Heat Transfer -- Correlation for heat transfer by natural convection from a horizontal pipe

  • Thread starter Tiberious
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  • #1
Tiberious
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Homework Statement



An appropriate correlation for heat transfer by natural convection from a horizontal pipe to the atmosphere is


Nu=0.53Gr^0.25 Pr^0.25

Where,

Gr= (αp^2 d^3 (T_1-T_f )g)/μ^2

And

Pr⁡〖= (C_p μ)/k〗

Show the above correlation can be simplified to

h ≈1.34((T_s-T_f)/d)^0.25 Wm^(-2) K^(-1)

When air has the values listed below

α=3.077∙10^(-3 ) K^(-1)
p=1.086 kg m^(-3)
C_p=1.0063 kj kg^(-1) K^(-1)
k=2.816∙10^(-5) kWm^(-1) K^(-1)
μ=1.962∙10^(-5) kg m^(-1) s^(-1)

The Attempt at a Solution




Inputting the Grashof and Prandtl number equations:

Nu=0.53〖(αp^2 d^3 (T_1-T_f )g)/μ^2 〗^0.25 〖(C_p μ)/k〗^0.25

As,

N_u= hd/k

Replacing this into our equation,

hd/k=0.53〖(αp^2 d^3 (T_1-T_f )g)/μ^2 〗^0.25 〖(C_p μ)/k〗^0.25



Rearranging,

h=k/d 0.53〖(αp^2 d^3 (T_1-T_f )g)/μ^2 〗^0.25 〖(C_p μ)/k〗^0.25

Distributing through the exponents,

Struggling to figure this one out. Any assistance is appreciated.

I know I have to Distribute though the terms. Just a little unsure in what order to carry our the operations.
 

Answers and Replies

  • #2
BvU
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Just a little unsure in what order to carry our the operations
Shouldn't matter ! So, please continue!
 
  • #3
Tiberious
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Thank-you for your hasty reply.

So, do I multiply each term by the power of 0.25. Apologies, I've been staring at this for ages and am at a loss.
 
  • #4
BvU
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You have ##\Delta T^{0.25}## already. ##(d^3)^{0.25}## and ## 1/d ## give you the expected power of ##d##. The remainder is numerical.
 
  • #5
Tiberious
73
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Does this seem correct as the next stage ? Or, do we multiply each term ?

h= (2.816∙10^(-5))/d 0.53〖(〖3.077∙10^(-3 )∙1.086〗^2 d^3 (T_s-T_f )9.81)/〖1.962∙10^(-5)〗^2 〗^0.25 〖(1.0063∙1.962∙10^(-5))/(2.816∙10^(-5) )〗^0.25
 
  • #6
BvU
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Try it and see where it takes you;
the end of the tunnel is that you want to end up at 1.34
 
  • #7
Tiberious
73
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Okay - I've found my way to the below. Now I'm stuck.

0.0044h= (2.816∙10^(-5))/d 0.53((0.2355)∙(1.042) d^0.75 (T_s-T_f )^0.25 1.7698)(0.915)
 
  • #8
BvU
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What is the problem ?
Lump all the numbers in one and I do hope you know what ##d^{0.75}/d## is ?
 
  • #9
Tiberious
73
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BvU - Thanks.

I got to the below eventually..

h=1.34 (T_s-T_f )^0.25/d^0.25

From this we can derive that,

h ≈1.34((T_s-T_f)/d)^0.25
 

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