Heat transfer coefficient for flow over a 2m x 3m flat plate with given drag force

[Peridot]

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 20^oC 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
Re_L=UL/v=7*3/1.5e-05=1.5e06 >5e05
x_cr=(Re_crv)/U=1.07

Nu_x,Laminar=0.332Pr^1/3Re_x^1/2
Nu_x,Turbulent=(0.0296Re^0.8Pr)/(1+2.11Re_x^-0.1(Pr-1))

Avg heat coeff
=(1/L){₀∫^x_crh_{x, Laminar}dx + _xcr∫³h_{x, Turbulent}dx}
= (1/L){₀∫^x_crNu_{x, Laminar}(k/x)dx + _xcr∫³Nu_{x, Turbulent}(k/x)dx}
= ...
=14.39

I did approximate 1+2.11Re_x^-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.

 

[LawrenceC]

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

 

[Peridot]

Thanks, I'll give it a shot again.

 

[Peridot]

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.

q_x/(ρcu_∞(T_∞-T_w))=τ_wx/(ρu_∞²)h_x/c=τ_wx/(u_∞)

τ_wx = (h_xu_∞)/(c)

∫_Aτ_wxdA = 0.86

∫_A(h_xu_∞)/(c)dA = 0.86

(u_∞/c)∫_Ah_xdA = 0.86

(u_∞/c)∫_Ah_xdA = 0.86

∫_Ah_xdA = (0.86c)/u_∞

avg h
= (1/A)∫_Ah_xdA
= (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]

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]

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.

St_x = Nu_x/(Re_xPr) = h_x/cpU

St_xPr^2/3= C_fx/2=т_wx/(pU²)

(h_x/cpU)Pr^2/3= т_wx(pU²)

h_x = (c/U)(Pr^-2/3)т_wx

∫_Ah_xdA = (c/U)(Pr^-2/3∫_Aт_wxdA

Avg heat coeff
=(1/A)∫_Ah_xdA
=(1/A)(c/U)(Pr^-2/3∫_Aт_wxdA

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

Avg heat coeff=12.86