Question

Consider atmospheric air at 20°C and a velocity of 30 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 130°C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 × 105, and 106 .

Answer 1

Answer:

$At 10^5 = 17270W/m$

$At 5*10^5 = 13127.84 W/m$

$At 10^6 = 8472.2 W/m$

Explanation:

We are given:

T_o = 20°C

Flat plate temp, T_s = 230°C

v = 30m/s

Let's find the mean:

$T_f = \frac{20+130}{2}$

= 75°C

Using air table at 75°C at 1 atm pressure, we have:

$Thermal conductivity, k = 29.66*10^-^3 W/m•K$

Prandtl number, Pr = 0.7087

$Viscosity, v= 20.82*10^-^6 m^2/s$

Calculating Reynolds number, we have:

$Re_L = \frac{VL}{v}$

$Re_L = \frac{30*1}{20.82*10^-^6}$

$= 1.44*10^6$

Since the Reynolds number is higher than the critical Reynolds number, we can say there is a turbulent flow.

Therefore we'll now use the formula:

$Nu_L = (0.037Re_L^0^.^8 - A)Pr^0^.^3^3$

$For Re_x = 10^5$

To get A using the formula, we have:

$A = 0.037(10^5) - 0.664(10^5)^0^.^5$

= 160.024

$To find Nu_L$

$Nu_L =(0.037(1.44*10^6)^0^.^8-160.024)(0.7087)^0^.^3^3$

= 2646.78

Connective heat transfer:

$h_L = \frac{Nu_L K}{L}$

$=\frac{(2646.78)(2966*10^-^3)}{1}$

$=78.50W/m^2•K$

Lets calc for heat transfer:

$q_1_0_^_5 = 2h_L L(T_s-T_o)$

= 2(78.50)(1)(130-20)

= 1720W/m

$For Re_x = 5*10^5$

$A = 0.037Re_x^0^.^8-0.664Re_x^0^.^5$

$= 0.037(5*10^5)^0^.^8-0.664(5*10^5)^0^.^5$

= 871.323

Calculating Nu_L we have:

$h_L =\frac{Nu_L K}{L}$

$= \frac{2011.88*29.66*10^-^3}{1}$

= 59.672W/m²•K

For the heat transfer:

$q_5_*_1_0_^5 = 2(59.672)(1)(130-20)$

= 13127.84W/m

$For Re_x = 10^6$

$A= 0.037(10^6)^0^.^8-0.664(10^6)^0^.^5$

= 1670.54

Calculating Nu_L:

$Nu_L = (0.037(1.44*10^6)^0^.^8-1670.54)(0.7087)^0^.^3^3$

= 1298.51

Calculating the convective heat transfer coefficient:

$h_L = \frac{Nu_L*k}{L}$

$\frac{(1298.51)(29.66*10^-^3)}{1}$

=38.51 W/m²•K

To calculte for rate of heat transfer, we have:

$q_1_0_^6=2h_LL(T_s-T_o)$

=2(38.51)(1)(130-20)

= 8472.2W/m