---
title: "MOC single reach, reservoir pipe (3 s)"
subject: "Computational Techniques in Civil Engineering"
chapter: 6
key: "semester_8/computational_techniques_in_civil_engineering:moc-reservoir-pipe-sin-3"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/moc-reservoir-pipe-sin-3
site: BCT Engineering
---

# MOC single reach, reservoir pipe (3 s)

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

- Asked in: 2070 Magh, 8 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-3
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-6

## Question

A pipe conveys water from a reservoir as shown in the figure. Take $f = 0.02$, $C = 1200$ m/s. The hydraulic grade line (HGL) at the reservoir is given as $H_{PA} = 100 + 3\sin(\pi t)$. The discharge at the downstream end is zero at all times. By using only one reach, compute the discharge from A and elevation of the hydraulic grade line at B at 3 sec using the discretized equation of the method of characteristics in the form of HGL and discharge.

[Figure: reservoir with water level 100 m above the pipe centreline at A; pipe AB of length $L = 600$ m and diameter $D = 400$ mm; a sketch of the $C^+$ and $C^-$ characteristics on the grid]

## Short answer

One reach with Δt=L/a=0.5 s, B_p=973.42 s/m², R_p=96.83 s²/m⁵. Upstream head is H_PA=100+3sin(πt), downstream Q=0. At t=3 s, discharge at A is 0.000 m³/s and HGL at B is 117.968 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 $\Delta t=L/a$:
$$
\Delta t=\frac{600}{1200}=0.50\ \text{s}
$$
$$
A_p=\frac{\pi}{4}(0.4)^2=0.12566\ \text{m}^2,\quad B_p=\frac{a}{gA_p}=973.42\ \text{s/m}^2,\quad R_p=\frac{f\,L}{2gDA_p^2}=96.83\ \text{s}^2/\text{m}^5
$$

#### 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):** $H_{PA}=100+3\sin(\pi t)$ 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 ($t=0.50$ s)
$H_{PA}=100+3\sin(0.50\pi)=103.0000$ m; $K_2=100-0+0=100$; $Q_{PA}=(103.0000-100)/973.42=0.00308$ 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) |
|---|---|---|---|
| 0.50 | 0.00308 | 103.000 | 100.000 |
| 1.00 | 0.00000 | 100.000 | 105.999 |
| 1.50 | -0.00924 | 97.000 | 100.000 |
| 2.00 | 0.00000 | 100.000 | 88.009 |
| 2.50 | 0.01540 | 103.000 | 100.000 |
| 3.00 | 0.00000 | 100.000 | 117.968 |

($Q_B=0$ at all times.)

**Answer:** at $t=3$ s, the discharge from A is $Q_A=0.00000$ m$^3$/s and the HGL elevation at B is $H_B=117.968$ m.

## Check yourself

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

1. For L = 600 m and C = 1200 m/s with one reach, what is dt?
   A. 0.5 s
   B. 2.0 s
   C. 0.25 s
   D. 1.0 s
   Answer: A. dt = L/a = 600/1200 = 0.50 s. (Key point: Time step)

2. Which equation gives the discharge at the reservoir node A?
   A. Q_PA = (H_PA - K1)/B_p
   B. Q_PA = 0
   C. Q_PA = (H_PA - K2)/B_p, with K2 = H_B - B_p Q_B + R_p Q_B|Q_B|
   D. Q_PA = K1 - K2
   Answer: C. The C- line from B gives K2 and, with H_PA known, Q_PA = (H_PA - K2)/B_p. (Key point: Reservoir node equation)

3. What is the condition at node B and how is H_B found?
   A. Q_B = 0 and H_PB = K1 from the C+ equation
   B. H_B = 100 m always
   C. H_B follows the reservoir sine wave
   D. Q_B = Q_A
   Answer: A. At the dead end Q_PB = 0, so H_PB = K1 = H_A + B_p Q_A - R_p Q_A|Q_A|. (Key point: Closed end node)

4. What are Q_A and H_B at t = 3 s?
   A. Q_A = 0.01540 m^3/s, H_B = 100 m
   B. Q_A = 0, H_B = 117.968 m
   C. Q_A = 0.00308 m^3/s, H_B = 103 m
   D. Q_A = 0, H_B = 88.009 m
   Answer: B. The time march gives Q_A = 0.00000 m^3/s and H_B = 117.968 m at 3 s. (Key point: Result at t = 3 s)

## Related in Computational Techniques in Civil Engineering

- [MOC single reach, reservoir pipe (2 s)](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Amoc-reservoir-pipe-sin-2) — 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