- #1
chuy
- 31
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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
Once the beaker fills with a little air with a volume [tex]V_0[/tex], 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
In order to calculate the work made by the buoyant force I made the following thing:
The buoyant force is:
[tex]B=V\rho g[/tex]
Like [tex]V=V(h)[/tex] (h is depth)
[tex]V=\frac{V_0p_0}{p}[/tex]([tex]V_0[/tex] is the volume of air that introduced the pump to the beaker. And [tex]p_p[/tex] is the presure of that deph)
The presure is:
[tex]p=\rho gh + p_{atm}[/tex]
Then
[tex]B=\frac{\rho gV_0p_0}{p}=\frac{\rho gV_0p_0}{\rho gh + p_{atm}}[/tex]
So, the work maked for the buoyant force is:
[tex]W_{B}=\int_0^H Bdx=\int_0^H \frac{\rho g p_0V_0}{\rho g(H-x) + p_{atm}}dx[/tex] ([tex]x[/tex] is measured from the bottom and [tex]H[/tex] is the deph of beaker in figure 1)
[tex]W_B=p_0V_0\ln(\rho gH + p_{atm})-p_0V_0\ln(p_{atm})[/tex]
[tex]W_B=p_0V_0\ln[\rho gH/p_{atm} + 1][/tex]
It's correctly?
By the way, how can I to calculate the work necessary to introduce the air volume [tex]V_0[/tex] to the beaker through the pump?
Note: I want to calculate the ideal case.
Greetings!
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
Once the beaker fills with a little air with a volume [tex]V_0[/tex], 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
In order to calculate the work made by the buoyant force I made the following thing:
The buoyant force is:
[tex]B=V\rho g[/tex]
Like [tex]V=V(h)[/tex] (h is depth)
[tex]V=\frac{V_0p_0}{p}[/tex]([tex]V_0[/tex] is the volume of air that introduced the pump to the beaker. And [tex]p_p[/tex] is the presure of that deph)
The presure is:
[tex]p=\rho gh + p_{atm}[/tex]
Then
[tex]B=\frac{\rho gV_0p_0}{p}=\frac{\rho gV_0p_0}{\rho gh + p_{atm}}[/tex]
So, the work maked for the buoyant force is:
[tex]W_{B}=\int_0^H Bdx=\int_0^H \frac{\rho g p_0V_0}{\rho g(H-x) + p_{atm}}dx[/tex] ([tex]x[/tex] is measured from the bottom and [tex]H[/tex] is the deph of beaker in figure 1)
[tex]W_B=p_0V_0\ln(\rho gH + p_{atm})-p_0V_0\ln(p_{atm})[/tex]
[tex]W_B=p_0V_0\ln[\rho gH/p_{atm} + 1][/tex]
It's correctly?
By the way, how can I to calculate the work necessary to introduce the air volume [tex]V_0[/tex] to the beaker through the pump?
Note: I want to calculate the ideal case.
Greetings!
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