--- title: "Economic diameter of penstock by graphical method" subject: "Hydropower Engineering" chapter: 6 key: "semester_7/hydropower_engineering:economic-diameter-penstock-graphical" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/economic-diameter-penstock-graphical site: BCT Engineering --- # Economic diameter of penstock by graphical method Hydropower Engineering · Chapter 6: Water Conveyance Structures - Asked in: 2079 Bhadra, 1+5 marks - 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_7/hydropower_engineering/economic-diameter-penstock-graphical - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-6 ## Question What is the economical diameter of a penstock? How do you determine the economic diameter by the graphical method? ## Short answer Economic diameter is the diameter minimizing total annual cost, made up of annual fixed charges on the penstock plus annual value of power lost in friction. Graphically, plot penstock cost (rising) and power-loss cost (falling) against diameter, sum them, and read the diameter at the minimum of the total curve. ## Answer #### Economic diameter The **economic diameter** of a penstock is the diameter for which the total annual cost (the annual fixed charges on the penstock plus the annual value of the power lost in friction) is the minimum. A small diameter is cheap but gives a high velocity and a high head loss; a large diameter is costly but gives a low head loss. #### Graphical method 1. Take the design discharge $Q$, the gross head and the penstock length. Select a range of trial diameters $d_1, d_2, \dots, d_n$. 2. For each diameter, find the velocity $v = 4Q/\pi d^2$ and the friction head loss $h_f = fLv^2/2gd$ (or use the Manning or Hazen-Williams formula). 3. Annual cost of the penstock (curve 1): the thickness $t = pd/2\sigma\eta$ gives the steel weight and cost; add fabrication, erection and anchors, and multiply by the annual charge rate (interest + depreciation + maintenance, say 8 to 12 %). 4. Annual cost of the power lost (curve 2): $P_{loss} = \gamma Q h_f \eta_{overall}$ (kW), converted to energy per year and multiplied by the tariff (cost per kW-year or per kWh). 5. Plot the diameter on the x-axis and the annual cost on the y-axis for both curves, then add them to get the total annual cost curve (curve 3). 6. The diameter at the lowest point of the total curve is the economic diameter (it is also where the slopes of curves 1 and 2 are equal and opposite). ```text Annual cost ^ | T T | T T | L T T P | L T T P | L T T T P | L : P | L : P | L : P | L : P | L P: | X : | P L: | P :L L | P : L L L | P : L L L L +-----------------+-----------------------> d d_economic P = annual cost of the penstock (rises with d) L = annual cost of the power lost in friction (falls with d) T = total annual cost = P + L (minimum at d_economic) ``` Note: since the cost of the power lost falls steeply with the diameter while the penstock cost rises, their sum has one clear minimum, which is read off the plot. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the economic diameter of a penstock? A. The diameter with the minimum total annual cost (fixed charges plus value of power lost in friction) B. The smallest diameter allowed by velocity limits C. The diameter with zero head loss D. The diameter giving maximum wall thickness Answer: A. It balances annual cost of the pipe against annual cost of friction loss. (Key point: Definition) 2. How do pipe cost and friction loss vary as diameter increases? A. Both rise B. Pipe cost rises while friction loss falls C. Both fall D. Pipe cost falls while friction loss rises Answer: B. Larger pipes cost more but lose less head, so the sum has a minimum. (Key point: Trend with diameter) 3. How is the economic diameter read from the graph? A. Take d where the two curves are parallel and far apart B. Take d at the highest point of the total curve C. Plot annual cost of pipe and of power loss against d, add them, and take d at the lowest point of the total curve D. Take d where the loss curve is zero Answer: C. The minimum of the summed curve is the economic diameter, where slopes of the two curves are equal and opposite. (Key point: Graphical procedure) 4. How is the annual cost of power loss obtained for each trial diameter? A. Wall thickness times steel price B. Velocity times pipe length C. Annual charge rate times the penstock cost D. Power lost = gamma Q hf x efficiency, converted to energy per year and multiplied by the tariff Answer: D. Loss in kW from gamma Q hf eta, times tariff, gives curve 2. 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