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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.     ...