Question A centrifugal water pump running at a speed w = 800 rpm has the following data for flow rate, Q, and pressure head, Ap. Q (ft/min) O 50 75 100 120 140 150 165 Ap (lbf/ft?) 7.54 7.29 6.85 6.12 4.80 3.03 2.38 1.23 The pressure head is a function of the flow rate, speed, impeller diameter (D=1 ft), and water density p. Plot the pressure head versus flow rate curve. Find the two parameters for this problem, and, from the above data, plot one against the other. By using Excel to perform a trendline analysis on this latter curve, generate and plot data for pressure head versus flow rate for impeller speeds of 600 rpm and 1200 rpm.

SOLUTION: The variables that are to be related are as follows:Flow rate is Q, pressure is \Delta p, diameter is D, density is \rho, and angular speed is \omega Dimensions of the variables are:\begin{array}{l}Q=\frac{\mathrm{L}^{3}}{\mathrm{t}} \\\Delta p=\frac{\mathrm{M}}{\mathrm{Lt}^{2}} \\D=\mathrm{L} \\\rho=\frac{\mathrm{M}}{\mathrm{L}^{3}} \\\omega=\frac{1}{\mathrm{t}}\end{array}Number of parameters is n=5Number of primary dimensions, m=3Consider repeating variables as \rho, D, and \omegaSo, number of repeating variables is r=3Calculate number of dimensionless groups:\begin{aligned}n-m & =5-3 \\& =2\end{aligned} Apply Buckingham pi theorem.Calculate \pi_{1} term:\pi_{1}=Q \rho^{a} D^{b} \omega^{c}So,\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{t}^{0}=\left(\frac{\mathrm{L}^{3}}{\mathrm{t}}\right)\left(\frac{\mathrm{M}}{\mathrm{L}^{3}}\right)^{a} \mathrm{~L}^{b}\left(\frac{1}{\mathrm{t}}\right)^{c}Compare M terms:a=0Compare L terms:\begin{array}{l}0=3-3 a+b \\b=-3\end{array}Compare t terms:\begin{array}{l}0=-1-c \\c=-1\end{array}Write \pi_{1} term:\begin{aligned}\pi_{1} & =Q \rho^{0} D^{-3} \omega^{-1} \\& =\frac{Q}{D^{3} \omega}\end{aligned} Calculate \pi_{2} term:\pi_{2}=\Delta p \rho^{a} D^{b} \omega^{c}So,\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{t}^{0}=\left(\frac{\mathrm{M}}{\mathrm{Lt}^{2}}\right)\left(\frac{\mathrm{M}}{\mathrm{L}^{3}}\right)^{a} \mathrm{~L}^{b}\left(\frac{1}{\mathrm{t}}\right)^{c}Compare M terms:\begin{array}{l}0=1+a \\a=-1\end{array}Compare \mathrm{L} terms:\begin{array}{l}0=-1-3 a+b \\b=-2\end{array}Compare t terms:\begin{array}{l}0=-2-c \\c=-2\end{array}Write \pi_{2} term:\begin{aligned}\pi_{2} & =\Delta p \rho^{-1} D^{-2} \omega^{-2} \\& =\frac{\Delta p}{\rho D^{2} \omega^{2}}\end{aligned} The two functions are related as \pi_{2}=f\left(\pi_{1}\right)So,\frac{\Delta p}{\rho D^{2} \omega^{2}}=f\left(\frac{Q}{D^{3} \omega}\right)From the data, it is assumed to be a parabolic expression.\frac{\Delta p}{\rho D^{2} \omega^{2}}=a+b\left(\frac{Q}{D^{3} \omega}\right)+c\left(\frac{Q}{D^{3} \omega}\right)^{2}Here,Density, \rho=1.94 \mathrm{slug} / \mathrm{ft}^{3}Speed,\begin{aligned}\omega & =800 \mathrm{rpm} \\& =(800 \times 2 \pi) / \mathrm{min} \\& =\left(\frac{800 \times 2 \pi}{3600}\right) / \mathrm{s}\end{aligned}Diameter, D=1 \mathrm{ft}Table shows the vales of flow rate and pressures. Plot the graph with \frac{\Delta p}{\rho D^{2} \omega^{2}} versus \frac{Q}{D^{3} \omega}.       P.S: If you are having any doubt, please comment here, I will surely reply you.Please provide your valuable feedback by a thumbsup. Your thumbsup is a result of our efforts. Your LIKES are very important for me so please LIKE….Thanks in advance! Please like my solution, its my sincere request, I am in really bad situation, your "upvote" helps me a lot.     ...