# Problem 06 - Bernoulli's Energy Theorem

**Problem 6**

As shown in Figure 4-03, the smaller pipe is cut off a short distance past the reducer so that the jet springs free into the air. Compute the pressure at 1 if Q = 5 cfs of water. D_{1} = 12 inches and D_{2} = 4 inches. Assume that the jet has the diameter D_{2}, that the pressure in the jet is atmospheric and that the loss of head from point 1 to point 2 is 5 ft of water.

**Solution 6**

$Q_1 = Q_2 = 5 \, \text{ ft}^3\text{/s}$

Velocity head

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

$\dfrac{{v_1}^2}{2g} = \dfrac{8(5^2)}{\pi^2(32.2)(1^4)} = 0.6293 \, \text{ ft}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(5^2)}{\pi^2(32.2)(4/12)^4} = 50.98 \, \text{ ft}$

Energy equation between 1 and 2

$E_1 - HL = E_2$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma_w} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma_w} + z_2$

$0.6293 + \dfrac{p_1}{\gamma_w} + 0 - 5 = 50.98 + 0 + 0$

$\dfrac{p_1}{\gamma_w} = 55.35 \, \text{ ft}$

$p_1 = 55.35\gamma_w = 55.35(62.4)$

$p_1 = 3453.88 \, \text{ psf} = 23.98 \, \text{ psi}$ answer

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