Is this Thermodynamic Cycle Possible and Reversible?

In summary: I'll have to defer to someone else on this, for I am not good with that theorem. Sorry.In summary, the conversation discusses determining whether a cycle violates the first law and whether it is reversible, irreversible, or not possible. The problem involves a heat engine producing 2000kW of power by taking 3000kW of heat from a 750C reservoir and rejecting 1000kW of heat to a 20C reservoir. The equation dU=Q-W is used to solve the problem, with the attempt being to sum the heat transfer and set dU equal to the energy produced. It is determined that the reasoning for the engine being irreversible due to heat transfer from a reservoir is not valid
  • #1
Boom100
4
0

Homework Statement


Determine Whether or not this cycle violates the first law and if it is reversible, irreversible, or not possible.

2000kW or power are produced by taking 3000 kW of heat transfer from a 750C reservior and rejecting 1000kW of heat to a 20C reservoir.


Homework Equations



dU=Q-W

The Attempt at a Solution



My attempt was to simply sum the heat transfer (3000-1000)kW = Q and set dU=2000kW (The energy produced, or the internal energy it gathered). I set W=0 because this is just heat transfer and there is no work being done.

So 2,000 does = 2,0000 so I'm thinking this is true. It seems a bit easy though... so I am thinking something is off and I might have to incorporate the temperatures.

I said this is irreversible because it is heat transfer from a reservoir. So, once the heat is transferred, both heat sources change temperatures (The cold gets hotter and the hot gets cooler) so they are tending towards an equilibrium temperature. This can not be reversed and heat can not be sent back.
 
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  • #2
.
Welcome to Physics Forums.

Boom100 said:

Homework Statement


Determine Whether or not this cycle violates the first law and if it is reversible, irreversible, or not possible.

2000kW or power are produced by taking 3000 kW of heat transfer from a 750C reservior and rejecting 1000kW of heat to a 20C reservoir.


Homework Equations



dU=Q-W

The Attempt at a Solution



My attempt was to simply sum the heat transfer (3000-1000)kW = Q and set dU=2000kW (The energy produced, or the internal energy it gathered). I set W=0 because this is just heat transfer and there is no work being done.

So 2,000 does = 2,0000 so I'm thinking this is true. It seems a bit easy though... so I am thinking something is off and I might have to incorporate the temperatures.

I said this is irreversible because it is heat transfer from a reservoir. So, once the heat is transferred, both heat sources change temperatures (The cold gets hotter and the hot gets cooler) so they are tending towards an equilibrium temperature. This can not be reversed and heat can not be sent back.
The "power produced" mentioned in the problem statement actually refers to the work done by this heat engine. That contradicts your thinking that W=0.
 
  • #3
Redbelly98 said:
.
Welcome to Physics Forums.


The "power produced" mentioned in the problem statement actually refers to the work done by this heat engine. That contradicts your thinking that W=0.

Ahhh Ok great, thanks Red. So with that thinking then- Since the engine is doing the work, that means that W=-2000, So dU=2000-2000=0. So 2000 still = 2000 So am i correct in saying that this means that this statement is valid? Is my reason correct for why it is irreversible as well?

Once again, thank you!
 
  • #4
I have asked around if any of the other Homework Helpers can help out on the question of reversible/irreversible/impossible. Meanwhile, I'll respond as I can.

Boom100 said:
I said this is irreversible because it is heat transfer from a reservoir. So, once the heat is transferred, both heat sources change temperatures (The cold gets hotter and the hot gets cooler) so they are tending towards an equilibrium temperature. This can not be reversed and heat can not be sent back.
But an ideal Carnot cycle also involves heat transfer from a reservoir, yet it is reversible.

Boom100 said:
Ahhh Ok great, thanks Red. So with that thinking then- Since the engine is doing the work, that means that W=-2000, So dU=2000-2000=0. So 2000 still = 2000 So am i correct in saying that this means that this statement is valid?
What statement are you referring to?
Is my reason correct for why it is irreversible as well?
I think not, for the reason I stated above on Carnot cycles.
 
  • #5
Redbelly98 said:
I have asked around if any of the other Homework Helpers can help out on the question of reversible/irreversible/impossible. Meanwhile, I'll respond as I can.


But an ideal Carnot cycle also involves heat transfer from a reservoir, yet it is reversible.


What statement are you referring to?

I think not, for the reason I stated above on Carnot cycles.

The statement I was referring to was the first law. So I was just seeing if that was proof enough that it doesn't violate the first law.

As far as the reversibility thing, I see what you mean. I am still pretty confused on this one.
 
  • #6
You want to compare the efficiency of the given engine to that of the Carnot engine and use Carnot's theorem to answer the question about whether the given engine is possible, and if so, if it's reversible or irreversible.
 

Related to Is this Thermodynamic Cycle Possible and Reversible?

1. What is a thermodynamics cycle?

A thermodynamics cycle is a series of thermodynamic processes that occur in a system and return the system to its initial state. It involves the transfer of energy between the system and its surroundings.

2. Why is understanding thermodynamics cycles important?

Understanding thermodynamics cycles is important because it allows us to analyze and predict the behavior of various systems, such as engines, power plants, and refrigeration systems. It also helps us optimize these systems for maximum efficiency and performance.

3. What are the four laws of thermodynamics?

The four laws of thermodynamics are: 1) The law of conservation of energy, stating that energy cannot be created or destroyed, only transferred or converted; 2) The law of entropy, stating that the total entropy of a closed system will always increase over time; 3) The law of absolute temperature, stating that the entropy of a perfect crystal at absolute zero temperature is zero; and 4) The law of conservation of mass and energy, stating that the total mass and energy in a closed system are constant.

4. What are some common types of thermodynamics cycles?

Some common types of thermodynamics cycles include the Carnot cycle, the Rankine cycle, the Otto cycle, and the Brayton cycle. These cycles are used in various systems, such as steam engines, internal combustion engines, and gas turbines.

5. How do you calculate the efficiency of a thermodynamics cycle?

The efficiency of a thermodynamics cycle can be calculated by dividing the work output of the cycle by the heat input. This is known as the thermal efficiency and is expressed as a percentage. The higher the efficiency, the more energy-efficient the system is.

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