (s / (ks * 0.09677)) + (t /

(ki * 0.09677)) + (s / (ks * 0.09677))))tinsul = (9 / 5) * (tinsul - 273) + 32tral = Format(tinsul, "#0.000")qvalue = Format(q, "#0.00")qval.Caption = qvalueresult.Caption = tralEnd Sub(back)

Code for Wall CalculationsPrivate Sub CommandButton1_Click()c = concretethickness.Valuet = feltthickness.Values = steelthickness.Valuekc = kcon.Valueki = kinsul.Valueks = ksteel.Valueq = (727.6 - 293.2) / ((1 / (5 * 1.162)) + (c / (kc * 1.162)) + (t / (ki

* 1.162)) + (0.0254 / ((28.5 * 10 ^ 3) * 1.162)) + (s / (ks * 1.162)) + (1 / (5 * 1.162)))

tinsul = 727.6 - (q * ((1 / (5 * 1.162)) + (c / (kc * 1.162)) + (t / (ki * 1.162)) + (0.0254 / ((28.5 * 10 ^ 3) * 1.162)) + (s / (ks * 1.162))))

tinsul = (9 / 5) * (tinsul - 273) + 32tral = Format(tinsul, "#0.000")qvalue = Format(q, "#0.00")qval.Caption = qvalueresult.Caption = tralEnd Sub

Displacement of BaseMax of 6.888e-003 in

Hardware Features Supply voltage- +5 V standard The clock- 20 MHz oscillator A/D converter- contains eight LCD drivers- enables microcontroller to

be connected directly to an LCD display Current sink/source capability- up to 25

mA per pin EEPROM

Hardware Features Continued RISC instruction set Harvard architecture-code and data

storage are on separate buses which allows code and data to be fetched simultaneously

All of the ports are bidirectional

Measurement Error Considerations

Calibration errors-result of offset and linearity errors

Electrical noise- thermocouples produce extremely low voltages

Thermal coupling-need a good contact with a measuring surface

Linear Approximation Continued

In order to obtain a 10 mV output from an amplifier we will need a gain of G=10mv/51.71uV=193

Safety Measures

Gas Valves Insulation Exterior (No sharp edges) Outdoor experiments Safe handling of propane

containers