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Eq APSRC

1) The document presents equations for calculating the temperature profile T of a cylindrical body exposed to convection. 2) Boundary conditions are defined relating the heat flux at the surface to the temperature difference with the surroundings. 3) Coefficients in the equations like A0, B0, A1, etc. are defined in terms of parameters like the Biot number and the radius ratio a/b.

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0% found this document useful (0 votes)
39 views1 page

Eq APSRC

1) The document presents equations for calculating the temperature profile T of a cylindrical body exposed to convection. 2) Boundary conditions are defined relating the heat flux at the surface to the temperature difference with the surroundings. 3) Coefficients in the equations like A0, B0, A1, etc. are defined in terms of parameters like the Biot number and the radius ratio a/b.

Uploaded by

aefontalvo
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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r ∞   r n  n 

X b
T̄ = A0 + B0 ln + An + Bn cos (nφ)
b n=1
b r
T − Tf αq”0 b
T̄ = k ⇒ T = Tf + T̄
αq”0 b k
∞ 
(  n  )
αq”0 b r X  r n b
T = Tf + A0 + B0 ln + An + Bn cos (nφ)
k b n=1
b r
∞ 
(  )
∂T αq”0 b B0 X An n−1 n −n−1
= + n nr − nBn b r cos (nφ)
∂r k r n=1
b
Boundary conditions
 ( ∞    )
αq”0 b X a n  a −n 
hf (T − Tf ) = B0 + n An − Bn cos (nφ)
k a n=1
b b

( )
αq”0 X
hext (T − Tamb ) = B0 + n [An − Bn ] cos (nφ)
k n=1
Where
( 2 amb )(1−Bi1 ln(a/b))
 1+πBi T̄

 π[Bi1 +Bi2 (1−Bi1 ln(a/b))]
 n=0
1
An = 2[g10 (1)+Bi2 g1 (1)]
n=1

2 cos(nπ/2)

n>1

π(1−n2 )[gn 0 (1)+Bi g (1)]
2 n
(
A0 1−Bi1Biln(a/b)
1
n=0
Bn = a
 2n n−Bi1
An b n+Bi1 n≥1
 a 2n  n − Bi 
1
gn (1) = 1 +
b n + Bi1
  a 2n  n − Bi 
0 1
gn (1) = n 1 +
b n + Bi1
Boundary conditions evaluated
      a −1      a −2  
αq”0 b a a 2
hf (T − Tf ) = B 0 + A1 − B1 cos (φ) + 2 A2 − B2 cos (2φ)
k a b b b b
αq”0
hext (T − Tamb ) = {B0 + [A1 − B1 ] cos (φ) + 2 [A2 − B2 ] cos (2φ)}
k
Where the coefficients are the following 
1 + πBi2 T̄amb (1 − Bi1 ln (a/b))
A0 =
π [Bi1 + Bi2 (1 − Bi1 ln (a/b))]
Bi1
B 0 = A0
1 − Bi1 ln (a/b)
1
A1 = 0
2 [g1 (1) + Bi2 g1 (1)]
 a 2  1 − Bi 
1
B 1 = A1
b 1 + Bi1
2
A2 =
3π [g20 (1) + Bi2 g2 (1)]
 a 4  2 − Bi 
1
B 2 = A2
b 2 + Bi1
 a 2  1 − Bi 
1
g1 (1) = 1 +
b 1 + Bi1
 a 2  1 − Bi 
1
g10 (1) = 1 +
b 1 + Bi1
 a 4  2 − Bi 
1
g2 (1) = 1 +
b 2 + Bi1
 a 4  2 − Bi 
1
g20 (1) = 2 + 2
b 2 + Bi1

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