## Related questions with answers

An emergency drain pump, shown in figure, should be able to pump $0.1 \mathrm{~m}^3 / \mathrm{s}$ of liquid water at $15^{\circ} \mathrm{C}, 10 \mathrm{~m}$ vertically up, delivering it with a velocity of $20 \mathrm{~m} / \mathrm{s}$. It is estimated that the pump, pipe, and nozzle have a combined isentropic efficiency expressed for the pump as $60 \%$. How much power is needed to drive the pump?

Solution

VerifiedWe are given following data for emergency drain pump:

$T=15\text{ C}$

$\Delta Z=10\text{ m}$

$v_2=20\frac{\text{ m}}{\text{ s}}$

$v_1=0$

$\dot V=0.1\frac{\text{ m}^3}{\text{ s}}$

$\eta=0.6$

From Saturated Water table B.1.1. corresponding to $T=15\text{ C}$ we can find specific volume:

$\nu=0.001001\frac{\text{ m}^3}{\text{ kg}}$

Calculating mass flow rate:

$\dot m=\dfrac{\dot V}{\nu}=\dfrac{0.1}{0.001001}=99.9\frac{\text{ kg}}{\text{ s}}$

Calculating ideal pump power:

$\begin{align*} \dot W_{id}&= \dfrac{1}{2}\cdot \dot m\cdot (v_2^2-v_1^2)+\dot m\cdot g\cdot \Delta Z\\\\ &= \dfrac{1}{2}\cdot 99.9\cdot (20^2-0)+99.9\cdot 9.81\cdot 10\\\\ &={29.78\text{ kW}} \end{align*}$

Calculating actual pump power:

$\dot W_{pump}=\dfrac{\dot W_{id}}{\eta}=\dfrac{29.78}{0.6}=\boxed{49.6\text{ kW}}$

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