The pump of a water distribution system is powered by a 20-kW electric motor whose efficiency is 90%. The water flow rate through the pump is 40 L/s. The diameters of the inlet and outlet pipes are the same and the elevation difference across the pump is negligible. If the absolute pressures at the inlet and outlet of the pump are measured to be 120 kPa and 320 kPa, respectively, determine (a) the mechanical efficiency of the pump, and (b) the temperature rise of water as it flows through the pump due to mechanical inefficiencies.

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.