Applying first law of thermo to a closed spring piston

In summary, a cylinder with a piston restrained by a linear spring (with a spring constant of 15 kN/m) contains 0.5 kg of saturated vapor water at 120ºC. Heat is transferred to the water causing the piston to rise. The final pressure reaches 500 kPa and the final temperature is approximately 875°C. The change in internal energy is 638.18 KJ, the work done against the spring is 29.1973 KJ, and the heat transfer is 617.475 KJ. The system undergoes a large increase in pressure and volume, resulting in positive values for energy change, work, and heat transfer.
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Homework Statement


A cylinder having a piston restrained by a linear spring (of spring constant 15 kN/m) contains 0.5 kg of saturated vapor water at 120ºC. Heat is transferred to the water causing piston to rise. If the piston’s cross sectional area is 0.05 m2 and the pressure varies linearly with volume until a final pressure of 500 kPa is reached,

1-Find the final temperature in the cylinder.

2-Calculate the amount of heat transfer.

3-Calculate the change in the internal energy of water.

4-Calculate the work is done against the spring.

5-Sketch the process on a T-v phase diagram.

Homework Equations


ΔU=Q+W
W=∫ PdV
F=kx , linear spring displacement equation
v=V/m
u=U/m

The Attempt at a Solution



So I'd like some fresh opinions to just verify that my approaches are sensible and I correctly understand everything here. Firstly - I'm imagining this system initially as a "saturated vapor of 100% steam quality".

Finding the final temperature

Using the given temperature I look in the steam tables for saturated water and find a specific volume of vg=.891 m3/kg. I multiply this by the mass
.5 kg to get an initial volume of .4459 m3. I then use the final pressure
and area of the piston face (to find force) with the spring equation to find that the spring is compressed by 1.666m. This times the piston area gives a newly added volume of .529 m3. I add this to the original volume to find new total volume. I find the new specific volume, and
look in the steam tables to verify that it is greater than saturation specific
volume of gas at the new final temperature, confirming the system ends
up as a superheated gas. I look in the appropriate steam table to locate
this new specific volume and final pressure, finding that the temperature is approximately 875 C after some extrapolation.

Finding the change in Internal Energy ΔU
Going back to the steam table for the initial state, I find that the specific internal energy ug (since it's 100% gas), is 2529.24 KJ/kg. Using the
mass, I find an initial internal energy U1 of 1264.6 KJ. Using the data
for the final state found earlier in the superheated region, I find a new specific internal energy of 3805.7 KJ/kg. Using the constant mass,
I find a final internal energy U2 of 1902.8 KJ. The change in internal
energy ΔU is therefore 638.18 KJ.


Finding the amount of Work done
I make a line graph of Pressure vs Volume, plotting the initial and final
states. I found the initial pressure by looking at the saturated steam
tables, finding it to be 198.5 KPa. Using the listed equation for finding
Work, I understand that the Work is equal to the area under the curve of
P vs V. However, because I don't have two endpoints and the function
is linear, instead of determining the function relating P and V and integrating it I just segment the area under the curve into a triangle and a
rectangle. I find the area of each and sum them, coming out to a Work of
20.7 KJ.
It's positive because the system has expanded and done work
against its surroundings. (left out all the numbers for sake of brevity/clarity)

Finding the heat transfer Q
To find Q, I rearrange the first law of thermodynamics to find that
Q=ΔU-W. Then simply use the values of ΔU and W from earlier to find
a Q of 617.475 KJ.

To sum up - I expected energy change, work, and heat transfer to be positive as the system underwent a large increase in pressure while increasing in volume. These numbers make sense to me, at least as far as their signs. If you see any glaring issues with my approach to the problem please let me know. Any and all advice is appreciated, thank you.
 
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I got a work of 29.1973 KJ

Area Under Curve=0.5(P1+P2)(V2-V1)=(0.5)(0.529-0.4454)(500-198.5)=29.1973 KJ
 
  • #3
RyanFrancis said:
I got a work of 29.1973 KJ

Area Under Curve=0.5(P1+P2)(V2-V1)=(0.5)(0.529-0.4454)(500-198.5)=29.1973 KJ

I think you're right. I think since I set the baseline of my pressure axis at 100 KPa, I used 98.5 as the height of the rectangle instead of 198.5. Thanks for catching that mistake!
 

FAQ: Applying first law of thermo to a closed spring piston

1. What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another.

2. How does the first law of thermodynamics apply to a closed spring piston?

The first law of thermodynamics can be applied to a closed spring piston by considering the energy input and output of the system. As the spring is compressed, work is done on the system, increasing its internal energy. When the spring is released, this energy is converted into mechanical work, such as moving a piston.

3. What is the role of entropy in the first law of thermodynamics?

Entropy is a measure of the disorder or randomness in a system. The first law of thermodynamics accounts for changes in the internal energy of a system, but it does not account for changes in entropy. This means that while energy is conserved, the overall disorder of a system can increase over time.

4. How can the first law of thermodynamics be used to calculate the work done by a closed spring piston?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat input minus the work done by the system. In the case of a closed spring piston, the work done by the system can be calculated by measuring the change in internal energy and subtracting the heat input.

5. What are the limitations of applying the first law of thermodynamics to a closed spring piston?

The first law of thermodynamics assumes that the system is in a state of equilibrium and that there is no energy lost to friction or other forms of energy transfer. In reality, there may be some energy lost to heat due to the compression and expansion of the spring, resulting in a less efficient conversion of energy.

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