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Question

An electrically driven pump operating at steady state draws water from a pond at a pressure of 1 bar and a rate of 50 kg/s and delivers the water at a pressure of 4 bar. There is no significant heat transfer with the surroundings, and changes in kinetic and potential energy can be neglected. The isentropic pump efficiency is 75%. Evaluating electricity at 8.5 cents per kWh,k W \cdot h, estimate the hourly cost of running the pump.

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We are given following data for electrically driven pump:

P1=1 barP_1=1\text{ bar}

P2=4 barP_2=4\text{ bar}

m˙=50 kg s\dot m=50\frac{\text{ kg}}{\text{ s}}

η=0.75\eta=0.75

From saturated water table A-3 corresponding to P1=1 barP_1=1\text{ bar} we can fin specific volume:

ν1=νf=1.0432 m3 kg\nu_1=\nu_f=1.0432\frac{\text{ m}^3}{\text{ kg}}

Calculating power required to drive a pump:

W˙=m˙ν2(P1P2)η=501.0432(400100)0.75=20.86 kW\begin{align*} \dot W&=\dfrac{\dot m\cdot \nu_2\cdot (P_1-P_2)}{\eta}=\dfrac{50\cdot 1.0432\cdot (400-100)}{0.75}\\\\ &=20.86\text{ kW} \end{align*}

Evaluating electricity is 8.5 cents kW h8.5\frac{\text{ cents}}{\text{ kW h}}. Calculating the cost in cents:

cost=20.86 kW1 h8.5 cent kW h=177 centcost=20.86\text{ kW}\cdot 1\text{ h}\cdot 8.5\frac{\text{ cent}}{\text{ kW h}}=\boxed{177\text{ cent}}

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