Discharge
$Q_1 = Q_2 = 0.03 \, \text{ m}^3\text{/s}$
Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$
$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.3^4)} = 0.0092 \, \text{ m}$
$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.1^4)} = 0.7436 \, \text{ m}$
Head loss
$HL = 0.05 \left( \dfrac{p_1}{\gamma_w} - \dfrac{p_2}{\gamma_w} \right) = 0.05 \left( \dfrac{p_1}{\gamma_w} - \dfrac{70}{\gamma_w} \right)$
$HL = \dfrac{0.05p_1}{\gamma_w} - \dfrac{3.5}{\gamma_w}$
Energy equation between 1 and 2
$E_1 - HL = E_2$
$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma_w} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma_w} + z_2$
$\left( 0.0092 + \dfrac{p_1}{\gamma_w} + 0 \right) - \left( \dfrac{0.05p_1}{\gamma_w} - \dfrac{3.5}{\gamma_w} \right) = \left( 0.7436 + \dfrac{70}{\gamma_w} + 0 \right)$
$\dfrac{0.95p_1}{\gamma_w} = 0.7436 - 0.0092 + \dfrac{70}{\gamma_w} - \dfrac{3.5}{\gamma_w}$
$\dfrac{0.95p_1}{\gamma_w} = 0.7344 + \dfrac{66.5}{\gamma_w}$
$0.95p_1 = 0.7344\gamma_w + 66.5$
$0.95p_1 = 0.7344(9.81) + 66.5$
$p_1 = 77.6 \text{ kPa}$ answer
