# Problem 16 - Bernoulli's Energy Theorem

**Problem 16**

A pump (Figure 4-07) takes water from a 200-mm suction pipe and delivers it to a 150-mm discharge pipe in which the velocity is 3.6 m/s. The pressure is -35 kPa at A in the suction pipe. The 150-mm pipe discharges horizontally into air at C. To what height h above B can the water be raised if B is 1.8 m above A and 20 hp is delivered to the pump? Assume that the pump operates at 70 percent efficiency and that the frictional loss in the pipe between A and C is 3 m.

**Solution 16**

$Q = v_BA_B$

$Q = 3.6 \, [ \, \frac{1}{4}\pi (0.15^2) \, ]$

$Q = 0.0636 \, \text{ m}^3\text{/s}$

Output power of the pump

$P_{output} = \text{Efficiency} \times P_{input}$

$P_{output} = 0.70P_{input} = 0.70(20)$

$P_{output} = 14 \text{ hp} (746 \text{ Watts} / 1 \text{ hp})$

$P_{output} = 10~444 \, \text{ Watts}$

Head Added

$P_{output} = Q\gamma HA$

$10~444 = 0.0636(9810)HA$

$HA = 16.74 \, \text{ m}$

Velocity heads

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

$\dfrac{{v_A}^2}{2g} = \dfrac{8(0.0636^2)}{\pi^2(9.81)(0.2^4)} = 0.2089 \, \text{ m}$

$\dfrac{{v_B}^2}{2g} = \dfrac{8(0.0636^2)}{\pi^2(9.81)(0.15^4)} = 0.6602 \, \text{ m}$

Energy equation from A to C

$E_A + HA - HL = E_C$

$\dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma_w} + z_A + HA - HL = \dfrac{{v_C}^2}{2g} + \dfrac{p_C}{\gamma} + z_C$

$0.2089 - \dfrac{35}{9.81} + 0 + 16.74 - 3 = 0.6602 + 0 + (1.8 + h)$

$h = 7.92 \, \text{ m}$ *answer*