Pressure in a closed system using an inclined manometer

[Dylan.Wallett]

Homework Statement

In the figure attached the pressure at A and B are the same 100 kPa. If water is introduced at A to increase p_A to 130 kPa find the new positions of the mercury menisci. The connecting tube has a diameter of 10 mm. Assume no change in Liquid density.

Homework Equations

P _A - P _B = ρ_merc g (Δ L) sin σ

The Attempt at a Solution

30 000 = (13600)(9.81) (Δ L) (0.26)
ΔL = 0.868
h = 0.225 m

Intuitively this answer should be wrong because if there is an increase in pressure at A it should transfer through out the system to create an equilibrium. There would be no change in the mercury menisci if the air did not compress.

I think I need to work out a change in air density to get the correct answer. I am also not sure why they mentioned the diameter of the connecting tube.

Would you mind helping me get on the correct track.

Thank you

 

[KataKoniK]

for your post! From what I understand, the system described in the problem is a U-tube manometer, with a liquid (mercury) on one side and air on the other. When water is introduced at A, it will cause the pressure at A to increase, but the pressure at B will remain the same. This will result in a pressure difference between A and B, causing the mercury levels to change.

To solve this problem, you can use the equation you mentioned: PA - PB = ρmerc g (ΔL) sin σ. However, in this case, the density of the liquid (ρmerc) will remain constant, as no change in liquid density is specified in the problem. The important thing to note is that the pressure difference (PA - PB) is now equal to 30 kPa, since PA has increased from 100 kPa to 130 kPa.

Now, let's consider the air on the other side of the U-tube. The air is compressed due to the increase in pressure at A, and this will result in a change in air density. To determine the new positions of the mercury menisci, we will need to consider the new pressure and density of the air at both points A and B.

To find the new air pressure at A, we can use the ideal gas law: PA = nRT/V, where n is the number of moles of air, R is the ideal gas constant, T is the temperature, and V is the volume. Since the volume of air at A is the same as before (the connecting tube diameter is given, so we can calculate the volume), and the temperature remains constant, the only variable that changes is the number of moles of air. This number will decrease due to the compression, but we don't know the exact amount. However, we do know that the pressure at A has increased by 30 kPa, so we can set up an equation:

130 kPa = (n - Δn)RT/V

where Δn is the change in the number of moles of air. We can rearrange this equation to solve for Δn:

Δn = n - (130 kPa V)/(RT)

Now, we can use this value of Δn to find the new air pressure at B. Since the pressure at B is the same as before (100 kPa), we can set up a similar equation:

100 k