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An emergency drain pump, shown in figure, should be able to pump 0.1 m3/s0.1 \mathrm{~m}^3 / \mathrm{s} of liquid water at 15C,10 m15^{\circ} \mathrm{C}, 10 \mathrm{~m} vertically up, delivering it with a velocity of 20 m/s20 \mathrm{~m} / \mathrm{s}. It is estimated that the pump, pipe, and nozzle have a combined isentropic efficiency expressed for the pump as 60%60 \%. How much power is needed to drive the pump?


Answered 2 years ago
Answered 2 years ago
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We are given following data for emergency drain pump:

T=15 CT=15\text{ C}

ΔZ=10 m\Delta Z=10\text{ m}

v2=20 m sv_2=20\frac{\text{ m}}{\text{ s}}


V˙=0.1 m3 s\dot V=0.1\frac{\text{ m}^3}{\text{ s}}


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

ν=0.001001 m3 kg\nu=0.001001\frac{\text{ m}^3}{\text{ kg}}

Calculating mass flow rate:

m˙=V˙ν=0.10.001001=99.9 kg s\dot m=\dfrac{\dot V}{\nu}=\dfrac{0.1}{0.001001}=99.9\frac{\text{ kg}}{\text{ s}}

Calculating ideal pump power:

W˙id=12m˙(v22v12)+m˙gΔZ=1299.9(2020)+99.99.8110=29.78 kW\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:

W˙pump=W˙idη=29.780.6=49.6 kW\dot W_{pump}=\dfrac{\dot W_{id}}{\eta}=\dfrac{29.78}{0.6}=\boxed{49.6\text{ kW}}

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