Question

Water flowing through a horizontal pipe of variable cross-section the rate of volume floor is 0.4 L per second the pressure is 60,000 Pa at a point where the area of cross-section is 5×10 power -5 m² the pressure at a point where the area of cross-section 10 power -4 is

Answer 1

Answer:

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Explanation:

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Answer 2

Given:-

Volume flow rate, Q=0.4lps=$0.4\times10^{-3}m^{3}/s$

Pressure at point 1, $P_{1}=60000Pa$

Area at point 1, $A_{1}=5\times10^{-5}m^{2}$

Area at point 2, $A_{2}=10^{-4}m^{2}$

Density of water, $\rho=1000kg/m^{3}$

To Find:-

Pressure at point 2.

Explanation:-

Since, water is flowing in a pipe therefore, flow rate will be same at every points.

$\Rightarrow Q=A_{1}\times v_{1}\\\\0.4\times10^{-4}=5\times10^{-5}\times v_{1}\Rightarrow v_{1}=0.8m/s\\\\Similarly,\\\\Q=A_{2}\times v_{2}\\\\\Rightarrow0.4\times10^{-4}=10^{-4}\times v_{2}\Rightarrow v_{2}=0.4m/s$

Now, apply bernoulli's equation between points 1&2:-

$\frac{P_{1}}{\rho g}+\frac{v_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\rho g}+\frac{v_{2}^{2}}{2g}+z_{2}\\\\\text{For horizontal pipe}, z_{1}=z_{2}\\\\\frac{60000}{1000}+\frac{0.8^{2}}{2}=\frac{P_{2}}{1000}+\frac{0.4^{2}}{2}\Rightarrow P_{2}=60240Pa$