With one reach of length L = 500 m and a = 1250 m/s, what is the time step and why?

dt = L/a = 0.40 s so characteristics pass through the nodes

dt = 1 s by choice, with interpolation

At node A (reservoir with known H_PA), which characteristic and unknown are used?

C- from B to find Q_PA, since H_PA is known

At the dead end B where Q = 0, how is H_PB found?

H_PB = K1 = H_A + B_p Q_A - R_p Q_A|Q_A| from the C+ equation

At t = 2 s, what are the discharge from A and the HGL at B?

Q_A = 0.00768 m^3/s, H_B = 93.846 m

Q_A = 0.00237 m^3/s, H_B = 100 m

Q_A = -0.00622 m^3/s, H_B = 102.351 m

2077 Chaitra · 6 marks

MOC single reach, reservoir pipe (2 s)

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.

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]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

MOC single reach, reservoir pipe (3 s) · Computational Techniques in Civil Engineering, chapter 6
MOC numerical: discharge and head at P (30 cm pipe) · Computational Techniques in Civil Engineering, chapter 6

All Computational Techniques in Civil Engineering chapter 6 questions →

MOC single reach, reservoir pipe (2 s)

Set-up

Equations

Initial condition (t=0t=0t=0)

First step (t=0.40t=0.40t=0.40 s)

Marching in time

Related questions

Initial condition ( $t=0$ )

First step ( $t=0.40$ s)