# Work made by the bouyant force

1. Oct 13, 2006

### chuy

Hi,

I need to know if I am making the calculations correctly. I thank much your aid.

Well, we have a deposit with water, with a beaker inside connected to an air pump in outside. The beaker is held to the bottom of the deposit by a pair of cords, as it is possible to be seen in the figure:

http://img169.imageshack.us/img169/1669/dibudx3.png [Broken]

Once the beaker fills with a little air with a volume $$V_0$$, we loosen it so that it can raise the surface (It appears 2). While the beaker raises, the pressure is reduced and the air volume increases (It appears 3).

http://img169.imageshack.us/img169/7906/dibu2wa0.png [Broken]

In order to calculate the work made by the buoyant force I made the following thing:

The buoyant force is:

$$B=V\rho g$$

Like $$V=V(h)$$ (h is depth)

$$V=\frac{V_0p_0}{p}$$($$V_0$$ is the volume of air that introduced the pump to the beaker. And $$p_p$$ is the presure of that deph)

The presure is:

$$p=\rho gh + p_{atm}$$

Then
$$B=\frac{\rho gV_0p_0}{p}=\frac{\rho gV_0p_0}{\rho gh + p_{atm}}$$

So, the work maked for the buoyant force is:

$$W_{B}=\int_0^H Bdx=\int_0^H \frac{\rho g p_0V_0}{\rho g(H-x) + p_{atm}}dx$$ ($$x$$ is measured from the bottom and $$H$$ is the deph of beaker in figure 1)

$$W_B=p_0V_0\ln(\rho gH + p_{atm})-p_0V_0\ln(p_{atm})$$

$$W_B=p_0V_0\ln[\rho gH/p_{atm} + 1]$$

It's correctly?

By the way, how can I to calculate the work necessary to introduce the air volume $$V_0$$ to the beaker through the pump?

Note: I want to calculate the ideal case.

Greetings!

Last edited by a moderator: May 2, 2017
2. Oct 13, 2006

### OlderDan

Your calcualtion looks OK to me. For the work needed to pump in the initial volume, assume the pump starts from a position with gas below the piston at pressure equal to the water pressure in the beaker. The piston then has to move against the gas in the cylinder to push some of it into the beaker. All the piston is doing is moving a volume Vo of gas at constant temperature and pressure.

3. Oct 14, 2006

### chuy

Ok. Then the work, maked for introduce the air will be:

$$W_{pump}=Fh=p_0Ah = p_0V_0 = (\rho g h + p_{atm}) V_0$$

http://img278.imageshack.us/img278/57/volhn8.png [Broken]

It's correctly?

Greetings.

Last edited by a moderator: May 2, 2017
4. Oct 14, 2006

### OlderDan

Looks good to me.

5. Oct 14, 2006

### chuy

Sure? Because if we reduced the work that we obtain from the system less the work which we put due to the pump we obtain an excess of energy:

$$W_B-W_{pump}=p_0V_0\ln[\rho gH/p_{atm} + 1]-p_0V_0=p_0V_0(\ln[\rho gH/p_{atm} + 1]-1)$$

If $$\ln[\rho gH/p_{atm} + 1]>1$$, then $$W_B>W_{pump}$$

Thanks!!

Last edited: Oct 14, 2006
6. Oct 14, 2006

### OlderDan

I'm pretty sure. I expect you could show that the work done by the water does not increase the energy of the system. When you pump air into the glass, you are displacing water from the glass into the tank, raising the level and potential energy of the water. As the glass rises, water is constantly moving from a higher level in the tank to a lower level, reducing the potential energy of the water. If the glass floats at the top, the water will be almost back to the way it was before the glass was filled with air, so it has given back potential energy as the glass has gained energy.

I haven't tried to prove that it all cancels out, but if you want to prove that it does not, you have to consider every energy change in the system.

7. Oct 14, 2006

### chuy

Then it means that the work that makes the force (F) that is applied on the handle of the pump:

http://img146.imageshack.us/img146/1650/dibu3tb5.png [Broken]

Is:

$$W = F\Delta y = V_0p_0+\Delta U_{watter}$$

Or:

$$W= F\Delta y= p_0V_0=\Delta U_{watter}$$ ??

Last edited by a moderator: May 2, 2017
8. Oct 15, 2006

### OlderDan

It does not mean $$W = F\Delta y = V_0p_0+\Delta U_{watter}$$

When you push the pump handle, you do work on the gas. I was careful in the beginning to say the gas was already at the pressure of the liquid at the end of the hose. All you are doing to the gas is moving some of it out of the cylinder into the beaker, which means no change in volume or temperature or pressure. How can you do work on a gas, and not change any of these things? Only by having the gas do the same amount of work on something else that you do on the gas. The gas in the first step is simply transferring the work you do into work done on the water to increase the potential energy of the water.

Think about changing your problem a little. Suppose the bubble of air you create in the tank completely fills a cross section of the tank. You create a complete disk layer of air in the tank. That disk will have the volume of the air you push out of the pump. The height of the disk will be the distance you move the pump handle, times the ratio of the area of the pump to the area of the tank. All the water above the disk of air is raised by this same distance. How much does the potential energy of the water increase? How does this compare to the work done by you pushing the pump handle?

The atmospheric pressure part might be a little tricky here. That part of the work calculation you did earlier for introducing the air into the beaker is not done by you. It is done by the air pushing down on the top of the pump piston. I want you do compare the work you do pushing the pistion to the potential energy increase of the water.

9. Oct 15, 2006

### chuy

The mass of watter above de disk is m, and x is the thickness of the disc, the increase of energy potencial is:

$$\Delta U_{water}=mgx=\rho Vgx=\rho AHgx$$ (depth of disk is H, and A area of disk).

But $$p=\rho gH$$, and $$V_0=Ax$$ Therefore:

$$\Delta U_{water}=pV_0$$.

Mmmm, since we had said, the work maked for introduce the air is:

$$W_{pump}= p_0V_0 = (\rho g H + p_{atm}) V_0$$

Thus:

$$W_{pump}= \Delta U_{water}+V_0p_{atm}$$

But as you mentioned:

$$W_{pump}=W_{hand}+W_{atmosfera}$$

Then,

$$W_{hand}=\Delta U_{water}$$

But the potencial energy of water, would not have to be equal to the work made by bouyant force?

$$W_B-\Delta U_{water}=p_0V_0\ln[\rho gH/p_{atm} + 1]-pV_0=(p+p_{atm})V_0\ln[\rho gH/p_{atm} + 1]-pV_0$$

Then:

$$W_B>\Delta U_{water}$$

I believe that there is something I am omitting

Greetings!

Last edited: Oct 15, 2006
10. Oct 15, 2006

### OlderDan

Probably so. This part of the process is more complex. The gas in the beaker is expanding, which is what made the first calculation you did much more complex than calculating the work done on a block of wood released from some depth. That means the gas was doing work during the process. To examine all the energy conversions, you would have to consider the energy change of the gas.

If the glass had been a block of wood with volume Vo, the buoyant force would have been constant, the work would have been the weight of a volume of water of Vo times the initial depth (the constant buoyant force), and if there were no energy losses to viscosity, the PE lost by the water would have been the work done on the block. The increase in PE of the block would be its weight times the depth of the water, which would be less than the decrease in PE of the water, which means the block would go flying out of the water with some initial kinietic energy. In reality, the flow of the block through viscous water is going to steal some of the energy, converting it into heat, slightly warming the water.

As I said earlier, if you want to show that your calculation is invalid based on energy conservation, you had better account for all of the energy. That is not simple to do.

11. Oct 16, 2006

### chuy

When the outer air enters the pump increases its pressure. A work is made on the gas. This change of energy also we must consider it?

Good night.

12. Oct 16, 2006

### OlderDan

You have to set up boundaries for every problem. Yes it would take work to build up pressure in the pump before any air ws pushed into the tank. But if the pump did not have that pressure, and you put the tube into the tank, water would run through the tube into the pump until the pressure was equalized. The initial condition I chose in my first post was carefully constructed to avoid this.

13. Oct 17, 2006

### chuy

Retaking that.

I can suppose constant temperature of the gas while the glass raises?

The work done by the gas on the water is:

$$W_{gas}=\int_{V_0}^{V_f}pdV$$

Like $$p=\frac{p_0V_0}{V}$$

$$W_g=\int_{V_0}^{V_f}\frac{p_0V_0}{V}pdV$$

$$W_g=p_0V_0\ln{\frac{V_f}{V_0}}$$

$$W_g=p_0V_0\ln{\frac{p_0}{p_f}}$$ ($$p_f=p_{atm}$$)

$$W_g=p_0V_0\ln{\frac{p_0}{p_{atm}}}$$

$$W_g=p_0V_0\ln{\frac{\rho g H+p_{atm}}{p_{atm}}}=p_0V_0\ln(\rho g H/p_{atm}+1)$$

Then, $$W_B=W_{g}$$ (The work made by buoyant force is the same than work made by gas on water)

But would not have to be:

$$W_B=W_g+\Delta U_{water}$$ ???

Greetings!