With only friction losses considered, how is the head loss in each pipe written in this solution?

h_f = rQ^2 with r = 8fL/(pi^2 g D^5)

h_f = rQ with r = 8fL/(pi^2 g D^5)

h_f = rQ^2 with r = 8fL/(pi^2 g D^2)

Why is it first tested whether B receives or supplies water by assuming H_D = 70 m?

Because with H_D = 70 m, pipe DB carries no flow, so comparing inflow from A with outflow to C shows whether B must take the surplus or supply the deficit

Because 70 m is the average of the three reservoir levels

Because the head at the junction must always equal the middle reservoir level

Because B has the smallest pipe so it always supplies

Inflow (0.0536) exceeds outflow to C (0.0155) when H_D = 70 m. What condition is then used to find the true H_D?

Q_AD = Q_DB + Q_DC, with H_D above 70 m

At the balanced junction head H_D = 95.9 m, which set of discharges is correct?

Q_AD = 37.2, Q_DB = 13.6, Q_DC = 23.6 l/s

Q_AD = 37.2, Q_DB = 23.6, Q_DC = 13.6 l/s

Q_AD = 23.6, Q_DB = 13.6, Q_DC = 37.2 l/s

Q_AD = 50.0, Q_DB = 13.6, Q_DC = 36.4 l/s

2068 Bhadra · 10 marks

Three reservoirs 120 m, 70 m and 50 m

Short answer

Resistances r=8fL/(π²gD⁵): AD=17409, DB=139276, DC=82627. Assume B receives; trial H_D balances Q_AD=Q_DB+Q_DC. Converges at H_D=95.9 m: Q_AD=37.2 L/s (A→D), Q_DB=13.6 L/s (D→B), Q_DC=23.6 L/s (D→C).

Reservoir A, water surface elevation 120 m is connected to reservoir B and C having surface elevation 70 m and 50 m respectively. A pipe line 150 mm diameter and 400 m long connects reservoir A to Junction D. Reservoir B and C are connected to Junction D by 75 mm diameter 100 m long and 100 mm diameter 250 m long pipeline respectively. Assuming friction factor f = 0.04 for all pipes, estimate the rate of flow for each pipe, neglecting minor head losses.

Data. $Z_A = 120$ m, $Z_B = 70$ m, $Z_C = 50$ m. Pipes: AD 400 m, 150 mm; DB 100 m, 75 mm; DC 250 m, 100 mm; $f = 0.04$ for all. Friction only: $h_f = rQ^2$ , $r = \dfrac{8fL}{\pi^2gD^5}$ .

Pipe	L (m)	D (m)	r (s²/m⁵)
AD	400	0.150	17409
DB	100	0.075	139276
DC	250	0.100	82627

Method

A is the highest reservoir and supplies water; C is the lowest and receives. Test whether B receives or supplies by assuming $H_D = 70$ m (B balanced): $Q_{AD} = \sqrt{50/17409} = 0.0536$ m³/s in, $Q_{DC} = \sqrt{20/82627} = 0.0155$ m³/s out. Inflow exceeds outflow, so the surplus goes to B: $H_D > 70$ m and $Q_{AD} = Q_{DB} + Q_{DC}$ .

$H_D$ (m)	$Q_{AD}$	$Q_{DB}$	$Q_{DC}$	$Q_{AD} - Q_{DB} - Q_{DC}$
70.5	0.05332	0.00189	0.01575	+0.03568
85	0.04484	0.01038	0.02058	+0.01388
90	0.04151	0.01198	0.02200	+0.00753
95.9	0.03721	0.01364	0.02357	+0.00000

The balance is at $H_D = 95.90$ m.

Discharges

Q_{AD} = \sqrt{\frac{120 - 95.90}{17409}} = 0.03721\ \text{m}^3/\text{s} = 37.21\ \text{l/s}

Q_{DB} = \sqrt{\frac{95.90 - 70}{139276}} = 0.01364\ \text{m}^3/\text{s} = 13.64\ \text{l/s}

Q_{DC} = \sqrt{\frac{95.90 - 50}{82627}} = 0.02357\ \text{m}^3/\text{s} = 23.57\ \text{l/s}

Check: $13.64 + 23.57 = 37.21$ l/s $= Q_{AD}$ .

Answer: $Q_{AD} = 37.2$ l/s (A to D), $Q_{DB} = 13.6$ l/s (D to B), $Q_{DC} = 23.6$ l/s (D to C); $H_D = 95.9$ m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Three reservoirs 120 m, 70 m and 50 m

Method

Discharges

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