Answer:
Input Power = 6.341 KW
Explanation:
First, we need to calculate enthalpy of the water at inlet and exit state.
At inlet, water is at 20° C and 100 KPa. Under these conditions from saturated water table:
Since the water is in compresses liquid state and the data is not available in compressed liquid chart. Therefore, we use approximation:
h₁ = hf at 20° C = 83.915 KJ/kg
s₁ = sf at 20° C = 0.2965 KJ/kg.k
At the exit state,
P₂ = 5 M Pa
s₂ = s₁ = 0.2965 K J / kg.k (Isentropic Process)
Since Sg at 5 M Pa is greater than s₂. Therefore, water is in compresses liquid state. Therefore, from compressed liquid property table:
h₂ = 88.94 KJ/kg
Now, the total work done by the pump can be calculated as:
Pump Work = W = (Mass Flow Rate)(h₂ - h₁)
W = (53 kg/min)(1 min/60 sec)(88.94 KJ/kg - 83.915 KJ/kg)
W = 4.438 KW
The efficiency of pump is given as:
efficiency = η = Pump Work/Input Power
Input Power = W/η
Input Power = 4.438 KW/0.7
<u>Input Power = 6.341 KW</u>
