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A small lawn sprinkler is shown. The sprinkler operates al a gage pressure of \(140 \mathrm{kPa}\). The total flow rate of water through the sprinkler is 4 L \(/\) min. Each jet discharges at \(17 \mathrm{m} / \mathrm{s}\) (relative to the sprinkler arm) in a direction inclined \(30^{\circ}\) above the horizontal. The sprinkler rotates about a vertical axis. Friction in the bearing causes a torque of \(0.18 \mathrm{N}\). \(\mathrm{m}\) opposing rotation. Evaluate the torque required to hold the sprinkler stationary.

Short Answer

Expert verified
The torque required to hold the sprinkler stationary is \(7.11\) N.m

Step by step solution

01

Convert Flow Rate to Standard Units

To have consistent units, convert the given flow rate from liters per minute to cubic meters per second. Use the fact that 1 L = \(10^{-3} \) m³ and 1 min = 60 s. Hence, the flow rate is \(4 \div 60 \times 10^{-3}\) m³/s = \(6.67 \times 10^{-5}\) m³/s.
02

Find the Mass Flow Rate

Now we calculate the mass flow rate. The formula for mass flow rate is the product of the fluid density and the flow rate. Assuming the density of water is \(1000 \) kg/m³, the mass flow rate is then \(1000 \times 6.67 \times 10^{-5}\) kg/s = \(0.067 \) kg/s.
03

Calculate the Angular Momentum

The angular momentum formula is mass flow rate times velocity times radius times sin(\(\theta\)), where \(\theta\) is the angle of inclination. From the properties of the sprinkler, we know each jet discharges at \(17\) m/s in a direction inclined \(30^{\circ}\) vertically with regard to the horizontal. Now, compute the angular momentum for each jet: \(0.067 \times 17 \times 30\sin(30^{\circ})\) = \(3.465\) N.m/s.
04

Calculate the total torque

The total torque required to hold the sprinkler stationary equals the sum of the frictional torque and the torque from the jet water. Since there are two jets of water, multiply the calculated angular momentum by 2. Then, add the frictional torque: \(2 \times 3.465 + 0.18\) = \(7.11\) N.m.

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