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scenario: you work for dr pepper bottling factory. the pump at the hydr…

Question

scenario: you work for dr pepper bottling factory. the pump at the hydraulic lift for the packing machine is running slow and is overheating. the speed of the system with the old pump is running at 72% its normal speed. the old pump is running at 1650 rpm, and has a d/rev of 7.3 cuin. maintenance found a replacement pump but with no plate or label on it, other than its a slightly smaller motor (6.8 cu in/rev). the supervisor wants to know about the replacement pump before installing it. use the old pump to determine the optimal operation (displacement and rpm) of the new pump. known: rpm = 1650 d/rev = 7.3 cuin effv = 72% find: d/min - q (gpm) known: d/rev = 6.8 cuin find: q (gpm) - rpm@100% - (hint: use q and effy of old pump) (hint: use q to find d/min, then d/rev to find rpm)

Explanation:

Step1: Calculate flow - rate of old pump at 100% efficiency

The formula for flow - rate $Q$ is $Q = \text{RPM}\times\text{D/rev}\times\text{Eff}$. For the old pump, $\text{RPM}=1650$, $\text{D/rev}=7.3$ cu in, and $\text{Eff}=0.72$. First, find the flow - rate at 100% efficiency. Let's assume the flow - rate at 100% efficiency of the old pump is $Q_{100\%}$. The actual flow - rate of the old pump $Q_{old}=1650\times7.3\times0.72$ cu in/min.
\[Q_{old}=1650\times7.3\times0.72 = 1650\times5.256=8662.4\] cu in/min. To convert to gpm (gallons per minute), since 1 gallon = 231 cu in, $Q_{old}=\frac{8662.4}{231}\approx37.5$ gpm.

Step2: Calculate RPM of new pump

The new pump has $\text{D/rev}=6.8$ cu in. We want to maintain the same flow - rate as the old pump's optimal (100% efficiency) flow - rate. Let the RPM of the new pump be $n$. We know that $Q = n\times\text{D/rev}\times1$ (at 100% efficiency). Since we want $Q$ to be the same as the old pump's optimal $Q$, and we know $Q$ (from Step 1) and $\text{D/rev}$ for the new pump. Rearranging the formula $n=\frac{Q}{\text{D/rev}}$. Substituting $Q = 8662.4$ cu in/min and $\text{D/rev}=6.8$ cu in, we get $n=\frac{8662.4}{6.8}\approx1274$ RPM.

Answer:

The RPM of the new pump is approximately 1274 RPM.