Head loss with enlarged central pipe length

Friction loss is found with Darcy's equation using the coefficient of friction $f$: $h_f = \dfrac{4fLV^2}{2gd}$ (Bansal convention, $f = 0.01$).

Case 1: Original pipe (5 cm, 5 m)

$$\begin{aligned} V_1 &= \frac{Q}{\frac{\pi}{4}d_1^2} = \frac{0.005}{\frac{\pi}{4}(0.05)^2} = 2.546\ \text{m/s} \\ h_f &= \frac{4 \times 0.01 \times 5 \times 2.546^2}{0.05 \times 2 \times 9.81} = 1.322\ \text{m} \end{aligned}$$

Case 2: Central 2 m replaced by 7.5 cm pipe

  5 cm, 1.5 m     7.5 cm, 2 m      5 cm, 1.5 m
 ==========|----------------------|==========
           ^ sudden enlargement   ^ sudden contraction

Velocity in the larger pipe:

$$V_2 = \frac{0.005}{\frac{\pi}{4}(0.075)^2} = 1.132\ \text{m/s}$$

Losses:

$$\begin{aligned} h_{f,5\,cm} &= \frac{4 \times 0.01 \times 3 \times 2.546^2}{0.05 \times 2 \times 9.81} = 0.793\ \text{m} \\ h_{f,7.5\,cm} &= \frac{4 \times 0.01 \times 2 \times 1.132^2}{0.075 \times 2 \times 9.81} = 0.070\ \text{m} \\ h_{e} &= \frac{(V_1 - V_2)^2}{2g} = \frac{(2.546-1.132)^2}{2 \times 9.81} = 0.102\ \text{m} \\ h_{c} &= 0.5\frac{V_1^2}{2g} = 0.5 \times \frac{2.546^2}{2 \times 9.81} = 0.165\ \text{m} \end{aligned}$$

Total loss:

$$h_L = 0.793 + 0.070 + 0.102 + 0.165 = 1.130\ \text{m}$$

Answer: Friction loss in the original pipe = 1.322 m. With the central 2 m replaced by a 7.5 cm pipe, total loss (friction + enlargement + contraction) = 1.130 m, a saving of about 0.19 m. The large pipe cuts friction a lot, but part of that saving is lost at the sudden enlargement and contraction.