# Sizing a pressurizer

So I can't find anything online to help me with this because there isnt very many resources for this. If anyone knows of a good resource that would be great or some pointers on how to get started on this.

I am working a problem trying to find the volume of a pressurizer... here is the problem

A PWR has a 8000 ft^3 primary loop at 2200 psi. Hot leg temp is 600F and cold leg is 540F and the pressurizer if 60% full of water. Calculate necessary pressurizer volume for no more than 1% system pressure rise for a 10F rise in primary loop temperatures. Ignore spray flow and heat added.

I AM NOT asking someone to work this for me, I simply want to know what equations can help me get started because the book "Powerplant Technology" does not cover this very well and I can't find any other resources...

gmax137
Use water properties (e.g., subcooled steam tables) to find out how much that 8000 ft expands as the temperature changes. When it expands, it flows up the surge line into the pressurizer, squeezing the steam bubble. That tends to raise the steam pressure, by how much?

how to you find the pressure rise in the pressurizer though?

i found the mass of steam and water in the pressurizer and multiplying them by their corresponding specific volumes I got a volume of 8000 ft^3 which is the volume of the primary loop, so obviously I did something wrong

Astronuc
Staff Emeritus
One is given "the pressurizer if 60% full of water", so the other 40% is steam. The pressurizer is at saturated conditions - initially at the hot leg conditions.

The primary loop should be subcooled liquid. The max pressure is at the discharge of the reactor coolant pump, will the lower pressure is at the intake or end of the cold leg.

It's not clear to me if one is to assume that the pressurizer temperature also increases by +10F when the loop temperature increases. Initially there would be a lag on the increase in temperature of the coolant in the pressurizer, and the steam would not effectively conduct heat.

When the coolant temperature rises, what happens to the liquid volume? What then happens to the steam in the pressurizer?

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when the temp rises the liquid volume should decrease because pressure increases and more steam is produced.

This is making more since now!! If I find the before and after specific volumes of the steam and water I can find how large the pressurizer needs to be so that it can handle a 1% pressure rise.

Xprops works so great in MathCad for stuff like this!

Thanks for the help!

gmax137
Well, just keep in mind that the pressurizer is initially at ~650F (saturated at 2200 psia) and you said the hot leg is at 600F, so: when the RCS heats up by 10F you will be pushing 600F water into the bottom of the 650F pressurizer. So... the overall pressurizer liquid doesn't heat up (in fact, in an actual reactor the heaters will come on if pressurizer level is rising, to try and keep it saturated).

NUCENG
Well, just keep in mind that the pressurizer is initially at ~650F (saturated at 2200 psia) and you said the hot leg is at 600F, so: when the RCS heats up by 10F you will be pushing 600F water into the bottom of the 650F pressurizer. So... the overall pressurizer liquid doesn't heat up (in fact, in an actual reactor the heaters will come on if pressurizer level is rising, to try and keep it saturated).

You are getting there. No heaters or spray in the pressurizer. The steam is compressed causing some condensation, releasing heat. The volume of the pressurizer is determined by the new equilibrium of a 1% pressure increase and a new saturation condition.

NUCENG
when the temp rises the liquid volume should decrease because pressure increases and more steam is produced.

This is making more since now!! If I find the before and after specific volumes of the steam and water I can find how large the pressurizer needs to be so that it can handle a 1% pressure rise.

Xprops works so great in MathCad for stuff like this!

Thanks for the help!

Method
• You know initial and final conditions for the system.
• Unknown final Pressurizer steam volume is Pressurizer volume * (1- fractional pressurizer ligud volume) and can be eliminated by substitution.
• Total system Mass is constant and is a function of the unknown pressurizer volume and initial fractional pressurizer liquid volume.
• Heat is added to the primary loop raising temperatures by 10 degrees.
• This creates a change in loop volume at 1% increased pressure that has to expand into the pressurizer.
• The pressurizer mass changes by the amount of inflow and is a function of the unknown pressurizer volume and the new fractional pressurizer volume.
• The total energy in the pressurizer is determined by the energy of the inflow, the increased temperature in the existing pressurizer liquid and the condensation of steam volume required to heat the water to the new saturation conditions. Again, this is a function of pressurizer volume and the new fractional liquid pressurizer volume.
• This results in three equations (total mass, final pressurizer mass, final pressurizer energy) in three unknowns (total Mass, Pressurizer Volume and Final fractional liquid Pressurizer volume) and solution is at hand.

Assumptions.
• Constant system mass.
• Constant system total volume. No change due to temperature or pressure changes.
• Non-mechanistic heating of loop volume by 10 degrees.
• No heaters, sprays or pump heat.
• Adiabatic conditions No heat loss or gain from system except the loop heating
.

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gmax137
NUCENG, I think you're also assuming thermal equilibrium within & between the liquid and vapor in the pressurizer, as well as assuming homogeneous (saturated) conditions there.

NUCENG
NUCENG, I think you're also assuming thermal equilibrium within & between the liquid and vapor in the pressurizer, as well as assuming homogeneous (saturated) conditions there.

You are right! Ah - the unrecognized assumption! And I also assumed elevation head and flow resistence were negligible. Thank you!

jim hardy
Gold Member
Dearly Missed
FWIW :

my old W 3-loop (2200 mwth) had a 1300 ft^3 pressurizer, to best of my recollection

i don't know how you'd size one.
presumably to accept the "swell" of primary system without lifting a primary side safety valve when secondary main steam valves slam shut and heat is constrained to go out secondary side safeties--- (Thot and T cold both correspond to saturation temperature of secondary safety valve pressure setpoint) ????
equally important, to maintain primary side pressure above SI setpoint when primary cools suddenly from that temperature to ~540F.
"in the ballpark" setpoints ( they're plant specific)
normal pressure" 2235 psi
primary safety : 2485 psi
secondary safety: 1200psi
SI setpoint: 1800 psi

old jim

Astronuc
Staff Emeritus
The sizing is determined by initial conditions and the requirement "Calculate necessary pressurizer volume for no more than 1% system pressure rise for a 10F rise in primary loop temperatures." One simply determines the system volume change with the increase in primary system temperature, which compresses the steam in the pressurizer.

We know that initially, the pressurizer is 0.6 water, so 0.4 steam, so the initial volume of steam is 0.4V, where V is the volume of the pressurizer.

The problem doesn't go into the fact that the pressure drop around the loop is something like 50 psid, and probably more. The max pressure is out of the RCP, while the min pressure is at the intake. I remember a 12 ft core has something like a 25 psid drop.

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