Heat engine that uses a rubber band (Thermodynamics)

In summary, the conversation discusses the calculation of efficiency for a heat engine using a rubber band as an example. The path from point a to b is isothermal and the values of T_a and T_b can be found using the equation of state and the diagram. The equations for the paths from a to b, b to c, and c to a are given, and the total work and heat absorbed are calculated. The efficiency is then derived and is found to depend on temperature. The conversation ends with a request for confirmation on the correctness of the efficiency calculation.
  • #1
Pedro Roman
10
0
Homework Statement
Find the efficiency of a three-step cycle heat engine that uses a rubber band (the cycle is shown in the figure), which idealized equation of state is J= aLT (J is the tension, L is the length of the rubber band per unit mass, a is a constant and T is the temperature). The specific heat capacity is a constant, c. Find the efficiency of this rubber band engine.
Relevant Equations
dQ=TdS=(dS/dT)dT+(dS/dL)dL
Screen Shot 2020-08-01 at 9.58.32 PM.png

Nota that the path a to b is isothermal since J is proportional to L, then we can find the value of T_a and T_b using the equation of state and the figure. We have,

\begin{equation}
J_0=\alpha L_0T_b
\end{equation}
or
\begin{equation}
T_b=T_a=\frac{J_0}{\alpha L_0}=T_0
\end{equation}

Also, by using the equation of state we find,

\begin{equation}
T_c=T_0/2
\end{equation}

Now, for the path a to b, we have that

\begin{equation}
\begin{split}
dQ&=T_0dS=T_0\left(\frac{\partial S}{\partial L}\right)_T dL\\
dQ&=-T_0\left(\frac{\partial J}{\partial T}\right)_L dL\\
dQ&=-\alpha LM T_0dL\\
\end{split}
\end{equation}
where M is the mass of the rubber band. Therefore,
\begin{equation}\Delta Q_{ab}=\frac{3L_0J_0M}{2}.\end{equation}

For b to c, we have
\begin{equation}
\begin{split}
dQ=& T\left(\frac{\partial S}{\partial T}\right)_LdT \\
+& T\left(\frac{\partial S}{\partial L}\right)_TdL\\
dQ=&cM dT-\alpha LMT dL\\
dQ=&cM dT-J_0 dL\\
\end{split}
\end{equation}
where I have used the equation of state. Now the change of heat is given by
\begin{equation}\Delta Q_{bc}=-\frac{McT_0}{2}-J_0L_0M.\end{equation}
Finally for c to a we have dL=0 and
\begin{equation}
\begin{split}
dQ=& T\left(\frac{\partial S}{\partial T}\right)_L dT \\
dQ=&cM dT\\
\end{split}
\end{equation}
Then, we find,
\begin{equation}
\Delta Q_{ca}=\frac{McT_0}{2}.
\end{equation}

The total work must be (since for a full cycle dU=0),
\begin{equation}
\begin{split}
\Delta W_{tot}&=\Delta Q_{tot}\\
\Delta W_{tot}&=\Delta Q_{ab}+\Delta Q_{bc}+\Delta Q_{ca}\\
\Delta W_{tot}&=\frac{J_0L_0M}{2}.
\end{split}
\end{equation}
The absorbed heat must be the heat changes that are positive (right?), then,
\begin{equation}
\begin{split}
\Delta Q_{abs}&=\Delta Q_{ab}+\Delta Q_{ca}\\
\Delta Q_{abs}&=\frac{3}{2}J_0L_0M+\frac{McT_0}{2}
\end{split}
\end{equation}
At last, we obtain the efficiency,
\begin{equation}
\begin{split}
\eta &=\frac{\Delta W_{tot}}{\Delta Q_{abs}}\\
\eta &=\frac{1}{3+\frac{cT_0}{J_0L_0}}.
\end{split}
\end{equation}
So, I would thank you if you could tell me if that efficiency is correct, I saw some pages solving the same problem that find the efficiency to be 1/3, so I am not sure about my calculation. Thanks in advance.
 
Last edited:
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  • #2
Not sure of the "work" this rubber band is doing or why anyone would call it a "heat engine". Never saw an efficiency calculation that took mass into account. Thermodynamic efficiency is proportional to delta T/T and your "efficiency" has no temperature dependence. Your math goes off of my screen.
 
  • Like
Likes Pedro Roman
  • #3
Let just say that we have a cycle as the one showed in the figure, which follows the following equation of state, J=aLT, and for that let us calculate the total work and the heat absorbed, in that sense one is able to compute the 'efficiency' of the 'heat engine'. This is just a textbook problem. The mass appears due to the fact that the heat capacity at constant length (c) and the length (L) have the following units [J/kg K] and [m/kg], respectively. As you said, the 'efficiency' must depend on the temperature, now I have written the efficiency in term of it. Also, I have split all equations to avoid they go off the screen. Thanks for your answer.
 

Related to Heat engine that uses a rubber band (Thermodynamics)

1. How does a heat engine using a rubber band work?

A heat engine using a rubber band works by converting thermal energy into mechanical energy. The rubber band is stretched and then allowed to contract, causing it to perform work. This process is repeated by heating and cooling the rubber band, resulting in a continuous conversion of heat into mechanical energy.

2. What is the principle behind a heat engine using a rubber band?

The principle behind a heat engine using a rubber band is based on the second law of thermodynamics, which states that heat will naturally flow from a hotter object to a cooler object. By utilizing this principle, the rubber band can convert heat into mechanical energy through its natural tendency to contract when heated and expand when cooled.

3. What materials are needed to build a heat engine using a rubber band?

To build a heat engine using a rubber band, you will need a rubber band, a heat source (such as a flame or hot water), a cooling source (such as ice or cold water), and a mechanism to stretch and release the rubber band (such as a pulley system).

4. Can a heat engine using a rubber band be used as a source of energy?

While a heat engine using a rubber band can convert thermal energy into mechanical energy, it is not a practical source of energy. The amount of energy produced is relatively small and the process is not efficient. It is primarily used as a demonstration of thermodynamic principles.

5. How does a heat engine using a rubber band differ from other heat engines?

A heat engine using a rubber band differs from other heat engines in that it does not use a traditional fuel source, such as gasoline or coal. Instead, it utilizes the natural properties of the rubber band to convert thermal energy into mechanical energy. It also operates on a much smaller scale and is not designed for practical use as a source of energy.

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