Thermodynamics Problem: Calculating Piston Heights in Connected Oil Tanks

[rbarngrover]

Two tanks filled with oil with density 860 kg/m3 are connected by a valve. Tank A has piston with 2.2 kg mass, cross sectional area of .02 m^3 and is suspended 1 m from the base of the tank. Tank B has a pistion mass of 1 kg, with cross sectional area .01 m^2 and is suspended .7m. when the valve is opened what are the final heights the two pistons are suspended.



I understand that Pressure of tank A has to equal Pressure tank B.

My main problem is i do not understand how i am suppose to correlate the two heights so i can solve for one of them. I tried have tank a height be x, then make tank b height be x-.3, but when u try and solve x cancels. Please would like to see how this is to go about being solved since most problems don't have the height's part of the problems.

 

[rbarngrover]

If need be i can draw and scan a diagram of the problem

 

[etudiant]

A diagram might help.
The statement that the piston(s) are 'suspended' at some height from the base of the tank(s) is not clear.
Does it mean the pistons form the upper bound of the cylinders, that their cross sectional areas are their respective cylinders area and that their weights add to the pressure in each cylinder? If so, the solution is pretty direct.
(Assume the Tank A cross sectional area is 0.02m^2 that was typoed to 0.02m^3.)

 

[rbarngrover]

Yes, that area would be the area of the respective cylinders and the pistons form upper bound of the tank so they add to pressure. I think it might help to say that the bases of each cylinder are on the same level if that helps.

before the tank valve is open the following equations determine pressure on each side of the valve...

Valve on side A: Pa = Po + Rho(g)(H) + (M(piston A)(g))/Area of a)
== ( 100000 + 860(9.81)(1) + 2.2(9.81)/.02 ) pascals

Valve on side B: Pb = Po + Rho(g)h + (M(piston b)(g))/Area of b
== ( 100000 + (860)(9.81)(.7) + (1)(9.81)/.01 ) PascalsI can see that once valve is opened that Po would not affect pressures since both are open to that atm, but after that i am kinda stuck with relating the two heights so i can solve for one.

 

[rbarngrover]

DIagram of problem

OK, I sketched a diagram that would represent the problem and have it as an attachment.

 

[Travis_King]

You have two pressures, one for A and one for B. When the valve is opened, will these pressures be equal or unequal?

How would you go about finding the final pressure in each cylinder?

 

[rbarngrover]

Once valve is opened the two tanks should go to equilibrium pressure wise. Intuitively the volume from tank a would rush into tank B till equilibrium is reached.

 

[Chestermiller]

Once valve is opened the two tanks should go to equilibrium pressure wise. Intuitively the volume from tank a would rush into tank B till equilibrium is reached.

The other constraint you need to match is that the total volume of liquid in the two tanks must remain constant. Let x = final height of liquid in tank A and y = final height of liquid in tank B. Write an equation in terms of x and y such that the final and initial volumes of liquid in the two tanks remains constant. Write a second equation in terms of x and y such that the final pressures at the bottom of the two tanks are equal to one another. Solve for x and y.

Chet

 

[rbarngrover]

I still not sure how this will work, But if i understand correctly...

Make tank A height x and Tank B height Y. Then Tank B = y = x - .3. is that what your talking about?

then Pressure tank A = x and Pressure tank B = y per say then x = y + difference between two pressures due to the pistons, such that x = y + 98.1? <--- I guess this comes from there being no liquid volume in the tanks

 

[rbarngrover]

well wait is this what u meant...

since total volume has to equal .027 m^3 i can write equations in terms of x and y as

.027 = (.02)x + (.01)y

then use pressures equal so that

rho(g)x +Ma(g)/Aa = Rho(g)y + Mb(g)/Ab

from these two equations solve?

 

[Chestermiller]

well wait is this what u meant...

since total volume has to equal .027 m^3 i can write equations in terms of x and y as

.027 = (.02)x + (.01)y

then use pressures equal so that

rho(g)x +Ma(g)/Aa = Rho(g)y + Mb(g)/Ab

from these two equations solve?

Yes. Right on! Nice job.

 

[rbarngrover]

ahh think I have it now with x = .896 m and y = .907 m. May i ask what made you think of coming up with two equations like that?

 

[Chestermiller]

ahh think I have it now with x = .896 m and y = .907 m. May i ask what made you think of coming up with two equations like that?

Just lots of modeling experience.

Chet

 

[Travis_King]

Two unknowns, you need two equations. You have a relationship between pressures, so you need to think of another relationship between the two. In this case, it's volume.