
#1
Jun1612, 12:20 AM

P: 4

A 2m x 3m flat plate is suspended in a room, and is subjected to air flow parallel to its surfaces along its 3mlong side. The free stream temperature and velocity of the air are 20^{o}C and 7m/s. the total drag force acting on the plate is measured to be 0.86N. Determine the average heat transfer coefficient for the plate. Answer:h=12.7W/m^{2}K
Properties of air at 300K(from my textbook ρ=1.1763 c=1.007e03 μ=1.862e05 v=1.5e05 Pr=0.717 k=2.614e02 Attempt at a solution Re_{L}=UL/v=7*3/1.5e05=1.5e06 >5e05 x_{cr}=(Re_{cr}v)/U=1.07 Nu_{x,Laminar}=0.332Pr^{1/3}Re_{x}^{1/2} Nu_{x,Turbulent}=(0.0296Re^{0.8}Pr)/(1+2.11Re_{x}^{0.1}(Pr1)) Avg heat coeff =(1/L){_{0}∫^{xcr}h_{x, Laminar}dx + _{xcr}∫^{3}h_{x, Turbulent}dx} = (1/L){_{0}∫^{xcr}Nu_{x, Laminar}(k/x)dx + _{xcr}∫^{3}Nu_{x, Turbulent}(k/x)dx} = ... =14.39 I did approximate 1+2.11Re_{x}^{0.1}(Pr1) to be equal to 1 in my working because I didn't know how to integrate it. Is that acceptable? And I don't know how to use the drag force provided in this question. :x I've been pondering over this question for a few days and I still can't get the answer please help check if my approach is correct. Thank you. 



#2
Jun1812, 02:10 PM

P: 1,195

Hint: The problem states the force on the plate. That implies they want you to approach it from a Reynold's Analogy standpoint.




#3
Jun1812, 07:04 PM

P: 4

Thanks, I'll give it a shot again.




#4
Jun1912, 06:03 AM

P: 4

Homework help:Heat transfer coefficient of flow over a flat plateq_{x}/(ρcu_{∞}(T_{∞}T_{w}))=τ_{wx}/(ρu_{∞}^{2}) h_{x}/c=τ_{wx}/(u_{∞}) τ_{wx} = (h_{x}u_{∞})/(c) ∫_{A}τ_{wx}dA = 0.86 ∫_{A}(h_{x}u_{∞})/(c)dA = 0.86 (u_{∞}/c)∫_{A}h_{x}dA = 0.86 (u_{∞}/c)∫_{A}h_{x}dA = 0.86 ∫_{A}h_{x}dA = (0.86c)/u_{∞} avg h = (1/A)∫_{A}h_{x}dA = (1/A) (0.86c)/u_{∞} Is this approach correct now? I subbed in the values of A=12, c = 1.007e03 and u_{∞}=7 but the answer I got is 10.3 instead of the given 12.7. I think the difference might be attributed to the value of c. Thanks in advance 



#5
Jun1912, 10:37 AM

P: 1,195

I started off with the basic relationship:
St*Pr^2/3 = Cf/2 I don't see the Prndtl number anywhere in your computations. 



#6
Jun1912, 09:32 PM

P: 4

I did as you told me and I got the answer.(at least reasonably close to it) ^^ Thanks a lot!(Sorry if I annoyed you with a lot of stupid mistakes and questions) Here is my working if anyone is interested. St_{x} = Nu_{x}/(Re_{x}Pr) = h_{x}/cpU St_{x}Pr^{2/3}= C_{fx}/2=т_{wx}/(pU^{2}) (h_{x}/cpU)Pr^{2/3}= т_{wx}(pU^{2}) h_{x} = (c/U)(Pr^{2/3})т_{wx} ∫_{A}h_{x}dA = (c/U)(Pr^{2/3}∫_{A}т_{wx}dA Avg heat coeff =(1/A)∫_{A}h_{x}dA =(1/A)(c/U)(Pr^{2/3}∫_{A}т_{wx}dA sub A=12, c=1.007e03, U=7, Pr=0.717, ∫_{A}т_{wx}dA = 0.86 Avg heat coeff=12.86 


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