Mercury Manometer

nasi112

Junior Member
Joined
Aug 23, 2020
Messages
206
5.jpg

I don't understand why my calculations are wrong. Can anyone help me?
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
7,458
It is interesting that you title this "mercury manometer" but your picture shows water, not mercury!
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
11,285
It is interesting that you title this "mercury manometer" but your picture shows water, not mercury!
It also shows mercury; apparently A is the interface between water and mercury.

View attachment 26044

I don't understand why my calculations are wrong. Can anyone help me?
The image is quite faded, and I don't want to strain my eyes. Can you explain your reasoning in words?
 

nasi112

Junior Member
Joined
Aug 23, 2020
Messages
206
Thanks a lot HallsofIvy and Dr.Peterson for giving me some of your valuable time by looking at this problem which perhaps is not for mathematicians.

Since the pressure at point \(\displaystyle A\) and \(\displaystyle B\) are equal, we can use this formula

\(\displaystyle P_o + \rho_w gh_w = P_o + \rho_m gh_m\)

where \(\displaystyle P_o\) is the atmospheric pressure, in this case it is the Air Pressure. It is not a big deal since it will be cancelled.

the equation will become

\(\displaystyle \rho_w h_w = \rho_m h_m\)

\(\displaystyle \rho_w (h + 0.61 + 0.61) = \rho_m h\)

All I have to do now is to plug the values of water density, mercury density, and solve for h.

\(\displaystyle 1000h + 1220 = 13600h\)

\(\displaystyle h = 0.0968 \ m\)

for (b) i am thinking of \(\displaystyle 2 \ psi = 13800 \ N/m^2\)

for (c) I have to repeat part (a) with a new density \(\displaystyle = 0.85 \cdot 1000\)
 
Top