Question

The intake in figure has cross-sectional area of $0.74{\text{m}}^2$ and water flow at $0.40\text{m/s}$. At the outlet, distance $D=180\text{m}$ below the intake, the cross sectional area is smaller than at the intake and the water flows out at $9.5\text{m/s}$ into equipment. What is the pressure difference between inlet and outlet?

• A. $3.14\text{MPa}$
• B. $2\text{MPa}$
• C. $1.71\text{MPa}$
• D. $6.2\text{MPa}$

Answer 1

The correct option is C $1.71\text{MPa}$

Given that
$\rho =1000{\text{kg/m}}^3$
$D=180\text{m}$
$v_1=0.40\text{m/s}$
$v_2=9.5\text{m/s}$
Now by using Bernoulli's equation
$\Delta P=P_2-P_1=\rho gD+\frac{1}{2}\rho (v_1^2-v_2^2)$
$\Rightarrow P_2-P_1=1000\times 9.8\times 180+\frac{1}{2}\times 1000\times ({0.40}^2-{9.5}^2)$
$\Rightarrow \Delta P=P_2-P_1=1718955\text{Pa}$
$\Delta P=P_2-P_1=1.71\text{MPa}$
Hence, option C is correct.