---
title: "Torque to hold two-armed sprinkler"
subject: "Fluid Mechanics"
chapter: 7
key: "semester_3/fluid_mechanics:two-armed-sprinkler-torque"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/two-armed-sprinkler-torque
site: BCT Engineering
---

# Torque to hold two-armed sprinkler

Fluid Mechanics · Chapter 7: Momentum principle and flow analysis

- Asked in: 2076 Chaitra, 6 marks
- Video: [Determine Torque required to hold the arm stationary, when water enter two armed sprinkler.|||||||||](https://www.youtube.com/watch?v=xD0tYNO5UiI) — Civil Engg Network
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/two-armed-sprinkler-torque
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-7

## Question

Water enters two armed sprinkler vertically at rate of 10 litre/sec, and leaves the nozzle horizontally. The diameter of both the nozzle is 12 mm. Calculate the torque required to hold the arm stationary. [Figure: sprinkler arms with $r_1$ = 60 cm and $r_2$ = 20 cm, $\theta$ = 60° at jet 1 and 60° at jet 2, shaft torque $T_{shaft}$ and angular speed $\omega$]

## Short answer

Torque from angular momentum: T = ρ q V cosθ (r1 + r2). With q=0.005 m³/s, V=44.21 m/s, cos60°=0.5, r1+r2=0.8 m, T = 88.4 N·m required to hold stationary.

## Answer

**Given:** total flow $Q = 10$ L/s $= 0.01$ m³/s entering vertically at the axis; two equal nozzles of $d = 12$ mm; $r_1 = 0.6$ m, $r_2 = 0.2$ m; the jets leave horizontally at $\theta = 60^\circ$ to the tangential direction (assumed from the figure). The arm is held stationary ($\omega = 0$), and both jets give torque in the same sense.

#### Jet velocity

The nozzles are equal, so each nozzle carries $q = Q/2 = 0.005$ m³/s:

$$
\begin{aligned}
a &= \frac{\pi}{4}(0.012)^2 = 1.131\times 10^{-4}\ \text{m}^2\\
V &= \frac{q}{a} = \frac{0.005}{1.131\times 10^{-4}} = 44.21\ \text{m/s}
\end{aligned}
$$

#### Torque

Water enters along the axis with zero angular momentum. The angular momentum leaving per second is $\rho q\,(r_1V\cos\theta + r_2V\cos\theta)$. The torque needed to hold the arm equals this rate of change of angular momentum:

$$
\begin{aligned}
T &= \rho\,q\,V\cos\theta\,(r_1 + r_2)\\
&= 1000\times 0.005\times 44.21\times\cos60^\circ\times(0.6 + 0.2)\\
&= 1000\times 0.005\times 44.21\times 0.5\times 0.8 = 88.4\ \text{N·m}
\end{aligned}
$$

Individually, jet 1 gives 66.3 N·m and jet 2 gives 22.1 N·m.

**Answer: torque required to hold the sprinkler stationary $\approx 88.4$ N·m.** (It is equal and opposite to the torque the jets exert on the arm. If $\theta$ were measured from the arm instead of from the tangential direction, replace $\cos60^\circ$ by $\sin60^\circ$.)

## Check yourself

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

1. How is the velocity of each jet found?
   A. V = 2Q/a
   B. V = sqrt(2gh)
   C. V = Q/a
   D. V = (Q/2)/a with a from d = 12 mm
   Answer: D. The two nozzles are equal so each carries 5 L/s; V = 44.2 m/s. (Key point: Flow per nozzle)

2. How is the holding torque obtained?
   A. Rate of change of angular momentum of the water
   B. Weight of the arm times radius
   C. Pressure times area
   D. Jet force times arm width
   Answer: A. Water enters with zero angular momentum and leaves with moment arms r1 and r2. (Key point: Torque principle)

3. What is the torque?
   A. About 88.4 N.m
   B. About 176.8 N.m
   C. About 66.3 N.m
   D. About 22.1 N.m
   Answer: A. T = 1000 x 0.005 x 44.21 x 0.5 x 0.8 = 88.4 N.m. (Key point: Result)

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