Question

In a horizontal pipeline of uniform area of cross-section, $2\times {10}^{-2}\text{J/kg}$ is the change in kinetic energy per $\text{kg}$ of the kerosene oil flowing through the pipe of length $1500\text{m}$. What will be the pressure drop between the end points of the pipe? (Take density of kerosene oil as $850{\text{kg/m}}^3)$

• A. $8{\text{N/m}}^2$
• B. $10{\text{N/m}}^2$
• C. $17{\text{N/m}}^2$
• D. $1.34{\text{N/m}}^2$

Answer 1

The correct option is C $17{\text{N/m}}^2$


Let $P_1$ and $P_2$ be the pressure at the inlet and outlet of the pipe respectively.
Let $v_1$ and $v_2$ be the velocities of oil at inlet and outlet of the pipe respectively.
Given, change in kinetic energy = $2\times {10}^{-2}\text{J/kg}$
According to the Bernoulli's theorem,
$P_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2[\because h_1=h_2]$
$\Rightarrow P_1-P_2=\frac{1}{2}\rho (v_2^2-v_1^2)$
Change in kinetic energy per $\text{kg}$ mass is given by
$\frac{1}{2}(v_2^2-v_1^2)=\frac{P_1-P_2}{\rho }$
$\Rightarrow 2\times {10}^{-2}=\frac{P_1-P_2}{\rho }\Rightarrow P_1-P_2=\rho \times (2\times {10}^{-2})$
$\Rightarrow P_1-P_2=850\times 2\times {10}^{-2}$
$\Rightarrow P_1-P_2=17{\text{N/m}}^2$
or $\Delta P=17{\text{N/m}}^2$