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Carnot Cycle Refrigerator - Thermal Battery Problem

  • Thread starter Kelsi_Jade
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  • #1
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The question is:
Consider a thermal battery - a device for storing mechanical energy. Use two large containers as heat reservoirs where one contains ice and water in equilibrium and the other contains steam and water in equilibrium. Both are held at pressure=1atm. Energy is stored by using a carnot cycle to transfer heat from the colder reservoir to the hotter - how much ice is created and how much vapor is created when 1kW*hr of energy is stored?

I know that a refrigerator is essentially a heat pump in reverse so we can consider a Carnot engine where heat is transferred from the cold reservoir. Work is done in order to do this, and the work for the refrigerator is negative.

Since I don't know the temperatures, and n=only the pressure of the systems and the energy stored, I know I can't use the COP or efficiency in the problem but I am unsure where to go next.

Any help is appreciated.
 

Answers and Replies

  • #2
DrClaude
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Since I don't know the temperatures
Really? Reread the problem carefully, they're hiding in there somewhere...
 
  • #3
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OK...so is it correct to say that if the ice and water are in equilibrium, T1 would be the temp where ice and water can exist together at 1 atm and T2 would be the temperature where water and steam exist in equilibrium at 1 atm?
So, after a little research: T1= ice point= 273K
T2= steam point= 373K

Next:
Is the 1 kW*hour (3600kJ) energy given the internal energy of the system if it is the "energy stored"?
which for an isobaric process would be: ΔU=ncpΔT
which can be found using the temperatures I just defined and where:
cpice=38.09 J/mol*K
cpsteam=37.47 J/mol*K

When the question asks for how much ice/steam, does that mean that I am to find the mols of ice/steam? Which I can find by rearranging the above equation?
 
  • #4
DrClaude
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OK...so is it correct to say that if the ice and water are in equilibrium, T1 would be the temp where ice and water can exist together at 1 atm and T2 would be the temperature where water and steam exist in equilibrium at 1 atm?
So, after a little research: T1= ice point= 273K
T2= steam point= 373K
Correct.

Next:
Is the 1 kW*hour (3600kJ) energy given the internal energy of the system if it is the "energy stored"?
I am not sure how to interpret this, by my take is that "energy stored" means energy that went into the hot reservoir, as this is where it can be kept for future use in a thermal engine.

When the question asks for how much ice/steam, does that mean that I am to find the mols of ice/steam?
Yes.

Which I can find by rearranging the above equation?
Yes, but you are missing one element.
 
  • #5
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Hmm...so, energy going into the hot reservoir would be the Heat energy Q? Where,
Q =mCpΔT =ncpΔT...

If that's true, how is the equation for ΔU= the equation for Q?
 
  • #6
DrClaude
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which for an isobaric process would be: ΔU=ncpΔT
I hadn't paid enough attention to what you wrote there. That equation for ΔU is not valid for a phase transformation. Look up "latent heat".

And I'll make more precise my previous hint: you have to use the fact that it is a Carnot engine that is transfering the heat.
 
  • #7
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OK, latent heat is the amount of energy in the form of heat (Q) required to completely effect a phase change of a unit of mass.
Latent heat=Q/m
Latent heat for water vapor:
L(T) = (2500.8 - 2.36 T + 0.0016 T^2 - 0.00006 T^3)J/g

Latent heat for ice:
L(T) = (2834.1 - 0.29 T - 0.004 T^2)J/g

On the right track?
 
  • #8
DrClaude
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OK, latent heat is the amount of energy in the form of heat (Q) required to completely effect a phase change of a unit of mass.
Latent heat=Q/m
Latent heat for water vapor:
L(T) = (2500.8 - 2.36 T + 0.0016 T^2 - 0.00006 T^3)J/g

Latent heat for ice:
L(T) = (2834.1 - 0.29 T - 0.004 T^2)J/g

On the right track?
That first formula is not valid at 100 °C. Wikipedia has a table with values you can use.

Otherwise, yes, you are on the right track.
 
  • #9
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That first formula is not valid at 100 °C.
Oh, I see that now. Thanks!
 
  • #10
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Ok, I may have confused myself a little going over the last bit. So, is the 3600kJ Q, or L(T)?
 
  • #11
DrClaude
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Ok, I may have confused myself a little going over the last bit. So, is the 3600kJ Q, or L(T)?
That would be Q going into the hot reservoir.
 

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