---
title: "Flow between reservoirs (15 cm and 25 cm pipe)"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:reservoirs-pipeline-15-25cm"
source: https://www.bctengineer.com/q/semester_6/hydro_power/reservoirs-pipeline-15-25cm
site: BCT Engineering
---

# Flow between reservoirs (15 cm and 25 cm pipe)

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2080 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/reservoirs-pipeline-15-25cm
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

Two reservoirs are connected by a pipeline which is 15 cm diameter for the first 6 m and 25 cm diameter in remaining 15 m length. The entrance and exit are sharp and change of section is sudden. If water surface in upper reservoir is 6 m above that of lower reservoir, calculate rate of flow. Take friction factor 0.04 for all pipes.

## Short answer

Apply Bernoulli equating the 6 m head to entry, friction, sudden enlargement and exit losses. Solving with continuity V₂=0.36V₁ and total loss coefficient 2.950 yields V₁=6.32 m/s, discharge Q=0.112 m³/s.

## Answer

Apply Bernoulli's equation between the two reservoir surfaces. The difference in levels equals the sum of all losses.

**Given:** pipe 1: $d_1 = 0.15$ m, $L_1 = 6$ m; pipe 2: $d_2 = 0.25$ m, $L_2 = 15$ m; $H = 6$ m; Darcy $f = 0.04$ ($h_f = fLV^2/2gd$).

```text
 ===== upper reservoir
   |  |                      
   |  | 6 m                   
   |  |______ 15 cm, 6 m ___
   |         |____ 25 cm, 15 m ______  
   ------------------------------- ===== lower
```

#### Continuity
$$
V_2 = V_1\left(\frac{d_1}{d_2}\right)^2 = V_1\left(\frac{0.15}{0.25}\right)^2 = 0.36\,V_1
$$

#### Losses in terms of $V_1^2/2g$
| Loss | Formula | Coefficient of $V_1^2/2g$ |
|---|---|---|
| Entrance (sharp) | $0.5V_1^2/2g$ | 0.5000 |
| Friction, pipe 1 | $f\frac{L_1}{d_1}\frac{V_1^2}{2g} = 0.04\times\frac{6}{0.15}$ | 1.6000 |
| Sudden enlargement | $\frac{(V_1 - V_2)^2}{2g} = (1 - 0.36)^2$ | 0.4096 |
| Friction, pipe 2 | $0.04\times\frac{15}{0.25}\times0.36^2$ | 0.3110 |
| Exit | $\frac{V_2^2}{2g} = 0.36^2$ | 0.1296 |
| **Total** | | **2.9502** |

#### Velocity and discharge
$$
\begin{aligned}
6 &= 2.9502\,\frac{V_1^2}{2g} \\
\frac{V_1^2}{2g} &= 2.0337\ \text{m} \\
V_1 &= \sqrt{2 \times 9.81 \times 2.0337} = 6.317\ \text{m/s} \\
V_2 &= 0.36 \times 6.317 = 2.274\ \text{m/s} \\
Q &= \frac{\pi}{4}(0.15)^2 \times 6.317 = 0.1116\ \text{m}^3/\text{s}
\end{aligned}
$$

Check: losses $= 1.017 + 3.254 + 0.833 + 0.633 + 0.264 = 6.00$ m.

**Answer:** rate of flow $Q \approx 0.112$ m³/s (about 112 L/s). (If 0.04 were the coefficient in $4fLV^2/2gd$, the friction terms would be four times larger and $Q$ smaller; the Darcy form is used here.)

## Check yourself

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

1. How does the 6 m level difference relate to the losses?
   A. It equals only the friction loss
   B. It equals the sum of all losses between the reservoir surfaces
   C. It equals the velocity head
   D. It equals the entry loss
   Answer: B. Bernoulli between the two surfaces (zero velocity, equal pressure) gives H = total losses. (Key point: Basic principle)

2. How is V2 related to V1?
   A. V2 = 0.36 V1
   B. V2 = 1.67 V1
   C. V2 = 0.6 V1
   D. V2 = V1
   Answer: A. V2 = V1 (d1/d2)^2 = (0.15/0.25)^2 V1. (Key point: Continuity)

3. Which list of losses is included?
   A. Friction only
   B. Enlargement only
   C. Entrance and exit only
   D. Entrance, friction in both pipes, sudden enlargement, exit
   Answer: D. Total coefficient is 2.9502 times V1^2/2g. (Key point: Losses included)

4. What is the rate of flow?
   A. About 0.0353 m^3/s
   B. About 0.112 m^3/s
   C. About 0.31 m^3/s
   D. About 1.12 m^3/s
   Answer: B. V1^2/2g = 6/2.9502 = 2.0337, V1 = 6.317 m/s, Q = 0.01767 x 6.317 = 0.1116 m^3/s. (Key point: Discharge)

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