Work done by a expanding and contracting wire over a Carnot cycle

In summary, the student is trying to calculate the work done over a Carnot cycle but is getting stuck. He has tried using the first law of thermodynamics and the equation of state, but is still having trouble. He has also tried using the work done over the isoterm curves and the work done over the adiabatic processes, but is still stuck. He has found that using the first law and the equation of state is enough to calculate the work done over the isotherm and adiabatic processes, but is having trouble with the adiabatic cycle.
  • #1
Uriel
16
0

Homework Statement



Basically I have the following problem. Given the equation of state [itex]\sigma = \frac{b}{T}\frac{L - L_0}{L_0}[/itex] where b and [itex]L_0[/itex] are positive constants, calculate the work done over a Carnot cycle by a wire with this equation of state.

Homework Equations

I already have the work done over the isoterm curves. But when I try to calculate the work on the adiabatic processes I just get stuck over and over again.

The Attempt at a Solution



I use the first law of thermodynamics

[itex] du = \delta Q + \delta W[/itex]

also we know that the work for this system is the following

[itex]\delta W = - \tau dL[/itex]

and in a book of basic thermodynamics I have found that

[itex]dT = -\frac{T}{c_\tau}\left( \frac{\partial L}{\partial T} \right)_\tau d\tau[/itex]

Any ideas?
 
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  • #2
Hi Uriel. Welcome to PF!

Uriel said:
I already have the work done over the isoterm curves. But when I try to calculate the work on the adiabatic processes I just get stuck over and over again.
Can you give more details about what you've tried?

Uriel said:
also we know that the work for this system is the following

[itex]\delta W = - \tau dL[/itex]
What is ##\tau##?
 
  • #3
DrClaude said:
Hi Uriel. Welcome to PF!


Can you give more details about what you've tried?


What is ##\tau##?

[itex] \tau [/itex] is the [itex] \sigma [/itex] in the equation of state, it's linear tension, I change the name of the variable because [itex] \sigma [/itex] might be confused by strains and that's not the case, I'm sorry for that.

Using the first law on the isotherm process we got that the internal energy doesn't change, therefore

[itex] d u = 0[/itex]

and using this [itex] \delta W = -\delta Q [/itex]

now we know that the work is computed with the integral [itex] \delta W = - \int \tau dL [/itex]
so it's really simple to calculate, for the case on the two isotherm processes.

My problem arises when I try to calculate the work performed over the adiabatic cycles. I have two options, integrate over the path of the adiabatic curve from one state to the other, or I need to know the internal energy of the system and using that plug in the temperature of each state and calculate the difference.
That is the awful part; given the equation [itex] dT = - \frac{T}{c_\tau} \left( \frac{\partial L}{\partial T} \right)_\tau d \tau [/itex]

I need to know L, but this is easy, [itex] L = \frac{L_0 T}{b}\tau + L_0 [/itex],

next I calculate the partial derivative of L with respect to T leaving [itex] \tau [/itex] constant.

So [itex]\left( \frac{\partial L}{\partial T} \right)_\tau = \frac{L_0 \tau}{b} [/itex]

then [itex]dT = - \frac{T}{c_\tau} \left( \frac{\partial L}{\partial T} \right)_\tau d \tau = - \frac{T}{c_\tau} \frac{L_0 \tau}{b} d \tau[/itex].

Reordering the terms in the differential equation we get

[itex] \frac{dT}{T} = - \frac{L_0}{c_\tau b} \tau d \tau [/itex]

solving the differential equation we get this

[itex] ln \left( \frac{T_f}{T_i} \right) = - \frac{L_0}{c_\tau b} \left( \frac{\tau_f^2 - \tau_i^2}{2} \right) [/itex]

And that's all I got, I don't have the slightest idea what to do next.
 
  • #4
In which book did you see that problem?
 
  • #5
Miroslava said:
In which book did you see that problem?

I'm not sure it was a book problem. I think the professor just made it up, but she was very fond of the book of Callen.
 

FAQ: Work done by a expanding and contracting wire over a Carnot cycle

1. What is work done by a expanding and contracting wire over a Carnot cycle?

The work done by a expanding and contracting wire over a Carnot cycle refers to the amount of energy transferred by the wire as it expands and contracts during a Carnot cycle. This energy transfer is typically measured in joules (J) and is a result of the expansion and contraction of the wire due to changes in temperature.

2. How is work done by a expanding and contracting wire calculated?

The work done by a expanding and contracting wire is calculated using the formula W = -∫Fdx, where W is the work done, F is the force applied, and dx is the displacement of the wire. This formula takes into account the change in force and displacement as the wire expands and contracts, resulting in a net work done over the entire cycle.

3. What factors affect the work done by a expanding and contracting wire over a Carnot cycle?

The work done by a expanding and contracting wire over a Carnot cycle is affected by several factors, including the initial length and temperature of the wire, the change in temperature during the cycle, and the properties of the wire such as its material and cross-sectional area. These factors determine the amount of expansion and contraction of the wire, which in turn affects the work done.

4. How does the work done by a expanding and contracting wire over a Carnot cycle relate to thermodynamics?

The work done by a expanding and contracting wire over a Carnot cycle is closely related to the principles of thermodynamics. The Carnot cycle is a theoretical model that follows the laws of thermodynamics and is used to analyze the efficiency of heat engines. The work done by the wire is a result of the transfer of heat energy, which is a fundamental concept in thermodynamics.

5. What are some practical applications of understanding the work done by a expanding and contracting wire over a Carnot cycle?

Understanding the work done by a expanding and contracting wire over a Carnot cycle has practical applications in various fields. For example, it can be applied in the design and optimization of heat engines, as well as in the development of materials with specific thermal properties. It can also be used in the study of thermodynamics and its applications in various industries, such as energy production and refrigeration systems.

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