Question

Water is flowing with a velocity of $5m/s$ in a horizontal pipe with cross-sectional area decreasing from $5\times 10^{-2}$ $m^2$ to $2\times 10^{-2}{m}^2$ at pressure $8\times 10^4$ pascal. What will be the pressure at smaller cross-section?

• A. $130kPa$
• B. $400kPa$
• C. $146kPa$
• D. $200kPa$

Answer 1

Answer: (C)

$a_1{v}_1=a_2{v}_2$
$\Rightarrow v_2=\frac{a_1{v}_1}{a_2}=\frac{5\times 10^{-2}\times 5}{2\times 10^{-2}}=12.5m/s$
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
$\Rightarrow P_2=P_1+\frac{1}{2}\rho \left(v_1^2-v_2^2\right)$
$P_2=8\times 10^4+\frac{1}{2}\times 10^3\left((5{)}^2-(12.5{)}^2\right)$
$P_2=8\times 10^4+\frac{1}{2}\times 10^3(25-156.25)$
$\Rightarrow P_2=8\times 10^4+66\times 10^3$
$P_2=146\times 10^3=146kPa$