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1 July, 03:18

Refrigerant 134a enters a horizontal pipe operating at steady state at 40oC, 300 kPaand a velocity of 40 m/s. At the exit, the temperature is 50oC and the pressure is 240 kPa. The pipe diameter is 0.04 m. Determine (a) the mass flow rate of the refrigerant, in kg/s, (b) the velocity at the exit, in m/s, and (c) the rate of heat transfer between the pipe and its surroundings, in kW

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  1. 1 July, 06:17
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    1. Mass flow rate = 0.621Kg/s

    2. Velocity = 52.2m/s

    3. Rate = + 6.82kW

    Explanation:

    Given

    Refrigerant = R134a

    Diameter, d = 0.04m

    T1 = 40°C

    P1 = 300kPa

    V1 = 40m/s

    T2 = 50°C

    P2 = 240kPa

    a.

    The specific volume of refrigerant R134a at T1 and P1 from Refrigerant table (A12) = v1 = 0.08089 m³/kg

    The specific volume of refrigerant R134a at T2 and P2 from Refrigerant table (A12) = v2 = 0.10562 m³/kg.

    Mass flow rate is calculated as follows ...

    = AV/v

    Where A = Area = πd²/4

    V = velocity =,40

    v = v1 = 0.08089 m³/kg.

    So. Mass flow rate = (π * 0.04²/4 * 40) / 0.0809

    Mass flow rate = 0.621Kg/s

    b.

    Calculating Velocity at the exit.

    This is given as;

    m * v2 / A

    0.621 * 0.10562 / (π * 0.04²/4)

    = 52.2m/s

    c. Calculating the rate of heat transfer between the pipe and its surroundings

    From energy equation, we have.

    Q - > m * [ (h2 - h1) + ((, V2² - V1²) / 2) ]

    Where h2 is the specific enthalpy for refrigerant at T1 and P1

    h2 = 234.05 Kj/Kg

    And h1 is the specific enthalpy for refrigerant at T2 and P2

    h2 = 244.47 Kj/Kg

    Rate = the specific enthalpy for refrigerant at T1 and P1

    h2 = 234.05 Kj/Kg

    Rate = 0.621[ (234.05 - 244.47) + (52.2² - 40²) / 2]

    Rate = 0.621[ (234.05 - 244.47) +

    562.42] - - - Convert the kW

    Rate = 0.621 ((244.47 - 234.05) +

    562.42/1000)

    Rate = + 6.82kW
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