# Two immiscible liquids in a container

• luciriv
In summary, the problem involves finding the height of an upper liquid in a container filled with two immiscible liquids. The upper liquid has a specific gravity of 0.45 and the lower liquid has a density of 61.78 lbf/ft^3. Using Imperial units, the equation to solve for the upper liquid's height is P = 62.4(SG)h_0 + rho(h_1), where P is the pressure exerted by the lower liquid at the bottom of the container. The value of 16.35 psi for P may be gauge pressure, so it should be converted to absolute pressure if needed. To find the height of the upper liquid, the equation h_1 = 5.
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.]

Last edited by a moderator:
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.

## 1. What are immiscible liquids?

Immiscible liquids are two or more liquids that are unable to mix together and form a homogeneous solution. This is due to differences in their molecular structures and intermolecular forces.

## 2. How can you tell if two liquids are immiscible?

If two liquids are immiscible, they will form distinct layers when mixed together in a container. This is because the molecules of each liquid are more attracted to each other than to the molecules of the other liquid.

## 3. What happens when two immiscible liquids are placed in a container?

When two immiscible liquids are placed in a container, they will form distinct layers due to their inability to mix together. The less dense liquid will float on top of the more dense liquid.

## 4. Can immiscible liquids be separated?

Yes, immiscible liquids can be separated through a process called liquid-liquid extraction. This involves adding a third solvent that is miscible with one of the liquids, causing the two immiscible liquids to separate and form distinct layers.

## 5. What are some examples of immiscible liquids?

Examples of immiscible liquids include oil and water, gasoline and water, and vinegar and oil. These liquids have different densities, polarities, and intermolecular forces, making them unable to mix together.

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