---
title: "Force on a reducing pipe bend"
subject: "Fluid Mechanics"
chapter: 4
key: "semester_4/fluid_mechanics_mechanical:reducing-bend-force-numerical"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/reducing-bend-force-numerical
site: BCT Engineering
---

# Force on a reducing pipe bend

Fluid Mechanics · Chapter 4: Basic Equations for Fluid Flow

- Asked in: year not given, 8 marks
- Video: [Force on a Pipe Bend - Fluid Momentum Example Problem](https://www.youtube.com/watch?v=zLPZtIBVJXY) — Brian Bernard
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/reducing-bend-force-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-4

## Question

A horizontal reducing pipe bend turns water through 45°. The diameter at the inlet is 300 mm and at the outlet is 150 mm. The discharge is 0.15 m$^3$/s and the gauge pressure at the inlet is 200 kPa. Neglecting energy losses, find the magnitude and direction of the force exerted by the water on the bend.

## Answer

#### Given data
$d_1 = 0.3\ \text{m}$, $d_2 = 0.15\ \text{m}$, $Q = 0.15\ \text{m}^3/\text{s}$, $p_1 = 200\ \text{kPa}$ (gauge), $\theta = 45^\circ$, $\rho = 1000\ \text{kg/m}^3$. Take $x$ along the inlet flow and $y$ perpendicular to it, towards the side to which the flow turns.

#### Areas and velocities
$$
\begin{aligned}
A_1 &= \frac{\pi}{4}(0.3)^2 = 0.07069\ \text{m}^2, & A_2 &= \frac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2 \\
V_1 &= \frac{0.15}{0.07069} = 2.122\ \text{m/s}, & V_2 &= \frac{0.15}{0.01767} = 8.488\ \text{m/s}
\end{aligned}
$$

#### Outlet pressure (Bernoulli, horizontal, no loss)
$$
p_2 = p_1 + \frac{\rho}{2}(V_1^2 - V_2^2) = 200\,000 + 500(4.503 - 72.05) = 166\,226\ \text{Pa}
$$

#### Pressure forces
$$
p_1A_1 = 200\,000 \times 0.07069 = 14\,137\ \text{N}, \qquad p_2A_2 = 166\,226 \times 0.01767 = 2937\ \text{N}
$$

#### Momentum equation (forces on the fluid)
Let $R_x$ and $R_y$ be the forces exerted by the bend on the fluid. $\rho Q = 150\ \text{kg/s}$.

In the $x$-direction:

$$
p_1A_1 - p_2A_2\cos45^\circ + R_x = \rho Q (V_2\cos45^\circ - V_1)
$$

$$
\begin{aligned}
14\,137 - 2077 + R_x &= 150(6.002 - 2.122) = 582 \\
R_x &= -11\,478\ \text{N}
\end{aligned}
$$

In the $y$-direction:

$$
\begin{aligned}
-p_2A_2\sin45^\circ + R_y &= \rho Q V_2\sin45^\circ \\
R_y &= 2077 + 150 \times 6.002 = 2977\ \text{N}
\end{aligned}
$$

#### Force of water on the bend
By Newton's third law, the force on the bend is equal and opposite:

$$
F_x = 11\,478\ \text{N}, \qquad F_y = -2977\ \text{N}
$$

$$
\begin{aligned}
F &= \sqrt{11\,478^2 + 2977^2} = 11\,858\ \text{N} \\
\alpha &= \tan^{-1}\frac{2977}{11\,478} = 14.5^\circ
\end{aligned}
$$

$F_x$ acts in the direction of the incoming flow, and $F_y$ acts away from the side to which the flow turns.

**Answer: $F = 11.86\ \text{kN}$ at $14.5^\circ$ to the inlet flow direction, tilted away from the side to which the water turns ($F_x = 11.48\ \text{kN}$, $F_y = 2.98\ \text{kN}$)**

## Check yourself

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

1. With Q = 0.15 m^3/s and d1 = 0.3 m, d2 = 0.15 m, the velocities V1 and V2 are about:
   A. 2.12 m/s and 8.49 m/s
   B. 8.49 m/s and 2.12 m/s
   C. 2.12 m/s and 4.24 m/s
   D. 0.5 m/s and 2.0 m/s
   Answer: A. V = Q/A: A1 = 0.07069 m^2 and A2 = 0.01767 m^2, so V1 = 2.122 and V2 = 8.488 m/s. (Key point: Velocities from continuity)

2. How is the outlet pressure p2 found?
   A. Bernoulli's equation for the horizontal bend, p2 = p1 + (rho/2)(V1^2 - V2^2)
   B. p2 = p1
   C. From the momentum equation
   D. p2 = p1 + (rho/2)(V2^2 - V1^2)
   Answer: A. The velocity rises, so pressure drops to about 166 kPa. (Key point: Outlet pressure from Bernoulli)

3. The x-momentum equation for the fluid in the bend is:
   A. p1 A1 - p2 A2 cos45 + Rx = rho Q (V2 cos45 - V1)
   B. p1 A1 + p2 A2 cos45 = rho Q V2
   C. p1 A1 - p2 A2 = rho Q (V2 - V1)
   D. Rx = rho Q (V1 - V2 cos45) only
   Answer: A. Pressure forces on the ends plus the bend's reaction equal the change in x-momentum flux. (Key point: Momentum equation in x)

4. How is the force of water on the bend obtained from Rx and Ry?
   A. Same as the force on the fluid
   B. Equal and opposite (Newton's third law), giving 11.86 kN
   C. Half of the reaction
   D. Add rho Q V1 to it
   Answer: B. R is the force of the bend on the water, so the force on the bend is -R; the resultant is sqrt(11478^2 + 2977^2) = 11.86 kN. (Key point: Force on the bend)

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