Head loss at sudden enlargement

Given: $D_1 = 0.15$ m, $D_2 = 0.225$ m, $Q = 0.05$ m³/s, horizontal pipe.

Velocities

$$\begin{aligned} A_1 &= \frac{\pi}{4}(0.15)^2 = 0.017671\ \text{m}^2, \quad V_1 = \frac{0.05}{0.017671} = 2.829\ \text{m/s} \\ A_2 &= \frac{\pi}{4}(0.225)^2 = 0.039761\ \text{m}^2, \quad V_2 = \frac{0.05}{0.039761} = 1.258\ \text{m/s} \end{aligned}$$

(i) Loss of head due to sudden expansion

$$h_e = \frac{(V_1 - V_2)^2}{2g} = \frac{(2.829 - 1.258)^2}{2 \times 9.81} = 0.1259\ \text{m}$$

(ii) Pressure difference (Bernoulli with loss, $z_1 = z_2$)

$$\frac{p_1}{\rho g} + \frac{V_1^2}{2g} = \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + h_e$$ $$\begin{aligned} \frac{p_2 - p_1}{\rho g} &= \frac{V_1^2 - V_2^2}{2g} - h_e = \frac{2.829^2 - 1.258^2}{19.62} - 0.1259 \\ &= 0.3274 - 0.1259 = 0.2015\ \text{m} \\ p_2 - p_1 &= 9810 \times 0.2015 = 1977\ \text{N/m}^2 \end{aligned}$$

(iii) Gradual enlargement without loss ($h_e = 0$)

$$p_2 - p_1 = \rho\frac{V_1^2 - V_2^2}{2} = 9810 \times 0.3274 = 3212\ \text{N/m}^2$$

Answer: (i) $h_e = 0.126$ m; (ii) pressure rises by $p_2 - p_1 \approx 1.98$ kN/m²; (iii) with a gradual, loss-free enlargement the rise would be $\approx 3.21$ kN/m².