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(A) According to Bernoulli's theorem for a horizontal tube:$P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2} \rho v_{2}^{2}$$P_{1} - P_{2} = \

Vedclass · Accepted Answer

$32$

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(A) According to Bernoulli's theorem for a horizontal tube:$P_{1} + \frac{1}{2} ho v_{1}^{2} = P_{2} + \frac{1}{2} ho v_{2}^{2}$$P_{1} - P_{2} = \frac{1}{2} ho (v_{2}^{2} - v_{1}^{2})$Given $P_{1} - P_{2} = h ho g$,where $h = 0.8\,m$.From the equation of continuity,$Q = A_{1}v_{1} = A_{2}v_{2}$,so $v_{1} = Q/A_{1}$ and $v_{2} = Q/A_{2}$.Substituting these into the Bernoulli equation:$h ho g = \frac{1}{2} ho \left[ \frac{Q^{2}}{A_{2}^{2}} - \frac{Q^{2}}{A_{1}^{2}} ight]$$2gh = Q^{2} \left[ \frac{A_{1}^{2} - A_{2}^{2}}{A_{1}^{2} A_{2}^{2}} ight]$$Q = A_{1} A_{2} \sqrt{\frac{2gh}{A_{1}^{2} - A_{2}^{2}}}$Diameters are $d_{1} = 0.3\,m$ and $d_{2} = 0.1\,m$,so radii are $r_{1} = 0.15\,m$ and $r_{2} = 0.05\,m$.Areas are $A_{1} = \pi (0.15)^{2} = 0.0225\pi\,m^{2}$ and $A_{2} = \pi (0.05)^{2} = 0.0025\pi\,m^{2}$.$Q = (0.0225\pi)(0.0025\pi) \sqrt{\frac{2 	imes 9.8 	imes 0.8}{(0.0225\pi)^{2} - (0.0025\pi)^{2}}}$Using $g \approx 10\,m/s^{2}$:$Q = (5.625 	imes 10^{-5} \pi^{2}) \sqrt{\frac{16}{\pi^{2}(5.0625 	imes 10^{-4} - 0.0625 	imes 10^{-4})}}$$Q = (5.625 	imes 10^{-5} \pi^{2}) \frac{4}{\pi \sqrt{5 	imes 10^{-4}}} = \frac{22.5 	imes 10^{-5} \pi}{0.02236} \approx 0.0316\,m^{3}/s = 31.6\,ltr/s \approx 32\,ltr/s$.

Water is flowing through a horizontal tube as shown in the figure. Its diameters at two points are $0.3\,m$ and $0.1\,m$ respectively. The pressure difference between these two points is equal to $0.8\,m$ of water column. Find the rate of flow of water in the tube in $ltr/s$.

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