--- title: "HGL and TEL for two-pipe system" subject: "Hydraulics" chapter: 1 key: "semester_4/hydraulics:hgl-tel-two-pipe-reservoir" source: https://www.bctengineer.com/q/semester_4/hydraulics/hgl-tel-two-pipe-reservoir site: BCT Engineering --- # HGL and TEL for two-pipe system Hydraulics · Chapter 1: Pipe flow - Asked in: 2076 Baisakh, 8 marks - Video: [Problem 1 on Total energy line and hydraulic gradient line / Pipe attached with tank](https://www.youtube.com/watch?v=AVb55pB_DAM) — Edutech Guru Engineering Learning-By Priyanka ma'am - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/hgl-tel-two-pipe-reservoir - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-1 ## Question Draw the hydraulic gradient line (HGL) and total energy line (TEL) for the following, considering all losses of head. The pipe is laid horizontal and discharges freely into the atmosphere. Take coefficient of friction f = 0.01. [Figure: a reservoir with water level 8 m above the pipe axis; a horizontal pipe of D1 = 0.15 m and length 25 m followed by a pipe of D2 = 0.3 m and length 15 m discharging to the atmosphere] ## Short answer Solving energy equation with all losses yields V1 = 7.46 m/s, V2 = 1.86 m/s. TEL drops from 8 m to 0.177 m at outlet; HGL ends at axis. In pipe 1 HGL is 2.83 m below TEL, in pipe 2 it is 0.177 m below; expansion causes HGL to rise from -0.97 m to 0.09 m. ## Answer **Given.** Reservoir level 8 m above the pipe axis (datum = pipe axis). Pipe 1: $D_1 = 0.15$ m, $L_1 = 25$ m. Pipe 2: $D_2 = 0.3$ m, $L_2 = 15$ m, $f = 0.01$. The pipe discharges freely, so the outlet velocity head stays as kinetic energy (no exit loss term, the HGL ends at the axis). #### Velocities Continuity: $V_2 = V_1\left(\dfrac{D_1}{D_2}\right)^2 = V_1/4$. Energy equation from the reservoir surface to the outlet (all losses): $$ H = 0.5\frac{V_1^2}{2g} + f\frac{L_1}{D_1}\frac{V_1^2}{2g} + \frac{(V_1-V_2)^2}{2g} + f\frac{L_2}{D_2}\frac{V_2^2}{2g} + \frac{V_2^2}{2g} $$ $$ 8 = \frac{V_1^2}{2g}\left[0.5 + 0.01\left(\frac{25}{0.15}\right) + (1-0.25)^2 + 0.01\left(\frac{15}{0.3}\right)(0.25)^2 + (0.25)^2\right] = 2.8229\,\frac{V_1^2}{2g} $$ $$ V_1 = \sqrt{\frac{2(9.81)(8)}{2.8229}} = 7.457\ \text{m/s},\qquad V_2 = 1.864\ \text{m/s} $$ Velocity heads: $V_1^2/2g = 2.834$ m, $V_2^2/2g = 0.177$ m. Discharge $= 131.8$ l/s. #### Head losses | Loss | Head (m) | |---|---| | Entrance, $0.5V_1^2/2g$ | 1.417 | | Friction in pipe 1 | 4.723 | | Sudden expansion, $(V_1-V_2)^2/2g$ | 1.594 | | Friction in pipe 2 | 0.089 | | Exit velocity head (kinetic) | 0.177 | | **Total** | 8.000 | The total equals the 8 m available head. #### Levels of TEL and HGL (m above the pipe axis) | Location | TEL | HGL = TEL $-V^2/2g$ | |---|---|---| | Reservoir surface | 8.000 | 8.000 | | Just inside pipe 1 (after entrance loss) | 6.583 | 3.749 | | End of pipe 1 (before expansion) | 1.860 | -0.974 | | Start of pipe 2 (after expansion) | 0.266 | 0.089 | | Outlet | 0.177 | 0.000 | ```text 8.0 ___ TEL | |\ 0.5V²/2g | | \__ TEL falls steeply in pipe 1 (small D) | | `-.__ | | HGL `-. | drop (V1-V2)²/2g | | (well below) |`.__ TEL falls gently in pipe 2 | | | `-.__ HGL 0 --|__|===[D1=0.15]===[D2=0.3]===> free outlet ``` Draw to scale: TEL: 8.0, 6.58 (entrance drop), 1.86 (end of pipe 1), 0.27 (after expansion drop), 0.18 (outlet). HGL is lower by 2.83 m in pipe 1 and by 0.18 m in pipe 2, and meets the pipe axis at the outlet. The HGL rises slightly at the expansion (from -0.97 to 0.09 m) because the velocity head falls. **Answer:** $V_1 = 7.46$ m/s, $V_2 = 1.86$ m/s, TEL ends 0.177 m above the axis at the outlet and the HGL ends at the axis. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which terms are included in the energy equation from the reservoir surface to the outlet? A. Entrance loss, friction in both pipes, expansion loss and outlet velocity head B. Only friction in both pipes C. Friction and expansion loss only, with zero outlet velocity head D. Entrance loss and contraction loss only Answer: A. The 8 m head is spent on entrance, both frictions, the expansion loss and the kinetic energy leaving the free outlet. (Key point: Energy equation) 2. How are V2 and V1 related for D1 = 0.15 m and D2 = 0.3 m? A. V2 = V1/2 B. V2 = 2V1 C. V2 = V1/4 D. V2 = V1 Answer: C. By continuity V2 = V1 (D1/D2)^2 = V1/4. (Key point: Velocity relation) 3. Where do the HGL and TEL end at the free outlet? A. Both at the pipe axis B. HGL at the pipe axis and TEL V2^2/2g above it C. TEL at the axis and HGL below the axis D. HGL at the reservoir level Answer: B. Gauge pressure is zero at a free discharge so the HGL meets the axis, and TEL is above it by the velocity head 0.177 m. (Key point: HGL at outlet) 4. Why does the HGL rise from -0.97 m to 0.09 m at the expansion even though energy is lost? A. The friction factor drops B. Water is added at the joint C. The velocity head falls, so pressure head recovers D. The reservoir level rises Answer: C. HGL = TEL - V^2/2g; V^2/2g falls from 2.83 m to 0.18 m, more than the TEL drop of the expansion loss. 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