Isothermal compression and adiabatic expansion

[pranj5]

I want to know if a specific amount of gas has been compressed isothermaly an then released to expand freely, how much of the energy that has been spent on compressing the gas can be recovered. As for example, 1 gm-mole of Nitrogen has been compressed to 1/4th of its initial volume from at 1 bara pressure. Now, if that amount has been released suddenly and used to rotate a turbine, then how much energy can be produced?
For simplicity, let's consider the whole process to be something totally ideal from start to end.

 

[Mech_Engineer]

You should lay out a cycle on an H-S or P-v diagram and then calculate the power output using the start and end states. It's impossible to answer your question without knowing something about what fluid is being used and its start and end states.

Take for example this tutorial on analyzing a Brayton cycle air turbine: http://web.mit.edu/16.unified/www/SPRING/propulsion/notes/node27.html
Rankine cycle: http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node65.html

 

[Nidum]

Now, if that amount has been released suddenly and used to rotate a turbine, then how much energy can be produced?

Certainly gives a new meaning to the term impulse turbine .

 

[pranj5]

You should lay out a cycle on an H-S or P-v diagram and then calculate the power output using the start and end states. It's impossible to answer your question without knowing something about what fluid is being used and its start and end states.

In the thread, I have clearly stated that it's 1 gm-mole of Nitrogen compressed from 1 bara pressure to 1/4th of its original volume isothermally. It can't be a cycle as the initial and final conditions aren't same.

 

[pranj5]

Certainly gives a new meaning to the term impulse turbine .

As far as I know, turbines or engines running with compressed air is now available in market.

 

[JBA]

Compressing the gas isothermally means that the heat of compression must be removed during the compression cycle. Determine the the temperature loss and you can determine the energy loss for your quantity of gas.

 

[pranj5]

As per wiki page on isothermal compression, the power consumption can be calculated by this way.

 

[russ_watters]

As per wiki page on isothermal compression, the power consumption can be calculated by this way.

The equation for the work required to isothermally compress your nitrogen is contained in that link. So please calculate it for us.

 

[pranj5]

If the temperature during compression is 27°C or 300°K and the compression ratio is 0.25 i.e. the gas/air has been compressed to 1/4th of its initial volume, then the power consumption would be 832 calorie. While if the same amount has bee expanded, then the power output would be 540 cal in ideal case.

 

[pranj5]

Sorry, I have made a mistake here. The degree of freedom for Nitrogen and other diatomic molecules are 5, while I have considered it 3. Therefore, during the expansion, the power released would be 900 cal.

 

[russ_watters]

If the temperature during compression is 27°C or 300°K and the compression ratio is 0.25 i.e. the gas/air has been compressed to 1/4th of its initial volume, then the power consumption would be 832 calorie.

I get 826. Close enough. And it's energy, not power.

While if the same amount has been expanded[snip]during the expansion, the power released would be 900 cal.

I get 486 cal. And since 900 is greater than 826, which would violate conservation of energy, it can't be right. Please post the equations/method you used to calculate it and we'll see if we can find the error.

 

[pranj5]

The formula given at wikipedia is W = αnRT₁{(V₂/V₁)^γ-1 -1}
Where, α is degrees of freedom divided by two and n is the number of moles of gas and T1 is the initial temperature. γ = (C_P/C_v) = 1.33 (for Nitrogen).
Like other diatomic gases, α for Nitrogen is 2.5 and for simplicity, we have considered n =1 i.e. just 1 gm-mole of Nitrogen is being expanded.
It has already been previously stated that (V₂/V₁) is 4. Therefore, γ-1 is 0.33 or just 1/3 and therefore (V₂/V₁)^γ-1 – 1 equals to 4^1/3 – 1 or roughly 1.6 – 1 or just 0.6.
Therefore, the power gained by the process is
= (2.5).2.300.(0.6) calorie = 900 calories
And, this isn't violation of law of conservation of energy. Let's the start the process from the expansion first. After being expanded, the temperature of the gas will fall low and it has to take heat from surrounding atmosphere to raise its temperature to atmospheric level again. That extra heat will convert into power during expansion after being isothermally compressed.

 

[russ_watters]

γ = (C_P/C_v) = 1.33 (for Nitrogen).

No, it's 7/5 = 1.4. Make that correction and you'll end up with about the same answer as you got for the compression, which isn't surprising because:

And, this isn't violation of law of conservation of energy. Let's the start the process from the expansion first. After being expanded, the temperature of the gas will fall low and it has to take heat from surrounding atmosphere to raise its temperature to atmospheric level again.

The title of the thread says "adiabatic expansion", which would mean it can't take heat from the atmosphere. The way you are defining the processes now, it is 100% reversible, making the work in and work out identical.

I calculated it with the 2nd process as originally stated (or, rather, without the 3rd process you added) and got 486 cal output.

 

[pranj5]

Well, as far as I can understand if γ-1 if 0.4 than 0.33, the value of (V₂/V₁)^γ-1 – 1 will be higher and that means the gross energy output will be higher.
Can you show me how you have calculated the output?

 

[russ_watters]

Well, as far as I can understand if γ-1 if 0.4 than 0.33, the value of (V₂/V₁)^γ-1 – 1 will be higher and that means the gross energy output will be higher.
Can you show me how you have calculated the output?

Your mixing and matching tripped me up and I think tripped you up too. Since I'm no longer sure what you processes are, I'm no longer sure I did the right calculation:

That equation is for adiabatic expansion. If you add heat, then it isn't adiabatic anymore and that equation doesn't work and you are misusing it. Specifically, you have the volume ratio wrong for adiabatic expansion: when it expands, it cools, so V₂ isn't 4 times V₁ (you have to calculate the value).

So which is it: do you want this to be adiabatic expansion or not?

 

[pranj5]

My process isn't a matter here but rather how much energy can be gained by adiabatic expansion of common diatomic gas like Nitrogen. I want to mean that if γ = 1.4, then γ-1 = 0.4 which is higher than 0.33 and that means the value of (V₂/V₁)^γ-1 – 1 would be higher and that simply means that the gross output will be higher. The process is nothing but adiabatic expansion of a diatomic gas like Nitrogen. The rest isn't necessary here.
If you want to consider the whole process, then let's start it this way. First, the gas has been isothermally compressed to 1/.4th of its volume at standard temperature and pressure from the condition of atmospheric pressure and 27°C or 300°K. The power consumption is already stated and both of us have to close figures here. Now, the gas has been released adiabatically and forced to rotate a machinery to produce power/electricity. Now, after producing power/electricity, the gas will come to the atmospheric pressure level but at far lower temperature and that means it has lost a good deal of its internal energy. In fact, the summation of this lost internal energy and the power consumption during isothermal compression exceeds the gain by adiabatic expansion and in short, it can be said that the conservation of energy isn't violated.
And kindly tell me how you have calculated the power gain during the adiabatic expansion.

 

[russ_watters]

My process isn't a matter here but rather how much energy can be gained by adiabatic expansion of common diatomic gas like Nitrogen...

Fine: adiabatic it is.

And kindly tell me how you have calculated the power gain during the adiabatic expansion.

You need to calculate V₃ first, because it isn't the same as V₁: the drop in temperature means your final volume is less than your initial volume. Then you can plug that number into your equation. That is the part you are missing.

Edit: the equation for that is TV^{Y -1}=Constant

 

[pranj5]

In the formula above, we can replace (V₂/V₁)^γ-1 with (P₁/P₂)^(γ-1)/γ and instead of 4, it will then become 1/6.4 and the outcome would be the same. And in that case, P₁ = P₃.
And, you haven't give me yet any calculation on how you have calculated the power gain.