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Homework help:Heat transfer coefficient of flow over a flat plate

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Peridot
#1
Jun16-12, 12:20 AM
P: 4
A 2-m x 3-m flat plate is suspended in a room, and is subjected to air flow parallel to its surfaces along its 3-m-long side. The free stream temperature and velocity of the air are 20oC 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/m2K



Properties of air at 300K(from my textbook
ρ=1.1763
c=1.007e03
μ=1.862e-05
v=1.5e-05
Pr=0.717
k=2.614e-02




Attempt at a solution
ReL=UL/v=7*3/1.5e-05=1.5e06 >5e05
xcr=(Recrv)/U=1.07

Nux,Laminar=0.332Pr1/3Rex1/2
Nux,Turbulent=(0.0296Re0.8Pr)/(1+2.11Rex-0.1(Pr-1))

Avg heat coeff
=(1/L){0xcrhx, Laminardx + xcr3hx, Turbulentdx}
= (1/L){0xcrNux, Laminar(k/x)dx + xcr3Nux, Turbulent(k/x)dx}
= ...
=14.39


I did approximate 1+2.11Rex-0.1(Pr-1) 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.
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LawrenceC
#2
Jun18-12, 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.
Peridot
#3
Jun18-12, 07:04 PM
P: 4
Thanks, I'll give it a shot again.

Peridot
#4
Jun19-12, 06:03 AM
P: 4
Homework help:Heat transfer coefficient of flow over a flat plate

Quote Quote by LawrenceC View Post
Hint: The problem states the force on the plate. That implies they want you to approach it from a Reynold's Analogy standpoint.
Okay thanks for the hint once again. I tried again using Reynold's Analogy.

qx/(ρcu(T-Tw))=τwx/(ρu2)


hx/c=τwx/(u)

τwx = (hxu)/(c)

AτwxdA = 0.86

A(hxu)/(c)dA = 0.86

(u/c)∫AhxdA = 0.86

(u/c)∫AhxdA = 0.86

AhxdA = (0.86c)/u

avg h
= (1/A)∫AhxdA
= (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
LawrenceC
#5
Jun19-12, 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.
Peridot
#6
Jun19-12, 09:32 PM
P: 4
Quote Quote by LawrenceC View Post
I started off with the basic relationship:

St*Pr^2/3 = Cf/2

I don't see the Prndtl number anywhere in your computations.
Oops, I misread my notes, the formula I used earlier was only for Pr=1 >.<

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.

Stx = Nux/(RexPr) = hx/cpU

StxPr2/3= Cfx/2=тwx/(pU2)

(hx/cpU)Pr2/3= тwx(pU2)

hx = (c/U)(Pr-2/3wx

AhxdA = (c/U)(Pr-2/3AтwxdA

Avg heat coeff
=(1/A)∫AhxdA
=(1/A)(c/U)(Pr-2/3AтwxdA

sub A=12, c=1.007e03, U=7, Pr=0.717, ∫AтwxdA = 0.86

Avg heat coeff=12.86


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