--- title: "Economic diameter of tunnel" subject: "Hydro Power" chapter: 5 key: "semester_6/hydro_power:economic-diameter-tunnel" source: https://www.bctengineer.com/q/semester_6/hydro_power/economic-diameter-tunnel site: BCT Engineering --- # Economic diameter of tunnel Hydro Power · Chapter 5: Components of Hydropower System - Asked in: 2075 Baisakh, 1+5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/economic-diameter-tunnel - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5 ## Question What is meant by economic diameter of tunnel? How do you work out it? ## Short answer The economic diameter minimizes total annual cost (construction plus value of energy lostdue to friction). It is found by expressing annual cost as a function of diameter, differentiating and solving, or by plotting total cost for trial diameters and selecting the minimum point. ## Answer The **economic diameter** of a tunnel (or penstock) is the diameter for which the **total annual cost**—the annual cost of construction (capital recovery + maintenance) plus the annual value of energy lost due to friction—is a **minimum**. - A larger diameter gives lower velocity and smaller head loss, so less energy is lost, but construction cost (excavation, lining) is higher. - A smaller diameter is cheaper to build but causes larger friction loss and loss of revenue every year. - The economic diameter is the best balance between these two opposing costs. #### Working it out 1. **Annual construction cost.** Cost of tunnel of length $L$ is roughly proportional to $D^2$ (excavation and lining). The annual cost is $$ \begin{aligned} C_1 = k_1 D^2 L \times (\text{capital recovery factor} + \text{O\&M rate}) \end{aligned} $$ 2. **Annual cost of energy lost.** Head loss by Manning (for a full circular section $R = D/4$): $$ \begin{aligned} h_f &= \frac{n^2 V^2 L}{R^{4/3}}, \quad V = \frac{4Q}{\pi D^2} \\ \Rightarrow h_f &= 10.29\,\frac{n^2 Q^2 L}{D^{16/3}} \end{aligned} $$ Power lost $P_L = 9.81\,Q\,h_f\,\eta$ kW; annual energy lost $= P_L \times 8760 \times$ plant factor; its value at the energy rate $r$ gives $$ C_2 = \frac{k_2}{D^{16/3}} $$ 3. **Minimise total cost.** $$ \begin{aligned} C &= C_1 + C_2 = k_1' D^2 + \frac{k_2}{D^{16/3}} \\ \frac{dC}{dD} &= 2k_1' D - \frac{16}{3}\frac{k_2}{D^{19/3}} = 0 \\ D_{eco} &= \left(\frac{8 k_2}{3 k_1'}\right)^{3/22} \end{aligned} $$ 4. **Graphical method (used in practice).** Assume several trial diameters, compute for each the annual construction cost and annual loss cost, and plot them against $D$. The diameter at the lowest point of the total cost curve is the economic diameter. ```text Annual cost ^ \ / construction | \ / cost | \ total cost / | \__ ___ / | \ ''-.-'' / | \ min /'. | '-..-' '-. energy loss cost | | '--.__ +--------------+---------------> D D_eco ``` 5. **Checks:** velocity within the usual range (about 2–3 m/s in concrete-lined tunnels), minimum size for construction (about 2–2.5 m), and practical round size. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The economic diameter of a tunnel is the diameter that gives A. minimum total annual cost (construction plus energy lost) B. minimum head loss C. minimum construction cost only D. maximum flow velocity Answer: A. It is the best balance between annual construction cost and annual value of friction energy loss. (Key point: Definition) 2. If the tunnel diameter is increased, what happens? A. Both friction loss and construction cost fall B. Friction loss falls but construction cost rises C. Friction loss rises and construction cost falls D. Both rise Answer: B. Larger D lowers velocity and head loss but raises excavation and lining cost. (Key point: Opposing costs) 3. In the derivation, how do the annual construction cost C1 and energy-loss cost C2 vary with D? A. C1 is proportional to 1/D^2 and C2 to D^(16/3) B. Both are proportional to D^2 C. C1 is proportional to D^2 and C2 to 1/D^(16/3) D. C1 is proportional to D^(16/3) and C2 to D^2 Answer: C. Construction cost goes with D^2; Manning head loss with Q^2/D^(16/3). (Key point: Cost relations) 4. How is the economic diameter found graphically? A. Take the diameter where construction cost is zero B. Take the diameter giving 5 m/s velocity C. Take the largest trial diameter D. Plot the total annual cost against trial diameters and take the lowest point Answer: D. The minimum of the total cost curve gives D_eco. 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