# Two immiscible liquids in a container

• luciriv

#### luciriv

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.

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##?

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.