Find Elevation at Point A - Math Homework

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Homework Help Overview

The discussion revolves around finding the elevation at point A in a fluid mechanics context, specifically involving pressures in connected tanks. Participants are analyzing pressures based on given specific gravities and atmospheric conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to calculate pressures at various heights in the tanks and are discussing the implications of specific gravity and atmospheric pressure. Questions arise regarding the interpretation of negative pressures and the correct application of density in calculations.

Discussion Status

There is active engagement with multiple interpretations being explored, particularly around the definitions and implications of specific gravity and pressure calculations. Some participants are providing guidance on how to approach the equations, while others are questioning assumptions and seeking clarification on units and definitions.

Contextual Notes

Participants note missing information regarding the densities of certain fluids and the implications of negative pressure readings in the context of sealed tanks. There is an emphasis on ensuring correct unit conversions and understanding the relationships between pressure, density, and height.

MrMechanic
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Homework Statement


Find the elevation at point A. in the figure shown.

Homework Equations



P = pgh

The Attempt at a Solution



So i Converted 0.21kg/cm^2 to Kn/m^2 and i got 20.6kN/m^2 or 20.6kPa

and -30cm Hg i got 40kN/m^2 or 40kPa
 

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I'm not sure I understand the diagram. It appears to be two tanks side by side, connected by a pipe at the bottom and sealed at the top. Pressures in the airspaces in the tanks are given. Right?
So, now calculate the pressures at 33m in each, then at 30m in each, etc.
 
Yes the pressure in the airspaces in the tanks are given.
I'll start with the left tank
P = Patm + (0.82)(9.81)(38-30)
is it correct?
 
MrMechanic said:
Yes the pressure in the airspaces in the tanks are given.
I'll start with the left tank
P = Patm + (0.82)(9.81)(38-30)
is it correct?
If by Patm you mean the pressure of the air in that tank, that term is ok. But watch out for the units in the other term. What does "specific gravity" mean exactly?
 
Specific gravity is the ratio of the density of a substance to the density. that's why it has no unit
yes If Patm =14.7 psi = 101.325kPa = 101.325kN/m^2
and
P = 101.325 + (0.82)(9.81(38-30)
P = 36.9714 kPa
What i don't know is how to do I form an equation where i can get the height (h)
 
MrMechanic said:
Specific gravity is the ratio of the density of a substance to the density. that's why it has no unit
OK, but you need to specify the reference substance. Did you mean
Specific gravity is the ratio of the density of a substance to the density of water.​
?
The density of that substance should therefore figure in the formula. As it stands you have (spec grav) * (acceleration) * ( height), which has dimension L2/T2.

What i don't know is how to do I form an equation where i can get the height (h)
If you continue to work out the pressure at each height, working down from the top, and put in an unknown h for the height you need, you eventually get to two expressions for the pressure right at at the bottom. They must be equal, of course.
 
is this correct?
(-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1
and i'll perform another on the other side? or should I set P1 = 0?
 
When I perform P1 = 0
I get h = 5.235
 
MrMechanic said:
is this correct?
(-40) + (0.82)(9.81)(38-30) + (1.5) (9.81) (h) - (9.81)(h+3) - 20.6 = P1
and i'll perform another on the other side? or should I set P1 = 0?
Two problems.
1. I had been ignoring the minus sign in "-30cm Hg". I don't understand how an absolute air pressure can be negative. These are sealed containers, so it shouldn't be relative to ambient pressure. Do you have any explanation for that?
2. You are ignoring my comment about density and units. If the s.g. is 0.82 relative to water, what is the density in kg/m3?
 
  • #10
The minus sign is not an absolute pressure i think.
The density of oil is not given. And also gasoline. In order to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3
 
  • #11
MrMechanic said:
to get their density you have to multiply their specific gravity to the density of water which is 9.81kg/m^3
No, that's where you are going wrong. The 9.81 is gravitational acceleration. That factor converts mass to weight. The density of water is much larger.
 
  • #12
Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry
 
  • #13
MrMechanic said:
Oh yeah sorry. It's not kg/m^3 ... It's 9.81 kN/m^3 forgot to change that sorry
Right, but answer my other question; what is the density of water in these units?
 

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