## Related questions with answers

Consider the piping system of the figure, with all the dimensions, parameters, minor loss coefficients, etc., of the earlier problem. The pump's performance follows a parabolic curve fit, $H_{\text {wailate }}=H_0-a \dot{V}^2$, where $H_0=19.8 \mathrm{~m}$ is the pump's shutoff head, and $a=0.00426 \mathrm{~m} /(\mathrm{Lpm})^2$ is a coefficient of the curve fit. Find the operating volume flow rate $\dot{V}$ in Lpm (liters per minute), and compare it with that of the earlier problem. Discuss.

Solution

VerifiedThe pump’s performance curve is given by:

$\begin{equation*} H_{\text{available}}=H_0-a \dot V^2 \end{equation*}$

where $H_0=19.8$ m is the pump’s shutoff head and $a= 0.00426$ m/(Lpm)$^2$ is a coefficient of the curve fit.

The coefficient $a$ can be expressed as:

$\begin{equation*} a=0.00426\:\frac{\text{m}}{\text{Lpm}^2\cdot\bigg(\frac{1}{1000}\cdot\frac{1}{60}\bigg)^2\:\frac{(\text{m}^3/\text{s})^2}{\text{Lpm}^2}}=1.533\cdot10^7\:\frac{\text{m}}{(\text{m}^3/\text{s})^2} \end{equation*}$

Therefore, the pump's performance curve is:

$\begin{equation*} H_{\text{available}}=19.6-1.533\cdot10^7\dot V^2 \end{equation*}$

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