Calculating work done by a Carnot engine

[Mimi Sanford]

"A Carnot engine operates using a heat source at 500 °C, and a heat sink at room temperature (20 °C). Suppose that as a heat source, you use the combustion of 100 cubic feet of natural gas at room temperature and pressure (e.g. in a fuel cell of some kind). Under ideal conditions, what is the largest amount of work that can be generated using this Carnot engine? (Use the ideal gas law, pV=nRT, with the gas constant R=8.3 J/(K mol).)"

I figured out that total work is equal to n * R * ln(v2/v1) * (Temp(hot) - Temp(cold)), but I have no idea how to solve for v2, especially since I assume the pressure is changing as well. Is there a way to solve for v2, or should I use a different equation? Thank you! (I also found that the efficiency is 62.1%, but I'm not sure if that's important for this question.)

 

[stockzahn]

I figured out that total work is equal to n * R * ln(v2/v1) * (Temp(hot) - Temp(cold)), but I have no idea how to solve for v2, especially since I assume the pressure is changing as well. Is there a way to solve for v2, or should I use a different equation? Thank you! (I also found that the efficiency is 62.1%, but I'm not sure if that's important for this question.)

That means that 62,1% of the (chemical) energy of the natural gas can be converted to work. If you now can calculate the energy contained in the natural gas, you find the work delivered by the machine. Do you have any information about the caloric value or the composition of the gas?

 

[Mimi Sanford]

That means that 62,1% of the (chemical) energy of the natural gas can be converted to work. If you now can calculate the energy contained in the natural gas, you find the work delivered by the machine. Do you have any information about the caloric value or the composition of the gas?

Sadly, no. That's all the info I was given.

 

[stockzahn]

Sadly, no. That's all the info I was given.

Sorry, I've also got my wires crossed right now. No heat capacities or assumptions about the number of atomes per molecule of the gas?

Regarding the work: I think it should read

$$-n R \left(T_h ln\frac{v2}{v1}+T_c ln\frac{v4}{v3}\right)$$

 

[Mimi Sanford]

Sorry, I've also got my wires crossed right now. No heat capacities or assumptions about the number of atomes per molecule of the gas?

Regarding the work: I think it should read

$$-n R \left(T_h ln\frac{v2}{v1}+T_c ln\frac{v4}{v3}\right)$$

Nope. Based on the ideal gas law, I believe there are 117.9 moles, but that's all I've got. Also, I simplified $$-n R \left(T_h ln\frac{v2}{v1}+T_c ln\frac{v4}{v3}\right)$$ because v2/v1 equals v3/v4 according to my textbook.

 

[stockzahn]

Maybe you can consider the gas mixture as air and assume an isentropic coefficient of 1.4? Would you have an idea how to proceed with this information?

 

[stockzahn]

Also, I simplified $$-n R \left(T_h ln\frac{v2}{v1}+T_c ln\frac{v4}{v3}\right)$$ because v2/v1 equals v3/v4 according to my textbook.

That's kind of a strange assumption, does your workbook say, if it can be treated as ideal or as perfect gas?

 

[Mimi Sanford]

I think so, cause it says to use the ideal gas law. There's a whole proof for the volume ratio assumption, but it's super long. Do you think I should calculate the work using a different formula? I just can't think of how to do so without volume. And as for your first question - no, we're not very far along in the course yet.

 

[stockzahn]

I think so, cause it says to use the ideal gas law. There's a whole proof for the volume ratio assumption, but it's super long. Do you think I should calculate the work using a different formula? I just can't think of how to do so without volume. And as for your first question - no, we're not very far along in the course yet.

I don't know the volume ratio assumption, I have to look that up.

Normally, to burn a gas, you need a certain amount of air for the combustion. In many cases it is allowed to treat the burning/burnt gas mixture as pure air - without making a large mistake. The so-called isentropic coefficient/exponent for air as ideal gas is $\kappa=1.4$. With this information (plus the ideal gas law) you can calculate the states of the four corners of your Carnot process - and consequently the volumes.

For an isentropic change of state

$$\frac{v_2}{v_1}=\left(\frac{p_1}{p_2}\right)^{1/\kappa}=\left(\frac{T_1}{T_2}\right)^{1/(\kappa-1)}$$

For an isothermal change of state

$$\frac{v_2}{v_1}=\frac{p_1}{p_2}$$ and $$T=const.$$

 

[Mimi Sanford]

I don't know the volume ratio assumption, I have to look that up.

Normally, to burn a gas, you need a certain amount of air for the combustion. In many cases it is allowed to treat the burning/burnt gas mixture as pure air - without making a large mistake. The so-called isentropic coefficient/exponent for air as ideal gas is $\kappa=1.4$. With this information (plus the ideal gas law) you can calculate the states of the four corners of your Carnot process - and consequently the volumes.

For an isentropic change of state

$$\frac{v_2}{v_1}=\left(\frac{p_1}{p_2}\right)^{1/\kappa}=\left(\frac{T_1}{T_2}\right)^{1/(\kappa-1)}$$

For an isothermal change of state

$$\frac{v_2}{v_1}=\frac{p_1}{p_2}$$ and $$T=const.$$

So would I take T1 to be 773 Kelvin, T2 to be 293 Kelvin, and kappa to be 1.4? And then solve for v2? Or am I mixing my variables up? I think I might be. Which volume would correspond to 100 ft^3?

 

[Mimi Sanford]

My Professor just posted this on his website, because apparently a lot of other people are confused, too:

"In the last question, the engine is burning up the methane. Figure out how much heat (energy) this burning produces, using the enthalpy of the reaction. That gives you a total heat input into your engine. You write that you already have figured out the efficiency. So then, with the total heat input, and the efficiency you have, you should be able to calculate the maximum work output."

I suppose we were meant to assume that the engine burns methane, which makes sense. Whoops. Thank you so much for your help!

 

[stockzahn]

So would I take T1 to be 773 Kelvin, T2 to be 293 Kelvin, and kappa to be 1.4? And then solve for v2?

I would define the state 1 as the initial state:
$T_1=293 K$
$p_1=10^5 Pa$
$V_1=2.83m^3$

About state 2 we know
$T_2=773 K$

$p_2$ and $V_2$ can be calculated by the correlations given in the previous post.

Which volume would correspond to 100 ft^3?

The $100 ft^3$ must be converted to $m^3$.

 

[stockzahn]

My Professor just posted this on his website, because apparently a lot of other people are confused, too:

"In the last question, the engine is burning up the methane. Figure out how much heat (energy) this burning produces, using the enthalpy of the reaction. That gives you a total heat input into your engine. You write that you already have figured out the efficiency. So then, with the total heat input, and the efficiency you have, you should be able to calculate the maximum work output."

I suppose we were meant to assume that the engine burns methane, which makes sense. Whoops. Thank you so much for your help!

Well, then easier than expected. Once you will have calculated that within the next minutes, there maybe is time to complete the calculation we just started.
Good night from CET!

 

[Chestermiller]

I would define the state 1 as the initial state:
$T_1=293 K$
$p_1=10^5 Pa$
$V_1=2.83m^3$

About state 2 we know
$T_2=773 K$

$p_2$ and $V_2$ can be calculated by the correlations given in the previous post.
The $100 ft^3$ must be converted to $m^3$.

The gas constant is 1.31443 atm ft^3/K.lb-mole. So it doesn't have to be converted to m^3. This can be used to get the number of lb-moles of natural gas.

 

[Chestermiller]

I might also mention that only a fraction of the heat of combustion of natural gas at 20 C will be available for the high temperature source, since, at most, the reaction products can only be cooled in the high temperature source to 500 C. Therefore, the amount of heat required to raise the temperature of the products (including the nitrogen associated with a stoichiometric mixture) to 500 C must be subtracted from the heat of combustion. Also, the lower heating value must be used in the calculation, since the water in the product mixture will not be condensed.