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Mechanical Properties of Fluids — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsMechanical Properties of Fluids3 Marks
Question
A building receives its water supply through an undergound pipe $2 \mathrm{~cm}$ in diameter at an absolute pressure of $4 \times 10^5 \mathrm{~Pa}$ and flow velocity $4 \mathrm{~m} / \mathrm{s}$. The pipe leading to higher floors is $1.5 \mathrm{~cm}$ in diameter. Find the flow velocity and pressure at the floor inlet $10 \mathrm{~m}$ above.
✓

Answer


Data : $\mathrm{d}_1=3 \mathrm{~cm}, \mathrm{p}_1=4 \times 10^5 \mathrm{~Pa}, \mathrm{v}_1=4 \mathrm{~m} / \mathrm{s}$,
$\mathrm{d}_2=2 \mathrm{~cm}, \mathrm{~h}_2-\mathrm{h}_1=10 \mathrm{~m}$
By continuity equation, the flow velocity at the higher floor inlet
$
v_2=v_1 \frac{A_1}{A_2}=v_1\left(\frac{d_1}{d_2}\right)^2=(4 \mathrm{~m} / \mathrm{s})\left(\frac{3 \mathrm{~cm}}{2 \mathrm{~cm}}\right)^2=9 \mathrm{~m} / \mathrm{s}
$
By Bernoulli's equation,
$
\begin{aligned}
p_1 & +\frac{1}{2} \rho v_1^2+\rho g h_1=p_2+\frac{1}{2} \rho v_2^2+\rho g h_2 \\
\therefore p_2 & =p_1-\frac{1}{2} \rho\left(v_2^2-v_1^2\right)-\rho g\left(h_2-h_1\right) \\
& =\left(4 \times 10^5\right)-\frac{1}{2}\left(10^3\right)\left(9^2-4^2\right)-\left(10^3\right)(10)(10) \\
& =(400-32.5-100) \times 10^3 \\
& =\mathbf{2 . 6 7 5} \times 10^5 \mathrm{~Pa}=\mathbf{2 6 7 . 6} \mathbf{k P a}
\end{aligned}
$

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