Two immiscible liquids in a container

AI Thread Summary
The discussion revolves around solving a problem involving two immiscible liquids in a container, focusing on calculating the height of the upper liquid. The user has set up the pressure equation using specific gravity and density but is unsure how to proceed. A moderator points out that using Imperial units simplifies the problem, as a pound mass equates to a pound force, eliminating the need for gravitational acceleration in calculations. The user clarifies the conversion of measurements and expresses confusion over the pressure value, suspecting it may not be accurate. Ultimately, the user confirms the relationship between the heights of the two liquids, leading to a clearer path for solving the problem.
luciriv
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I really need a help with this exercise:
A ##1.75##-m-high container has two immiscible liquids stacked on top of each other. The upper liquid has specific gravity ##SG = 0.45## and the other has density ##\rho = 61.78\, lbf/ft^3##. If the pressure exerted by the lower liquid at the bottom is ##16.35## psi, find the height of the upper liquid in the container.​

To solve it, I have supposed that ##h_{0}, h_{1}## are the heights of the upper and lower liquids, so I have this equation:
$$SG \cdot \alpha \cdot g \cdot h_{0} + \rho \cdot g \cdot h_{1} = P,$$
where ##\alpha = 1000\,kg/m^3##, ##g = 9.8\, m/s^2## and ##P## is the pressure at the bottom of the container. But I don't know how to proceed from this point. Converting units is not a problem for me. Any hint or help is welcome.

[Moderator's note: moved from a technical forum.]
 
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This problem is really easy to solve using Imperial units. What makes it so easy is that a 1 lb mass weighs 1 lb force. You don't even need to use g to do this problem.
 
Chestermiller said:
This problem is really easy to solve using Imperial units. What makes it so easy is that a 1 lb mass weighs 1 lb force. You don't even need to use g to do this problem.
I don't get it. Could you please explain me a little bit more? I was trying to follow the method used in standard textbooks.
 
Using Imperial units, $$P=\alpha (SG) h_0+\rho h_1$$ where ##\alpha = 62.4 lb/ft^3##. If ho and h1 are in feet, then P is in lb/ft^2. And there are 144 in^2 in 1 ft^2.

1.75 m = 175 cm = 69.9 in = 5.74 ft.

Also, that 16.35 psi sounds very fishy, unless it is absolute pressure. In that case, the gauge pressure would be 1.65 psi. 1.75 m of fluid is not enough to cause 16.35 psi gauge.
 
OK. Now I get it. To find the height ##h_{0}##, do I have to write ##h_{1} = 5.74 - h_0##? Or maybe ##h_{0} < 5.74\, ft##?
 
luciriv said:
OK. Now I get it. To find the height ##h_{0}##, do I have to write ##h_{1} = 5.74 - h_0##? Or maybe ##h_{0} < 5.74\, ft##?
##h_{1} = 5.74 - h_0##
 
Thank you very much for all your help.
 
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