Problem 14 - Bernoulli's Energy Theorem | Fluid Mechanics and Hydraulics Review at MATHalino

Problem 14
Water discharges through an orifice in the side of a large tank shown in Figure 4-06. The orifice is circular in cross section and 50 mm in diameter. The jet is the same diameter as the orifice. The liquid is water, and the surface elevation is maintained at a height h of 3.8 m above the center of the jet. Compute the discharge: (a) neglecting loss of head; (b) considering the loss of head to be 10 percent of h.

Solution 14

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$h = 3.8 \, \text{ m}$

Neglecting head lost
$E_1 = E_3$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 = \dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3$

$0 + 0 + 3.8 = \dfrac{8Q^2}{\pi^2g{D_3}^4} + 0 + 0$

$\dfrac{8Q^2}{\pi^2(9.81)(0.05^4)} = 3.8$

$Q = 0.01695 \, \text{ m}^3\text{/s} = 16.95 \, \text{ L/s}$ answer

Considering head lost
$HL = 0.10h = 0.10(3.8)$

$HL = 0.38 \, \text{ m}$

$E_1 - HL = E_3$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 - HL = \dfrac{{v_3}^2}{2g} + \dfrac{p_3}{\gamma} + z_3$

$(0 + 0 + 3.8) - 0.38 = \dfrac{8Q^2}{\pi^2g{D_3}^4} + 0 + 0$

$\dfrac{8Q^2}{\pi^2(9.81)(0.05^4)} = 3.42$

$Q = 0.01608 \, \text{ m}^3\text{/s} = 16.08 \, \text{ L/s}$ answer

Problem 14 - Bernoulli's Energy Theorem

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