Question

Water is flowing with a velocity of $5 m / s$ 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. $130 kPa$
• B. $400 kPa$
• C. $146 kPa$
• D. $200 kPa$

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}}=12.5 m / 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) $
$\Rightarrow P_{2}=8 \times 10^{4}+66 \times 10^{3}$
$P _{2}=146 \times 10^{3}=146 kPa$