Oil volume required to rise the piston

[Camille]

Homework Statement

The piston shown weighs 11 lbf. In its initial position, the piston is restrained from moving to the bottom of the cylinder by means of the metal stop. Assuming there is neither friction nor leakage between piston and cylinder, what volume of oil (S = 0.85) would have to be added to the 1-in. tube to cause the piston to rise 1 in. from its initial position?

Homework Equations

$p = \frac{dF}{dA}$

And the basic differential equation of the fluid statics:

$\frac{dp}{dz} = - γ$

Also the density or the specific weight of the oil is:

$ρ = 1000 \frac{kg}{m^3} \cdot 0.85 = 850 \frac{kg}{m^3}$

$γ = ρ \cdot g = 8.336 \frac{kN}{m^3}$

The Attempt at a Solution

Firstly, I have calculated what volume of oil is needed just to keep the piston "a little" above the stop, ie. to hold just the weight of the piston:

$V_{oil p} = \frac{W_p}{A_p} \cdot \frac{A_o}{ρ \cdot g}$ ,

where W_p is the weight of the piston,
A_p is the c-s area of the cylinder
A_o is the c-s area of the tube.

This turns out to be:

$V_{oil p} = 22.39 in^3$.

Next, I have imagined, that when the piston is 1 in above the orignal level, we have to add even more oil to hold the weight of this 1-in thick layer of oil. This gave me in addition the volume of:

$V_{oil o} = 3.14 in^3$.

So altogether the total amount of oil needed calculated by me was:

$V_{oil} = 22.39 in^3 + 3.14 in^3 = 25.53 in^3$.

The answer should be $V_{oil} = 35.7 in^3$.

I see the hole in my reasoning, because the volume of oil in the 1-in layer has to be taken from somewhere, so it's either the old oil that was in the cylinder or the old + the new oil added. However, I don't know how to solve it.

Any help will be appreciated!

 

[nasu]

For one thing, have you included the additional volume in the large cylinder?
And the 3.14 in^3 volume looks suspect. How did you get that?

 

[Camille]

Do you mean the additional volume of the 1-in layer in the cylinder? Yes. It's weight is:

$W_1 = π \cdot (4 in)^2 \cdot 1 in \cdot ρ \cdot g = 6.87 N$

Then we have:

$\frac{V_{oil o}}{A_o} = \frac{W_1}{A_p}$

And solving it I got $V_{oil o} = 3.14 in^3$

 

[Chestermiller]

The volume of a 1" layer in the large cylinder is π(4)²(1)/4 = 4π in³. This is the extra volume that had to be added, over and above the 22.39.

 

[Camille]

Okay, I finally got it right. The first mistake was that 4 in and 1 in are diameters, not radii...

Here's how one can look at it:

Now, obviously the "additional" oil that must be poured is the whole blue one, so both on the left and right.

$V_{oil} = V_1 + V_2$

We know that, because the oil that is hatched in black has still the same volume as it had in the beginning. Now we calculate the equality of pressures at the line of equal pressures (marked green):

Left = Right

$p_L = p_R$

$p_L = \frac{W_p + W_{V_1}}{A_L}$

$p_R = \frac{W_{V_2}}{A_R}$

$\frac{W_p + W_{V_1}}{A_L} = \frac{W_{V_2}}{A_R}$

The areas (now correct...):

$A_L = π \cdot (\frac{4in}{2})^2 = 12.57 in^2$

$A_R = π \cdot (\frac{1in}{2})^2 = 0.79 in^2$

And the volume of the 1-in layer of oil:

$\boxed{V_1 = A_L \cdot 1 in = 12.57 in^3}$

And the weight of it:

$W_{V_1} = V_1 \cdot ρ \cdot g = 0.39 lbf$

We calculate the weight of the oil V_2:

$\frac{W_p + W_{V_1}}{A_L} \cdot A_R = W_{V_2}$

$W_{V_2} = \frac{11 lbf + 0.39 lbf}{12.57 in^2} \cdot 0.79 in^2 = 0.72 lbf$

So the volume of it is:

$\boxed{V_2 = \frac{W_{V_2}}{ρ \cdot g} = 23.17 in^3}$

Summing two volumes we get:

$V_{oil} = V_1 + V_2 = 12.57 in^3 + 23.17 in^3 = 35.74 in^3$