# Conceptual thermodynamics problem about ammonia executing a Carnot cycle

• Andrew1234
In summary: For example, in process 2 to 3, the gas may also be turned into liquid by heating it, and then turned back into gas by cooling it. In this process, there is work being done on both the gas and liquid phases. However, the value of delta work is still zero because the temperature and pressure are constant.
Andrew1234
Homework Statement
Explain why the work done in process 2-3 is zero.
Relevant Equations
ΔH = ΔQ-ΔW

Is there a mathematical explanation for why the work done in the condenser (in process 2 to 3) is zero? I am aware that ammonia does not expand or compress in the condenser, only changes phase, but without knowing that the process takes place in a condenser and only considering the graph, according to which temperature is constant for process 2-3, how can it be seen that no work is done in this process?

In process 2-3 the temperature and pressure are both constant. Because the cycle is reversible, Q = ∫Tds.

Substituting this in the energy balance, -m'(Δh)+∫Tds = ΔW. Because temperature is constant, ΔW = T(Δs)-m'(Δh).

W = TΔs - ΔH = T(sf-sg) + m(hg - hf)

It is not clear to me why this quantity is zero for process 2-3.

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Andrew1234 said:
Homework Statement:: Explain why the work done in process 2-3 is zero.
Relevant Equations:: ΔH = ΔQ-ΔW

View attachment 259982

Is there a mathematical explanation for why the work done in the condenser (in process 2 to 3) is zero? I am aware that ammonia does not expand or compress in the condenser, only changes phase, but without knowing that the process takes place in a condenser and only considering the graph, according to which temperature is constant for process 2-3, how can it be seen that no work is done in this process?

In process 2-3 the temperature and pressure are both constant. Because the cycle is reversible, Q = ∫Tds.

Substituting this in the energy balance, -m'(Δh)+∫Tds = ΔW. Because temperature is constant, ΔW = T(Δs)-m'(Δh).

W = TΔs - ΔH = T(sf-sg) + m(hg - hf)

It is not clear to me why this quantity is zero for process 2-3.
There is no way to know unless, as in the case of a condenser, the device has no shaft that does shaft work. In other words, you need to know something about how your device works.

Thank you for clarifying this concept

Both processes, 2 to 3 and 4 to 1, have only transfer of heat.
The value of delta work in the equation ΔH = ΔQ-ΔW is always zero for these processes.
In real systems, there are sub-cooling and overheating portions besides pure change of phase in these processes.

## 1. What is a Carnot cycle?

A Carnot cycle is a theoretical thermodynamic cycle that consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. It is used as a model to understand the maximum efficiency that can be achieved by a heat engine.

## 2. How does ammonia play a role in a Carnot cycle?

Ammonia is used as the working fluid in a Carnot cycle. It is a common choice because it has a high boiling point and is easily available. It also has a large temperature range in which it can undergo phase change, making it efficient for use in a heat engine.

## 3. What is the purpose of a conceptual thermodynamics problem about ammonia executing a Carnot cycle?

The purpose of such a problem is to apply the principles of thermodynamics to a real-world scenario. It allows scientists to analyze and understand the efficiency and performance of a Carnot cycle using ammonia as the working fluid.

## 4. What are some key factors that affect the efficiency of a Carnot cycle using ammonia?

The efficiency of a Carnot cycle using ammonia is affected by the temperature difference between the hot and cold reservoirs, the pressure at which the cycle is operated, and the properties of the ammonia such as its specific heat and boiling point.

## 5. What are the limitations of a Carnot cycle using ammonia?

One major limitation of this type of cycle is that it is only a theoretical model and cannot be achieved in practice. Additionally, the efficiency of the cycle is limited by the temperature difference between the hot and cold reservoirs, and the presence of any irreversibilities in the system.

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