The document describes a simulation of heat transfer into a copper ball heated by a torch. It derives equations to model the temperature change over time considering convection and optionally radiation heat transfer. The simulation is performed in MATLAB considering first only convection, then convection and radiation. The results show that including radiation lowers the final steady state temperature from 152°C to 131°C and decreases the settling time from 4540 seconds to 3340 seconds.

1. Michael Davis
1000781532
8-2-2014
6.26 Thermal Simulation
A copperBall isheatedwitha small torchin ambientairas shownbelow.
Considerthe heatinputandthe convectionheatTransfer.Derive a differential equationforthe
temperature of the ball.
𝑄𝑖𝑛 − 𝑄 𝑐𝑜𝑛𝑣 = 𝑚𝐶 𝑝 𝑇̇𝑖;𝑤ℎ𝑒𝑟𝑒 𝑚𝐶 𝑝 𝑇̇𝑖 = 𝑇ℎ𝑒𝑟𝑚𝑎𝑙 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑎𝑛𝑐𝑒
𝑄 𝑐𝑜𝑛𝑣 = ℎ𝐴 𝑠( 𝑇𝑖 − 𝑇∞)
[ 𝜏𝐷 + 1] 𝑇𝑖 = 𝑇∞ +
𝑄𝑖𝑛
ℎ𝐴 𝑠
; 𝑤ℎ𝑒𝑟𝑒 𝜏 =
𝑚𝐶 𝑝
ℎ𝐴 𝑠
The heat inputis200 watts.The diameterof the copperball is100 mm.The convectioncoefficientis50
Watts/m2
°C.The ambientTemperature is25°C. Basedon thislinearsystemmodel,whatare the steady-
state temperature andthe settlingtime?
Material Propertiesof copper@25°C:
𝜌 = 8933
𝑘𝑔
𝑚3 ; 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑉𝑜𝑙𝑢𝑚𝑒 =
4
3
𝜋(
𝑑
2
)
3
𝑚3
𝑚 = 𝜌𝑉 = 4.68 𝑘𝑔; 𝑚𝑎𝑠𝑠
𝐶 𝑝 = 385
𝐽
𝑘𝑔 ∙ 𝐾
; 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 ℎ𝑒𝑎𝑡 𝑎𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝐴 𝑠 = 𝜋𝑑2 = 0.3142 𝑚2; 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
h = convectioncoefficientof air
Under steadystate the transient 𝑇̇𝑖 = 0,therefore:
𝑇𝑖 = 𝑇∞ +
𝑄𝑖𝑛
ℎ𝐴 𝑠
= 152.3°𝐶
The settlingtime isapproximately4𝜏 𝑤ℎ𝑒𝑟𝑒 𝜏 =
𝑚𝐶 𝑝
ℎ 𝐴 𝑠
:
4𝜏 = 4588 𝑠𝑒𝑐𝑜𝑛𝑑𝑠
Ti
Qconv
T∞
Qin

2. Part 1 Discussion
My initial considerationsof the situationpresented,includedconsiderationthatwe learnedinMaterial
PropertiesClassthatCopperhasa highthermal andelectrical conductivity.Withthatinmind,Ithought
of the size of the material andthe shape,basicallysomethinglike aball bearing.The situationthatIdo
not fullygraspisthe heatintothe iteminwatts. Due to lack of experience andnopractical equivalent
to the 200 watt amountof energyrate in mindotherthana realizationthata 100 watt lightbulbwould
be verywarm ina relativelyshortperiodof time.Myinitial guessforthe steadystate conditionwas2
minutes. These resultswere surprising.
Basedon thistemperature,doyouthinkyoushouldhave included radiation?
Initiallythe guessforconsideringradiationisyesandthiswasbasedona recentprojectcompletedfor
Heat Transferclasswhere we were heatingwaterwithwarmairanddeterminingthe lengthof the pipe
neededtosuccessfullyreachaspecifiedoutputtemperature. Inthisprojectitbecame apparentthat
the waterhas a grosslyhigherheattransfercoefficientascomparedtothe air and wasthe dominant
propertyto considerwhenbalancingthe heatrate intothe control volume.Since the convection
coefficient of airisrelativelylow,thiscouldallow radiationtoplayasignificantrole inthe processof
heatlossto the environment.
Nowconsiderthe radiationterminyourmodelingequationsandderiveastate-space modelforinternal
temperature.
𝑄𝑖𝑛 − 𝑄 𝑐𝑜𝑛𝑣 − 𝑄 𝑟𝑎𝑑 = 𝑚𝐶 𝑝 𝑇̇𝑖
𝑄 𝑐𝑜𝑛𝑣 = ℎ𝐴 𝑠( 𝑇𝑖 − 𝑇∞)
𝑄 𝑟𝑎𝑑 = 𝐹𝑒 𝐹𝑣 𝜎𝐴 𝑠( 𝑇𝑖
4
− 𝑇∞
4
);
𝑊ℎ𝑒𝑟𝑒 𝐹𝑒 = 𝑒𝑚𝑖𝑠𝑠𝑖𝑣𝑖𝑡𝑦 = 1 𝑎𝑛𝑑 𝐹𝑣 = 𝑣𝑖𝑒𝑤 𝑓𝑎𝑐𝑡𝑜𝑟 = 1 𝑎𝑛𝑑
𝜎 = 𝑡ℎ𝑒 𝑆𝑡𝑒𝑓𝑎𝑛 − 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 56.68 × 10−9
𝑤𝑎𝑡𝑡
𝑚2°𝐾4
𝑈 ≜ 𝑄𝑖𝑛
𝑥(1) ≜ 𝑇𝑖
𝑥̇(1) ≜ 𝑇̇𝑖 =
1
𝑚𝐶 𝑝
[ 𝑈 − ℎ𝐴 𝑠( 𝑥(1) − 𝑇∞) − 𝜎𝐴 𝑠(𝑥(1)4 − 𝑇∞
4
)]
Ti
Qconv
T∞
Qin
Qrad

3. Performa MATLAB simulationof the systemneglectingradiation.
Performa MATLAB simulationof the systemconsideringradiation.
Plotthe two inthe same graph. Expressthe internal temperature indegreesCelsius(notKelvin).
Compare the difference inthe final temperature andthe settlingtime.
MATLAB CODE
clc;
clear all;
%Main Program for Dynamic Simulation
n = 1; %order of the system
x0 = zeros (n,1); %reserves x0
x0(1) = 25+273; %initial conditions
Tinit = 0; %initial time
Tfinal =8000; %final time
Tspan = [Tinit,Tfinal];
%integrate
[t,x] = ode45(@template_hx1,Tspan,x0);
%plot output
n = 1; %order of the system
y0 = zeros (n,1); %reserves x0
y0(1) = 25+273; %initial conditions
Tinit = 0; %initial time
Tfinal =8000; %final time
Tspan = [Tinit,Tfinal];
%integrate
[z,y] = ode45(@template_hx2,Tspan,y0);
%plot output
figure
plot(t,x(:,1)-273,'--');hold on
plot(z,y(:,1)-273,'-o');hold on
grid
legend('Convection','Convection and Radiation')
legend('Location','Southeast')
xlabel('Time (s)')
ylabel ('Internal Temperature, T_i^{circ}C','fontsize',16)
Dx = zeros (1,1);
Tin=25+273;
Q=200;
h=50;
d=(100/1000);
As=pi*d^(2);
V=(4/3)*pi*(d/2)^(3);
cp = 385;
p=8933;
m=p*V;
tau = (m*cp)/(h*As);

4. Dx(1)=(1/tau)*(Tin+(Q/(h*As))-x(1));
function Dy=template_hx2(z,y)
Dy = zeros (1,1);
Tin=25+273;
Q=200;
hc=50;
d=(100/1000);
As=pi*d^(2);
V=(4/3)*pi*(d/2)^(3);
cp = 385;
p=8933;
m=p*V;
s=56.68*10^(-9);
Dy(1)=(1/(m*cp))*(Q -(hc*As*(y(1)-Tin))-(s*As*(y(1)^4-Tin^4)));
PLOT:
Lookingat the MATLAB createdarrays for eachfunctionplottedthe (4tau,0.98*T) positionwas
obtained. Forthe convectiononlycase the pointisat (4540 s , 149°C) and for the convectionand
radiationcase the correspondingpointis(3340s , 128°C). Tfinal forthe convectiononlycase was152°C
and Tfinal forthe convectionandradiationcase was 131°C. Comparingthese results itisapparentthat
withconvectionandradiationthatthe final temperature isloweranditreachessteadystate faster. This
resultiseasiertoobtainif youconsiderthe equivalentradiationheattransfercoefficienththerefore
0 1000 2000 3000 4000 5000 6000 7000 8000
20
40
60
80
100
120
140
160
Time (s)
InternalTemperature,Ti

C
Convection
Convection and Radiation

5. linearizingthe convectionandradiationscenario. Consideringthishvalue the equationforthe
convectionandradiationcase is:
[ 𝜏𝐷 + 1] 𝑇𝑖 = 𝑇∞ +
𝑄𝑖𝑛
(ℎ 𝑐𝑜𝑛𝑣 + ℎ 𝑟𝑎𝑑)𝐴𝑠
; 𝑤ℎ𝑒𝑟𝑒 𝜏 =
𝑚𝐶 𝑝
(ℎ 𝑐𝑜𝑛𝑣 + ℎ 𝑟𝑎𝑑)𝐴𝑠
Basicallyitiseasyto see that the final steadystate temperaturewillvaryfromthe convectiononlycase
by the size of the radiationheattransferconvectioncoefficientrelationandthisisthe same forthe time
constantfor the system.