Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 2 (Q.No. 1)
1.
Which area is used in case of heat flow by conduction through a cylinder ?
Discussion:
5 comments Page 1 of 1.
Ghori Mayur said:
1 decade ago
Why using this area? any reason behind it ?
Arpit said:
10 years ago
Just go through the derivation, and you will know the reason.
K.Rama Krishna said:
8 years ago
The formula of heat transfer by conduction in cylindrical coordinate is:
q=2*π*k*L*(T1-T2)/ln(r2/r1).
So, the area used in case of heat flow by conduction through a cylinder is the logarithmic mean area.
q=2*π*k*L*(T1-T2)/ln(r2/r1).
So, the area used in case of heat flow by conduction through a cylinder is the logarithmic mean area.
Jaat boy said:
6 years ago
Yes due to radius ratio.
Gemedo geleto said:
2 years ago
Q = -KA dT/dx ---> 1
But for cylinder;
A=2*π *r*L----------> 2
and dx = dr---------->3
Substitute 2 and 3 into 1 and integrate;
Q=(T1-T2)/((1/2pkL)*ln(r2/r1)----> LMTD
=> we use Ac= 2prL.
But for cylinder;
A=2*π *r*L----------> 2
and dx = dr---------->3
Substitute 2 and 3 into 1 and integrate;
Q=(T1-T2)/((1/2pkL)*ln(r2/r1)----> LMTD
=> we use Ac= 2prL.
(7)
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