Question on Vapor Compression Cycle.

[ashishvinayak]

For convenience, let's assume the following process to be ideal.

Process 1-2 Isenthalpic compression
Process 2-3 Condensation
Process 3-4 Throttling
Process 4-1 Evaporation

The refrigerant is dry-saturated at the end of compression.

Here's what I don't understand:
During process 2-3, if we refer P-h and T-S graphs we see that both the pressure and temperature remain constant during the condensation process. However, there is a constant pressure heat rejection at the same time. How is that possible? How is heat rejected without change in temperature?

I've uploaded a reference photo. Please help.Photo

 

[Jupiter6]

For convenience, let's assume the following process to be ideal.

Process 1-2 Isenthalpic compression
Process 2-3 Condensation
Process 3-4 Throttling
Process 4-1 Evaporation

The refrigerant is dry-saturated at the end of compression.

Here's what I don't understand:
During process 2-3, if we refer P-h and T-S graphs we see that both the pressure and temperature remain constant during the condensation process. However, there is a constant pressure heat rejection at the same time. How is that possible? How is heat rejected without change in temperature?

I've uploaded a reference photo. Please help.Photo

In reality, it's not. Of course the temp will drop between the inlet and the outlet. But what they're trying to show you here is an idealized cycle. You can dump heat and keep constant temp in an ideal condenser. As soon as you get into realistic non-ideal systems you'll get a better grasp on it.

It's been a long time but I suspect the Profs and Thermo books teach it this way so you put the total compressor work in. The "s" in the T-s diagram will realistically veer to the right (pos x). Later on, that's how you will figure out your compressor efficiency.

 

[ashishvinayak]

You can dump heat and keep constant temp in an ideal condenser. As soon as you get into realistic non-ideal systems you'll get a better grasp on it.

This is what I want to understand. How is it possible to dump heat without reduction of temperature?

 

[Jupiter6]

This is what I want to understand. How is it possible to dump heat without reduction of temperature?

I suppose you can think of heat as an amount of energy and temperature as the intensity of that energy. The only requirement to dump heat is a difference in temps between masses.

If you have a cup of water at 100 degrees F and a barrel of water at 100 degrees F the barrel will contain more heat. So if you drop an ice cube in the cup, the water temp will drop while adding an ice cube to the barrel won't change anything even though both the cup and barrel gave up the same amount of heat to melt the ice cube.

In the case of the condenser, the phase of the refrigerant has changed from a vapor to liquid. So the latent heat of vaporization is given up with the change.

 

[ashishvinayak]

Brilliant. Perfect explanation. Couldn't ask for anything more!

 

[SteamKing]

Condensation means there is a phase change in the working fluid, from the vapor phase to the liquid phase. Phase changes are usually isothermal, i.e., they occur with no change in temperature.

For example, when water freezes, the liquid at 0 C turns to a solid at 0 C, once the heat of fusion is removed from the liquid.

 

[pukb]

its a phase change process. latent heat is what that is given away as heat.

 

[sanka]

As others have posted, during condensation a phase change occurs as vapour changes to liquid. Firstly, let's take a step back; during boiling/evaporation of a liquid, energy is added (again at constant temperature). This energy is known as "Enthalpy of Vaporization" or "Latent Heat". It is a form of potential energy as it is the energy used to overcome the intermolecular forces that exist in a liquid state. As it is potential energy being added, no temperature change will occur as kinetic energy is responsible for a fluids temperature.
During condensation, as the vapor returns to the liquid state, the fluid will reject the energy that was previously added, which was potential energy, resulting in no change in temperature.

This might seem confusing at first but I always seem to understand a process much better from a molecular level