---
title: "MOC single reach, reservoir pipe (2 s)"
subject: "Computational Techniques in Civil Engineering"
chapter: 6
key: "semester_8/computational_techniques_in_civil_engineering:moc-reservoir-pipe-sin-2"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-reservoir-pipe-sin-2
site: BCT Engineering
---

# MOC single reach, reservoir pipe (2 s)

Computational Techniques in Civil Engineering · Chapter 6: Method of Characteristics

- Asked in: 2077 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-reservoir-pipe-sin-2
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-6

## Question

The figure below shows a pipe conveying water from a reservoir. The HGL at the reservoir is given as $H_{PA} = 100 + 2\sin(\pi t)$. The discharge at the downstream end is zero at all times. By using only one reach, compute the discharge from A and elevation of HGL at B at 2 seconds using the discretized equation of the MOC in the form of head and discharge. Take $f = 0.02$ and $c = 1250$ m/s.

[Figure: reservoir with water level 100 m above the pipe centreline at A; pipe AB of length $L = 500$ m and diameter $D = 450$ mm]

## Short answer

Use one reach, dt=L/a=0.4 s, Bp=801.17 s/m^2, Rp=44.78 s^2/m^5. A is reservoir with H_A=100+2 sin(pi t), B is dead end Q_B=0. Marching gives at t=2 s Q_A=0.00768 m^3/s and H_B=93.846 m.

## Answer

#### Set-up
One reach means two nodes: **A** at the reservoir ($x=0$) and **B** at the dead end ($x=L$). The time step follows from $\Delta x=L$ and $\Delta t=L/a$:
$$
\Delta t=\frac{500}{1250}=0.40\ \text{s}
$$
$$
A_p=\frac{\pi}{4}(0.45)^2=0.15904\ \text{m}^2,\quad B_p=\frac{a}{gA_p}=801.17\ \text{s/m}^2,\quad R_p=\frac{f\,L}{2gDA_p^2}=44.78\ \text{s}^2/\text{m}^5
$$

#### Equations
$C^+$ (from A at the old time to B) and $C^-$ (from B at the old time to A):
$$
H_{P}=H_A-B_p(Q_{P}-Q_A)-R_pQ_A|Q_A|,\qquad H_{P}=H_B+B_p(Q_{P}-Q_B)+R_pQ_B|Q_B|
$$
- **Node A (reservoir):** $H_{PA}=100+2\sin(\pi t)$ is known; use $C^-$ from B:
$$
Q_{PA}=\frac{H_{PA}-K_2}{B_p},\qquad K_2=H_B-B_pQ_B+R_pQ_B|Q_B|
$$
- **Node B (closed end):** $Q_{PB}=0$; use $C^+$ from A:
$$
H_{PB}=K_1=H_A+B_pQ_A-R_pQ_A|Q_A|
$$

#### Initial condition ($t=0$)
No flow and static head: $Q_A=Q_B=0$, $H_A=H_B=100$ m.

#### First step ($t=0.40$ s)
$H_{PA}=100+2\sin(0.40\pi)=101.9021$ m; $K_2=100-0+0=100$; $Q_{PA}=(101.9021-100)/801.17=0.00237$ m$^3$/s; $K_1=100$, so $H_{PB}=100$ m.

#### Marching in time
Repeating the two boundary calculations at every step (each uses the values of the previous step):

| t (s) | $Q_A$ (m$^3$/s) | $H_A$ (m) | $H_B$ (m) |
|---|---|---|---|
| 0.40 | 0.00237 | 101.902 | 100.000 |
| 0.80 | 0.00147 | 101.176 | 103.804 |
| 1.20 | -0.00622 | 98.824 | 102.351 |
| 1.60 | -0.00531 | 98.098 | 93.847 |
| 2.00 | 0.00768 | 100.000 | 93.846 |

($Q_B=0$ at all times.)

**Answer:** at $t=2$ s, the discharge from A is $Q_A=0.00768$ m$^3$/s and the HGL elevation at B is $H_B=93.846$ m.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With one reach of length L = 500 m and a = 1250 m/s, what is the time step and why?
   A. dt = L/a = 0.40 s so characteristics pass through the nodes
   B. dt = a/L = 2.5 s
   C. dt = 1 s by choice, with interpolation
   D. dt = L/(2a) = 0.2 s
   Answer: A. Taking dx = L and dt = dx/a puts the characteristic feet exactly on the nodes. (Key point: Time step for one reach)

2. At node A (reservoir with known H_PA), which characteristic and unknown are used?
   A. C+ from B to find H_PA
   B. C- from B to find Q_PA, since H_PA is known
   C. Neither, Q_PA is zero
   D. C+ from A to find Q_PA
   Answer: B. At the upstream node only the C- equation arrives from the pipe; with H_PA given, Q_PA = (H_PA - K2)/B_p. (Key point: Reservoir boundary uses C-)

3. At the dead end B where Q = 0, how is H_PB found?
   A. H_PB = K2
   B. H_PB = H_PA
   C. H_PB = K1 = H_A + B_p Q_A - R_p Q_A|Q_A| from the C+ equation
   D. H_PB = 100 m always
   Answer: C. With Q_PB = 0 the C+ equation from A gives H_PB = K1. (Key point: Closed-end boundary)

4. At t = 2 s, what are the discharge from A and the HGL at B?
   A. Q_A = 0.00768 m^3/s, H_B = 93.846 m
   B. Q_A = 0, H_B = 100 m
   C. Q_A = 0.00237 m^3/s, H_B = 100 m
   D. Q_A = -0.00622 m^3/s, H_B = 102.351 m
   Answer: A. Marching in steps of 0.4 s to t = 2 s gives Q_A = 0.00768 m^3/s and H_B = 93.846 m. (Key point: Result at t = 2 s)

## Related in Computational Techniques in Civil Engineering

- [MOC single reach, reservoir pipe (3 s)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-reservoir-pipe-sin-3) — Computational Techniques in Civil Engineering, chapter 6
- [MOC numerical: discharge and head at P (30 cm pipe)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-pipe-discharge-head-pc-30cm) — Computational Techniques in Civil Engineering, chapter 6