Work done as air expands in piston with spring

In summary, the problem involves warm air in a piston-cylinder assembly cooling from an initial volume of .003 m^3 to a final volume of .002m^3. A spring exerts a force that varies from 900N to 0N during the process, and there is no friction. The initial and final pressures were found to be 150 kPa and 100 kPa respectively. When calculating the work done, the assumption was made that the magnitude of the work done by the spring is equal to the work done by the air. However, this is not entirely correct and the correct method involves calculating the pressure as a function of the distance the spring is compressed and then integrating to find the work done. The final answer
  • #1
Giuseppe
42
0
There's a problem in my thermo class that I think I have the right answer for, but the value I get doesn't seem to make sense. The problem reads:

Warm air is contained in a piston-cyliner assembly oriented horizontally. The air cools slowly from an intial volume of .003 m^3 to a final volume of .002m^3. During the process, a spring exerts a force that varies linearly from an intial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is .018m^2. Friction between the piston and the cylinder wall can be neglected. For the air, determine the intial and final pressures, in kPa, and the work in kJ.

So I split the system up into two states.

State 1: V1 = .003 m^3 Fspring = 900N
State 2: v2 = .002 m^3 Fspring = 0N

I found the pressures for each state using a force balance equation and the relationship between force and pressure.

I came up with:

Fair = Fspring + Farm
(Pair)(Apiston) = Fspring + (Pair)(Apiston)

Using the values for State 1 and 2 I came up with 150 kPa for State 1 and 100 kPa for state 2.

Here's where I run into problems. Calculating the work.

First off I make the assumption that the magnitude of the work done as the spring moves is equal to the magnitude of the work the air does during this change.

Therefore, I used 1/2kx^2 for work done. I found the distance the spring moved using (V2-V1)/Apiston = delta X

Using this x value I went to Fspring = 900 = kx and found k.

From here I used evaluated 1/2kx^2 and I came up with an answer of 25Joules.

My question is, is the answer I getting make sense physically, because it seems like a really small answer, also are the assumptions I am making for work done correct, or is there another way I should go about this.
 
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  • #2
Giuseppe said:
There's a problem in my thermo class that I think I have the right answer for, but the value I get doesn't seem to make sense. The problem reads:

Warm air is contained in a piston-cyliner assembly oriented horizontally. The air cools slowly from an intial volume of .003 m^3 to a final volume of .002m^3. During the process, a spring exerts a force that varies linearly from an intial value of 900N to a final value of zero. The atmospheric pressure is 100kPa, and the area of the piston face is .018m^2. Friction between the piston and the cylinder wall can be neglected. For the air, determine the intial and final pressures, in kPa, and the work in kJ.

So I split the system up into two states.

State 1: V1 = .003 m^3 Fspring = 900N
State 2: v2 = .002 m^3 Fspring = 0N

I found the pressures for each state using a force balance equation and the relationship between force and pressure.

I came up with:

Fair = Fspring + Farm
(Pair)(Apiston) = Fspring + (Pair)(Apiston)

Using the values for State 1 and 2 I came up with 150 kPa for State 1 and 100 kPa for state 2.

That's correct.
Giuseppe said:
Here's where I run into problems. Calculating the work.

First off I make the assumption that the magnitude of the work done as the spring moves is equal to the magnitude of the work the air does during this change.

Therefore, I used 1/2kx^2 for work done. I found the distance the spring moved using (V2-V1)/Apiston = delta X

Using this x value I went to Fspring = 900 = kx and found k.

From here I used evaluated 1/2kx^2 and I came up with an answer of 25Joules.

My question is, is the answer I getting make sense physically, because it seems like a really small answer, also are the assumptions I am making for work done correct, or is there another way I should go about this.

Now, while calculating the work done, you have to be careful. First of all, what work is the question asking for? I'll assume that the air in the cylinder is the system, and the work done on the system (by the surroundings, ie the atmosphere and the spring) in this process is what is required.

The only work done on the system is expansion work, which is [tex] \int P dv [/tex]. So, let us say that the spring is compressed by a distance 'x'. Can you calculate the pressure in the cylinder as a function of 'x'? Note that this process has to be reversible, which means that the net force on the piston is always 0.

Once you do this, you will be able to calculate the work done, as 'x' goes from the initial value, to 0. Also, watch your signs when you calculate the final answer.
 
Last edited:
  • #3


I would say that your approach and assumptions are generally correct. However, there are a few factors that could be affecting your final answer.

First, it's important to make sure that all of your units are consistent. In your calculation for work, you used the distance in meters and the force in Newtons, but you should also convert the spring constant k from N/m to N/m^2 to match the units of pressure. This would result in a larger value for k and therefore a larger value for work.

Second, it's possible that there are other forms of energy involved in this process that you may not have accounted for, such as thermal energy. This could affect the overall work done by the system and could explain the small value you obtained.

Lastly, it's always a good idea to double check your calculations and make sure you haven't made any errors. It's possible that a simple mistake in your calculations could result in a significantly different answer.

Overall, your approach seems sound and it's possible that the small value you obtained is due to one of these factors. I would recommend checking your units and double checking your calculations to see if there are any errors. If you still have concerns, it may be helpful to discuss your approach with your instructor or a classmate to see if they have any insights.
 
  • #4


I would suggest double-checking your calculations and assumptions to ensure that your answer is accurate. It is possible that there may be a mistake in your calculations or a factor that you may have overlooked.

In terms of the physical sense of your answer, it is possible that the work done by the air in this system is relatively small. This could be due to the fact that the air is only expanding a small amount and the force of the spring is decreasing linearly. It may be helpful to compare your calculated work value to other similar systems to see if it is within a reasonable range.

Additionally, it is important to consider the assumptions you are making in your calculations. While it is reasonable to assume that the work done by the air is equal to the work done by the spring, it is always good to double-check and consider any potential factors that may affect the accuracy of this assumption.

Overall, it is important to carefully evaluate your calculations and assumptions to ensure that your answer is accurate and makes sense physically. If you are still unsure, it may be helpful to seek guidance from a professor or colleague.
 

1. What is work done as air expands in a piston with spring?

Work done as air expands in a piston with spring is the amount of energy required to push the air molecules against the resistance of the piston and the spring. This work is typically measured in joules (J) and is a result of the force applied and the distance the air expands.

2. How is work calculated in this scenario?

The work done as air expands in a piston with spring can be calculated using the formula W = F * d, where W is work, F is the force applied, and d is the distance the air expands. This calculation assumes that the force remains constant throughout the expansion process.

3. What factors affect the amount of work done in this process?

The amount of work done as air expands in a piston with spring is affected by several factors, including the initial pressure of the air, the volume of the piston, and the spring constant of the spring. The greater the pressure and volume, and the stiffer the spring, the more work will be required to expand the air.

4. How does work done in this scenario relate to the first law of thermodynamics?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted. In the case of work done as air expands in a piston with spring, the energy put into expanding the air is converted into potential energy stored in the compressed spring. This potential energy can then be released and converted back into work when the air is allowed to contract.

5. What are some real-world applications of this concept?

The concept of work done as air expands in a piston with spring has applications in various fields, such as engineering, physics, and thermodynamics. It is used in the design of engines, air compressors, and other mechanical devices. It also plays a crucial role in understanding and optimizing the efficiency of heat engines, such as car engines and power plants.

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