Two immiscible liquids in a container

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Discussion Overview

The discussion revolves around a physics exercise involving two immiscible liquids in a container, focusing on calculating the height of the upper liquid based on given pressures and properties. The scope includes mathematical reasoning and technical explanations related to fluid mechanics and pressure calculations using both Imperial and metric units.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the problem and sets up an equation involving specific gravity and density to find the height of the upper liquid.
  • Another participant claims the problem is straightforward using Imperial units, stating that a 1 lb mass equals 1 lb force, thus eliminating the need for gravitational acceleration in calculations.
  • A subsequent post seeks clarification on the reasoning behind the simplicity of the Imperial unit approach, indicating confusion with standard textbook methods.
  • Further clarification is provided, suggesting a specific equation for pressure in Imperial units and converting the height of the container from meters to feet, while questioning the validity of the given pressure value.
  • Participants discuss the relationship between the heights of the two liquids, with one proposing an equation to express the height of the lower liquid in terms of the height of the upper liquid.

Areas of Agreement / Disagreement

Participants express differing views on the approach to solving the problem, with some favoring the use of Imperial units and others adhering to standard textbook methods. There is no consensus on the interpretation of the pressure value provided or the implications for the problem's solution.

Contextual Notes

There are unresolved questions regarding the pressure value's nature (gauge vs. absolute) and its implications for the calculations. Additionally, the conversion between units and the assumptions made about the properties of the liquids are not fully explored.

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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