In a stream line steady flow, two points A and B on a stream line are 1 m apart and the flow velocity varies uniformly from 2 m/s to 5 m/s. What is the acceleration of fluid at B?

Concept:

For flow along a stream line acceleration is given as

If V = f(s, t)

Then, \(dV = \frac{{\partial V}}{{\partial s}}ds + \frac{{\partial V}}{{\partial t}}dt\)

\(a = \frac{{dV}}{{dt}} = \;\frac{{\partial V}}{{\partial s}} \times \frac{{ds}}{{dt}} + \frac{{\partial V}}{{\partial t}}\) 

For steady flow \(\frac{{\partial V}}{{\partial t}} = 0\)

Then \(a = \frac{{\partial V}}{{\partial s}} \times \frac{{ds}}{{dt}}\) 

Since V = f(s) only for steady flow therefore \(\frac{{\partial v}}{{\partial s}} = \frac{{dv}}{{ds}}\)

Therefore \(a = V \times \frac{{dV}}{{ds}}\)

Calculation:

Given, V_A = 2 m/s, V_B = 5 m/s, and distance s = 1 m

\(\frac{{dV}}{{ds}} = \frac{{\left( {5 - 2} \right)}}{1} = 3\)

So acceleration of fluid at B is

\({a_B} = {V_B} \times \frac{{dV}}{{ds}} = 5 \times 3 = 15\)