Estimating Waste Heat Rejected from PWRs

  • Thread starter bizarro186
  • Start date
  • Tags
    Heat
In summary: Actually, some of the German PWRs are around 35.7 to 36.4% efficient based on some innovation in turbine design by Siemens.Hello,Thank you everyone for your replies. I am a mechanical engineering undergraduate student with little nuclear engineering knowledge. For one of my heat transfer projects, I plan on investigating the amount of useful heat energy that can be recovered from the waste heat rejected from a nuclear reactor, which could then perhaps be used to power an adsorption desalination plant. Unfortunately, I'm not entirely sure how I can go about doing this. Theoretically, how does one calculate the amount of heat energy available from the water in the condensers? Surely it can't
  • #1
bizarro186
3
0
Hello everyone,

I was wondering how one can go about accurately estimating the amount of the waste heat rejected from a PWR and the temperatures of said waste heat. Thank you very much.

Regards,
Jon
 
Engineering news on Phys.org
  • #2
By knowing the thermal output, the efficiency and the maximum power producing temperature.
 
  • #3
And the waste heat is rejected through the condenser, so its temperature would be controlled by the temperature of the river / lake / ocean / cooling tower water running through the condenser tubes.
 
  • #4
bizarro186 said:
Hello everyone,

I was wondering how one can go about accurately estimating the amount of the waste heat rejected from a PWR and the temperatures of said waste heat. Thank you very much.

Regards,
Jon
More or less as others have indicated. We can know mass flow rate and temperature in various systems of the plant, so we can determine the temperature (enthalpy drop) through the condenser in the secondary cooling system, or through the steam generator.

We can also estimate from the MWt (thermal power) and MWe (electrical). For PWRs (steam Rankine cycle), the efficiency varies from 32 to ~37%, so one rejects 63 to 68% of the nuclear thermal energy directly do the environment.
 
  • #5
... and to add further we can not further utilize this 63~ 68 % energy because temperature rise in condenser water is small. Secondly we can not further increase PWR theoretical efficiency from around 33% because steam it produce is saturated.
 
  • #6
bizarro186 said:
accurately estimating... the temperatures of said waste heat.

From the temperature of the ultimate heat sink.

Condenser temperature will be greater than cooling water temperature by the ΔT across the tubes... a few degrees.

Great Lakes water in wintertime or an evaporative cooling tower up north would be probably not much above freezing
while shallow seawater in S Florida in peak of summer , well, I've seen 98 degF.
Saturation temperature at absolute pressure of 1 inch of mercury is about 80 degf.
One may not be able to get condenser pressure much lower 1" due to practical limits of the vacuum pumps used to extract the little bit of air that leaks in.

Our condenser warmed its cooling water by about thirteen degF .
 
  • #7
Muti said:
... and to add further we can not further utilize this 63~ 68 % energy because temperature rise in condenser water is small. Secondly we can not further increase PWR theoretical efficiency from around 33% because steam it produce is saturated.
Actually, some of the German PWRs are around 35.7 to 36.4% efficient based on some innovation in turbine design by Siemens.
 
  • #8
Hello,

Thank you everyone for your replies. I am a mechanical engineering undergraduate student with little nuclear engineering knowledge. For one of my heat transfer projects, I plan on investigating the amount of useful heat energy that can be recovered from the waste heat rejected from a nuclear reactor, which could then perhaps be used to power an adsorption desalination plant. Unfortunately, I'm not entirely sure how I can go about doing this. Theoretically, how does one calculate the amount of heat energy available from the water in the condensers? Surely it can't just be Qr(Reactor thermal power) - Pe(Electric power output) = Qw(Heat dissipation)?

Thank you once again.

Regards,
Jon
 
Last edited:
  • #9
bizarro186 said:
Hello,

Thank you everyone for your replies. I am a mechanical engineering undergraduate student with little nuclear engineering knowledge. For one of my heat transfer projects, I plan on investigating the amount of useful heat energy that can be recovered from the waste heat rejected from a nuclear reactor, which could then perhaps be used to power an adsorption desalination plant. Unfortunately, I'm not entirely sure how I can go about doing this. Theoretically, how does one calculate the amount of heat energy available from the water in the condensers? Surely it can't just be Qr(Reactor thermal power) - Pe(Electric power output) = Qw(Heat dissipation)?

Thank you once again.

Regards,
Jon

That would be a decent starting point. I mean, really there are going to be ambient losses, small leaks, and there is going to be steam which is used to operate steam powered equipment which may have heat removed using other cooling water systems. But yea you can start with your Qw there.

I've seen 1000-1100 MWe plants (~3480 reactor MWth) with 2 waterboxes at about 600,000 gpm have a ~30 degF temperature rise across the condenser. That should give you a starting point. A 3 waterbox plant had 20-25 degF over a similar sized condenser. Both BWR series plants.
 
  • Like
Likes mheslep
  • #10
I see, thank you.

Does anyone know how I can go about calculating the temperature, pressure, and mass flow rate of the water within the condenser of a PWR just as it comes in contact with the secondary circuit? I cannot seem to find this anywhere.

Cheers :)
 
Last edited:
  • #11
Saturation temperature for steam at 27 inches of mercury vacuum (about 2-3 psia) wouldn't be a bad assumption. Large steam turbines cannot operate at low load with less than 25 inches hg of vacuum, and high load less than 24 inches continuously. Typical vacuum is 27 inches Mercury +/- 1. Again that's just approximate numbers for the couple large bwr units I've seen.

As for mass flow, it's going to be roughly equal to your reactor or steam generator steam output. Most nuclear plants boil their water at saturation conditions, so figure out how many pounds of water it takes to remove 3400 MWth from the core. Bwr pressure is in the 950-1050 range. Not sure what steam generator pressure for PWRs is.
 
Last edited:
  • #12
Hiddencamper said:
[ ... ] Not sure what steam generator pressure for PWRs is.
PSAT for TAVG?
 
  • #13
Doug Huffman said:
PSAT for TAVG?
minus the temperature difference across tubes..

Around 770 - 800 psig.
 
  • #14
jim hardy said:
minus the temperature difference across tubes..
Really? A measurable tube Delta_T?
 
  • #15
Our design was full power TAVG= 574F, steam temp = 516F, 770 psi.
Feedwater comes in at ~430 to 432F.

At shutdown of course Delta T across tubes is zero, so Tave =547 and steam pressure = 1005.

We had u-tubes.
B&W PWR plant has once through tubes, and I'm told.could make ~ 2 degrees superheat.
 
  • #16
Muti said:
... and to add further we can not further utilize this 63~ 68 % energy because temperature rise in condenser water is small. ...
The rejected low temperature heat can not be economically used in more conversion cycles, but it can certainly be used as is, that is for space/water heating in district heating applications. The current remote location of large nuclear plants makes this untenable, but the potential of small modular reactors and their much increased safety margins is that they'll be located close to the load and then allow the waste heat to be used for space heat. I don't know of any better method to eventually decarbonize heating.
 
  • #17
Hiddencamper said:
Saturation temperature for steam at 27 inches of mercury vacuum (about 2-3 psia) wouldn't be a bad assumption. Large steam turbines cannot operate at low load with less than 25 inches hg of vacuum, and high load less than 24 inches continuously. Typical vacuum is 27 inches Mercury +/- 1. Again that's just approximate numbers for the couple large bwr units I've seen.

As for mass flow, it's going to be roughly equal to your reactor or steam generator steam output. Most nuclear plants boil their water at saturation conditions, so figure out how many pounds of water it takes to remove 3400 MWth from the core. BWR pressure is in the 950-1050 range. Not sure what steam generator pressure for PWRs is.
I believe PWR steam pressure is about 900 to 1000 psi, for modern high duty plants. Alstom's Arabelle 1750 MW turbine for the EPR has a main steam pressure of 75 bar (at 290 C) or 1088 psia. Condenser pressure is 46 mbar.

http://www.alstom.com/Global/Power/Resources/Documents/Brochures/flamanville-france-nuclear-power-plant-wallchart.pdf?epslanguage=en-GB

http://www.power-technology.com/projects/flamanvillenuclear/
 
Last edited:
  • #18
mheslep said:
The rejected low temperature heat can not be economically used in more conversion cycles, but it can certainly be used as is, that is for space/water heating in district heating applications. The current remote location of large nuclear plants makes this untenable, but the potential of small modular reactors and their much increased safety margins is that they'll be located close to the load and then allow the waste heat to be used for space heat. I don't know of any better method to eventually decarbonize heating.
Such as the Pius reactor proposed by ABB atom in the 80s.. If it weren't for Chernobyl, one of those would have been built in Helsinki.
Pius was an intrimsicly safe reactor specifically designed for district heating. Heating water is what reactors are best at.

https://www.euronuclear.org/e-news/e-news-17/nps-kth.htm

Whoops, ABBAS called the district heating version SECURE. PIUS make both epheat and electricity.
 

1. What is waste heat rejected from PWRs?

Waste heat rejected from PWRs refers to the excess heat that is produced during the nuclear fission process in Pressurized Water Reactors (PWRs). This heat is generated as a byproduct of the nuclear reactions and needs to be removed from the reactor core to prevent overheating and potential damage to the reactor.

2. Why is it important to estimate waste heat rejected from PWRs?

Estimating waste heat rejected from PWRs is important for several reasons. First, it helps in the design and operation of the reactor cooling systems to ensure that the excess heat is effectively removed. Additionally, accurate estimation of waste heat can aid in the overall efficiency and performance of the reactor, as well as in the evaluation of potential environmental impacts.

3. How is waste heat rejected from PWRs estimated?

The estimation of waste heat rejected from PWRs involves complex calculations and modeling based on various factors such as reactor power, coolant flow rate, and reactor design. The most commonly used method is the heat balance method, which takes into account all heat inputs and outputs to the reactor core.

4. What factors can affect the amount of waste heat rejected from PWRs?

Several factors can affect the amount of waste heat rejected from PWRs. These include the reactor power level, coolant flow rate, reactor design and operational parameters, as well as environmental conditions such as ambient temperature and humidity. Any changes in these factors can impact the overall amount of waste heat rejected from the reactor.

5. How is waste heat rejected from PWRs managed?

Waste heat rejected from PWRs is typically managed through various cooling systems, such as the primary, secondary, and tertiary cooling systems. These systems use different heat exchange processes to remove the excess heat and dissipate it into the environment. Some PWRs also utilize waste heat for other purposes, such as district heating or desalination.

Similar threads

  • Nuclear Engineering
Replies
7
Views
2K
  • Thermodynamics
Replies
5
Views
1K
  • Nuclear Engineering
Replies
4
Views
1K
Replies
14
Views
1K
Replies
10
Views
2K
  • Thermodynamics
Replies
15
Views
1K
Replies
9
Views
736
  • Calculus and Beyond Homework Help
Replies
7
Views
837
  • Introductory Physics Homework Help
Replies
3
Views
965
Replies
9
Views
150
Back
Top