--- title: "Economic diameter of penstock" subject: "Hydro Power" chapter: 5 key: "semester_6/hydro_power:penstock-economic-diameter" source: https://www.bctengineer.com/q/semester_6/hydro_power/penstock-economic-diameter site: BCT Engineering --- # Economic diameter of penstock Hydro Power · Chapter 5: Components of Hydropower System - Asked in: 2077 Chaitra, 8 marks; 2079 Chaitra, 4 marks; 2075 Bhadra, 1+4 marks; 2072 Asoj, 3+3 marks; 2070 Magh, 3 marks (asked 5 times) - Video: [Unit:5 | Explain the analytical method to determine the economic diameter of Penstock Pipe | TU,PU](https://www.youtube.com/watch?v=uhi6LPl1JdA) — Prashant YT - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/penstock-economic-diameter - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-5 ## Question What is economic diameter of penstock? How do you determine the economic diameter of a penstock? ## Short answer Economic diameter minimizes total annual cost (penstock + energy loss). It is determined by computing penstock annualized cost and energy loss value for trial diameters, plotting total cost, and selecting the minimum. ## Answer #### Economic diameter The **economic diameter** of a penstock is the diameter for which the **total annual cost** (annual cost of the penstock + annual value of energy lost in friction) is **minimum**. - A **larger diameter** gives lower velocity and lower head loss (more energy), but costs more steel, excavation, supports and transport. - A **smaller diameter** is cheaper to build, but head loss $h_f = \dfrac{8fLQ^2}{\pi^2 g D^5}$ is high, so energy is lost every year. ```text Cost |T T | T T | T T | T min T | T T * T T C |E : C | E : C | E : C | E C: | C E | C : E E E +----------:-------------> D D_econ T = total, C = penstock cost, E = value of energy lost ``` #### Method of determination (analytical/graphical) 1. **Fix data:** design discharge Q, gross head H, penstock length L, friction factor f, energy rate, interest rate, life, steel price. 2. **Assume trial diameters** $D_1, D_2, \dots$ (e.g. velocity 2–6 m/s). 3. **Penstock cost:** for each D, find shell thickness $t = \dfrac{pD}{2\sigma\eta_j}$ + corrosion allowance (p includes water hammer), steel weight $= \pi D t L\rho_s$, then capital cost (steel + fabrication + supports + excavation). Annual cost = capital × capital recovery factor + O&M. 4. **Energy loss cost:** $h_f = \dfrac{8fLQ^2}{\pi^2 g D^5}$; power lost $= 9.81Qh_f\eta$ kW; annual energy lost $= $ power × hours of operation (× plant factor); its value = energy × tariff. 5. **Total annual cost** = penstock annual cost + value of lost energy. 6. **Plot** the three curves against D; the D at the **lowest total cost** is the economic diameter. Analytically, set $\dfrac{d(\text{total cost})}{dD} = 0$. 7. Round up to a standard size and check velocity, water hammer, and that head loss is usually 2–5% of gross head. #### Empirical formulas (for preliminary design) With P = rated power in kW, H = head in m, D in m: - **Sarkaria (1979):** $D = 0.62\dfrac{P^{0.35}}{H^{0.65}}$ - **USBR:** $D = 0.176\left(\dfrac{P}{H}\right)^{0.466}$ - **Fahlbusch (1987):** $D = 0.52\,H^{-0.17}\left(\dfrac{P}{H}\right)^{0.43}$ - Velocity method: $D = \sqrt{4Q/(\pi V)}$ with economic velocity about 3–5 m/s. These give a first estimate, which is refined by the cost analysis above. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The economic diameter of a penstock is the diameter for which: A. Total annual cost (penstock cost + value of energy lost) is minimum B. Steel weight is minimum C. Head loss is zero D. Velocity is exactly 1 m/s Answer: A. Economic diameter minimises the sum of annual penstock cost and the annual value of energy lost in friction. (Key point: Definition of economic diameter) 2. If a smaller penstock diameter is chosen, what happens? A. Both cost and head loss rise B. Cost rises and head loss falls C. Cost falls but head loss and energy loss rise D. Both cost and head loss fall Answer: C. A smaller pipe is cheaper but velocity rises and h_f (proportional to 1/D^5) rises, so more energy is lost each year. (Key point: Trade-off with diameter) 3. Which sequence correctly describes finding the economic diameter? A. Assume trial diameters, compute annual penstock cost and annual energy-loss value, add them, and pick the D with lowest total B. Maximise the energy lost and choose the largest D C. Choose D so that head loss equals gross head D. Pick one diameter, check velocity, finish Answer: A. Trial diameters are costed for both pipe and lost energy; the minimum of the total curve (or d(total)/dD = 0) gives D. (Key point: Steps of determination) 4. Which is the velocity-method estimate of penstock diameter? A. D = pD/(2 sigma) B. D = 0.62 P^0.35 / H^0.65 only C. D = sqrt(4Q/(pi V)) D. D = 8fLQ^2 / (pi^2 g) Answer: C. From continuity Q = (pi D^2/4)V, so D = sqrt(4Q/(pi V)), with economic velocity about 3-5 m/s. 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