Most economic penstock diameter (cost numerical)

Given

$Q = 8\ \text{m}^3/\text{s}$, $h = 25$ m, $\eta = 0.8$.

• Pipe cost per metre: $C = 250\,h\,d^2 = 250(25)d^2 = 6250\,d^2$ (US$)
• Annual fixed charge: $0.08 \times 6250\,d^2 = 500\,d^2$ (US$/m/year)

Head loss and power loss per metre

$$h_f = \frac{0.025\,Q^2}{12.1\,d^5} = \frac{0.025 \times 64}{12.1\,d^5} = \frac{0.1322}{d^5}\ \text{m/m}$$

Power lost per metre length:

$$P = \gamma Q h_f \eta = 9.81 \times 8 \times 0.8 \times \frac{0.1322}{d^5} = \frac{8.302}{d^5}\ \text{kW}$$

Annual cost of lost power at US$500 per kW per year:

$$C_{loss} = 500 \times \frac{8.302}{d^5} = \frac{4151}{d^5}$$

Total annual cost and minimum

$$C_T = 500\,d^2 + \frac{4151}{d^5}$$ $$\frac{dC_T}{dd} = 1000\,d - \frac{20755}{d^6} = 0 \;\Rightarrow\; d^7 = 20.755$$ $$d = (20.755)^{1/7} = 1.542\ \text{m}$$

Check: at $d = 1.542$ m the fixed charge is US$1189 and the power-loss cost is US$476, total US$1665 per metre per year.

Answer: most economic diameter $\approx 1.54$ m.