--- title: "Flow between reservoirs (20 cm and 30 cm pipe)" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:reservoirs-pipeline-20-30cm" source: https://www.bctengineer.com/q/semester_6/hydro_power/reservoirs-pipeline-20-30cm site: BCT Engineering --- # Flow between reservoirs (20 cm and 30 cm pipe) Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2070 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/reservoirs-pipeline-20-30cm - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question Two reservoirs are connected by a pipe line which is 20 cm in diameter for the first 5 m and 30 cm in diameter for the remaining 20 m. The entrance and exit are sharp and change of section is sudden. If the water surface in the upper reservoir is 6 m above that in the lower reservoir, calculate head losses and rate of flow. Take f = 0.01 for all cases. ## Short answer With coefficient of friction f=0.01, friction loss is 4fLV²/2gd. Summing all loss coefficients gives 2.533 equated to 6 m yields V₁=6.82 m/s, discharge Q=0.214 m³/s, and individual losses as tabulated. ## Answer The total difference in water level (6 m) is used up by all the losses in the pipe: entry loss, friction in both pipes, sudden enlargement loss and exit loss. Since $f = 0.01$ is a coefficient of friction (R.K. Bansal convention), friction loss is $h_f = \dfrac{4fLV^2}{2gd}$. #### Data - Pipe 1: $d_1 = 0.20$ m, $L_1 = 5$ m; Pipe 2: $d_2 = 0.30$ m, $L_2 = 20$ m - $H = 6$ m, $f = 0.01$, $g = 9.81$ m/s² ```text Upper res. ~~~~~~~~~| H = 6 m |==== 20 cm ====|=========== 30 cm ==========| | 5 m | 20 m |~~~~~ Lower res. ``` #### Velocity relation By continuity, $V_2 = V_1\left(\dfrac{d_1}{d_2}\right)^2 = V_1\left(\dfrac{0.2}{0.3}\right)^2 = 0.4444\,V_1$ #### Losses in terms of $V_1^2/2g$ | Loss | Formula | Coefficient of $V_1^2/2g$ | |---|---|---| | Entrance (sharp) | $0.5V_1^2/2g$ | 0.5 | | Friction, pipe 1 | $4fL_1V_1^2/(2g d_1)$ | $4(0.01)(5)/0.2 = 1.0$ | | Sudden enlargement | $(V_1-V_2)^2/2g$ | $(1-0.4444)^2 = 0.3086$ | | Friction, pipe 2 | $4fL_2V_2^2/(2g d_2)$ | $\frac{4(0.01)(20)}{0.3}(0.4444)^2 = 0.5267$ | | Exit | $V_2^2/2g$ | $(0.4444)^2 = 0.1975$ | | **Total** | | **2.5329** | #### Velocity and discharge $$ \begin{aligned} 6 &= 2.5329\,\frac{V_1^2}{2g} \\ V_1 &= \sqrt{\frac{2 \times 9.81 \times 6}{2.5329}} = 6.817\ \text{m/s} \\ V_2 &= 0.4444 \times 6.817 = 3.030\ \text{m/s} \\ Q &= \frac{\pi}{4}(0.2)^2 \times 6.817 = 0.2142\ \text{m}^3/\text{s} \end{aligned} $$ #### Individual head losses | Loss | Value (m) | |---|---| | Entrance | 1.184 | | Friction in 20 cm pipe | 2.369 | | Sudden enlargement | 0.731 | | Friction in 30 cm pipe | 1.248 | | Exit | 0.468 | | **Total** | **6.000** | **Answer:** Discharge $Q \approx 0.214$ m³/s (214 L/s); the losses are as tabulated, adding up to the 6 m head difference. (If $f = 0.01$ were taken as the Darcy factor, $h_f = fLV^2/2gd$, the friction terms would be four times smaller and Q would be larger; the coefficient-of-friction reading is the usual one for $f = 0.01$.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the 6 m difference in reservoir levels equal in this problem? A. Only the friction loss in both pipes B. The sum of entry, friction, enlargement and exit losses C. Only the velocity head in the 30 cm pipe D. The entry loss plus the exit loss only Answer: B. The whole level difference is used up by all losses: entry, friction in both pipes, sudden enlargement and exit. (Key point: Energy balance) 2. How is the velocity in the 30 cm pipe related to that in the 20 cm pipe? A. V2 = V1 x (0.3/0.2)^2 B. V2 = V1 x (0.2/0.3) C. V2 = V1 x (0.2/0.3)^2 = 0.444 V1 D. V2 = V1 Answer: C. Continuity gives A1V1 = A2V2, so V2 = V1(d1/d2)^2 = 0.4444 V1. (Key point: Continuity) 3. Which expression gives the loss at the junction where the pipe widens from 20 cm to 30 cm? A. 0.5 V1^2/2g B. (V1 - V2)^2/2g C. V2^2/2g D. (V1 + V2)^2/2g Answer: B. Sudden enlargement loss is (V1 - V2)^2/2g, which is 0.3086 V1^2/2g here. (Key point: Sudden enlargement loss) 4. After summing the coefficients (total 2.5329 of V1^2/2g), how is V1 and then Q found? A. Set 6 = 2.5329 V1^2/2g, solve V1 = 6.82 m/s, then Q = area of 20 cm pipe x V1 B. Set 6 = V1^2/2g, then multiply by the 30 cm area C. Divide 6 by 2.5329 to get V1 directly D. Use the exit loss alone to find V2, then Q = A1 V2 Answer: A. Equate total loss to 6 m, solve V1 = 6.817 m/s and Q = (pi/4)(0.2)^2 x V1 = 0.214 m^3/s. 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