Answer:
a) ηpump = 44.44%
b) ΔT = 0.059°C
Explanation:
Given
- Power input to the electric motor: Wmotor = 20 kW
- Electric motor efficiency: ηmotor = 90%
- Water flow rate: V = 40 l/s
- Pressure at the pump inlet: Pint = 120 kPa
- Pressure at the pump outlet: Pout = 320 kPa
- C = 4.186 kJ/Kg*°C
Required:
a) Determine the mechanical efficiency of the pump.
b) Determine the temperature rise of water as it flows through the pump due to mechanical inefficiencies.
Assumptions:
- Steady state operation.
- The elevation difference across the pump is negligible.
Solution:
Mass flow rate of the water could be defined as the following:
m = V/v ⇒ m = 0.04/0.001 = 40 Kg/s
The power supplied to the fluid is obtained from the First Law of Thermodynamics for open system.
Wfluid = m*(Pout - Pin)*v
⇒ Wfluid = (40 Kg/s)*(320 kPa - 120 kPa)*0.001 = 8 kW
The shaft power could be defined as the following
Wshaft = ηmotor*Wmotor
⇒ Wshaft = 0.9*20 = 18 kW
The mechanical efficiency of the pump could be defined as the following:
ηpump = Wfluid/Wshaft
⇒ ηpump = 8 kW/18 kW = 0.44
⇒ ηpump = 44.44%
Of the 18 kW mechanical power supplied by the pump, only 8 kW is imparted to the fluid as mechanical energy. The remaining 10 kW is converted to thermal energy due to frictional effects, and this "lost" mechanical energy manifests itself as a heating effect in the fluid,
Emech,loss = Wshaft - Wfluid = 18 kW - 8 kW = 10 kW
The temperature rise of water due to this mechanical inefficiency is determined from the thermal energy balance,
Emech,loss = m*(u₂- u₁) = m*C*ΔT.
Solving for ΔT,
ΔT = Emech,loss/(m*C)
⇒ ΔT = 10 kW/(40 Kg/s*4.186 kJ/Kg*°C) = 0.059°C
Therefore, the water will experience a temperature rise of 0.059°C due to mechanical inefficiency, which is very small, as it flows through the pump.
