---
title: "Network: unknown discharges and head losses"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:network-find-unknown-discharges-head-losses"
source: https://www.bctengineer.com/q/semester_4/hydraulics/network-find-unknown-discharges-head-losses
site: BCT Engineering
---

# Network: unknown discharges and head losses

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2077 Chaitra, 4+4 marks; 2071 Bhadra, 4 marks (asked 2 times)
- Video: [Minor Losses Example Problems - with Major Losses and Moody Diagram](https://www.youtube.com/watch?v=7meVgDi257Y) — Brian Bernard
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/network-find-unknown-discharges-head-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

Figure below shows a network in which Q and $h_f$ refer to discharges and pressure drops respectively. Subscripts 1, 2, 3, 4 and 5 designate respective values in pipe length AB, BC, CD, DA and AC. Subscripts A, B, C and D designate discharges entering or leaving the junction points A, B, C and D respectively. By sticking to the values given in the figure find the following discharges $Q_B$, $Q_2$, $Q_4$ and $Q_5$ and the pressure drops $h_{f4}$ and $h_{f5}$ and give these computed values at their respective places on a neat sketch of the network along with flow directions. [Figure: square network ABCD with diagonal AC; $Q_A$ = 20 leaves at A; $Q_1$ = 30, $h_{f1}$ = 60 in AB; $Q_B$ = ? leaves at B; $Q_2$ = ?, $h_{f2}$ = 40 in BC; $Q_C$ = 30 leaves at C; $Q_3$ = 40, $h_{f3}$ = 120 in CD; $Q_D$ = 100 enters at D; $Q_4$ = ?, $h_{f4}$ = ? in DA; $Q_5$ = ?, $h_{f5}$ = ? in AC]

## Short answer

Applying continuity at nodes and equating head losses along alternative paths yields: Q_B = 50, Q_2 = 20 (C to B), Q_4 = 60 (D to A), Q_5 = 10 (A to C); head losses h_{f4} = 100, h_{f5} = 20.

## Answer

**Method.** Apply continuity at every node, then use the fact that the head (pressure) drop between any two nodes is the same along every path joining them. External flows: $Q_A = 20$ out, $Q_C = 30$ out, $Q_D = 100$ in; $Q_B$ is unknown (out).

#### Directions of the known flows
Water enters the network only at D (100 units), so D has the highest head in the network and every pipe connected to D carries water away from it. The flow in CD is therefore from **D to C**, and the flow in DA is from **D to A**. The flow $Q_1 = 30$ in AB is taken from **A to B**, as shown in the figure (A is fed from D).

#### Discharges
- Node D (100 in): $Q_3 + Q_4 = 100 \Rightarrow Q_4 = 100 - 40 = \mathbf{60}$ (D to A)
- Node A: inflow $Q_4 = 60$; outflow $= Q_A + Q_1 + Q_5 = 20 + 30 + Q_5 \Rightarrow Q_5 = 60 - 50 = \mathbf{10}$ (A to C)
- Node C: inflow $Q_3 + Q_5 + Q_2 = 40 + 10 + Q_2$; outflow $Q_C = 30 \Rightarrow Q_2 = 30 - 50 = -20$, i.e. $Q_2 = \mathbf{20}$ flowing **from C to B**
- Node B: inflow $Q_1 + Q_2 = 30 + 20 = 50$, so the outflow is $Q_B = \mathbf{50}$

Check for the whole network: inflow 100 = outflows $20 + 50 + 30 = 100$.

#### Head losses (pressure drops)
Let $H$ denote the head at a node.
- Path A to B to C: $H_A - H_B = h_{f1} = 60$. The flow in BC is from C to B, so $H_C - H_B = h_{f2} = 40$. Hence
$$
H_A - H_C = 60 - 40 = 20 \;\Rightarrow\; h_{f5} = \mathbf{20}
$$
- Path D to C: $H_D - H_C = h_{f3} = 120$. Path D to A to C: $H_D - H_C = h_{f4} + h_{f5}$, so
$$
h_{f4} = 120 - 20 = \mathbf{100}
$$
(Check with the square law $h_f \propto Q^2$: $h_{f1}/Q_1^2 = 0.0667$, $h_{f2}/Q_2^2 = 0.1$, $h_{f3}/Q_3^2 = 0.075$, $h_{f4}/Q_4^2 = 0.0278$, $h_{f5}/Q_5^2 = 0.2$; all the resistances are positive, so the values are consistent.)

```text
            Q1=30 (hf1=60)
   Qa=20 <- A ----------> B -> QB=50
            |\             ^
 Q4=60 (100)| \ Q5=10 (20) |  Q2=20 (hf2=40)
            |  \           |
   Qd=100 ->D ----------> C -> QC=30
            Q3=40 (hf3=120)
```
Arrows: D to A (Q4), A to B (Q1), A to C (Q5, diagonal), C to B (Q2), D to C (Q3).

**Answer:** $Q_B = 50$, $Q_2 = 20$ (C to B), $Q_4 = 60$ (D to A), $Q_5 = 10$ (A to C); $h_{f4} = 100$, $h_{f5} = 20$.

## Check yourself

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

1. How is Q4 found?
   A. Node D: Q3 + Q4 = 100, so Q4 = 60
   B. Node A: Q4 = Q_A
   C. Node C
   D. From head loss
   Answer: A. 100 enters at D and 40 leaves via CD, so 60 goes to A. (Key point: Continuity at nodes)

2. At node A, with inflow 60 and outflows 20 and 30, what is Q5?
   A. 10 from A to C
   B. 30 from A to C
   C. 50
   D. 10 from C to A
   Answer: A. 60 - 20 - 30 = 10 leaving along AC. (Key point: Q5 at node A)

3. How is h_f5 found?
   A. Heads along path A-B-C: H_A - H_C = 60 - 40 = 20
   B. h_f3 - h_f1
   C. Sum of all losses
   D. Q5 squared
   Answer: A. Head difference between two nodes is the same along any path; flow in BC is from C to B, so 60 - 40. (Key point: Head loss around path)

4. Given h_f3 = 120 and h_f5 = 20, what is h_f4?
   A. 140
   B. 100
   C. 120
   D. 60
   Answer: B. Path DA + AC equals DC: h_f4 + 20 = 120. (Key point: h_f4)

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