Question A liquid is flowing in steady state flow through a 75 mm pipe. The local velocity varies with distance from the pipe axis as shown in following table. Calculate (a) Average velocity (b) Kinetic energy correction factor (c) Momentum correction factor. Use numerical technique. Local velocity #, m/s Distance from pipe axis, mm Local velocity , m/s Distance from pipe axis, mm 1.042 1.033 1.019 0.996 0.978 0.955 0 3.75 7.50 11.25 15.00 18.75 0.919 0.864 0.809 0.699 0.507 0 22.50 26.25 30.00 33.75 35.625 37.50

Step 1 of 3(a)Write the equation for average velocity as follows:\begin{aligned}\tilde{F} & =\frac{1}{S} \int_{T} u d S \\& =\frac{1}{\pi r_{0}^{2}} \int_{0}^{2} u d\left(\pi r^{2}\right) \\& =\frac{\pi}{\pi r_{0}^{2}} \int_{0} u 2 \pi r d r\end{aligned}The average velocity expression it,P=\frac{2}{r_{r}^{5}} \int_{0}^{e} \omega rNow, calculate the values of ur as follows:Draw a plot between or vi r and calculate the area under the curve as follows:r Vs ruArea under the curve \left(\int_{0}^{\text {ur }}\right. ur \left.d\right)= no. of boxes \times \mathrm{X} axis units \times \mathrm{Y} axis units\begin{array}{l}=23 \times 0.005 \mathrm{~m} \times 0.005 \mathrm{~m}^{2} / \mathrm{s} \\=5.75 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\end{array}Substitute the corresponding values in equation (1) to ealculate the average velocity.\begin{aligned}\tilde{F}^{F} & =\frac{2}{r_{s}^{T}} \int_{i}^{r} u r d r \\& =\frac{2}{(0.0375 \mathrm{~m})^{2}} \times 5.75 \times 10^{-\mathrm{At}} \mathrm{m}^{3} / \mathrm{s} \\& =0.81 \frac{\mathrm{m}^{3} / \mathrm{s}}{\mathrm{m}^{7}} \\& =0.81 \frac{\mathrm{m}}{\mathrm{s}}\end{aligned}Therefore, the average velocity is 0.81 \mathrm{~m} / \mathrm{s}.Step 2 of 3(b)Write the equation for kirietic energy correction factor as follows:\begin{aligned}a & =\frac{\int w^{3} d S}{p^{3} S} \\& =\frac{\int_{u^{3}} d\left(\pi r^{2}\right)}{F^{3} \pi r_{w}^{2}} \\& =\frac{\pi u^{3} 2 r d r}{P^{3} \pi r_{*}^{2}}\end{aligned}The kirietic energy correction foctor is,\alpha=\frac{2}{p^{3} r_{-}^{2}} \int_{i}^{\alpha} u^{3} r d rCalculate the values of u^{3} r from the data as follows:Draw a plot between u^{3} r vs r and calculate the atea under the curve as follows:r vs rus\text { Area under the curve } \begin{aligned}\left(\int_{0} u^{\prime} r d r\right) & =\text { no. of boxs } \times X \text { axis units } \times Y \text { axis units } \\& =43 \times 0.005 \mathrm{~m} \times 0.002 \mathrm{~m}^{2} / \mathrm{s}^{3} \\& =4.3 \times 10^{-4} \mathrm{~m}^{4} / \mathrm{s}^{3}\end{aligned}Substitute the corresponding values in equation (2) to calculate the kinetic friction factor.\begin{array}{l}\alpha=\frac{2}{P^{3} r^{2}} \int_{-}^{r} u^{3} r d r \\=\frac{2}{(0.81 \mathrm{~m} / \mathrm{s})^{1}(0.0375 \mathrm{~m})^{2}} \times 4.3 \times 10^{-64} \mathrm{~m}^{2} / \mathrm{s}^{2} \\=1.15 \\\end{array}Therefore, the kinetic energy correction factor is \mathbf{1 . 1 5}Step 3 of 3(C)Write the expression of momentum correction factor as follows:\beta=\frac{i / S}{\left.\rho\right|^{i}}Substitute the value of M / S in the above equation.\begin{array}{l}\beta=\frac{\left(\frac{\rho \int_{0}^{2} d S}{S}\right)}{\rho v^{n}} \\=\frac{\int u^{2} d\left(\pi r^{2}\right)}{\Gamma^{3} \pi r^{2}} \\=\frac{\pi \int v^{2} \cdot 2 r d r}{F r r_{w}^{2}} \\=\frac{2}{\sqrt{2}^{2} r^{2}} \int_{0}^{2} u^{2} r d r \\\end{array}The momentum correction factor expression is,\rho=\frac{2}{r^{2} r^{2}} \int_{0}^{2} u^{2} r d-(3)Calculate the values of u^{2} r from the dota as follows:Draw a plot between u^{2} r vs r and calculate the area under the curve as follows:r vs run 2Area under the curve \left(\int_{0}^{r} s^{2} r d r\right)= No. of boxes \times x- axis units \times y- axis units\begin{array}{l}=19 \times 0.005 \mathrm{~m} \times 0.005 \mathrm{~m}^{3} / \mathrm{s}^{\mathrm{t}} \\=4.75 \times 10^{-24} \mathrm{~m}^{1} / \mathrm{s}^{2}\end{array}Substitute the corresponding values in equation (3) to calculate the momentum correction factor as follows:\begin{aligned}\beta & =\frac{2}{\bar{V}^{2} r_{w}^{2}} \int_{0}^{r_{\tilde{2}}} u^{2} r d r \\& =\frac{2}{(0.81 \mathrm{~m} / \mathrm{s})^{2}(0.0375 \mathrm{~m})^{2}}\left(4.75 \times 10^{-04} \mathrm{~m}^{3} / \mathrm{s}^{2}\right) \\& =1.02\end{aligned}Thus, the momentum correction factor is 1.02 .I HOPE THAT YOU HAVE UNDERSTOOD THE PROBLEM. IF NOT, FEEL FREE TO ASK ME. BUT PLEASE GIVE ME A LIKE(UPVOTE) ud83dudc4d   * STAY HOME - STAY SAFE * ...