Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume the following phase diagram $T-H$ of water

enter image description here

At a nuclear engineering course, I was told that in order to increase the performance of a pressurized water reactor, one has to increase the pressure.

How such a reactor works is seen on the next image

enter image description here

In the steam generator the is saturated water. As fusion takes place, very hot water moves towards the steam generator where it transfers its thermal energy to the saturated water. The saturated water becomes hotter and finally it vaporizes. The steam moves the turbine and electric energy is produced.

I cannot understand the following: Why the increment of pressure in the reactor vessel, will transfer more energy in the steam generator making the thermal energy of the steam larger?

share|cite|improve this question
I think statement "in order to increase the performance of a pressurized water reactor, one has to increase the pressure" is not clear. The key function of Pressurizer is to maintain stability of system between parts with different pressure and to monitor level of water. If pressurizer is not configured well it main lead to overload or thermal shock, of some parts. I don't know any fusion reactors where it used as active element of system. Maybe it's related to some changes in other part of system? – Sigrlami May 29 '13 at 7:04
We do have some nuclear power guys who frequent the site. Your best bet is to wait for one of the to notice this questions. – dmckee May 29 '13 at 14:09
up vote 1 down vote accepted

I'm not a nuclear engineer, or a power systems engineer but lets try to work the question as simply as possible.

  1. You want the maximum efficiency out of the turbine.
  2. That is a large fraction of the thermodynamic (Carnot) limit which is set by the temperature of steam entering the turbine and the design of the turbine. The only parameter you can change dynamically here is the temperature of the input steam (and you want it to be as hot as possible).
  3. The steam can be no hotter than the water leaving the core of the reactor, and will actually be a bit less because of latent heat.
  4. Assuming that you want to maintain the core-cooling loop liquid the whole way (I believe that you do), the only way to raise its temperature is to increase it's pressure (because this increases the boiling point, which is your upper limit).

So, to maintain high power generation efficiency you must maintain high pressure on the core cooling system.

You need to keep in mind that there are two separate loops at work here. One is in contact with the core and potentially contaminated, so it is kept carefully inside the containment vessel. The other one is nominally clean and actually powers the turbine. You can see this in the figure you posted: one is colored in oranges, yellows and reds; the other colored in shades of blue (and actually a third water system that provides the cold sink for the back of the turbine also colored blue, but it plays only a passive roll here).

Steam is generated in the clean loop by exchanging heat with the hot liquid water in the dirty loop. Latent heat only comes into play on the clean side. I don't know if that system is actively pressurized or not, but there will be some back pressure in the system from the turbine.

share|cite|improve this answer
Thank you very much for your answer! The only part that bothers me is the latent heat. In the diagram for higher pressures the latent heat is smaller that in lower pressures, isn't that right? So this contradicts the part that higher pressure means higher Temperature... – Thanos May 29 '13 at 14:10
Well, I ignored that wrinkle, but it does not contradict anything. I will edit a bit. – dmckee May 29 '13 at 14:15

Its simple - with more pressure, you've got more particles to carry the energy per $m^3$. Liquid is better carrier of temperature than gas, because it has better Thermal conductivity and with pressure, you can prevent the water becoming steam.
That is the difference between inner and outer circle on your image. The outer circle on the other side, profits from big difference in pressure.

share|cite|improve this answer
Thank you very much for your answer! The thing is, that to increase the performance you have to have a high thermal energy steam in order to produce more energy via the turbine. So, you are saying that, the higher the pressure in the primary loop, the greater the heat transfer in the pressurizer, the higher the steam energy. If you consider the diagram, the higher the pressure, the lower the latent heat... That's the part I don't get! – Thanos May 29 '13 at 7:19
Well, I pretty much used the common sense to answer. I may be even wrong - make sure you ask your teacher and correct me in such case. Anyway, I think with enormous pressure, the power of friction must be taken into account. – Tomáš Zato May 29 '13 at 7:23
The thing is that I have asked him, but he is an engineer! I am not sure that he understands the thermodynamics beneath that. He just said something like what you answered, which is pretty much what I was thinking of. But the latent heat crossed my mind... – Thanos May 29 '13 at 7:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.