Question 7 of 13: Jet Force on a Plate vs Static Nozzle Pressure Force

Question 7: Jet Force on a Plate vs Static Nozzle Pressure Force (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) The force on the plate from the free jet; (b) the static pressure force when the plate blocks the nozzle; are they equal?

Approach. With friction and turbulence neglected, the exit velocity follows from Torricelli's law using the available head. The free-jet force on a flat plate perpendicular to the flow equals the rate of momentum destruction, $F=\rho QV$; the blocked-nozzle force is simply the static pressure (from the same head) acting over the nozzle area.

1. Jet velocity and flow rate. $$V=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}\qquad A=\frac{\pi}{4}(0.030)^2=7.069\times10^{-4}\ \text{m}^2$$ $$Q=AV=(7.069\times10^{-4})(6.264)=4.43\times10^{-3}\ \text{m}^3/\text{s}$$
2. Part (a) — force of the free jet on the plate. A flat plate perpendicular to the jet destroys all of the jet's momentum in the flow direction (the deflected flow leaves parallel to the plate, carrying no residual momentum in the original direction): $$F_{jet}=\rho QV=(1000)(4.43\times10^{-3})(6.264)=\boxed{27.7\ \text{N}}$$
3. Part (b) — static force with the nozzle blocked. With no flow, the pressure everywhere in the nozzle equals the full static head pressure, $$p=\rho gh=(1000)(9.81)(2)=19{,}620\ \text{Pa}\qquad F_{static}=pA=(19{,}620)(7.069\times10^{-4})=\boxed{13.9\ \text{N}}$$
4. Comparison. $$\frac{F_{jet}}{F_{static}}=\frac{\rho AV^2}{\rho ghA}=\frac{V^2}{gh}=\frac{2gh}{gh}=2$$ $$\boxed{\text{The forces are NOT the same: } F_{jet}=2F_{static}}$$