Adiabatic compressed air and energy calculations

  • #26
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Yes.
It seems our formulas are different. I mean the structure of them. But either ways they give the same answer.
 
  • #27
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Can we please do part C?. I know this is very bad and irritating for you. I am sorry for the trouble.
 
  • #28
haruspex
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Can we please do part C?. I know this is very bad and irritating for you. I am sorry for the trouble.
No, it's fine, I'm just trying to get you to be clearer and more careful in writing out your work, as much for your benefit as mine.

For c, you need to know the specific heat of the steel.
If the final temperature is θ, what is the gain in internal energy of the cylinder and what is the internal energy loss of the gas?
 
  • #29
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The specific heat of steel is 448 Kg-1 K-1. The density of steel is 7900 Kg m-3. I'm sorry I do not understand what you are asking
 
  • #30
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Doesn't the gas gain energy during the compression due to the work done?
 
  • #31
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The teacher also said I could assume that the final temperature is close to the initial temperature, so that all the of the energy in the hot gas is transferred to the steel
 
  • #32
haruspex
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Doesn't the gas gain energy during the compression due to the work done?
Yes, but we've done that part. We are now interested in what happens as heat from the compressed gas is transferred to the steel.
 
  • #33
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The steel and the compressed gas come into thermal equilibrium. Does that mean due to a difference in temperature heat transfer occurs?. If so, How do I calculate the increase in the the temperature of the steel if i do not know the initial temperature of the steel
 
  • #34
haruspex
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The teacher also said I could assume that the final temperature is close to the initial temperature, so that all the of the energy in the hot gas is transferred to the steel
Ok. I would have said it a bit differently... The specific heat of the gas is very small compared to that of the cylinder, so nearly all the gained internal energy will finish up in the steel.
The steel and the compressed gas come into thermal equilibrium. Does that mean due to a difference in temperature heat transfer occurs?
Yes. The steel will heat up and the gas cool down until they reach the same temperature. It does not say so, but you need to ignore any heat loss from the cylinder to the surroundings.
How do I calculate the increase in the the temperature of the steel if i do not know the initial temperature of the steel
You can assume it started at the same temperature as the gas did, 18C.
 
  • #35
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I see. How do I calculate the increase in temperature of the steel. What formulas would I use?.
I am sorry this is my first doing this sort of problem. Could you show me an example?
 
  • #36
haruspex
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I see. How do I calculate the increase in temperature of the steel. What formulas would I use?.
I am sorry this is my first doing this sort of problem. Could you show me an example?
You know the heat energy going into it, its volume, density and specific heat. If you don't know an equation connecting those with temperature change, look up specific heat.
 
  • #40
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The energy lost by the gas + the energy gained by the steel cylinder = 0 Since the energy is conserved.
The mass of the steel cylinder = Volume x density = (π(0.01)^2 - π(0.009)^2)*0.6 = 3.58 x 10 (-5)* 7900 = 0.283 Kg
MsCs(Tf - 18) + MgCg(Tf- 18) = 0
 
  • #41
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I do not know the mass of air, or it's specific heat.
So would it be correct if i said the energy.
Also I do not know what to do from here.
 
  • #42
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(π(0.01)^2 - π(0.009)^2)*
The 2cm is the internal diameter, not the external diameter.
MgCg(Tf- 18)
No, the gas is cooling from the high temperature found previously to Tf. But anyway, we established that you can just write that the heat transferred to the steel is the whole of the internal energy gained by the gas during compression, and you already calculated that.
 
  • #43
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Do you mean to use the internal energy that was calculated in part b?
 
  • #44
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Density of steel = π(0.011)^2 - π(0.01)^2 * 0.6 = 3.96 x10(-5) * 7900 = 0.313 Kg
 
  • #45
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(0.313)(448)(Tf - 18) = 59.5 J
Tf - 291.15 = 59.5/(0.313*448)
Tf = 291.57 K
Therefore there is an increase of 0.42 degrees
 
  • #46
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I think this is way it should be done. I see that you are offline. Thank you very much for you help. This might be wrong, but it's too late now the homework is due tomorrow. If there is any mistakes with please just list them so i can take care of it late
 
  • #47
haruspex
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But anyway, we established that you can just write that the heat transferred to the steel is the whole of the internal energy gained by the gas during compression, and you already calculated that.
whoops! That's wrong.
Even though the gas cools, it is still at a higher pressure than it started. This means it retains some of the work done.
You need to calculate the pressure when it returns to about 18C.
 
  • #48
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I am so confused. Could you please give show me how it meant to be done?
 
  • #49
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Excuse me, Mr.Haruspex it's getting quite late for me could you please show me what you mean?.
 
  • #50
haruspex
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Excuse me, Mr.Haruspex it's getting quite late for me could you please show me what you mean?.
Sorry, but I'm not that expert in this area myself, and I don't want to lead you astray. I need to check some things before I give a firm answer. I understand that this will be too late for you tonight.
 

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