Question

In a horizontal pipeline of uniform cross section the pressure falls by $8N/m^2$ between two points separated by 1 km. If oil of density $800kg/m^3$ flows through the pipe, find the change in KE per kg of oil at these points.

• A. ${10}^{-2}J/kg$
• B. ${10}^{-3}J/kg$
• C. ${10}^{-4}J/kg$
• D. ${10}^{-1}J/kg$

Answer 1

The correct option is A ${10}^{-2}J/kg$

Applying Bernoulli's principle on a horizontal pipe, we have
$P_1+\frac{1}{2}\rho {v_1}^2=P_2+\frac{1}{2}\rho {v_2}^2\Rightarrow \frac{1}{2}\rho {v_2}^2-\frac{1}{2}\rho {v_1}^2=P_1-P_2\Rightarrow \frac{1}{2}{v_2}^2-\frac{1}{2}{v_1}^2=\frac{P_1-P_2}{\rho }\Rightarrow \text{Change in kinetic energy per unit mass}(\Delta K.E.)=\frac{P_1-P_2}{\rho }\Rightarrow \Delta K.E.=\frac{8}{800}={10}^{-2}J/Kg$