Respuesta :
Answer:
[tex]\dfrac{\bar{h}}{h}=\dfrac{5}{4}[/tex]
Explanation:
Given that
[tex]Nu_x=0.035Re_x^{0.8} Pr^{1/3}[/tex]
We know that
Rex=ρvx/μ
So
[tex]Nu_x=0.035Re_x^{0.8} Pr^{1/3}[/tex]
[tex]Nu_x=0.035\times\left(\dfrac{\rho vx}{\mu}\right)^{0.8}Pr^{1/3}[/tex]
All other quantities are constant only x is a variable in the above equation .so lets take all other quantities as a constant C
[tex]Nu_x=C.x^{0.8}=C.x^{4/5}[/tex]
We also know that
Nux=hx/K
[tex]C.x^{4/5}=\dfrac{hx}{k}[/tex]
m is the constant
[tex]h=mx^{-1/5}[/tex]
This is local heat transfer coefficient
The average value of h given as
[tex]\bar{h}=\dfrac{\int_{0}^{L}hdx}{L}[/tex]
[tex]\bar{h}=\dfrac{5m}{4}\times\dfrac{L^{4/5}}{L}[/tex]
[tex]\bar{h}=\dfrac{5m}{4}L^{-1/5}[/tex] ---------1
The value of local heat transfer coefficient at x=L
[tex]h=mx^{-1/5}[/tex]
[tex]h=mL^{-1/5}[/tex] -----------2
From 1 and 2 we can say that
[tex]\dfrac{\bar{h}}{h}=\dfrac{5}{4}[/tex]
