Water flows in a horizontal tube as shown in | vedclass

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Question

$\mathrm{A}_{1} \mathrm{v}_{1}=\mathrm{A}_{2} \mathrm{v}_{2}$

or $\mathrm{v}_{2}=\left(\frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}\right) \mathrm{v}_{1}$

Vedclass · Accepted Answer

$\frac{6}{{\sqrt {10} }} 	imes {10^{ - 3}}\,{m^3}/s$

Vedclass · Answer

$\mathrm{A}_{1} \mathrm{v}_{1}=\mathrm{A}_{2} \mathrm{v}_{2}$

or $\mathrm{v}_{2}=\left(\frac{\mathrm{A}_{1}}{\mathrm{A}_{2}}ight) \mathrm{v}_{1}$     $...(i)$

from bernoulli's equation for horizontal flow,

$\mathrm{P}_{1}+\frac{1}{2} ho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} ho \mathrm{v}_{2}^{2}$

$\mathrm{P}_{1}+\frac{1}{2} ho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} ho\left[\frac{\mathrm{A}_{1}^{2}}{\mathrm{A}_{2}^{2}}ight] \mathrm{v}_{1}^{2}$

or $\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2} ho \mathrm{v}_{1}^{2}\left[\frac{\mathrm{A}_{1}^{2}}{\mathrm{A}_{2}^{2}}-1ight]$

but $\mathrm{P}_{1}-\mathrm{P}_{2}=600 \mathrm{N} / \mathrm{m}^{2}$

$600=\frac{1}{2} ho v_{1}^{2}\left[\frac{A_{1}^{2}}{A_{2}^{2}}-1ight]$

$\mathrm{v}_{1}=\sqrt{\frac{1200}{ho}}\left[\frac{\mathrm{A}_{1}^{2}}{\mathrm{A}_{2}^{2}}-1ight]^{-1 / 2}$

$\mathrm{Q}=\mathrm{A}_{1} \mathrm{v}_{1}=\mathrm{A}_{2} \mathrm{v}_{2}$

$=\frac{6}{\sqrt{10}} 	imes 10^{-3} \mathrm{m}^{3} / \mathrm{s}$

Water flows in a horizontal tube as shown in figure. The pressure of water changes by $600\, N/ m^2$ between $A$ and $B$ where the area of crosssection are $30\, cm^2$ and $15\, cm^2$ respectively. Find the rate of flow of water through the tube.

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