Answer:
16.21 kW
Explanation:
Solution
Given that,
The velocity of wind = 55 km/hr
The length of the wall L = 10m
The height of the wall w = 4m
The surface temperature at wall Ts = 12° C
Temperature of air T∞ = 5°C
Now,
The properties of the air at atm and average film temperature =( 12 + 5)/2 = 8.5°C, which is taken from the air table properties.
k= 0.02428 W/m°C
v= 1.413 *10 ^⁻5
Pr =0.7340
Now,
Recall Reynolds number when air flow parallel to 10 m side
[ 55 * 1000/3600) m/s (10 m)/1.413 *10^⁻5 m²/s
Rel =1.081 * 10⁷
This value is greater than Reynolds number.
The nusselt number is computed as follows:
Nu =hL/k
(0.037Rel^0.08 - 871)Pr^1/3
Nu =1.336 * 10 ^4
The heat transfer coefficient is
h = k/L Nu
= 0.2428 W/m°C /10 m (1.336 * 10 ^4)
h = 32.43 W/m°C
The heat transfer area of surface,
As = 40 m²
= ( 4 m) (10 m)
As = 40 m²
The rate of heat transfer is determined as follows:
Q = hAs( Ts - T∞)
= (32.43 W/m²°C) (40 m) (12 - 5)°C
=9081 W
Q = 9.08 kW
When the velocity is doubled,
let say V = 110km/hr
The Reynolds number is
Rel = VL/v
= [110 * 100/3600) m/s] (10 m)/ 1.413 *10^⁻5 m²/s
Rel = 2.163 * 10 ^7
This value is greater for critical Reynolds number
The nusselt number is computed as follows:
Nu =hL/k
(0.037Rel^0.08 - 871)Pr^1/3
[0.037 ( 2.163 * 10 ^7)^0.08 - 871] (0.7340)^1/3
Nu =2.344 * 10^4
The heat transfer coefficient is
h = k/L Nu
= 0.2428 W/m°C /10 m (2.384 * 10 ^4)
h= 57.88 W/m²°C
The heat transfer area of surface,
As = wL
= ( 10 m) (4 m)
As = 40 m²
he rate of heat transfer is determined as follows:
Q = hAs( Ts - T∞)
= (57.88 W/m²°C) (40 m²) (12 - 5)°C
= 16,207 W
= 16.21 kW
