A lawn sprinkler with two nozzles 5mm in diameter each at 0.2m and 0.15m radii is connected across a tap capable of discharging 6 litres/min. The nozzles discharge water upwards and outwards from the plane of rotation. What torque will sprinkler exert on the hand if held stationary, and at what angular velocity will it rotate free?

Fluid Mechanics Problem Solution

Problem Statement

Given Data

Diameter of nozzle (d) 5 mm = 0.005 m

Area of nozzle (A) π/4 × (0.005)² = 1.9635×10^-5 m²

Radius to nozzle A (r_A) 0.2 m

Radius to nozzle B (r_B) 0.15 m

Total discharge rate 6 litres/min

Discharge per nozzle (Q_A = Q_B) 3 litres/min = 0.00005 m³/s

Density of water (ρ) 1000 kg/m³

Solution Approach

To determine the torque and angular velocity of the sprinkler, we’ll apply momentum principles. We need to:

Calculate the relative velocity of water at each nozzle
Determine the force exerted by the water jets
Calculate the resulting torque when the sprinkler is held stationary
Apply the principle of zero net moment for free rotation

Calculations

Part 1: Torque Required to Hold the Sprinkler Stationary

Step 1: Calculate the relative velocity of water at each nozzle:

V_A = V_B = Q / A = 0.00005 m³/s / 1.9635×10^-5 m² = 2.54 m/s

Step 2: Calculate the vertical components of the relative velocities:

V_YA = 2.54 × Cos(30°) = 2.2 m/s

V_YB = 2.54 × Cos(45°) = 1.8 m/s

Step 3: Calculate the torque exerted by water on the sprinkler:

T = ρQ_AV_YAr_A + ρQ_BV_YBr_B

T = 1000 × 0.00005 × 2.2 × 0.2 + 1000 × 0.00005 × 1.8 × 0.15

T = 0.022 + 0.0135 = 0.0355 Nm

Torque required to hold the sprinkler stationary = 0.0355 Nm

Part 2: Angular Velocity of Free Rotation

Step 1: For free rotation, the final moment of momentum must be zero:

V_1a = V_YA – r_Aω = 2.2 – 0.2ω

V_2a = V_YB – r_Bω = 1.8 – 0.15ω

Step 2: For zero net moment:

ρQ_AV_1ar_A + ρQ_BV_2ar_B = 0

V_1ar_A = -V_2ar_B

Step 3: Solving the equation:

(2.2 – 0.2ω) × 0.2 = -(1.8 – 0.15ω) × 0.15

0.44 – 0.04ω = -0.27 + 0.0225ω

0.44 + 0.27 = 0.04ω + 0.0225ω

0.71 = 0.0625ω

ω = 11.36 rad/s

Step 4: Converting angular velocity to RPM:

N = ω × 60 / (2π) = 11.36 × 60 / (2π) ≈ 109 RPM

Angular Velocity of Free Rotation = 11.36 rad/s = 109 RPM

Detailed Explanation

Working Principle of Lawn Sprinkler

A lawn sprinkler operates on the principle of momentum conservation and reaction force. When water is forced through the nozzles, it creates a reaction force in the opposite direction, which causes the sprinkler to rotate.

Torque Generation Mechanism

The torque is generated due to the momentum change of water as it exits the nozzles. The water enters the system with zero angular momentum and exits with a significant tangential component. According to Newton’s third law, this creates a reaction torque on the sprinkler arm.

Free Rotation Condition

During free rotation, the sprinkler reaches an equilibrium angular velocity. At this velocity, the absolute tangential momentum of the water leaving the nozzles becomes zero relative to the ground, meaning all the momentum is directed radially. This occurs because the tangential velocity of the nozzles exactly compensates for the tangential component of the water’s velocity relative to the nozzles.

Practical Applications

The principles demonstrated in this problem have several important applications:

Design of efficient irrigation systems
Understanding of reaction turbines in hydropower generation
Development of rotating spray systems for fire protection
Design of propulsion systems that use the reaction principle

Factors Affecting Performance

Several factors can affect the performance of a lawn sprinkler:

Water pressure: Higher pressure increases discharge rate and torque
Nozzle design: Shape and size affect the velocity and direction of water jets
Friction in bearings: Reduces the actual rotation speed below theoretical values
Distribution of nozzles: Affects balance and uniformity of coverage
Wind conditions: External forces can affect spray pattern and rotation

Analysis of Results

The calculated torque of 0.0355 Nm might seem small, but it’s significant enough to cause rotation. The free rotation speed of 109 RPM is typical for a lawn sprinkler and ensures adequate coverage while preventing excessive water drift. This analysis demonstrates the elegant application of conservation principles in everyday devices.

The understanding of fluid mechanics principles demonstrated in this problem is fundamental in the design of numerous hydraulic and irrigation systems, showing how engineering physics principles can be applied to create practical solutions for everyday needs.