---
title: "Hydrodynamic force on bend-nozzle assembly"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:vertical-bend-nozzle-hydrodynamic-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/vertical-bend-nozzle-hydrodynamic-force
site: BCT Engineering
---

# Hydrodynamic force on bend-nozzle assembly

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2075 Asoj, 8 marks
- Video: [Forces on Pipe Bend | Fluid Mechanics](https://www.youtube.com/watch?v=AXSz1yxLJM8) — Smart Civil Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/vertical-bend-nozzle-hydrodynamic-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

Water flows into atmosphere through a vertical bend nozzle assembly as shown in figure below. The pipe diameter is 10 cm and nozzle exit diameter is 5 cm. The rate of flow of water is 2400 lpm. The interior volume of the assembly is 18.2 litres. The head loss in the bend is $0.5\frac{v^2}{2g}$ and in the nozzle it is $2\frac{v^2}{2g}$, where V is the velocity of water in the pipe. Compute the hydrodynamic force on the system. [Figure: vertical pipe of 10 cm diameter turning through a bend to a horizontal nozzle with exit diameter 5 cm; 5 m vertical distance marked]

## Short answer

Find velocities from Q=0.04 m³/s: V1=5.09 m/s, V2=20.37 m/s. Inlet pressure from Bernoulli with losses: p1=276 kPa. Momentum analysis gives hydrodynamic force components: opposite to jet 815 N, upward 2193 N; resultant 2.34 kN at 69.6° above horizontal.

## Answer

**Given:** pipe $d_1 = 0.10$ m, nozzle exit $d_2 = 0.05$ m, $Q = 2400$ L/min $= 0.04$ m³/s, volume of assembly $= 18.2$ L $= 0.0182$ m³, head loss in bend $0.5V^2/2g$ and in nozzle $2V^2/2g$ ($V$ = velocity in the pipe).

**Assumed from the figure:** water flows upward in the vertical pipe, passes a bend and leaves through a horizontal nozzle into the atmosphere, 5 m above the inlet section.

```text
   exit ====>  (nozzle, d = 5 cm)
   |
   |  5 m
   |
   ^  inlet 1 (pipe d = 10 cm, flow up)
```

#### Velocities

$$
\begin{aligned}
A_1 &= 7.854\times 10^{-3}\ \text{m}^2,\quad A_2 = 1.9635\times 10^{-3}\ \text{m}^2\\
V_1 &= 5.093\ \text{m/s},\quad V_2 = 20.372\ \text{m/s}
\end{aligned}
$$

#### Inlet pressure (Bernoulli with losses)

$$
\frac{p_1}{\gamma} + \frac{V_1^2}{2g} = 0 + \frac{V_2^2}{2g} + 5 + (0.5 + 2)\frac{V_1^2}{2g}
$$

$$
\begin{aligned}
\frac{p_1}{\gamma} &= 21.15 + 5 + 2.5(1.322) - 1.322 = 28.13\ \text{m}\\
p_1 &= 276\,009\ \text{Pa} \approx 276\ \text{kPa (gauge)}
\end{aligned}
$$

(using $V_2^2/2g = 21.15$ m, $V_1^2/2g = 1.322$ m).

#### Momentum equation

- $p_1A_1 = 276\,009\times 7.854\times 10^{-3} = 2168\ \text{N}$ (up, on the inlet face)
- $W = 9810\times 0.0182 = 178.5\ \text{N}$ (down)
- $\rho Q = 40$ kg/s. Inlet velocity $= (0, +V_1)$; exit velocity $= (+V_2, 0)$.

Forces on the fluid ($R$ = force of the assembly on the fluid):

$$
\begin{aligned}
x:\ \ R_x &= \rho Q\,V_2 = 40\times 20.372 = 815\ \text{N}\\
y:\ \ p_1A_1 - W + R_y &= \rho Q\,(0 - V_1)\\
2168 - 178.5 + R_y &= -203.7\\
R_y &= -2193\ \text{N}
\end{aligned}
$$

The hydrodynamic force exerted **by the water on the assembly** is the opposite of $R$:

$$
F_x = -815\ \text{N},\qquad F_y = +2193\ \text{N}
$$

$$
F = \sqrt{815^2 + 2193^2} = 2339\ \text{N},\qquad \theta = 110.4^\circ\ \text{from the +x direction (the direction of the exit jet)}
$$

**Answer: the hydrodynamic force on the system $\approx 2.34$ kN, directed upward and backwards (against the jet), at $69.6^\circ$ above the horizontal, with components 0.815 kN opposite to the jet and 2.19 kN upward.**

## Check yourself

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

1. How is the inlet pressure found?
   A. Hydrostatic only
   B. Bernoulli between inlet and exit with elevation 5 m and losses (0.5 + 2)V1^2/2g
   C. Continuity
   D. Atmospheric
   Answer: B. Both losses are based on pipe velocity V1. (Key point: Inlet pressure)

2. What are the pipe and exit velocities?
   A. 5.09 and 5.09
   B. 5.09 m/s and 20.37 m/s
   C. 10.2 and 40.7
   D. 2.04 and 8.15
   Answer: B. Q = 0.04 m3/s over 7.854e-3 and 1.9635e-3 m^2. (Key point: Velocities)

3. What determines the x-reaction on the fluid?
   A. Weight
   B. Zero
   C. p1 A1
   D. rho Q V2, because the fluid enters with zero x-momentum and leaves horizontally
   Answer: D. Rx = 40 x 20.372 = 815 N. (Key point: Momentum x direction)

4. What is the force of water on the assembly?
   A. About 4.7 kN forward
   B. About 2.2 kN downward
   C. About 2.34 kN, upward and backward
   D. About 0.8 kN forward
   Answer: C. Fx = -815 N (against the jet) and Fy = +2193 N upward. (Key point: Force on assembly)

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