# Problem 09 - Bernoulli's Energy Theorem

**Problem 9**

The diameter of a pipe carrying water changes gradually from 150 mm at A to 450 mm at B. A is 4.5 m lower than B. What will be the difference in pressure, in kPa, between A and B, when 0.176 m^{3}/s is flowing, loss of energy is being neglected.

**Solution 9**

Velocity heads

$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$

$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.176^2)}{\pi^2(9.81)(0.15^4)} = 5.0557 \, \text{ m}$

$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.176^2)}{\pi^2(9.81)(0.45^4)} = 0.0624 \, \text{ m}$

Neglecting head loss

$E_A = E_B$

$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A = \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B$

$5.0557 + \dfrac{p_A}{\gamma} + 0 = 0.0624 + \dfrac{p_B}{\gamma} + 4.5$

$\dfrac{p_B}{\gamma} - \dfrac{p_A}{\gamma} = 0.4933 \, \text{ m}$

$p_B - p_A = 0.4933\gamma$

$p_B - p_A = 0.4933(9.81)$

$p_B - p_A = 4.839 \, \text{ kPa}$ *answer*