Find Change in height in a Manometer

[nick.falconer]

Homework Statement

Question 5

Figure 5.1
Figure 5.1 shows a nozzle with a jet of liquid 30 mm in diameter with a density of 850 kg.m-3
hitting a plate that has hole of 10 mm diameter in it. The jet of water requires a force of 50 N to
hold it in place.
(a) Assuming frictionless flow and an even distribution of flow up and down the plate for
deflected water, calculate the velocity of the water hitting the plate.
[10 marks]
(b) The manometer shown is filled with mercury with a density of 13,570 kg.m-3. calculate
the value of h that will be measured in this situation. (If you have not answered part (a)
then use a velocity of 10 m.s-1.)
[10 marks]

Homework Equations

Bernoulli's Equation
P₁ + pgh₁ + 1/2u₁² = P₂ + pgh₂ + 1/2u₂²

The Attempt at a Solution

I have worked out that the velocity in part a is 8.601 m/s

But I'm having trouble with part b.

Using Bernoulli's Equation I can remove the pgh part as I can set the datum in the middle causing these to go to zero. Also knowing that the force on the plate is 50N then using P=F/A I can work out that P₂ = 70.7kPa. That is as far as I can get. I believe that the P₁ and P₂ can't be equal otherwise the height difference in the manometer will be zero. So can I assume that the velocity is constant in the two parts? Because if so I believe I can work it out from there, otherwise I'm stumped.

Thanks
Nick

 

[Lavabug]

Use the equation of hydrostatic/manometric pressure:

dP = pgdy
integrating
P = P0 + pgh

You already know the pressure in both points in the tube (one on the left necessarily has the higher pressure), just solve for h.

 

[nick.falconer]

How do I get the pressure on the left?

Thanks

 

[Lavabug]

You've got the velocity on both sides right? You know the gravitational potential energy remains constant, so cross those terms out of bernoulli's equation. and solve for P1.

If you don't have the velocity on the left already, use the discharge equation ("volume goes in = volume goes out"): v1S1 = v2S2. Where S1 and S2 are the cross sections of the tube in point 1 and point 2, v is the velocity in respective points. Your cross sections are piR^2, and you've got the diameter to both tubes, so that's how you'd find v1 in case you haven't calculated it yet.

 

[paranormal]

Hello Nick, it seems like you have made some good progress on part a of the question. For part b, you are correct in using Bernoulli's equation to solve for the height difference in the manometer. However, you should also consider the pressure at point 1 (P1) in the manometer. Since the density of mercury is much higher than water, the pressure at point 1 will be much greater than the pressure at point 2 (P2). This pressure difference will cause the mercury to rise in the manometer, giving a non-zero height difference. Therefore, you can use the equation P1 + pgh1 + 1/2u12 = P2 + pgh2 + 1/2u22 to solve for the height difference, h.

Additionally, it is important to note that the velocity is not constant in the two parts. The velocity at point 1 (u1) will be higher than the velocity at point 2 (u2) due to the decrease in cross-sectional area of the nozzle. This change in velocity will also affect the pressure at each point, leading to a non-zero height difference in the manometer.

I hope this helps you in solving part b of the question. Good luck!