Which equation gives the discharge at the reservoir node A?

Q_PA = (H_PA - K2)/B_p, with K2 = H_B - B_p Q_B + R_p Q_B|Q_B|

What is the condition at node B and how is H_B found?

Q_B = 0 and H_PB = K1 from the C+ equation

H_B follows the reservoir sine wave

What are Q_A and H_B at t = 3 s?

Q_A = 0.01540 m^3/s, H_B = 100 m

Q_A = 0.00308 m^3/s, H_B = 103 m

2070 Magh · 8 marks

MOC single reach, reservoir pipe (3 s)

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.

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]

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

MOC single reach, reservoir pipe (2 s) · Computational Techniques in Civil Engineering, chapter 6
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MOC single reach, reservoir pipe (3 s)

Set-up

Equations

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

First step (t=0.50t=0.50t=0.50 s)

Marching in time

Related questions

Initial condition ( $t=0$ )

First step ( $t=0.50$ s)