Carnot Engine: A Firebox at 750K, Efficiency 60.0%: Part (a) & (b) Explained

[Gear300]

A firebox is at 750K, and the ambient temperature is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150J/.600 = 250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that some other heat engine S could have efficiency 70.0%.

(a) Find the energy input and wasted energy output of energy S as it does 150J of work.
(b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated.

Part (a) was done relatively easily but I have no idea what situation part (b) is bringing up (its a problem in conceptualizing what they're asking). How is engine S operating with the Carnot engine?

 

[alphysicist]

A firebox is at 750K, and the ambient temperature is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150J/.600 = 250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that some other heat engine S could have efficiency 70.0%.

(a) Find the energy input and wasted energy output of energy S as it does 150J of work.
(b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated.

Part (a) was done relatively easily but I have no idea what situation part (b) is bringing up (its a problem in conceptualizing what they're asking). How is engine S operating with the Carnot engine?

I believe it means to operate the Carnot engine as a heat pump, letting the output of work by engine S be the input of work to the Carnot heat pump. After following the process, do you see how Clausius's statement is violated?

 

[Gear300]

I believe it means to operate the Carnot engine as a heat pump, letting the output of work by engine S be the input of work to the Carnot heat pump. After following the process, do you see how Clausius's statement is violated?

Somewhat...the work being done on the heat pump is internal to the system, whereas the Clausius statement does not allow for such cyclic processes to continue constantly without external work being applied to the system. However, if the work done by engine S is applied to the heat pump to bring the exhaust energy back to the hot reservoir (while ignoring friction, conduction, or any unnecessary dissipation of energy), then how and where in the process is the limitation mentioned in the Clausius statement produced?

 

[alphysicist]

Somewhat...the work being done on the heat pump is internal to the system, whereas the Clausius statement does not allow for such cyclic processes to continue constantly without external work being applied to the system. However, if the work done by engine S is applied to the heat pump to bring the exhaust energy back to the hot reservoir (while ignoring friction, conduction, or any unnecessary dissipation of energy), then how and where in the process is the limitation mentioned in the Clausius statement produced?

Let the engine S produce 150 J of work, and the Carnot engine use 150 J of work. What's the total (net) change in this closed system? In other words, what has been accomplished?

 

[Gear300]

Let the engine S produce 150 J of work, and the Carnot engine use 150 J of work. What's the total (net) change in this closed system? In other words, what has been accomplished?

Hmmm...if I'm thinking in the right direction...then there is no net work done throughout the process...I see...and that violates the Clausius statement...heh, well that's something.

Also, if its not too much of a bother...in part (b)...whats the difference between the first find and the second find. Both of them are -35J (according to the answer given). I know the second is exhaust energy...what is the first one supposed to be?

 

[alphysicist]

Hmmm...if I'm thinking in the right direction...then there is no net work done throughout the process...I see...and that violates the Clausius statement...heh, well that's something.

Well, I think you need a bit more. No work has been done, true, but what has been accomplished? What has occurred due to the two of them running?

What is the Clausius statement? If you consider what it states, it might be easier to focus on the relevant happening.

Also, if its not too much of a bother...in part (b)...whats the difference between the first find and the second find. Both of them are -35J (according to the answer given). I know the second is exhaust energy...what is the first one supposed to be?

I'm a bit confused by the way you're using the word "find" here, but from the number you give, it appears that you are looking at the input/output effects of the total combined machine. This relates to answering my questions above.

 

[Gear300]

Well, I think you need a bit more. No work has been done, true, but what has been accomplished? What has occurred due to the two of them running?
What is the Clausius statement? If you consider what it states, it might be easier to focus on the relevant happening.

Would this fulfill it?: There is a cyclical process that is now continuous and working by itself...and so the entropy change for one cycle is 0. The Clausius statement indicates that such an accomplishment is not possible.
To stray a little...why is the Clausius statement true? I'm not denying it...I'm just asking how it is true. Apparently, according to the statement, even if a system did not hold any friction or any other dissipation of energy, a limitation will still exist and external work is needed. What is the source of the limitation?

 

[alphysicist]

Would this fulfill it?: There is a cyclical process that is now continuous and working by itself...and so the entropy change for one cycle is 0. The Clausius statement indicates that such an accomplishment is not possible.

What is the Clausius statement? For example the Kelvin statement of the second law goes something like "No process can convert heat completely into work." What is the Clausius statement of the second law?

(I'd rather not just write it out here, because it kind of gives away this problem. If you can't find it in your text you can find it on the web.)


Also, why do you say the entropy change is zero? Try calculating it for a single cycle. What do you get?

 

[Gear300]

What is the Clausius statement? For example the Kelvin statement of the second law goes something like "No process can convert heat completely into work." What is the Clausius statement of the second law?

Also, why do you say the entropy change is zero? Try calculating it for a single cycle. What do you get?

According to text, the Clausius statement states that "it is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work." Wouldn't that be somewhat similar to my explanation...or am I still missing something?

The reason why I stated that the entropy of one cycle would be zero would be that I was assuming that the entire process would be reversible...although, looking back at things, my reasoning in that was off.

 

[alphysicist]

According to text, the Clausius statement states that "it is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work." Wouldn't that be somewhat similar to my explanation...or am I still missing something?

When the problem directly references the Clausius statement, I think you would need to constuct your answer in the language of the Clausius statement. You have found that the total effect of the combined device is to transfer 35 J (or something like that) of heat from the cold reservoir to the hot reservoir, and there is no other effect. That specifically is what the Clausius statement says is impossible.

You could also think about the problem in terms of entropy, but in my opinion the problem was specifically wanting you to think about what Clausius had to say.

The reason why I stated that the entropy of one cycle would be zero would be that I was assuming that the entire process would be reversible...although, looking back at things, my reasoning in that was off.

What did you get for the change in entropy?

 

[Gear300]

What did you get for the change in entropy?

Sorry it took so long to respond...it took me a while to get the question back in front of me. I understand the problem now. I wasn't conceptualizing the process right. Apparently there is a net transfer of heat energy of 35.7J from the cold reservoir to the hot reservoir without any net external work, which violates the Clausius statement.

Since you brought up the matter, wouldn't I need to know the medium I'm using (the type of gas mixture, amount, etc...) to solve for the change in entropy (considering that the heat energy shifted from a colder temperature to a higher temperature)?

 

[alphysicist]

Sorry it took so long to respond...it took me a while to get the question back in front of me. I understand the problem now. I wasn't conceptualizing the process right. Apparently there is a net transfer of heat energy of 35.7J from the cold reservoir to the hot reservoir without any net external work, which violates the Clausius statement.

Since you brought up the matter, wouldn't I need to know the medium I'm using (the type of gas mixture, amount, etc...) to solve for the change in entropy (considering that the heat energy shifted from a colder temperature to a higher temperature)?

Since you know that the temperatures of the reservoirs is constant, and you know the heat transfer to and from each reservoir, you can simplify

$$\Delta S = \int \frac{dQ}{T}$$

directly for each reservoir.

 

[Gear300]

Since you know that the temperatures of the reservoirs is constant, and you know the heat transfer to and from each reservoir, you can simplify

$$\Delta S = \int \frac{dQ}{T}$$

directly for each reservoir.

Oh...I see...and the sum would give total change in entropy for the system of the 2 reservoirs, right?

 

[alphysicist]

Oh...I see...and the sum would give total change in entropy for the system of the 2 reservoirs, right?

That sounds right; and once you calculate the number, it should give more evidence as to why that combined device is impossible.

 

[Gear300]

I see, I see...the entropy is apparently decreasing for this kind of device, making it impractical...thanks for the help.

 

[stewartcs]

I see, I see...the entropy is apparently decreasing for this kind of device, making it impractical...thanks for the help.

Just wanted to point out that the entropy change of a system may very well decrease due to heat transfer from the system for a reversible process. It is the entropy generated in the system that may never decrease. It can only remain zero in the case of a reversible process. Hence, if the entropy transfer is negative (i.e. out of the system), the system entropy will decrease.

CS

 

[Andrew Mason]

Just wanted to point out that the entropy change of a system may very well decrease due to heat transfer from the system for a reversible process. It is the entropy generated in the system that may never decrease. It can only remain zero in the case of a reversible process. Hence, if the entropy transfer is negative (i.e. out of the system), the system entropy will decrease.

CS

This is a bit confusing. It is not clear what you mean by "system".

A consequence of the second law is that the entropy of the universe cannot decrease in any process. If the system includes the reservoirs, then it is not correct to say that the entropy of the system can decrease (since this is a closed system with no external work being supplied).

It is also unclear what you mean by "entropy generated in the system". Work is generated and heat flows in these thermodynamic processes. As a result of heat flowing over one full cycle, the entropy of each of the reservoirs changes (by dQ/T where dQ is the heat flow into/out of the reservoir and T is the temperature of the reservoir). The sum total entropy of those changes cannot be less than 0 if no external work is being supplied.

AM

 

[stewartcs]

This is a bit confusing. It is not clear what you mean by "system".

A consequence of the second law is that the entropy of the universe cannot decrease in any process. If the system includes the reservoirs, then it is not correct to say that the entropy of the system can decrease (since this is a closed system with no external work being supplied).

It is also unclear what you mean by "entropy generated in the system". Work is generated and heat flows in these thermodynamic processes. As a result of heat flowing over one full cycle, the entropy of each of the reservoirs changes (by dQ/T where dQ is the heat flow into/out of the reservoir and T is the temperature of the reservoir). The sum total entropy of those changes cannot be less than 0 if no external work is being supplied.

AM

The increase of entropy principle does not imply that entropy of a system cannot decrease. The entropy change of a system can be negative during a process, but entropy generation cannot. If the process were reversible, then the entropy generation would be zero. However, since all processes in the real world are irreversible, it is said that the entropy of the universe is always increasing. That was my point to the OP.

Here's an example: Take rigid tank and some refrigerant as the system. If heat is transferred out of the system, the entropy change of the refrigerant will be negative.

By entropy generation I mean irreversibilities such as friction, mixing, chemical reactions, expansion, etc.

Entropy generation and entropy transfer are not the same. Entropy transfer can occur by two mechanisms: heat transfer and mass flow. There is no entropy transfer associated with energy transfer as work.

CS

 

[Andrew Mason]

The increase of entropy principle does not imply that entropy of a system cannot decrease. The entropy change of a system can be negative during a process, but entropy generation cannot.

Entropy describes a thermodynamic state, not a quantity of something physical. So it is very confusing to speak about entropy generation. One does not speak about generating temperature, for example.

If the process were reversible, then the entropy generation would be zero. However, since all processes in the real world are irreversible, it is said that the entropy of the universe is always increasing. That was my point to the OP.

Here's an example: Take rigid tank and some refrigerant as the system. If heat is transferred out of the system, the entropy change of the refrigerant will be negative.

This is not exactly a thermodynamic system. You need some place to deliver the heat to. The system has to include the hot reservoir.

The entropy change of the refrigerant is always 0 after one complete cycle because it returns to its initial thermodynamic state. The entropy of the cold reservoir is negative in a refrigeration cycle simply because heat flows out of it (dQ<0 so dQ/T < 0). This is the case whether the cycle is reversible or irreversible.

By entropy generation I mean irreversibilities such as friction, mixing, chemical reactions, expansion, etc.

This is, again, confusing. If I take a Carnot engine system (engine + reservoirs) and I do work that is dissipated by friction externally to the system, is the change in entropy of the system not still zero?

Entropy generation and entropy transfer are not the same. Entropy transfer can occur by two mechanisms: heat transfer and mass flow. There is no entropy transfer associated with energy transfer as work.

If entropy is a state function, how is it transferred or generated?

There can be an entropy change associated with the transfer of work. Consider work done against friction force where the heat generated is transferred to a reservoir: the entropy of the reservoir will increase (by dQ/T = W/T).

AM

 

[stewartcs]

Entropy describes a thermodynamic state, not a quantity of something physical. So it is very confusing to speak about entropy generation. One does not speak about generating temperature, for example.

Entropy generation is a well known term used in thermodynamic texts. It is not a term I coined by any means.

This is not exactly a thermodynamic system. You need some place to deliver the heat to. The system has to include the hot reservoir.

Of course is it a thermodynamic system. Thermodynamics is the study of energy. Let’s put some numbers in and see what happens to the entropy change of this system. Say the tank has 5kg of R134a initially at 20 °C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa.

Since the tank is rigid $$v_2 = v_1$$. It is a closed system since no mass crosses the system boundary during the process. The entropy change of a substance is simply the difference between the entropy values at the final and initial states. Since the specific volume remains constant during this process the properties of the refrigerant at both states are:

State 1:

$$P_1 = 140$$ kPa
$$T_1 = 20$$ °C

Given these two intensive independent properties, one can see from a property table for R134a the following:

$$s_1 = 1.0624 \frac{kJ}{kg \cdot K}$$
$$v_1 = 0.16544 \frac{m^3}{kg}$$

State 2:

$$P_2 = 100$$ kPa
$$v_2 = v_1$$

Using the property table again one finds:

$$V_f = 0.0007259 \frac{m^3}{kg}$$
$$V_g = 0.19252 \frac{m^3}{kg}$$

From these the quality at state 2 is found to be 0.859.

Thus,

$$s_2 = s_f + x_2 \cdot s_{fg} = 0.07188 + (0.89)(0.87995) = 0.8278 \frac{kJ}{kg \cdot K}$$

The entropy change of the refrigerant during this process is:

$$\Delta{S} = m({s_2} - {s_1}) = (5)(0.8278 – 1.0624) \frac{kJ}{kg \cdot K}$$
EDIT: For some reason latex is not being displayed properly here. Delta S should equal m(s_2 - s_1).

= -1.173 kJ/K

Notice the negative sign. This indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation that cannot be negative. This is why one must know the difference between the general term “entropy” and entropy generation. They are not the same. Equating the terms is more confusing, not less.

BTW, this is not a cyclic device as you seem to think (in my example).

The entropy of the cold reservoir is negative in a refrigeration cycle simply because heat flows out of it (dQ<0 so dQ/T < 0). This is the case whether the cycle is reversible or irreversible.

This was my point all along which you seemed to have been saying wasn’t possible. The stipulation about being reversible simply points out that the entropy generation is zero for a reversible process.

This is, again, confusing. If I take a Carnot engine system (engine + reservoirs) and I do work that is dissipated by friction externally to the system, is the change in entropy of the system not still zero?

Friction is an irreversibility of a system. If friction is involved, the entropy generation is positive, not zero.

If entropy is a state function, how is it transferred or generated?

Entropy generation is a measure of the magnitudes of the irreversibilities present during a process.

There can be an entropy change associated with the transfer of work. Consider work done against friction force where the heat generated is transferred to a reservoir: the entropy of the reservoir will increase (by dQ/T = W/T).

AM

Being an organized form of energy, work is free of disorder or randomness and thus free of entropy. The work done against a frictional force results in an entropy change due to the irreversibility (i.e. the friction). It is not a result of the work itself, rather it is the fact that the organized energy (work) is being degraded to a less useful form (heat). Hence, the entropy generation is increased.

CS

 

[Andrew Mason]

Entropy generation is a well known term used in thermodynamic texts. It is not a term I coined by any means.

Lots of texts use the term "disorder" which is also very confusing if not incorrect. I am just pointing out that the term "entropy generation" adds unnecessary confusion to the study of the second law.

Of course is it a thermodynamic system.

You can call it a system. But it is incomplete in the sense that there is no mechanism by which the gas in the tank is cooled. How/why does the heat leave the tank?

The entropy change of the refrigerant during this process is:

$$\Delta{S} = m({s_2} - {s_1}) = (5)(0.8278 – 1.0624) \frac{kJ}{kg \cdot K}$$
EDIT: For some reason latex is not being displayed properly here. Delta S should equal m(s_2 - s_1).

= -1.173 kJ/K

Notice the negative sign. This indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation that cannot be negative. This is why one must know the difference between the general term “entropy” and entropy generation. They are not the same. Equating the terms is more confusing, not less.

BTW, this is not a cyclic device as you seem to think (in my example).

This was my point all along which you seemed to have been saying wasn’t possible. The stipulation about being reversible simply points out that the entropy generation is zero for a reversible process.

You used the term "refrigerant" which suggests that it is involved in absorbing heat from a cold reservoir and delivering it to a hot reservoir.

Of course this is not a violation of the second law. The entropy of a body can decrease by some amount - the second law merely requires the entropy of the rest of the universe to increase by a greater amount.

Friction is an irreversibility of a system. If friction is involved, the entropy generation is positive, not zero.

The entropy of a closed system consisting of a Carnot engine operating between two reservoirs at different temperatures can be 0 (or arbitrarily close to 0). The output of that engine is used to do work against a gravitational force or a frictional force. The entropy change of the Carnot engine thermodynamic system does not depend on what kind of work the engine does.

Entropy generation is a measure of the magnitudes of the irreversibilities present during a process.

All I am saying is that this terminology is very confusing. Entropy is difficult for a student to understand, in part because of a lot of loose terminology. "Entropy generation" is one such term.

The only concept that needs to be used is "increase in entropy". Entropy increases because energy disperses from a more concentrated or useable form to a less concentrated or useable form. Entropy is not "generated" any more than temperature is "generated", although it may be a term that is useful to the engineer.

You can have a reversible thermodynamic process doing work against a frictional force - in which case the thermodynamic system experiences a zero change of entropy. Or you can have a non-reversible thermodynamic process working against a gravitational force - in which case there is a positive change in entropy of the thermodynamic system. It is important to distinguish the thermodynamic system itself from the work done by the system.

Being an organized form of energy, work is free of disorder or randomness and thus free of entropy. The work done against a frictional force results in an entropy change due to the irreversibility (i.e. the friction). It is not a result of the work itself, rather it is the fact that the organized energy (work) is being degraded to a less useful form (heat). Hence, the entropy generation is increased.

CS

There is no such distinction between energy and work. Work is something that a body of matter does. Energy is the capacity or ability to do work. The statement: "Being an organized form of energy, work is free of disorder or randomness and thus free of entropy." is erroneous and, at best, confusing.

AM

 

[stewartcs]

Entropy increases because energy disperses from a more concentrated or useable form to a less concentrated or useable form.

"More concentrated or useable" is no more of a better definition than is "disorganzied". You even contradict yourself later with the following quote:

The statement: "Being an organized form of energy, work is free of disorder or randomness and thus free of entropy." is erroneous and, at best, confusing.

You can call it a system. But it is incomplete in the sense that there is no mechanism by which the gas in the tank is cooled. How/why does the heat leave the tank?

Heat leaves the tank simply because heat flows naturally from a higher temp to a lower temp. The obvious implication is that the refrigerant was at a higher temp than the surroundings. Hence, heat is rejected from the refrigerant to the surroundings which yielded the negative entropy change.

Thermodynamic systems have no requirement that they be cyclic.

You used the term "refrigerant" which suggests that it is involved in absorbing heat from a cold reservoir and delivering it to a hot reservoir.

Of course this is not a violation of the second law. The entropy of a body can decrease by some amount - the second law merely requires the entropy of the rest of the universe to increase by a greater amount.

The increase of entropy principle specifies that the system be isolated in order for the entropy of the universe to increase (or stay the same if the process is reversible). If it is not an isolated system (i.e. not adiabatic and closed) this does not necessarily hold true.

The entropy of a closed system consisting of a Carnot engine operating between two reservoirs at different temperatures can be 0 (or arbitrarily close to 0). The output of that engine is used to do work against a gravitational force or a frictional force. The entropy change of the Carnot engine thermodynamic system does not depend on what kind of work the engine does.

It has nothing to do with the work involved. The entropy generated is due to irreversibilities in the process.

All I am saying is that this terminology is very confusing. Entropy is difficult for a student to understand, in part because of a lot of loose terminology. "Entropy generation" is one such term.

The only concept that needs to be used is "increase in entropy". Entropy increases because energy disperses from a more concentrated or useable form to a less concentrated or useable form. Entropy is not "generated" any more than temperature is "generated", although it may be a term that is useful to the engineer.

“Increase in entropy” does not explain how an entropy change can be negative. If all one assumes is that entropy is always increasing, then the results, as previously shown, could not possibly exist…when in fact they do.

You can have a reversible thermodynamic process doing work against a frictional force - in which case the thermodynamic system experiences a zero change of entropy. Or you can have a non-reversible thermodynamic process working against a gravitational force - in which case there is a positive change in entropy of the thermodynamic system. It is important to distinguish the thermodynamic system itself from the work done by the system.

If you have a reversible process doing work against a frictional force and you add enough heat the entropy change will not be zero.

There is no such distinction between energy and work. Work is something that a body of matter does. Energy is the capacity or ability to do work. The statement: "Being an organized form of energy, work is free of disorder or randomness and thus free of entropy." is erroneous and, at best, confusing.

AM

There is no entropy transfer associated with energy transfer as work. Here is a simple example: Take a rotating shaft that has a cable connected to it and then to a weight (ignoring all friction – i.e. irreversibilities). If energy is used in the form of shaft work to raise the weight, no entropy is produced. The reason why no entropy is produced is because the weight may be lower and the process reversed with no loss in the quality of the energy. Hence, no entropy transfer is associated with energy transfer as work.

CS

 

[Andrew Mason]

"More concentrated or useable" is no more of a better definition than is "disorganzied". You even contradict yourself later with the following quote:

. Disorder is not a correct description of entropy. This should illustrate the problem:

If heat spontaneously flows from a small hot body to a large cool body, the thermal energy in the hot body is no longer concentrated in the hot body but is dispersed over both bodies. The hot body is cooler, which means that at the molecular level it has less random motion than before. The cool body is slightly warmer.

Now if one equates entropy with disorder or randomness, you have a problem because the energy in the hot body was much more random before it cooled and the energy in the cool body is only slightly more random. How can you say that there has been an increase in disorder?

If you are going to equate entropy with disorder you have to define disorder in a very particular way. If you define disorder as the number of possible physical microstates that a body can have for a given thermodynamic state you might be on the right track, but this is a rather esoteric way to explain entropy.

Heat leaves the tank simply because heat flows naturally from a higher temp to a lower temp. The obvious implication is that the refrigerant was at a higher temp than the surroundings. Hence, heat is rejected from the refrigerant to the surroundings which yielded the negative entropy change.

But it increases the entropy of the surroundings by a greater amount.

Thermodynamic systems have no requirement that they be cyclic.

I never said they did. I thought you were trying to describe a refrigerator with your use of the term "refrigerant".

The increase of entropy principle specifies that the system be isolated in order for the entropy of the universe to increase (or stay the same if the process is reversible). If it is not an isolated system (i.e. not adiabatic and closed) this does not necessarily hold true.

I am not sure what you mean. The second law says that the entropy of the universe can never decrease in any thermodynamic process. This is true whether the thermodynamic system that carries on the thermodynamic process is isolated or not.

If the system is isolated, (where there is no interaction with the surroundings) the entropy of the system cannot decrease - ie. it must increase or, if it is a reversible process, remain unchanged. If it is not isolated, the entropy of the system can decrease by some amount but the entropy of its surroundings must increase by an equal or greater amount.

“Increase in entropy” does not explain how an entropy change can be negative. If all one assumes is that entropy is always increasing, then the results, as previously shown, could not possibly exist…when in fact they do.

I was using "increase in entropy" to replace "entropy generation".

But you raise another point. Entropy does not exist in the abstract. It is a property of body of matter. When one speaks about entropy, one has to specify the body to which the entropy relates.

Any flow of heat out of a body results in a negative entropy change for that body. Any flow of heat into a body creates a positive entropy change for that body. What the second law says is that the entropy change of the system + surroundings will never be negative.

If you have a reversible process doing work against a frictional force and you add enough heat the entropy change will not be zero.

If you have a reversible thermodynamic system (Carnot engine) doing work against a frictional force, the entropy of the system + surroundings cannot be zero. This is because heat has to flow into some body as a result of the that work, so the entropy of that body must increase.

There is no entropy transfer associated with energy transfer as work. Here is a simple example: Take a rotating shaft that has a cable connected to it and then to a weight (ignoring all friction – i.e. irreversibilities). If energy is used in the form of shaft work to raise the weight, no entropy is produced. The reason why no entropy is produced is because the weight may be lower and the process reversed with no loss in the quality of the energy. Hence, no entropy transfer is associated with energy transfer as work.

Your terminology is confusing. Work is simply the application of a force over a distance. The ability to do that work is energy. eg: A body traveling at a speed v has the ability to do .5mv^2 of work. Now, if I have that body expend its kinetic energy by raising itself in a gravitational field, it does work but it does not transfer its kinetic energy into work. It transfers it into gravitational potential energy. If the body expends its kinetic energy by applying a frictional force over a distance, it transfers its kinetic energy into thermal energy (kinetic energy of molecules in the body that provides the friction.

AM

 

[stewartcs]

. Disorder is not a correct description of entropy. This should illustrate the problem:

If heat spontaneously flows from a small hot body to a large cool body, the thermal energy in the hot body is no longer concentrated in the hot body but is dispersed over both bodies. The hot body is cooler, which means that at the molecular level it has less random motion than before. The cool body is slightly warmer.

Now if one equates entropy with disorder or randomness, you have a problem because the energy in the hot body was much more random before it cooled and the energy in the cool body is only slightly more random. How can you say that there has been an increase in disorder?

The second law requires that the entropy of the cooler body be increased by a greater amount than was lost by the hot body. The hot body will have less entropy or disorganization, however, the cooler body’s entropy or disorganization is increase by a greater amount. They are not proportional.

If you are going to equate entropy with disorder you have to define disorder in a very particular way. If you define disorder as the number of possible physical microstates that a body can have for a given thermodynamic state you might be on the right track, but this is a rather esoteric way to explain entropy.

That is how entropy is defined in statistical thermodynamics (i.e. Boltzmann’s equation).

http://en.wikipedia.org/wiki/Boltzmann's_entropy_formula

But it increases the entropy of the surroundings by a greater amount.

I’ve always agreed with that. My point originally to the OP was a caveat on obtaining a negative value for entropy change!

I am not sure what you mean. The second law says that the entropy of the universe can never decrease in any thermodynamic process. This is true whether the thermodynamic system that carries on the thermodynamic process is isolated or not.

If the system is isolated, (where there is no interaction with the surroundings) the entropy of the system cannot decrease - ie. it must increase or, if it is a reversible process, remain unchanged. If it is not isolated, the entropy of the system can decrease by some amount but the entropy of its surroundings must increase by an equal or greater amount.

The net change when including the surroundings (i.e. the universe) will increase or stay the same, I agree with that.

I was using "increase in entropy" to replace "entropy generation".

But you raise another point. Entropy does not exist in the abstract. It is a property of body of matter. When one speaks about entropy, one has to specify the body to which the entropy relates.

Any flow of heat out of a body results in a negative entropy change for that body. Any flow of heat into a body creates a positive entropy change for that body. What the second law says is that the entropy change of the system + surroundings will never be negative.

I agree with that as well.

If you have a reversible thermodynamic system (Carnot engine) doing work against a frictional force, the entropy of the system + surroundings cannot be zero. This is because heat has to flow into some body as a result of the that work, so the entropy of that body must increase.

That’s what my quote said. I was responding to what you posted which seemed to say the opposite:

You can have a reversible thermodynamic process doing work against a frictional force - in which case the thermodynamic system experiences a zero change of entropy.

Your terminology is confusing. Work is simply the application of a force over a distance. The ability to do that work is energy. eg: A body traveling at a speed v has the ability to do .5mv^2 of work. Now, if I have that body expend its kinetic energy by raising itself in a gravitational field, it does work but it does not transfer its kinetic energy into work. It transfers it into gravitational potential energy. If the body expends its kinetic energy by applying a frictional force over a distance, it transfers its kinetic energy into thermal energy (kinetic energy of molecules in the body that provides the friction.

AM

That is exactly what I said just phrased differently. The transfer from kinetic energy to potential energy does not involve any entropy transfer unless friction (or other irreversibility) is involved in the process. If friction is involved part of the energy that went into the work is transformed into heat. This fact is what causes the entropy to increase (i.e. the friction).

CS

 

[Andrew Mason]

The second law requires that the entropy of the cooler body be increased by a greater amount than was lost by the hot body. The hot body will have less entropy or disorganization, however, the cooler body’s entropy or disorganization is increase by a greater amount. They are not proportional.

My point is that there is no obvious overall increase in disorder there, as the term "disorder" is used in normal speech. You can say the final result has more disorder but what does it mean? How is it measured?

Suppose the hot body was 10 grams of steam at 100 C and the cold body was a 10000 kg block of ice at -1 C. The random translational motion of the molecules in the steam disappears and all the molecules form crystal ice, which has very little disorder (no translational motion at all). How can you say that the ice block now has more disorder than the steam + ice?

If you say that the greatest amount of "disorder" or "randomness" occurs when equilibrium is reached, then the second law merely says that all bodies tend toward equilibrium. But this is a special way of defining disorder or randomness. It is not how we normally think of disorder, as my illustration of the steam and ice shows.

That is how entropy is defined in statistical thermodynamics (i.e. Boltzmann’s equation).

http://en.wikipedia.org/wiki/Boltzmann's_entropy_formula

And my point is that this is not what students automatically think of when they encounter the word "disorder".

That’s what my quote said. I was responding to what you posted which seemed to say the opposite.

No. I said the entropy of the reversible thermodynamic system does not change. The increase in entropy is in the surroundings. This is because there is an increase in the entropy of the body on which work is done (the work produced from the system).

That is exactly what I said just phrased differently. The transfer from kinetic energy to potential energy does not involve any entropy transfer unless friction (or other irreversibility) is involved in the process. If friction is involved part of the energy that went into the work is transformed into heat. This fact is what causes the entropy to increase (i.e. the friction).

I agree. Although there are other reasons (other than friction losses) that result in an increase in entropy.

Stewart, we don't disagree on the physics here - just the terminology.

AM

 

[stewartcs]

My point is that there is no obvious overall increase in disorder there, as the term "disorder" is used in normal speech. You can say the final result has more disorder but what does it mean? How is it measured?

Suppose the hot body was 10 grams of steam at 100 C and the cold body was a 10000 kg block of ice at -1 C. The random translational motion of the molecules in the steam disappears and all the molecules form crystal ice, which has very little disorder (no translational motion at all). How can you say that the ice block now has more disorder than the steam + ice?

If you say that the greatest amount of "disorder" or "randomness" occurs when equilibrium is reached, then the second law merely says that all bodies tend toward equilibrium. But this is a special way of defining disorder or randomness. It is not how we normally think of disorder, as my illustration of the steam and ice shows.

My point was that the incremental increase in the entropy of the cooler body is what is greater.

Stewart, we don't disagree on the physics here - just the terminology.

AM

I too believe we ultimately agree on the physics of it.

Chalk another one up to semantics.

CS

 

[RonL]

Well CS, AM,

I'm glad you both agree, but am disappointed that the thread might stop.

CS almost stated what is in my mind that I just can't bring into clear focus. (entropy generation)

If a tank is submerged in water (let's say the ocean) and a refrigeration design is used to absorb heat from the water, as I understand, the entropy decreases as heat moves from the ocean into the tank. In order to keep the cycle repeating, an exact amount of heat has to be disposed of, in some fashion.(an increases of entropy)

If that is correct, then any method of keeping this balance exact, would in fact have a net energy of 0.

If this is right, I have something else in mind, if not I hope someone will help clear it up.

Ron

 

[stewartcs]

Well CS, AM,

I'm glad you both agree, but am disappointed that the thread might stop.

CS almost stated what is in my mind that I just can't bring into clear focus. (entropy generation)

If a tank is submerged in water (let's say the ocean) and a refrigeration design is used to absorb heat from the water, as I understand, the entropy decreases as heat moves from the ocean into the tank. In order to keep the cycle repeating, an exact amount of heat has to be disposed of, in some fashion.(an increases of entropy)

If that is correct, then any method of keeping this balance exact, would in fact have a net energy of 0.

If this is right, I have something else in mind, if not I hope someone will help clear it up.

Ron

I’m not exactly sure what you are describing. However, if what I am picturing is correct (a tank full of refrigerant sitting in the ocean) then the ocean must be at a higher temperature than the refrigerant for heat to flow naturally without a work input. If this is the case, as soon as the two equalize (the ocean and refrigerant) no heat will flow an no work could be produced. If you could somehow remove the heat transferred into the refrigerant from the ocean (again assuming the ocean was at a higher temperature) then you could continue having a work output. However, to remove the heat from the refrigerant would require another lower temperature sink or a work input to reject it to a higher source. If you have another sink then the same process of having a lower temperature sink to reject the heat to would have to happen over and over and over again to continue getting any work output…which is not possible let alone practical. So you are only left with the option of a work input to remove the thermal energy from the refrigerant that was transferred from the ocean. Otherwise you would violate the second law and have a PMM2.

Hence the only method to continue the cycle would be to have a work input which of course requires energy.

Does that help?

CS

 

[RonL]

I’m not exactly sure what you are describing. However, if what I am picturing is correct (a tank full of refrigerant sitting in the ocean) then the ocean must be at a higher temperature than the refrigerant for heat to flow naturally without a work input. If this is the case, as soon as the two equalize (the ocean and refrigerant) no heat will flow an no work could be produced. If you could somehow remove the heat transferred into the refrigerant from the ocean (again assuming the ocean was at a higher temperature) then you could continue having a work output. However, to remove the heat from the refrigerant would require another lower temperature sink or a work input to reject it to a higher source. If you have another sink then the same process of having a lower temperature sink to reject the heat to would have to happen over and over and over again to continue getting any work output…which is not possible let alone practical. So you are only left with the option of a work input to remove the thermal energy from the refrigerant that was transferred from the ocean. Otherwise you would violate the second law and have a PMM2.

Hence the only method to continue the cycle would be to have a work input which of course requires energy.

Does that help?

CS

Thanks CS, yes it does.

As I mentioned in another thread, (Heat Pump and COP) a tank long enough and large enough in diameter, to draw in around around 5 HP of heat energy, might not be so large, but for a start, let's say that is enough.

What I can't grasp in my mind, is how it is not possible to design a power unit that can have in store (many ways) a mass of energy, far in excess of the 5 HP that moves through the system. The thrust of a propeller returns the energy to the ocean. There are no energy losses associated with the thermal energy moving from the water, into the refrigerant, and the frictional losses of the tank moving through the water, and the propeller working against the water, should be a negative value producing a positive effect, forward movement, even though the work is net 0.

I think it should be understood that a large amount of external energy would be needed to start the process, but once a state of imbalance is established, the process should remain at a stable rate of energy flow in and out. With the power unit insulated and inside the cold tank, all losses that are associated with the (internal) unit in motion, are recycled over and over.

I think copper is still the best transfer metal, (the highest K value) and even at todays prices, the most affordable. Using electric heat elements to establish a high temperature of around 1250 F. degrees and a refrigerant operating in the -50 F. degrees range, the ocean water can serve as both the heat source and the heat sink for the system cycle.

There are so many ways to design the internal power unit, that I would hope the need of a detailed description can slide for now. A cycle, and the time needed for the amount of thermal transfer to take place, will have many variables, I just hope what I have said makes a little sense in regards to this thread, and what you and Andrew have talked about.



Ron

 

[stewartcs]

What I can't grasp in my mind, is how it is not possible to design a power unit that can have in store (many ways) a mass of energy, far in excess of the 5 HP that moves through the system. The thrust of a propeller returns the energy to the ocean. There are no energy losses associated with the thermal energy moving from the water, into the refrigerant, and the frictional losses of the tank moving through the water, and the propeller working against the water, should be a negative value producing a positive effect, forward movement, even though the work is net 0.

Even though the first law may be satisfied, the second law is not. When the propeller returns energy to the ocean it is of a lower quality than it was originally (i.e. entropy is increased), so the amount of work that can be extracted due to heat transfer is reduced.

It's probably better to look at an ideal Rankine cycle and then an actual power cycle to see why this can't be done. Those are clearly defined systems in normal Thermodynamic terms and well understood. Ultimately, you won't get a perpetual cycle for some or all of the reasons listed by AM and I throughout this thread.

CS