HEAT TRANSFER CASE:  I want to know what temperature in (°F) the cylinder will have inside. It's a heat transfer problem. what is T2 ? HEAT TRANSFER They gave me an answer all squashed together that i can't make sense of it. If you could help me makes sense of it thank you!

Elements Of Electromagnetics
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Author:Sadiku, Matthew N. O.
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HEAT TRANSFER

CASE:  I want to know what temperature in (°F) the cylinder will have inside. It's a heat transfer problem.

what is T2 ?

HEAT TRANSFER

They gave me an answer all squashed together that i can't make sense of it. If you could help me makes sense of it thank you!

Step 1
Given that,
Inner radius, ri=0.55 in,Outer radius, ro=0.7 in,Length, L=5.7 in,Thermal conductivity, k=15 W/m.K,
1 W/m.K=6.9334 Btu.in/h.ft2°F
Step 2
The conduction heat transfer from the steel cylinder will be equal to the convection heat transfer,
Qcond=Qconv2+kLlnroriT2-T1=h2kLT1-To2äkLlnroriT2-T1=h2¬LT1-
ToLet h of air = 18 Btu/h.ft2.°F,klnroriT2-T1=hT1-To6.9334×15ln0.700.55T2-75.2=1875.2-68431.25T2-
75.2=129.6T2=75.5 °F,T2=6.71+75.2T2=81.91 °F,
Transcribed Image Text:Step 1 Given that, Inner radius, ri=0.55 in,Outer radius, ro=0.7 in,Length, L=5.7 in,Thermal conductivity, k=15 W/m.K, 1 W/m.K=6.9334 Btu.in/h.ft2°F Step 2 The conduction heat transfer from the steel cylinder will be equal to the convection heat transfer, Qcond=Qconv2+kLlnroriT2-T1=h2kLT1-To2äkLlnroriT2-T1=h2¬LT1- ToLet h of air = 18 Btu/h.ft2.°F,klnroriT2-T1=hT1-To6.9334×15ln0.700.55T2-75.2=1875.2-68431.25T2- 75.2=129.6T2=75.5 °F,T2=6.71+75.2T2=81.91 °F,
MATERIAL DATA
Stainless steel
value of
conductivity: 15
w/mk
12
No
ri: 0.55 in
ro:0.7 in
lenght: 5.7 in
the temperature
touching the
T₁ cylinder outside is
75.2 Fº
Ambient
temeperature 68 Fº
Target heating time:
20 s
Transcribed Image Text:MATERIAL DATA Stainless steel value of conductivity: 15 w/mk 12 No ri: 0.55 in ro:0.7 in lenght: 5.7 in the temperature touching the T₁ cylinder outside is 75.2 Fº Ambient temeperature 68 Fº Target heating time: 20 s
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