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