Problem 19 - Bernoulli's Energy Theorem

Part (a): Discharge, Q = 12 L/s
$Q = 0.012 \, \text{ m}^3\text{/s}$
 

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

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.012^2)}{\pi^2(9.81)(0.15^4)} = 0.0235 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.012^2)}{\pi^2(9.81)(0.1^4)} = 0.119 \, \text{ m}$
 

Head lost
$HL_{A-1} = 3 \times \dfrac{{v_1}^2}{2g} = 3(0.0235) = 0.0705 \, \text{ m}$

$HL_{2-B} = 20 \times \dfrac{{v_2}^2}{2g} = 20(0.119) = 2.38 \, \text{ m}$
 

Energy equation between A and B
$E_A - HL_{A-1} + HA - HL_{2-B} = E_B$

$\left( \dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A \right) - HL_{A-1} + HA - HL_{2-B} = \left( \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B \right)$

$(0 + 0 + 0) - 0.0705 + HA - 2.38 = (0 + 0 + 72)$

$HA = 74.4505 \, \text{ m}$
 

Output power of the pump
$P = Q \gamma HA = 0.012(9810)(74.4505)$

$P = 8764.31 \text{Watts} \times (1 \text{hp} / 746 \text{Watts})$

$P = 11.75 \, \text{ hp}$           answer
 

Energy equation between A and 1
$E_A - HL_{A-1} = E_1$

$\left( \dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A \right) - HL_{A-1} = \left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right)$

$(0 + 0 + 0) - 0.0705 = \left( 0.0235 + \dfrac{p_1}{\gamma} - 6 \right)$

$\dfrac{p_1}{\gamma} = 5.906 \, \text{ m}$           answer
 

Energy equation between 2 and B
$E_2 - HL_{2-B} = E_B$

$\left( \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 \right) - HL_{2-B} = \left( \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B \right)$

$\left( 0.119 + \dfrac{p_2}{\gamma} - 6 \right) - 2.38 = (0 + 0 + 72)$

$\dfrac{p_2}{\gamma} = 80.261 \, \text{ m}$           answer
 

Checking
Energy equation between 1 and 2
$E_1 + HA = E_2$

$\left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right) + HA = \left( \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 \right)$

$(0.0235 + 5.906 - 6) + 74.4505 = (0.119 + 80.261 - 6)$

$74.38 = 74.38$       (Check!)
 

Part (b): Discharge, Q = 36 L/s
$Q = 0.036 \, \text{ m}^3\text{/s}$
 

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

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.036^2)}{\pi^2(9.81)(0.15^4)} = 0.2115 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.036^2)}{\pi^2(9.81)(0.1^4)} = 1.0708 \, \text{ m}$
 

Head lost
$HL_{A-1} = 3 \times \dfrac{{v_1}^2}{2g} = 3(0.2115) = 0.6345 \, \text{ m}$

$HL_{2-B} = 20 \times \dfrac{{v_2}^2}{2g} = 20(1.0708) = 21.416 \, \text{ m}$
 

Energy equation between A and B
$E_A - HL_{A-1} + HA - HL_{2-B} = E_B$

$\left( \dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A \right) - HL_{A-1} + HA - HL_{2-B} = \left( \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B \right)$

$(0 + 0 + 0) - 0.6345 + HA - 21.416 = (0 + 0 + 72)$

$HA = 94.0505 \, \text{ m}$
 

Output power of the pump
$P = Q \gamma HA = 0.036(9810)(94.0505)$

$P = 33~214.87 \text{Watts} \times (1 \text{hp} / 746 \text{Watts})$

$P = 44.52 \, \text{ hp}$           answer
 

Energy equation between A and 1
$E_A - HL_{A-1} = E_1$

$\left( \dfrac{{v_A}^2}{2g} + \dfrac{p_A}{\gamma} + z_A \right) - HL_{A-1} = \left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right)$

$(0 + 0 + 0) - 0.6345 = \left( 0.2115 + \dfrac{p_1}{\gamma} - 6 \right)$

$\dfrac{p_1}{\gamma} = 5.154 \, \text{ m}$           answer
 

Energy equation between 2 and B
$E_2 - HL_{2-B} = E_B$

$\left( \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 \right) - HL_{2-B} = \left( \dfrac{{v_B}^2}{2g} + \dfrac{p_B}{\gamma} + z_B \right)$

$\left( 1.0708 + \dfrac{p_2}{\gamma} - 6 \right) - 21.416 = (0 + 0 + 72)$

$\dfrac{p_2}{\gamma} = 98.3452 \, \text{ m}$           answer
 

Checking
Energy equation between 1 and 2
$E_1 + HA = E_2$

$\left( \dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 \right) + HA = \left( \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2 \right)$

$(0.2115 + 5.154 - 6) + 94.0505 = (1.0708 + 98.3452 - 6)$

$93.416 = 93.416$       (Check!)