Help with thermodynamics problem

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The discussion revolves around calculating the optimal intermediate pressure for a two-stage compressor compressing air from 1 atm to 25 atm to minimize work. The key equation mentioned, P = sqrt(P1*P2), is derived from the principles of adiabatic compression and work minimization. Participants emphasize the importance of using the adiabatic condition and considering air as an ideal gas for accurate calculations. The conversation also touches on the necessity of removing heat after the first compression stage to achieve efficiency. Recommendations for further reading on thermodynamics and compression processes are requested to deepen understanding.
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Homework Statement


Hi I would like you guys to help me with this problem

Air is to be compressed steadily and isentropically from 1 atm to 25 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be
(a) 3 atm (b) 5 atm (c) 8 atm (d) 10 atm (e) 13 atm

Homework Equations



(a) 3 atm (b) 5 atm (c) 8 atm (d) 10 atm (e) 13 atm

P=sqrt(P1*P2)

The Attempt at a Solution



I found that this pressure can be calculated with the square root of the product of those two pressures, however, i want to know where that equation comes from

thanks!
 
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cmon, help me, please, how do you obtain that equation?
 
Deathcrush said:

Homework Statement


Hi I would like you guys to help me with this problem

Air is to be compressed steadily and isentropically from 1 atm to 25 atm by a two-stage compressor. To minimize the total compression work, the intermediate pressure between the two stages must be
(a) 3 atm (b) 5 atm (c) 8 atm (d) 10 atm (e) 13 atm

Homework Equations



(a) 3 atm (b) 5 atm (c) 8 atm (d) 10 atm (e) 13 atm

P=sqrt(P1*P2)
!
If there is any savings of work by doing it in two stages, the heat from the first compression has to be removed. So compress it to 1/5 of original volume (adiabatic), remove the heat and then adiabatically compress it further to 1/5 of this compressed state.

You should be able to show that if you do it any other way (eg. compress 1/2 then 1/12.5, or vice versa) you will end up doing more work. Write out the mathematical expression for work for the first compression to V1 + work for the second compression to Vf. You have to use the adiabatic condition (T and V) for each compression with a reduction in temperature to initial temperature after the first compression. When is the work a minimum?

AM
 
yeah, well, what i am interested in , is in finding how can I get that equation listed above, I had thought in deriving the equation for the sum of work in each phase, and finding the minimum value, however i can't set the whole thing to get that equation. Can I consider the air as an ideal gas? or how else can I get an equation P to integrate
 
Deathcrush said:
yeah, well, what i am interested in , is in finding how can I get that equation listed above, I had thought in deriving the equation for the sum of work in each phase, and finding the minimum value, however i can't set the whole thing to get that equation. Can I consider the air as an ideal gas? or how else can I get an equation P to integrate
Use the adiabatic condition for the first stage from V0 to V1:

T_0V_0^{(\gamma-1)} = T_1V_1^{(\gamma-1)}

Since it is adiabatic, the work done is equal to the change in internal energy: W = nC_v\Delta T

Do the same for the second stage. Then show that the minimum value for total Work is when V2/V1 = V1/V0 (ie. V_1 = \sqrt{V_2V_0}).

AM
 
I think I'm getting it, I would thank you if you could recommend me some reading about this
 
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Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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