---
title: "Valve openings for equal discharge to B and C"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:three-reservoir-valve-openings-equal-discharge"
source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-valve-openings-equal-discharge
site: BCT Engineering
---

# Valve openings for equal discharge to B and C

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2081 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-valve-openings-equal-discharge
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Determine the percentage opening $n_1$ and $n_2$ of valve $V_1$ and $V_2$ of the three reservoir system as shown in figure below to supply equal amount of discharge in reservoirs B and C from A. [Figure: reservoir A at 110 m; reservoir B at 95 m; reservoir C at 90 m; junction D at 100 m with $P_D$ = 39.24 kN/m²; pipe AD to the junction, pipe BD with valve $V_1$ and pipe CD with valve $V_2$]

| Pipe | L (m) | D (m) | f |
|---|---|---|---|
| AD | 1000 | 0.25 | 0.02 |
| BD | 800 | 0.20 | 0.02 |
| CD | 1200 | 0.20 | 0.02 |

| Valve | Percentage opening | Resistance (r) |
|---|---|---|
| $V_1$ (BD) | $n_1$ | $(6000/n_1)^2$ |
| $V_2$ (CD) | $n_2$ | $(8000/n_2)^2$ |

## Short answer

From p_D = 39.24 kN/m², H_D = 104 m; pipe resistances give Q_AD = 59.55 L/s and Q_B = Q_C = 29.77 L/s. Available head balances valve resistances: n₁ ≈ 77.3%, n₂ ≈ 81.7%.

## Answer

**Data.** A = 110 m, B = 95 m, C = 90 m; junction D at elevation 100 m with $p_D = 39.24$ kN/m². Friction only, $h_f = rQ^2$, $r = \dfrac{8fL}{\pi^2gD^5}$ for the pipes and the valve resistances given.

#### Piezometric head at D
$$
H_D = z_D + \frac{p_D}{\gamma} = 100 + \frac{39.24}{9.81} = 100 + 4.00 = 104.00\ \text{m}
$$
Since $H_D = 104$ m is below A (110 m) and above B (95 m) and C (90 m), pipe AD supplies D, and D supplies both B and C.

#### Pipe resistances (s²/m⁵)
| Pipe | L (m) | D (m) | r |
|---|---|---|---|
| AD | 1000 | 0.25 | 1692.2 |
| BD | 800 | 0.20 | 4131.3 |
| CD | 1200 | 0.20 | 6197.0 |

#### Discharges
$$
Q_{AD} = \sqrt{\frac{110 - 104}{1692.2}} = 0.05955\ \text{m}^3/\text{s} = 59.55\ \text{l/s}
$$
Equal discharge to B and C: $Q_B = Q_C = \dfrac{Q_{AD}}{2} = 0.02977$ m³/s ($29.77$ l/s).

#### Valve $V_1$ in pipe DB
Head available from D to B: $104 - 95 = 9$ m $= (r_{BD} + r_{V1})Q_B^2$:
$$
r_{V1} = \frac{9}{0.02977^2} - 4131.3 = 6021.8 = \left(\frac{6000}{n_1}\right)^2 \;\Rightarrow\; n_1 = \frac{6000}{\sqrt{6021.8}} = 77.3\ \%
$$

#### Valve $V_2$ in pipe DC
Head available from D to C: $104 - 90 = 14$ m $= (r_{CD} + r_{V2})Q_C^2$:
$$
r_{V2} = \frac{14}{0.02977^2} - 6197.0 = 9596.8 = \left(\frac{8000}{n_2}\right)^2 \;\Rightarrow\; n_2 = \frac{8000}{\sqrt{9596.8}} = 81.7\ \%
$$

**Answer:** $n_1 \approx 77.3\ \%$ and $n_2 \approx 81.7\ \%$ (each with $Q = 29.8$ l/s to B and to C).

## Check yourself

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

1. How is H_D obtained?
   A. z_D + p_D/gamma = 100 + 4 = 104 m
   B. z_D = 100 m
   C. p_D/gamma = 4 m
   D. Average of reservoir levels
   Answer: A. Piezometric head equals elevation plus pressure head. (Key point: Piezometric head at D)

2. How is Q_AD determined?
   A. Q = sqrt((110 - 104)/r_AD)
   B. Q = sqrt((110 - 95)/r_AD)
   C. Q = sqrt(104/r_AD)
   D. Q from valve resistance
   Answer: A. Pipe AD has head loss H_A - H_D = 6 m with r_AD. (Key point: Flow in AD)

3. What does equal discharge to B and C imply?
   A. Each gets Q_AD/2
   B. Valves must be equal
   C. Q_B = 2 Q_C
   D. Valves fully open
   Answer: A. Continuity at D: Q_AD = Q_B + Q_C with Q_B = Q_C. (Key point: Equal discharge)

4. How is r_V1 obtained?
   A. From (D to B head)/Q_B^2 minus pipe resistance r_BD, then equate to (6000/n1)^2
   B. From the whole head 110 - 95
   C. From pipe resistance only
   D. Equal to r_V2
   Answer: A. The available head 9 m = (r_BD + r_V1) Q_B^2, then n1 follows from the valve law. (Key point: Valve resistance)

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