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Pumpby Yegor
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#1
Feb905, 03:47 AM

P: 149

Air pump is pumping out a gas from a vessel (vessel's volume is V).
After each cycle i it pumps out dV. Gas is ideal. T=Const. After how much cycles pressure P will decrease k times? Here is my work. PV = nRT Initially n=n(0) After each cycle n decreases (1dV/V) times. So, after x cycles n(x)=n(0)(1dV/V)^x k=p(0)/p(x)=n(0)/n(x) n(x)/n(0)=(1dV/V)^x So x =ln(1/k)/ln(1dV/V) But the answer is x=ln(k)/ln(1+dV/V) As I understand, they assume that after each cycle n decreases 1/(1+dV/V) times. In this way i really receive x=ln(k)/ln(1+dV/V) I don't understand it. Help me please. 


#2
Feb905, 01:05 PM

P: 149

Has anybody any ideas?



#3
Feb905, 01:36 PM

P: 333

They have done some mathematics with x =ln(1/k)/ln(1dV/V) to arrive at a more nice looking answer.
x= ln (k) / ln (1 dV/V) = ln (k) / ln [(1dV/V)] ^1 now, [(1dV/V)] ^1 = 1 + dV/V (approximately) so, x = ln (k) / ln [(1+dV/V) 


#4
Feb905, 02:05 PM

P: 120

Pump
Each cycle corresponds to a T=const. process. (in another version of this problem the pumping process is very fast and a the transformation would correspond to an adiabatic process  all you have to do is to add the exponent [tex]\gamma[/tex] to the equations below). After the first cycle the air volume increases from V to V+dV and the pressure decreases at [tex]p_1[/tex]. You have
[tex]p_0\cdot V=p_1\cdot (V+dV)[/tex] After the first cycle, the pump takes out the air from dV, and you get again [tex]p_1=p_0 \frac{V}{V+dv}[/tex] In the second cycle, [tex]p_1\cdot V=p_2\cdot (V+dV)[/tex] and [tex]p_2=p_1 \frac{V}{V+dv}=p_0 \frac{V^2}{(V+dv)^2}[/tex] and so on..... [tex]p_n=p_{n1} \frac{V}{V+dv}=p_0 \frac{V^n}{(V+dv)^n}[/tex] 


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