Work done in Compressing a fluid in a piston (Bulk Modulus)‏

[shivkumarsing]

Homework Statement

I am trying to solve one theoretical problem.
If I had an Iron piston cylinder with radius of 1 cm, such that it
contains 1 liter of water(therefore about 318 cm tall), ( say Bulk
Modulus of water is 1 GPascal), How much work(in joules) will be
required to depress the piston by 5 cm ( approx 15 ml compression, =
1.5%).

Homework Equations

I don't know which formulas to use.
Cylinder area = pi x r x r
clylinder vol = area x displacement

The Attempt at a Solution

Not sure

 

[blochwave]

http://hyperphysics.phy-astr.gsu.edu/hbase/permot3.html

So basically you know B(it's 2.2 GPa in fact)you know the initial and final volumes, so you can find delta V/V(the geometry is only a little different from the picture since you use cylinders)and I guess we assume the water fills the cylinder so the initial pressure is 0, so you'll find basically what the pressure the cylinder has to apply is. P=F/A, so multiply that (should be huge)pressure by the area of the cylinder pushing on the water, and you have the force you have to apply to compress it, multiply by the distance you compress it for the work, make sure you get the units right

I think

 

[Shooting Star]

The Attempt at a Solution

Not sure

Have you done calculus?

 

[blochwave]

You sure you need calculus? I'm trying to decide what assumptions I made in dumping the differentials for the finite deltas. I guess that the change in pressure with respect to the change in volume is constant as you compress it, and I'm not sure how unrealistic an assumption that is

 

[shivkumarsing]

Work done

Thanks for your response. Using your steps, the work done,
P = F/A, F = PXA, and Work done = F x Distance
therefore, Workdone = PXAXdistance = P X volume change.

Bulk modulus gives P required per unit area. Here in lies my problem. What area should I choose. Surface Area of the total cylinder or Surface area of the piston?
Based on the area, I choose, I can multipy the Bulk modulus with the area and get the pressure. After that it is easy.

 

[blochwave]

Yah, I actually think it's the whole cylinder, not just the moving surface. You could imagine the water bursting out the sides violently as the piston lowered if the walls weren't as strong, that's the idea, that the water is applying pressure everywhere as it compresses

 

[Shooting Star]

You sure you need calculus? I'm trying to decide what assumptions I made in dumping the differentials for the finite deltas. I guess that the change in pressure with respect to the change in volume is constant as you compress it, and I'm not sure how unrealistic an assumption that is

(Sorry for the late reply.)

Calculus gives you the exact and simple answer, as long as B is constant. (For varying B, it'll be difficult to integrate.) But let’s think of the approximations you have suggested.

Work done W = Integral PdV.

So you simply want to put delta P = final P = P2, and V=constant= initial volume, and P1 = 0. Perhaps V = (V1+V2)/2 will make for a better approximation.

The problem may be that P changes enormously. What should be taken as the average value of P? Simply equating W=P*delta_V may or may not yield a realistic answer.

It's best to compare with the exact answer arrived at by integrating. Will post it very soon.

 

[Shooting Star]

(Sorry again, forgot to post it.)

From defn, B = -VdP/dV, which on integration between V1 to V2 gives,

P2 = Bln(V1/V2) + P1, if B is constant.

Work done = Integral(from V1 to V2) PdV = (P2V2-P1V1) + B(V2-V1). You can get rid of P2 using the previous relationship.

Now we can see how best to approximate without using calculus.