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16 May, 09:16

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

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  1. 16 May, 11:21
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    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
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