---
title: "Three reservoirs with 250 mm concrete pipes"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:three-reservoir-concrete-pipes-flow-rates"
source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-concrete-pipes-flow-rates
site: BCT Engineering
---

# Three reservoirs with 250 mm concrete pipes

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2072 Asoj, 10 marks; 2070 Magh, 10 marks (asked 2 times)
- Video: [Three Reservoir Problem|Type-III|Hydraulics BE Civil Purbanchal University TU KU PoU 4th Sem.](https://www.youtube.com/watch?v=n1N7XYVOvGs) — Civil Engineering Solution Channel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reservoir-concrete-pipes-flow-rates
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

For the three reservoir system of figure below $Z_1$ = 29 m, $L_1$ = 80 m, $Z_2$ = 129 m, $L_2$ = 150 m, $Z_3$ = 69 m and $L_3$ = 110 m. All pipes are 250 mm diameter concrete with roughness height 0.5 mm. Compute the flow rates. Take $\nu = 1.02\times10^{-6}$ m²/s. You are not allowed to use the Moody's chart. [Figure: three reservoirs with water levels $Z_1$, $Z_2$, $Z_3$ connected by pipes of lengths $L_1$, $L_2$, $L_3$ to a common junction a]

## Short answer

Solve Colebrook-White for f and iterate H_J for continuity: Q_2 = 450.4 L/s (in), Q_3 = 47.9 L/s (in), Q_1 = 498.3 L/s (out); piezometric head at junction ≈ 68.5 m.

## Answer

**Data.** $Z_1 = 29$ m, $Z_2 = 129$ m, $Z_3 = 69$ m; $L_1 = 80$ m, $L_2 = 150$ m, $L_3 = 110$ m; $D = 0.25$ m, $k = 0.5$ mm, $\nu = 1.02\times10^{-6}$ m²/s. The friction factor is found from the **Colebrook-White equation** (no Moody chart), neglecting minor losses and velocity heads at the reservoirs.

$$
\frac{k}{D} = \frac{0.5}{250} = 0.0020,\qquad \frac{1}{\sqrt f} = -2\log_{10}\left(\frac{k/D}{3.7} + \frac{2.51}{Re\sqrt f}\right)
$$
A starting value comes from the fully rough formula $\dfrac{1}{\sqrt f} = 2\log_{10}\dfrac{D}{k} + 1.14$, giving $f_0 = 0.0234$.

#### Method
Let $H_J$ be the piezometric head at the junction. For pipe $i$ the head loss is $|Z_i - H_J| = f_i\dfrac{L_i}{D}\dfrac{V_i^2}{2g}$, so $V_i = \sqrt{\dfrac{2gD|Z_i - H_J|}{f_iL_i}}$, with $f_i$ updated from $Re_i = V_iD/\nu$. The correct $H_J$ satisfies continuity at the junction: $\sum Q = 0$.

Reservoir 2 is the highest (129 m) and reservoir 1 the lowest (29 m), so the junction head lies between 29 and 129 m. If $H_J > 69$ m, reservoir 3 receives water; if $H_J < 69$ m, it also feeds the junction. Trying values (each with Colebrook for $f$):

- $H_J = 69$ m: $Q_1 \approx 0.50$ m³/s out, $Q_2 \approx 0.45$ m³/s in, $Q_3 = 0$. Outflow > inflow, so $H_J$ must be lowered below 69 m, and reservoir 3 then supplies the junction.
- Further trials just below 69 m give the balance at $H_J \approx 68.48$ m.

#### Final solution ($H_J = 68.48$ m)
| Pipe | Z (m) | L (m) | $Z - H_J$ (m) | f | V (m/s) | Re | Q (l/s) |
|---|---|---|---|---|---|---|---|
| 1 | 29 | 80 | -39.48 | 0.02350 | 10.151 | 2.488e+06 | -498.3 |
| 2 | 129 | 150 | +60.52 | 0.02350 | 9.176 | 2.249e+06 | +450.4 |
| 3 | 69 | 110 | +0.52 | 0.02417 | 0.975 | 2.390e+05 | +47.9 |

(Positive $Q$ = into the junction; negative = out of the junction.) Check: $450.4 + 47.9 = 498.3$ l/s in, and $498.3$ l/s out.

**Answer:** Reservoir 2 delivers $Q_2 = 450.4$ l/s and reservoir 3 delivers $Q_3 = 47.9$ l/s to the junction; both flow on to reservoir 1 with $Q_1 = 498.3$ l/s. Junction piezometric head $H_J = 68.48$ m (about 68.5 m).

## Check yourself

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

1. How is f found without a Moody chart?
   A. Colebrook-White iteration, starting from the fully rough value
   B. f = 64/Re
   C. Assume f = 0.02
   D. Blasius equation
   Answer: A. Colebrook-White is solved iteratively, with the rough-pipe formula giving a starting f. (Key point: Friction factor without Moody chart)

2. Why does H_J lie between 29 m and 129 m?
   A. It is between the lowest and highest reservoir levels
   B. It equals Z3
   C. It is above all
   D. It is below all
   Answer: A. The junction head must be between the highest and lowest reservoir levels. (Key point: Junction head range)

3. What defines the correct H_J?
   A. Sum of discharges at the junction is zero (continuity)
   B. Equal flows in all pipes
   C. Equal friction factors
   D. H_J = Z3
   Answer: A. The trial H_J is adjusted until inflow equals outflow. (Key point: Condition to find H_J)

4. Since H_J = 68.48 m is just below Z3 = 69 m, what does reservoir 3 do?
   A. Receives water
   B. Supplies a small flow (47.9 l/s) to the junction
   C. Is isolated
   D. Supplies all flow
   Answer: B. Z3 slightly exceeds H_J, so water flows from reservoir 3 to the junction; with reservoir 2 it feeds reservoir 1. (Key point: Role of reservoir 3)

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