Calculating this gauge pressure reading

AI Thread Summary
To calculate gauge pressure using a U-tube manometer, it's essential to understand that the pressure gauge indicates the static pressure at the liquid's height, influenced by the trapped air in the tank. Establishing equations for both sides of the manometer based on the common pressure of the trapped air is crucial for solving the problem. Height x, which was initially considered a variable, is ultimately deemed irrelevant in the calculations. The final gauge pressure was determined to be 33.5 kPa after equating the pressures on both sides. This approach effectively clarifies the relationship between the trapped air pressure and the liquid levels.
Bolter
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Homework Statement
See below
Relevant Equations
pressure = rho x g x h
specific weight = rho x g
Hi everyone!
How do I go about solving this problem?

Screenshot 2020-09-30 at 13.04.21.png

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I tried working out the gauge pressure using this but I have a few unknowns which won't make this possible such as what is the length of x which I labelled in the figure

IMG_5299.JPG


Any help would be appreciated! Thanks
 
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The U-tube manometer is telling you the pressure of the volume of air trapped in the tank.
The pressure gauge should be telling you the static pressure at that height of liquid, which surface is being pressured down by the trapped air.
 
Lnewqban said:
The U-tube manometer is telling you the pressure of the volume of air trapped in the tank.
The pressure gauge should be telling you the static pressure at that height of liquid, which surface is being pressured down by the trapped air.

Ok I get that, but I am not entirely sure how I can form an equation with that statement. Do I begin off by setting up an expression on the left and right side of the figure then equate them?
 
The trapped pocket of air in the tank is simultaneously in contact with the surface of both liquids, having a unique pressure.
I would establish relations for each side around that common pressure.

I believe that height x is irrelevant, do you agree?
 
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Lnewqban said:
The trapped pocket of air in the tank is simultaneously in contact with the surface of both liquids, having a unique pressure.
I would establish relations for each side around that common pressure.

I believe that height x is irrelevant, do you agree?

Thanks I got the right answer from making relations on both sides and then equalling those 2. Gauge pressure turned out to be 33.5 kPa

IMG_5300.JPG
 
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You are welcome. :smile:
Your result is correct.
 
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