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!
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
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Question
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,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fd72755-fbcd-4bf5-b9e3-e08fe5879c40%2Fbb72308d-fb91-4cab-8ff3-b303d7ebb7e3%2Fbvsrou7_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9fd72755-fbcd-4bf5-b9e3-e08fe5879c40%2Fbb72308d-fb91-4cab-8ff3-b303d7ebb7e3%2Fb116u3d_processed.png&w=3840&q=75)
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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