--- title: "Pressure-time diagram 1200 m from reservoir" subject: "Hydraulics" chapter: 4 key: "semester_4/hydraulics:pressure-time-diagram-1200m-from-reservoir" source: https://www.bctengineer.com/q/semester_4/hydraulics/pressure-time-diagram-1200m-from-reservoir site: BCT Engineering --- # Pressure-time diagram 1200 m from reservoir Hydraulics · Chapter 4: Unsteady flow in pipes - Asked in: 2071 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/pressure-time-diagram-1200m-from-reservoir - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-4 ## Question Water is flowing from a reservoir in a pipe of 600 mm diameter, 3000 m long and 6 mm thick at a velocity of 3.5 m/s. Assuming the value of bulk modulus of elasticity for water as 2.06 GPa, modulus of elasticity for pipe material 206 GPa and velocity of pressure wave 1400 m/s. Draw pressure-time diagram at location 1200 m from reservoir if the valve located at the end of the pipe is closed in 1 second. ## Short answer At 1200 m from reservoir (1800 m from valve), rapid closure yields ΔH = cV/g = 499 m. Pressure is H0 until 1.29 s, rises trapezoidally to H0+499 m by 2.29–3.0 s, falls to H0 by 4.0 s, then H0−499 m from 5.57 to 7.29 s, repeating every 8.57 s. ## Answer Closure time $t_c=1$ s is shorter than the pipe period $2L/c=2\times3000/1400=4.29$ s, so the closure is rapid and the full Joukowsky pressure develops. The pressure at the section is then a trapezoidal wave that repeats every $4L/c$. #### Given data $D=0.6$ m, $L=3000$ m, $v=3.5$ m/s, $c=1400$ m/s (given), $t_c=1$ s, section at $x=1200$ m from the reservoir, i.e. 1800 m from the valve. The static head $H_0$ is not given, so pressure is plotted as $H_0\pm\Delta H$. (Note: the given $K$ and $E$ would give $c\approx1015$ m/s for a thin-walled elastic pipe; the stated $c=1400$ m/s is used as instructed.) #### Pressure rise $$ \Delta p=\rho c v=1000\times1400\times3.5=4.90\ \text{MPa},\qquad \Delta H=\frac{cv}{g}=\frac{1400\times3.5}{9.81}=499.5\ \text{m} $$ #### Wave travel times - Pipe period: $4L/c=4\times3000/1400=8.571$ s. - Wave from valve reaches the section: $(L-x)/c=1800/1400=1.286$ s. - Wave from valve reaches the reservoir: $L/c=2.143$ s, and returns as a negative wave. #### Pressure history at the section (1200 m from reservoir) | Time (s) | Event | Pressure | |---|---|---| | 0 – 1.29 | Wave not yet arrived | $H_0$ | | 1.29 – 2.29 | Positive wave arrives (rises over 1 s) | rises to $H_0+499$ m | | 2.29 – 3.00 | Positive wave passing | $H_0+499$ m | | 3.00 – 4.00 | Negative wave reflected from reservoir arrives | falls to $H_0$ | | 4.00 – 5.57 | Net effect zero | $H_0$ | | 5.57 – 6.57 | Negative wave reflected from closed valve arrives | falls to $H_0-499$ m | | 6.57 – 7.29 | Negative wave passing | $H_0-499$ m | | 7.29 – 8.29 | Positive wave reflected from reservoir arrives | rises to $H_0$ | | 8.29 – 9.86 | Cycle repeats from 8.57 s | $H_0$ | #### Pressure-time diagram (friction neglected) ```text H0+dH| ___ | //// \\\\ H0 |_____ ______ _______ | \\\\\ //// H0-dH| __ +|---|---|---|---|---|---|---|---|---|--- 0 1 2 3 4 5 6 7 8 9 t(s) ``` dH = 499 m. Sloping parts last 1 s (equal to the closure time). In practice friction damps the wave, and the low-pressure part cannot fall below vapour pressure. **Answer: peak pressure rise = 4.90 MPa (499 m of water); the section sees +499 m from 1.29 s to 3.00 s (falling to $H_0$ by 4.00 s), then −499 m from 5.57 s to 7.29 s, repeating every 8.57 s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is full Joukowsky pressure developed when 2L/c = 4.29 s and t_c = 1 s? A. Because the pipe is steel B. Because the section is far from the valve C. Because closure is shorter than 2L/c, so it is rapid D. Because velocity is above 3 m/s Answer: C. Closure faster than 2L/c is complete before reflected relief arrives, so dp = rho c v. (Key point: Rapid closure) 2. What is the head rise dH? A. L v / g t_c = 1070 m B. c/g = 142.7 m C. v^2/2g = 0.62 m D. cv/g = 499.5 m Answer: D. dH = 1400 x 3.5 / 9.81 = 499.5 m. (Key point: Pressure rise) 3. When does the positive wave from the valve first reach the section 1200 m from the reservoir? A. 4.29 s B. 0.86 s C. 2.14 s D. 1.29 s Answer: D. Distance from valve is 1800 m, so t = 1800/1400 = 1.29 s. (Key point: First arrival time) 4. How often does the pressure pattern repeat? A. Every L/c = 2.14 s B. Every 2L/c = 4.29 s C. Every t_c = 1 s D. Every 4L/c = 8.57 s Answer: D. One full cycle of wave travel is 4L/c. 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