--- title: "Head loss with enlarged central pipe length" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:pipe-replaced-central-length" source: https://www.bctengineer.com/q/semester_6/hydro_power/pipe-replaced-central-length site: BCT Engineering --- # Head loss with enlarged central pipe length Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2073 Bhadra, 4+6 marks - Video: [Laminar and Turbulent flow (Problems on Minor Losses ) Lecture 10 by PSS](https://www.youtube.com/watch?v=KowbfFf4X04) — prashant shedbalkar - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/pipe-replaced-central-length - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question A pipe 5 cm diameter is 5 m long and carries a discharge of 0.005 m³/s. Find the loss of head due to friction. The central 2 m length of the pipe is replaced by a pipe 7.5 cm diameter, the changes of section being sudden. Determine the loss of head due to these alternatives considering all losses. Take f = 0.01 for all pipes and contraction loss coefficient = 0.5. ## Short answer Original uniform pipe friction loss = 1.322 m. With central 2 m replaced by 7.5 cm pipe, total loss (friction + sudden enlargement + contraction) = 1.130 m. ## Answer Friction loss is found with Darcy's equation using the coefficient of friction $f$: $h_f = \dfrac{4fLV^2}{2gd}$ (Bansal convention, $f = 0.01$). #### Case 1: Original pipe (5 cm, 5 m) $$ \begin{aligned} V_1 &= \frac{Q}{\frac{\pi}{4}d_1^2} = \frac{0.005}{\frac{\pi}{4}(0.05)^2} = 2.546\ \text{m/s} \\ h_f &= \frac{4 \times 0.01 \times 5 \times 2.546^2}{0.05 \times 2 \times 9.81} = 1.322\ \text{m} \end{aligned} $$ #### Case 2: Central 2 m replaced by 7.5 cm pipe ```text 5 cm, 1.5 m 7.5 cm, 2 m 5 cm, 1.5 m ==========|----------------------|========== ^ sudden enlargement ^ sudden contraction ``` Velocity in the larger pipe: $$ V_2 = \frac{0.005}{\frac{\pi}{4}(0.075)^2} = 1.132\ \text{m/s} $$ Losses: $$ \begin{aligned} h_{f,5\,cm} &= \frac{4 \times 0.01 \times 3 \times 2.546^2}{0.05 \times 2 \times 9.81} = 0.793\ \text{m} \\ h_{f,7.5\,cm} &= \frac{4 \times 0.01 \times 2 \times 1.132^2}{0.075 \times 2 \times 9.81} = 0.070\ \text{m} \\ h_{e} &= \frac{(V_1 - V_2)^2}{2g} = \frac{(2.546-1.132)^2}{2 \times 9.81} = 0.102\ \text{m} \\ h_{c} &= 0.5\frac{V_1^2}{2g} = 0.5 \times \frac{2.546^2}{2 \times 9.81} = 0.165\ \text{m} \end{aligned} $$ Total loss: $$ h_L = 0.793 + 0.070 + 0.102 + 0.165 = 1.130\ \text{m} $$ | Arrangement | Total head loss | |---|---| | Uniform 5 cm pipe | 1.322 m | | With 7.5 cm central length | 1.130 m | **Answer:** Friction loss in the original pipe = 1.322 m. With the central 2 m replaced by a 7.5 cm pipe, total loss (friction + enlargement + contraction) = 1.130 m, a saving of about 0.19 m. The large pipe cuts friction a lot, but part of that saving is lost at the sudden enlargement and contraction. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With f = 0.01 taken as coefficient of friction, which friction loss formula is used? A. hf = fLV^2/(2gd) B. hf = 4fLV^2/(2gd) C. hf = fLV/(2gd) D. hf = 4fLV/(gd) Answer: B. The Bansal convention uses hf = 4fLV^2/(2gd). (Key point: Friction formula) 2. When the central 2 m of the 5 m pipe is replaced, what are the lengths of the 5 cm pipe left in the line? A. 2 m in total B. 5 m in total C. 3 m in total (1.5 m on each side) D. 1.5 m in total Answer: C. 5 m minus the 2 m replaced leaves 3 m of 5 cm pipe, 1.5 m on each side. (Key point: Pipe arrangement) 3. Which minor losses are added in the modified pipe? A. Sudden enlargement and sudden contraction B. Only the exit loss C. Only the entrance loss D. Bend and valve losses Answer: A. Flow goes from 5 cm to 7.5 cm (enlargement, (V1-V2)^2/2g) and back (contraction, 0.5 V1^2/2g). (Key point: Minor losses) 4. How does the total loss with the 7.5 cm central length compare with the uniform pipe? A. Larger, because of two extra minor losses B. Equal, since the discharge is the same C. Smaller (1.130 m vs 1.322 m) because friction falls more than the minor losses add D. Zero friction loss in the central part Answer: C. The larger pipe cuts friction a lot, but part of the saving is lost at the enlargement and contraction; net saving is about 0.19 m. 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