Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to another horiz

Water is flowing into a factory in a horizontal pipe with a radius of 0.0183 m at ground level. This pipe is then connected to another horizontal pipe with a radius of 0.0420 m on a floor of the factory that is 12.6 m higher. The connection is made with a vertical section of pipe and an expansion joint. Determine the volume flow rate that will keep the pressure in the two horizontal pipes the same.

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  1. Answer:

    [tex]0.0168 m^3/s[/tex]

    Explanation:

    We are given that

    [tex]r_1=0.0183 m[/tex]

    [tex]h_1=0[/tex]

    [tex]r_2=0.0420 m[/tex]

    [tex]h_2=12.6 m[/tex]

    Let [tex]P_1=P_2=P[/tex]

    By using Bernoulli theorem

    [tex]P+\frac{1}{2}\rho v^2_1+\rho gh_1=P+\frac{1}{2}\rho v^2_2+\rho gh_2[/tex]

    [tex]\frac{1}{2}\rho v^2_1+\rho gh_1=\frac{1}{2}\rho v^2_2+\rho gh_2[/tex]

    [tex]v^2_1+2gh_1=v^2_2+2gh_2[/tex]

    [tex]A_1v_1=A_2v_2[/tex]

    [tex]v_1=\frac{A_2v_2}{A_1}[/tex]

    [tex](\frac{A_2}{A_1})^2v^2_2+2g\times 0=v^2_2+2\times 9.8\times 12.6[/tex]

    [tex](\frac{\pi r^2_2}{\pi r^2_1})^2v^2_2-v^2_2=246.96[/tex]

    [tex]v^2_2((\frac{r^2_2}{r^2_1})^2-1)=246.96[/tex]

    [tex]v^2_2=246.96\frac{r^4_1}{r^2_4-r^4_1}[/tex]

    [tex]v_2=\sqrt{246.96\frac{r^4_1}{r^4_2-r^4_1}}[/tex]

    [tex]v_2=\sqrt{246.96\times \frac{(0.0183)^4}{(0.042)^4-(0.0183)^4}}[/tex]

    [tex]v_2=3.038 m/s[/tex]

    Volume flow rate =[tex]A_2v_2[/tex]

    Volume flow rate =[tex]\pi r^2_2v_2=\pi (0.042)^2\times 3.038=0.0168 m^3/s[/tex]

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