Question
A 3.5 hp pump delivers 1140 lbf of ethylene glycol at $20^{\circ} \mathrm{C}$ in 12 seconds, against a head of $17 \mathrm{ft}$. Calculate the efficiency of the pump.
Step 1
First, we need to find the power input to the pump. We know that the pump has a power of 3.5 hp, so we need to convert this to watts. We can use the conversion factor 1 hp = 746 watts. Power input = 3.5 hp * 746 watts/hp = 2611 watts Show more…
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Water at $15^{\circ} \mathrm{C}$ is pumped at a rate of $20 \mathrm{~L} / \mathrm{s}$ using a $5 \mathrm{~kW}$ pump. If the efficiency of the pump is $80 \%,$ what is the head added to the water as it passes through the pump?
The efficiency $\eta$ of a pump is defined as the (dimensionless) ratio of the power developed by the flow to the power required to drive the pump: $$\eta=\frac{Q \Delta p}{\text { input power }}$$ where $Q$ is the volume rate of flow and $\Delta p$ is the pressure rise produced by the pump. Suppose that a certain pump develops a pressure rise of 35 lbf/in $^{2}$ when its flow rate is $40 \mathrm{L} / \mathrm{s} .$ If the input power is $16 \mathrm{hp},$ what is the efficiency?
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