--- title: "Optimization of hydraulic tunnel diameter" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:tunnel-optimization-diameter" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/tunnel-optimization-diameter site: BCT Engineering --- # Optimization of hydraulic tunnel diameter Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2082 Bhadra, 4 marks; 2082 Baishakh, 3 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/tunnel-optimization-diameter - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question How do you carry out the optimization of a hydraulic tunnel (how can the diameter of a power tunnel in a hydropower project be optimized using cost-benefit analysis)? Explain. ## Short answer Select trial diameters; compute annualized pipe cost and annual cost of lost energy (hf ∝ Q²L/D⁵). Total cost minimum gives optimum; for cost Cc = k D^m, Ce = a D⁻⁵, Dopt = [5a/(m·k·CRF)]^(1/(m+5)). Check velocity 2.5–4 m/s. ## Answer The diameter of a power tunnel is chosen so that the **net benefit** (value of the energy saved minus the cost of the tunnel) is highest. A small diameter is cheap but gives a high velocity, a high head loss and less energy; a large diameter costs more but loses less head. #### Step by step procedure 1. **Select several trial diameters** $D_1, D_2, \dots$ (e.g. 3.0, 3.5, 4.0, 4.5, 5.0 m) for the design flow $Q$ and the tunnel length $L$. 2. **Head loss** for each: $h_f=\dfrac{f\,L}{D}\dfrac{V^2}{2g}$ (or Manning), with $V=\dfrac{4Q}{\pi D^2}$, so $h_f\propto\dfrac{Q^2L}{D^5}$. Add the minor losses. 3. **Power and energy loss**: $P_{loss}=\rho g Q h_f\eta$, and the annual energy loss is $E_{loss}=P_{loss}\times8760\times PF$ (PF = plant factor; the variation of $Q$ over the flow-duration curve should be included). 4. **Annual cost of the lost energy**: $C_E=E_{loss}\times\text{tariff}$ (loss of income). Add the cost of the lost capacity. 5. **Capital cost** of the tunnel for each diameter: excavation, support, lining and grouting, usually of the form $C_c=k\,D^{m}$ (with $m$ about 1.2-2). Convert it to an **annual cost**: $C_A=C_c\times CRF+\text{O\&M}$, where CRF is the capital recovery factor for the given interest rate and life. 6. **Total annual cost**: $C_T(D)=C_A(D)+C_E(D)$. 7. **Optimum diameter**: the diameter at which $C_T$ is minimum, i.e. where $\dfrac{dC_A}{dD}+\dfrac{dC_E}{dD}=0$; or the diameter at which the **incremental benefit/cost ratio** (extra energy value / extra annual cost) just exceeds 1. Plot $C_A$, $C_E$ and $C_T$ against $D$ and read off the minimum. If $C_c=kD^{m}$ and $C_E=a\,D^{-5}$, then $$ \frac{d}{dD}\left(k\,CRF\,D^{m}+aD^{-5}\right)=0\ \Rightarrow\ D_{opt}=\left(\frac{5a}{m\,k\,CRF}\right)^{\frac{1}{m+5}} $$ 8. **Check** the practical limits: velocity (about 2.5-4 m/s for lined pressure tunnels, less for unlined ones), the minimum construction diameter (about 2.5-3 m for TBM or drill and blast), geology and surge conditions. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the optimum tunnel diameter exist? A. Small D is cheap but has high head and energy loss; large D costs more but loses less B. Large D is both cheaper and loses less C. Small D reduces velocity and loss D. Diameter has no effect on loss Answer: A. Capital cost rises and energy loss falls with D, so total annual cost has a minimum. (Key point: Trade-off in diameter) 2. How does head loss vary with diameter for given Q and L? A. h_f is proportional to D B. h_f is proportional to Q^2 L / D^5 C. h_f is proportional to 1/D^2 only D. h_f is independent of Q Answer: B. With V = 4Q/(pi D^2), h_f = f L V^2/(2 g D) is proportional to Q^2 L / D^5. (Key point: Head loss dependence) 3. What is minimised to select the diameter? A. Capital cost alone B. C_T = annualised capital cost (capital x CRF + O&M) + annual cost of lost energy C. Energy loss alone D. The velocity Answer: B. Capital cost is converted to annual cost via CRF and added to the value of lost energy; the minimum C_T gives the optimum D. (Key point: Total annual cost) 4. What defines the optimum diameter mathematically? A. dC_A/dD + dC_E/dD = 0, or incremental benefit/cost ratio just exceeding 1 B. C_A = 0 C. D where velocity is zero D. D where C_E is maximum Answer: A. At the minimum of total cost the slope is zero; equivalently extra benefit equals extra cost (incremental B/C about 1). (Key point: Optimality condition) ## Related in Hydropower Engineering - [Optimize capacity by marginal cost-benefit](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Amarginal-cost-benefit-installed-capacity) — Hydropower Engineering, chapter 3 - [Hydraulic design of tunnel; alignment criteria](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Atunnel-hydraulic-design-alignment-selection) — Hydropower Engineering, chapter 6 - [Hydraulic design of surge tank and pressure tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-tank-hydraulic-design-pressurised-tunnel) — Hydropower Engineering, chapter 6 - [Economic diameter of penstock (analysis)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aeconomic-diameter-penstock-mathematical) — Hydropower Engineering, chapter 6, asked 2× - [Most economic penstock diameter (cost numerical)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Apenstock-economic-diameter-cost-numerical) — Hydropower Engineering, chapter 6 - [Surge tank: 7.5 m³/s through 2.5 m tunnel](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Asurge-numerical-7-5cumec-2-5m) — Hydropower Engineering, chapter 6 ## Related in other subjects - [Project proposal; cost benefit analysis for road](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Aproposal-cba-road-project) — Project Engineering, chapter 2 - [Socio-economic survey and social cost benefit analysis](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Asocio-economic-survey-scba) — Project Engineering, chapter 2, asked 3× - [Short note: Cost-benefit analysis](https://www.bctengineer.com/question.md?key=semester_7%2Fproject_engineering%3Acost-benefit-analysis-note) — Project Engineering, chapter 2 - [Economic diameter of tunnel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aeconomic-diameter-tunnel) — Hydro Power, chapter 5