Question

A coolant fluid at ${30}^{\circ }$C flows over a heated flat plate maintained at a constant temperature of ${100}^{\circ }$C. The boundary layer temperature distribution at a given location on the plate may be approximated as T= 30+70 exp(-y) where y (in m) is the distance normal to the plate and T is in${}^{\circ }$C. If thermal conductivity of the fluid is 1.0 W/mK, the local convective heat transfer coefficient (in $W/m^{2}k$) at the location will be

• A. 0.2
• B. 1
• C. 5
• D. 10

Answer 1

The correct option is B 1


At y=0; $q_{cond}=q_{conv}$

$-k_f{\begin{array}{cc}& \frac{\partial T}{\partial y}\end{array}|}_{y=0}=h\Delta T=h(T_s-T_{\infty })$

${\begin{array}{cc}& -\left(1\right)\frac{\partial }{\partial y}[30+70e^{-y}]\end{array}|}_{y=0}=h(100-30)$

${\begin{array}{cc}& -[0+70(-1\left)e^{-y}\right]\end{array}|}_{y=0}=h\left(70\right)$

$70e^{-0}=h\left(70\right)$

or $h=1W/m^{2}K$

Alternative solution

Given data:

$T_{\infty }={30}^{\circ }C$

$T_s={100}^{\circ }C$

$k_t=1W/mK$

T=30 + 70exp(-y)

Differentiating w.r.t. y, we get

$\frac{dT}{dy}=-70e^{-y}$

At y=0

${\left(\frac{dT}{dy}\right)}_{y=0}=-70$

We know that local convective heat transfer coefficient:

$h_x=\frac{-k_f{\left(\frac{dT}{dy}\right)}_{y=0}}{T_s-T\infty }=\frac{-1\times (-70)}{100-30}=1W/m^{2}K$