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Multidisciplinary Engineering Senior Design Project 05424 High Temperature Pizza Oven 2005 Critical Design Review May 13, 2005. Project Sponsor: Abraham Fansey / VP Office of Finance and Administration Team Members: Izudin Cemer – Electrical Engineering Adam George – Mechanical Engineering

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Multidisciplinary Engineering Senior DesignProject 05424High Temperature Pizza Oven2005 Critical Design ReviewMay 13, 2005

Project Sponsor:

Abraham Fansey / VP Office of Finance and Administration

Team Members:

Izudin Cemer – Electrical Engineering

Adam George – Mechanical Engineering

Nathan Mellenthien – Mechanical Engineering

Derek Stallard – Mechanical Engineering

Team Mentor:

Dr. Satish Kandlikar

Kate Gleason College of Engineering

Rochester Institute of Technology

mission statement
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
Design Process
  • Define problem
  • Data collection/Research
  • Concept development/Brainstorming
  • Feasibility assessment
  • Performance objectives & specifications
  • Analysis & synthesis
  • Prototype detailed design
key requirements critical parameters
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
Major Design Challenges
  • Time
    • 20 weeks
  • Money
    • Budget
  • Suppliers
    • Reliability
  • Manpower
performance specifications
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
Analysis of Design
  • Thermal Analysis
  • Mechanical Analysis
  • Electrical Analysis
thermal analysis
Thermal Analysis
  • Thermal Model
  • Pizza Heat Transfer Methods
  • Heat loss
  • Heat generation
thermal model elements
Thermal Model Elements

Flue

Dome

Flame

Door

Pizza

Stone Deck

IR Burner

pizza heat transfer model
Pizza Heat Transfer Model

Radiation from dome

Convection from air

Pizza

Conduction from stone

conduction detail
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
radiation detail
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
convection detail
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
pizza heat transfer summary
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
heat loss elements
Heat Loss Elements

Heat Loss Through Flue

Heat Loss Through Dome

Dome

Flue

Heat Loss to Pizza

Heat Loss Through Door

Flame

Door

Pizza

Stone Deck

IR Burner

heat loss summary
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
Heat Generation
  • Mass rate of propane required
  • Preheat conditions
mass rate of propane
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
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%
design of prototype1
Design of Prototype
  • Total weight of dome = 495 lbs
design of prototype2
Design of Prototype
  • Oven base support
  • Constructed of 3”x3”x3/8” angle iron
  • Total height = 30”
mechanical analysis
Mechanical Analysis
  • Using COSMOS finite element analysis
    • Top load of 600 lbs
      • Concrete dome
    • Lower load of 50 lbs
      • Deck, deck support, shaft, etc.
strain
Strain

Max of 3.807e-005

mechanical design concerns
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
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
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
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
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/Disadvantages of a Thermocouple
  • Advantages
    • Wide operating temp. range
    • Low cost
  • Disadvantages
    • Non-linear
    • Low sensitivity
    • Reference junction compensation required
    • Subject to electrical noise
thermocouples
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
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
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
linear approximation
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 Conclusion
  • Electrical Block Diagram
results
Desired Outcomes

Cooking Time < 5 min.

Dome Temp. = x °C

Deck Temp. = x °C

Budget < $3000.00

Rotating Deck

Exterior Temp< 49 °C

Actual Outcomes

Cooking Time=

Dome Temp=

Deck Temp=

Budget=

Rotating Deck

Exterior Temp=

Results
trial 1
Trial 1
  • Toven=232.2 °C
  • Mi=.805 kg
  • Mf=.715 kg
  • T=11 minutes
  • Heat=2034 kJ
trial 2
Trial 2
  • Toven=260 °C
  • Mi=.655 kg
  • Mf=.585 kg
  • T=10 minutes
  • Heat=1582 kJ
trial 3
Trial 3
  • Toven=287.8 °C
  • Mi=.800 kg
  • Mf=.720 kg
  • T=8 minutes
  • Heat=1808 kJ

(back)

determination of thermal conductivity
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)

experimentation
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
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
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

Back

heat transfer through door closed
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
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

heat loss through conduction closed door and wall
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 calculations
Code for Wall Calculations

Private Sub CommandButton1_Click()

s = steelthickness.Value

t = insulationthickness.Value

ks = ksteel.Value

ki = kinsul.Value

q = (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) + 32

tral = Format(tinsul, "#0.000")

qvalue = Format(q, "#0.00")

qval.Caption = qvalue

result.Caption = tral

End Sub

(back)

code for wall calculations1
Code for Wall Calculations

Private Sub CommandButton1_Click()

c = concretethickness.Value

t = feltthickness.Value

s = steelthickness.Value

kc = kcon.Value

ki = kinsul.Value

ks = ksteel.Value

q = (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) + 32

tral = Format(tinsul, "#0.000")

qvalue = Format(q, "#0.00")

qval.Caption = qvalue

result.Caption = tral

End Sub

displacement of base
Displacement of Base

Max of 6.888e-003 in

hardware features
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
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
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
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
Safety Measures
  • Gas Valves
  • Insulation
  • Exterior (No sharp edges)
  • Outdoor experiments
  • Safe handling of propane containers
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