Water flows in a horizontal pipe one end of which is closed by a … - Vidyadip

Question

Water flows in a horizontal pipe one end of which is closed by a value and the reading in the pressure meter on the pipe is $3.5 \times 10^5$ Newton $/ m ^2$.
When the value on the pipe is opened the reading on the pressure meter becomes $3.0 \times 10^5 Newton / m ^2$. Calculate the velocity of water flowing in the pipe.

✓

Answer

From Bernoulli's principle
$
\begin{aligned}
P_1+\frac{1}{2} \rho v_1^2 & =P_2+\frac{1}{2} \rho v_2^2 \\
\text { or } \frac{1}{2} \rho\left(v_2^2-v_1^2\right) & =P_1-P_2
\end{aligned}
$
Here, $v _1=0$ (Initially the value is closed so velocity of water will be zero.)
$
v_2^2=\frac{2}{\rho}\left(P_1-P_2\right)
$
For water
$
\begin{aligned}
\rho & =1 \times 10^3 kg / m^3 \\
P_1 & =3.5 \times 10^5 N / m^2 \\
P_2 & =3.0 \times 10^5 N / m^2 \\
v_2^2 & =\frac{2}{1 \times 10^3}(3.5-3.0) \times 10^5 \\
& =2 \times 0.5 \times 10^2=100 \\
v_2 & =\sqrt{100}=10 m / sec
\end{aligned}
$

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