On a cold winter day, wind at 55 km/hr is blowing parallel to a 4-m high and 10-m long wall of a house. If the air outside is at 5o C and the surface temperature of the wall is 12o C, find the rate of heat loss from the wall by convection

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