# Fluid food is pumped through a pipeline with a decrease in the pipe diameter. The velocity in the...

## Question:

Fluid food is pumped through a pipeline with a decrease in the pipe diameter. The velocity in the first pipe in 1.6 m/s and the velocity in the second pipe is 120% of that value. Assuming turbulent flow, what is the change in kinetic energy of fluid (in J/kg) in going from the first pipe to the second pipe?

## Kinetic Energy:

The term kinetic energy can be defined as when the object is moving, and then the energy exhibits are said to be kinetic energy. The kinetic energy can be calculated or measured in Joules.

Given data:

• The velocity in the first pipe is {eq}{v_1} = 1.6\,{\rm{m/s}} {/eq}
• The velocity in the second pipe is {eq}{v_2} = 120\% \times {v_1} = 1.2 \times 1.6 = 1.92\,{\rm{m/s}} {/eq}

The expression for the decrease in the kinetic energy per unit mass is given by

{eq}\Delta K.e = \dfrac{1}{2}\left[ {{v_2}^2 - {v_1}^2} \right] {/eq}

Substituting the values in the above equation as,

{eq}\begin{align*} \Delta K.e &= \dfrac{1}{2}\left[ {{v_2}^2 - {v_1}^2} \right]\\ \Delta K.e &= \dfrac{1}{2}\left[ {{{\left( {1.92} \right)}^2} - {{\left( {1.6} \right)}^2}} \right]\\ \Delta K.e &= 0.5632\,{\rm{J/kg}} \end{align*} {/eq}

Thus the change in the kinetic energy per unit mass is {eq}\Delta K.e = 0.5632\,{\rm{J/kg}} {/eq} 