Pressure Drop Calculation in a Pipe

Question

Calculate the drop in pressure head when the cross-section of a pipe through which water is flowing is reduced by 10%. Consider the flow to be laminar.

Solution

When the cross-section of a pipe is reduced by 10%, we can calculate the pressure head drop by using Bernoulli’s equation and the continuity equation for steady, laminar flow.

Step 1: Bernoulli’s Equation

Bernoulli's equation for a horizontal pipe is:

\( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \)

Where:

\( P_1 \) and \( P_2 \) are the pressures at points 1 and 2, respectively
\( v_1 \) and \( v_2 \) are the velocities at points 1 and 2
\( \rho \) is the density of the fluid (water in this case)

Step 2: Continuity Equation

The continuity equation relates the flow rates at two points:

\( A_1 v_1 = A_2 v_2 \)

If the cross-sectional area is reduced by 10%, then:

\( A_2 = 0.9 A_1 \)

Hence, the velocity at point 2 is:

\( v_2 = \frac{v_1}{0.9} = \frac{10}{9} v_1 \)

Step 3: Substitute into Bernoulli’s Equation

Substituting the velocity relation into Bernoulli’s equation:

\( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho \left( \frac{10}{9} v_1 \right)^2 \)

Simplifying:

\( P_1 - P_2 = \frac{1}{2} \rho v_1^2 \left( \frac{100}{81} - 1 \right) \)

\( P_1 - P_2 = \frac{1}{2} \rho v_1^2 \left( \frac{19}{81} \right) \)

Step 4: Convert Pressure Difference to Pressure Head

Pressure head \( h \) is given by:

\( h = \frac{P_1 - P_2}{\rho g} \)

Substituting the pressure difference:

\( h = \frac{1}{2} \frac{ \rho v_1^2 \left( \frac{19}{81} \right)}{\rho g} = \frac{v_1^2}{2g} \times \frac{19}{81} \)

Final Expression for Pressure Head Drop

The drop in pressure head due to the reduction in cross-sectional area is:

\( h = \frac{19}{162} \times \frac{v_1^2}{g} \)

This formula gives the pressure head drop as a function of the initial velocity \( v_1 \) and gravitational acceleration \( g \).