Find Elevation at Point A - Math Homework

  • Thread starter Thread starter MrMechanic
  • Start date Start date
Click For Summary
SUMMARY

The discussion revolves around calculating the elevation at point A in a fluid mechanics problem involving two tanks connected by a pipe. The participants utilize the equation P = pgh, converting pressures from kg/cm² to kN/m² and discussing specific gravity in relation to density. Key calculations include determining pressures at various heights and addressing the significance of specific gravity and density in the equations. The final calculations lead to a height (h) of approximately 5.235 meters based on the pressure equilibrium principle.

PREREQUISITES
  • Understanding of fluid mechanics principles, specifically hydrostatic pressure.
  • Familiarity with the equation P = pgh and its application in pressure calculations.
  • Knowledge of specific gravity and its relationship to density.
  • Ability to convert units between kg/cm² and kN/m².
NEXT STEPS
  • Learn about hydrostatic pressure calculations in fluid mechanics.
  • Study the concept of specific gravity and its implications in fluid density.
  • Explore unit conversions in fluid mechanics, particularly between different pressure units.
  • Investigate the principles of pressure equilibrium in connected fluid systems.
USEFUL FOR

Students and professionals in engineering, particularly those studying fluid mechanics, as well as anyone involved in solving practical problems related to pressure and elevation in fluid systems.

MrMechanic
Messages
46
Reaction score
0

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
 

Attachments

  • Untitled.png
    Untitled.png
    3.5 KB · Views: 573
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
14
Views
3K