A compressor operates at steady state with Refrigerant 134a as the working fluid. The refrigerant enters at 0.24 MPa, 0°C, with a volumetric flow rate of 0.64 m/min. The diameters of the inlet and exit pipes are 3 and 1.5 cm, respectively. At the exit, the pressure is 1.0 MPa and the temperature is 50°C. If the magnitude of the heat transfer rate from the compressor to its surroundings is 5% of the compressor power input, determine the power input, in kW.

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  • Answer posted by: rejkjavik
    Answer:POWER INPUT = 82.989 KWExplanation:For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kgFor pressure P =1  MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kgHeat loss Q = 0.05wInlet diameter = 3 cm exit diamter = 1.5  cmvolume of tank will be v = area * velocityvelocity at inlet[tex] = \frac{0.64\60 m/s}{ \frac{\pi}{4} (3\times 10^{-2})^2} = 15.09  m/s[/tex]velocity at outlet[tex] = \frac{0.64\60 m/s}{ \frac{\pi}{4} (1.5\times 10^{-2})^2} = 60.36  m/s[/tex]steady flow energy equation[tex]E_{IN} = E_{OUT}[/tex][tex]h_1 + \frac{v_1^2}{2g} +wc = h_2 + \frac{v_2^2}{2g} + 0.05wc[/tex][tex]248.89 + \frac{15.09^2}{2} + wc = 280.18 + \frac{60.36^2}{2} + 0.05 wc[/tex]solving wc = 1830.64  kJ/kgwc in KWHwe know that[tex] wc = \dot m wc[/tex]        [tex]\dot m = 4.25 kg/m3 \times (0.64/60) m^3/s[/tex][tex]\dot m = 0.04533  kg/s[/tex][tex]wc = 0.04533 \times 1830.64 = 82.989 kW[/tex]

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