Question

Water enters through one end of a horizontal pipe of non-uniform cross-section with a velocity of $0.4\text{m/s}$ and leaves the other end with a velocity $0.6\text{m/s}$. The pressure of water at the entry point is $1500{\text{N/m}}^2$. What is the pressure at the exit point? (Density of water $=1.0\times {10}^3{\text{kg/m}}^3)$

• A. $1000{\text{N/m}}^2$
• B. $1200{\text{N/m}}^2$
• C. $1500{\text{N/m}}^2$
• D. $1400{\text{N/m}}^2$

Answer 1

The correct option is D $1400{\text{N/m}}^2$

Given: $P_1=1500{\text{N/m}}^2$
$v_1=0.4\text{m/s},v_2=0.6\text{m/s},P_2=?$

According to Bernoulli's theorem,
$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$
$\Rightarrow P_2=P_1-\frac{1}{2}\rho (V_2^2-V_1^2)$

$P_2=1500-\frac{1}{2}\times 1\times {10}^3\left[\right(0.6{)}^2-\left(0.4{)}^2\right]$
$P_2=1500-100=1400{\text{N/m}}^2$

Hence, $\left(D\right)$ is the correct answer.

Why this Q? Bernoulli's theorem states that the total energy (pressure energy, kinetic energy, gravitational potential energy) per unit volume is constant at all points along a streamline. $p+\frac{1}{2}\rho v^2+\rho gh=\text{constant or}\frac{p}{\rho g}+\frac{1}{2}\frac{v^2}{g}+h=\text{constant}$