Discharge

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

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