Bernoulli’s Equation -2 ( Practice Problems )

Here in this second part of the Bernoulli’s Equation, we discuss the Bernoulli effect and the practice problems in Physics of fluids. This is very important topic in Physics at all levels of study from class 11 physics to JEE Physics, NEET Physics and BSc Physics.

Practice Problems – Bernoulli’s Equation

Here we provide one practice problem which requires good understanding of the concepts of the subject.

Problem 1

Find the following for the fluid flow diagram given the below figure. The fluid is water and flow is steady state.

1. The mean fluid velocities at sections 1-1 and 2-2
2. The diameter of pipe d at section 1-1
3. The diameter of the pipe at section 2-2

Solution

Here the fluid is water. Water is an in-compressible fluid . Hence density remains constant at all sections of flow.

For normal conditions , the value of density of water is assumed as ${\mathit{\rho }}_1$= ${\mathit{\rho }}_2$ = $\mathit{\rho }$ = $1000\mathrm{kg}/m^3$

Gravitational acceleration = $g=9.81m/s^2$

Ar section 1-1 , let datum head $H_1$ =0 m,

At section 2-2 , $H_2=30\times \tan {30}^0$ = 17.32 m

$p_1$ = 2.5 bar = $2.5\times {10}^5\mathrm{Pa}$

$p_2$ = 4.5 bar = $4.5\times {10}^5\mathrm{Pa}$

Given that steady state flow through pipe = $Q=100m^3/s$

Application of continuity Equation

From conservation of mass principle and continuity equation,

Flow rate = Product of Cross sectional area and velocity of low = Constant.

$Q=A_1{V}_1=A_2{V}_2$

At section 1-1 , the diameter of pipe = d m

At section 2-2 , the diameter of pipe = 3.5 d m

$\frac{\mathit{\pi }}{4}{d}^2{V}_1=\frac{\mathit{\pi }}{4}(3.5d{)}^2{V}_2=100m^3/s$

From above, $V_2=\frac{V_1}{12.25}m/s$

Application of Bernoulli’s Equation

The Bernoulli’s Equation is given by,

$\frac{p_1}{{\mathit{\rho }}_{1}g}+\frac{V_1{}^2}{2g}+H_1=\frac{p_2}{{\mathit{\rho }}_{2}g}+\frac{V_2{}^2}{2g}+H_2$

By substituting the values obtained from above calculations in Bernoulli’s equation , we get,

$\frac{2.5\times {10}^5}{1000\times 9.81}+\frac{V_1{}^2}{2\times 9.81}+0=\frac{4.5\times {10}^5}{1000\times 9.81}+\frac{V_1{}^2}{12.{25}^2\times 2\times 9.81}+17.32$

From above, we get

(i) $V_1=27.29m/s$

$V_2=\frac{27.29}{12.25}=2.23m/s$

(ii) From Continuity equation, $\frac{\mathit{\pi }}{4}{d}^2{V}_1$ = $100m^3/s$

As $V_1=27.29m/s$ , by substituting this value in above equation we get,

$\begin{array}{c}{d}_1=\sqrt{4.665}=2.16m\\ d_2=3.5\times 2.16=7.56m\end{array}$

As above the problem is solved in full.

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