Problem 20 - Bernoulli's Energy Theorem

Discharge
$Q = 0.7 \, \text{ m}^3\text{/s}$
 

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

$\dfrac{{v_1}^2}{2g} = \dfrac{{v_2}^2}{2g} = \dfrac{8(0.7^2)}{\pi^2(9.81)(0.6^4)} = 0.3124 \, \text{ m}$
 

Head lost
$HL_{A-1} = 5 \times \dfrac{{v_1}^2}{2g} = 5(0.3124) = 1.562 \, \text{ m}$

$HL_{2-B} = 0.2 \times \dfrac{{v_2}^2}{2g} = 0.2(0.3124) = 0.0625 \, \text{ m}$
 

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

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

$(0 + 0 + 60) - 1.562 - HE - 0.0625 = (0 + 0 + 0)$

$HE = 58.3755 \, \text{ m}$
 

Power given up by the water to the turbine
$P = Q\gamma HE = 0.7(9810)(58.3755)$

$P = 400~864.56 \text{ Watts } \times (1 \text{ hp } / 746 \text{ Watts } )$

$P = 537.35 \, \text{ hp}$           answer
 

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

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

$(0 + 0 + 60) - 1.562 = \left( 0.3124 + \dfrac{p_1}{\gamma} + 4.5 \right)$

$\dfrac{p_1}{\gamma} = 53.6256 \, \text{ m}$           answer
 

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

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

$\left( 0.3124 + \dfrac{p_2}{\gamma} + 4.5 \right) - 0.0625 = (0 + 0 + 0)$

$\dfrac{p_2}{\gamma} = -4.7499 \, \text{ m}$           answer
 

Checking
Energy equation between 1 and 2
$E_1 - H_E = E_2$

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

$(0.3124 + 53.6256 + 4.5) - 58.3755 = (0.3124 - 4.7499 + 4.5)$

$0.0625 = 0.0625$       (Check!)