Respuesta :
Answer:
the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW
the isentropic pump efficiency is 78%
Explanation:
Given that;
m = 1.2 kg/sec
T = 50 degree Celsius { Vf = 0.001012 m^3/kg}
P1 = 1.5 Mpa
P2 = 15 Mpa
W-actual = 21 kw
W reversible = m*Vf (p2 - p1)
= 1.2 * 0.001012 * ( 15*10^3 - 1.5*10^3)
= 1.2 * 0.001012 * 13500
= 16.39 kW
the work required by a reversible pump operating with the same conditions, in kW is 16.39 kW
Isentropic Pump efficiency = W-reversible / W-actual
= 16.39 / 21 = 0.78
= 78%
the isentropic pump efficiency is 78%
A) The work required by a reversible pump operating with the same conditions, in kW is; 16.39 kW
B) The Isentropic pump efficiency for the given work condition is;
η_p ≈ 78%
Isentropic Efficiency
We are given;
- Mass transfer; m' = 1.2 kg/sec
- Temperature; T = 50 °C
- Pressure at entry; P₁ = 1.5 Mpa = 1500 kPa
- Pressure at exit; P₂ = 15 Mpa = 15000 kPa
- Work required by pump; W_actual = 21 kw
From steam tables online, at a temperature of 50 °C, the specific volume of saturated liquid is;
υ_f = 0.001012 m³/kg
A) Formula for work required by the reversible pump is;
W_reversible = m' × υ_f × (P₂ - P₁)
W_reversible = 1.2 × 0.001012 × ( 15000 - 1500)
W_reversible = 1.2 × 0.001012 × 13500
W_reversible ≈ 16.39 kW
B) The formula for the isentropic pump efficiency is;
η_p = (W_reversible/W_actual) * 100%
η_p = (16.39/21) * 100%
η_p = 78.05%
η_p ≈ 78%
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