Answer:
POWER INPUT = 82.989 KW
Explanation:
For pressure P =0.24 MPa and T_1 = 0 DEGREE, Enthalapy _1 = 248.89 kJ/kg
For pressure P =1 MPa and T_1 = 50 DEGREE, Enthalapy _1 = 280.19 kJ/kg
Heat loss Q = 0.05w
Inlet diameter = 3 cm
exit diamter = 1.5 cm
volume of tank will be v = area * velocity
velocity 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/kg
wc in KWH
we 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]
