Question

L2 Two identical rectangular plates, A and B, of dimensions LixLa are at the same temperature. Both plates are placed parallel to a moving fluid. Plate A is oriented with side Li facing the flow while side L2 of plate B faces the flow → If the flow over plate A is laminar, what is it for B? If the heat transfer rate from plate A is 435 W, what is the rate B ? a) b) from plate L1

Answer 1

a)

Reynolds number for flow over a plate is given as,

$Re= \frac{\rho vL}{\mu}$, where 'L' is the characteristic length which is equal to the dimension of the plate along the flow.

For plate A, $Re_A= \frac{\rho vL_2}{\mu}$

Also, since flow is laminar,

$Re_A<5\times 10^5$

$\Rightarrow \frac{\rho vL_2}{\mu}<5\times 10^5$

Now, $Re_B= \frac{\rho vL_1}{\mu}$

$\because \: L1<L2$

$\therefore Re_B<Re_A<5\times 10^5$

Thus, the flow over plate B will be laminar also.

b)

Heat transfer rate from the plate will be given as,

$\dot{Q}=hA\Delta T$

For plates A and B, heat transfer area, 'A' and temperature difference, $\Delta T$ will be same.

Now, average heat transfer coefficient for the plate under laminar flow is given as,

$h= 2\times h_{trailing\: edge}$

i.e.; $h_A= 2\times h_{x=L2}$ and $h_B= 2\times h_{x=L1}$

where, $Nu_x=\frac{h_x x}{k}=0.332Re_x^{1/2}Pr^{1/2}$

$\Rightarrow h_x\propto \frac{1}{\sqrt{x}}$

For the plates A and B, all other quantities will be same except the characteristic length 'x'.

Thus, $\frac{h_A}{h_B}=\sqrt{\frac{L_1}{L_2}}$

$\Rightarrow \frac{Q_A}{Q_B}=\frac{h_A A \Delta T}{h_B A \Delta T}=\sqrt{\frac{L_1}{L_2}}$

Q_{​​​​​​A} is known. Knowing ratio of the two dimensions of the plate, Q_{​​​​​​B} can be easily calculated.