Question

Water enters a pipe with inlet diameter of $1.5\text{cm}$ at an absolute pressure of $5\times {10}^5\text{Pa}$. The outlet of the pipe is $1\text{cm}$ in diameter and is at height of $6\text{m}$ above the inlet. When the flow speed at the inlet pipe is $2\text{m/s}$. Find the flow speed $\left(\text{m/s}\right)$ and pressure $\left(\text{Bar}\right)$ at the outlet of the pipe. (Take $g=10{\text{m/s}}^2$)

• A. $4.5\text{m/s},4.3\times {10}^5\text{Pa}$
• B. $9\text{m/s},4.3\times {10}^5\text{Pa}$
• C. $4.5\text{m/s},8.6\times {10}^5\text{Pa}$
• D. $9\text{m/s},4.3\times {10}^5\text{Pa}$

Answer 1

The correct option is A $4.5\text{m/s},4.3\times {10}^5\text{Pa}$

Given, $v_1=2\text{m/s}$
$P_1=5\times {10}^5\text{Pa}.$
Let the velocity of outlet be $v_2$
Let the absolute pressure at the outlet be $P_2$
From equation of continuity,
$A_1{v}_1=A_2{v}_2$
$\frac{\pi }{4}{d}_1^2\times v_1=\frac{\pi }{4}{d}_2^2\times v_2$
(given $d_1=1.5\text{cm},d_2=1\text{cm}$)
$v_2={\left(\frac{d_1}{d_2}\right)}^2\times v_1={\left(\frac{1.5}{1}\right)}^2\times 2$
$\Rightarrow v_2=4.5\text{m/s}$
Applying Bernoulli's Equation ,
$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$
$\Rightarrow P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)+\rho g(h_2-h_1)$
$\Rightarrow 5\times {10}^5-P_2=\frac{1}{2}\times 1000\times ({4.5}^2-2^2)+1000\times 10\times \left(6\right)$
$(\because h_2-h_1=6\text{m})$
$\Rightarrow P_2=5\times {10}^5-6.812\times {10}^4$
$\Rightarrow P_2\approx 4.3\times {10}^5\text{Pa}$