Pressure Drop???

[Adnoh97]

Does Anyone know the pressure drop (delta P) around the primary loop of a PWR using standard operating conditions?? Or how I can calculate the pressure drop??  I am currently in a class "Intro to NUC Reactor Sys" and we're being asked these type questions with no book or other source of help for calculations. Any info will be appreciated. ThANKS!

[Already Gone]

Well, that depends on the size of piping, the flowrate, the power of the pump... etc.  But, considering that the primary loop is a closed system, the pressure drop almost has to be zero.  Of course, there will be a pressure drop across the core, and some headloss in the piping, but that has to be equal to the pressure increase across the RCP.  Otherwise the pressure would constantly be rising or falling.

So, think Big Picture.  There has to be a pressure drop across the core and through the S'G's, or there would be no flow.  There has to be a pressure rise across the RCP, otherwise there would be no flow.  Those will always offset each other, because the flow is the cause for the dP.  Get it yet?  The pressure drop across the entire loop is always zero.

[thenuttyneutron]

I think about an 80 pound drop is a good number.  The core is about 50 pounds I think.  There is a small difference in pressure drop depending on the loop configuration and things like pressureizer, let down, and makeup taps.

I can double check this by looking at the data from my plant.  If my gut feeling is off, I will correct it.

[Fermi2]

Uh don't you have the size of the core plate and the height of the fuel bundles?

Mike

[Adnoh97]

No, my teacher didn't give us a height or any other variable. We haven't even done a single sample problem yet. He just wants us to find out the change in pressure around the primary loop. He elaborated by saying the pressure drop around the primary loop= the pressure drop across the reactor + the pressure drop across the steam generator.   And by the way, THANK YOU VERY MUCH to the ppl who answered above.

[wlrun3@aol.com]

   ...is the instructor looking for a pressure versus temperature calculation...

   ...if so the temperature changes on either side of the core and steam generators are easily had...

   ...broadzilla knows them without having to look them up...

   ...what is your school and program...

[S3GLMS]

If you are given the KW of the running pump in the loop at the steady state conditions you can back calculate the total pressure drop (Pump Head), based on the Pump and motor performance curves.  You can assume some industry standards for Motor efficiency = .94, and Pump efficiency = .83.  You need to have the correct Specific Gravity of the fluid pumped typically .95 for the hot water loop.  Also you need to know the Flow in the loop in Gallons per minute ( You can convert to this from a LBM/Sec number if that is given).

Then use the Equation:     Pump Brake Horsepower =  Flow(GPM) * Pump Head (In Feet) * Specific Gravity
                                                                                           3960 * Pump Efficiency

     From This you can then use the relationship that:   KW (pump) = Pump Brake Horsepower *.7457
                                                                                                      Motor Efficiency

By relating these two equations you solve for the one unknown which is:     pump head (In Feet).

You do not need to know all of the other specifics of the loop to obtain this overall number.  It includes pipe frictional losses as well as all of the other height variables and component through put losses. It does not quantify what each component headloss is only the over all total for the loop.  These numbers should be available from the system you have as basic operating parameters not related to the S/g's or the reactor fuel channels etc.  This is basic fluid flow centrifugal pump laws. 

If you don not have the pump information given such as flow or KW then you have no way to solve anything and why would you want if you do not have even basic operating data to work with.  It gives you no hard background to make system assessments.

[Already Gone]

Wow!  You have gotten some good knowledge from some of the best minds in nuclear power... and from me too.

Let us know how you do with this question.  And what the final answer turns out to be.

[Adnoh97]

We went to the professor and asked for some specifics. He gave us a table in the book to use. We have the flow in GPM, the pump efficiency, the KW of the pump, and the motor efficiency. I'll do the calculation and let you guys know the answer. Thanks again for all of the help.

[thenuttyneutron]

I looked today at my plant data and it is about a 150 psid difference between the suction and discharge of the RCP.

Plan parameters at 100% Rx Power are:

2155 PSIG RCS
144 MMPH flow rate
606 Thot
558 Tcold
582 Tave (Tave does not change on my plant after 28.5% power)

[Adnoh97]

I solved the problem using the equations listed above. Here are my Known Values:
Primary Coolant Pumps for a 1200 MW Plant (number of pumps = 4 pumps)
Flow (GPM) = 95,650
Specific Gravity = .95
Pump efficiency = .83
Motor efficiency = .94
I substituted the second equation into the first equation and solved for the Pump Head (in ft).
My answer was 54,715.41 (would the units be ft??)
Does this answer seem right?
Should this number be divided by 4???
Sorry for all the questions, but I keep getting bad grades on these homework assignments and I need to learn this stuff.
By the way I attend USC (Columbia) and the class is Intro to Nuclear Reactor systems.

[S3GLMS]

you need to divide total flow by 4.  then you need the KW of the individual pump involved in one loop.  In a Westinghouse plant  4 loop 1200 MW Electric gross power output,  this should be about 650 - 700 KW.  When you multiply all of this out, you should have:

Pump Head = 3,960* (.83) * (.94) * 650 KW
                          13,678 GPM * (.95) * (.7457)

                = 2,008,234.8       
                          9,690.3

                = 207.24 Feet (Water column)        Note 2.26 Feet of Water Column = 1 PSI

    Thus     207.24 Feet/2.26 Feet/PSI  =  91.69 PSI Differential head across the pump


    Each plant has a different KW number for the pump based on normal load conditions, temperature and pressure.  But this is very realistic for the 4 loop Westinghouse design.  I have worked on the 1275 MW 4 loop plants and pump KW loads were as high as 800 KW depending on coolant temp and core fuel status of life.

[S3GLMS]

I am updating the last post to use the flow that you provided of 95,650 GPM which is divided then by 4 to get the individual loop flow.  Without you providing the KW for the loop Pump, I still used the 650 KW based on my own experience.


Pump Head = 3,960* (.83) * (.94) * 650 KW
                  23,912.5 GPM * (.95) * (.7457)

                = 2,008,234.8        
                     16,940

                = 118.5 Feet (Water column)        Note 2.26 Feet of Water Column = 1 PSI

    Thus     118.5 Feet/2.26 Feet/PSI  =  52.46 PSI Differential head across the pump 

This would be the number that reflects the conditions given by the system parameters, if the pump KW was 700 or 600 KW very little would change the range would still be in the 50 to 60 PSI differential at this flow rate condition all of the rest of the physical attributes are based on the Physics of water.

[thenuttyneutron]

I solved the problem using the equations listed above. Here are my Known Values:
Primary Coolant Pumps for a 1200 MW Plant (number of pumps = 4 pumps)
Flow (GPM) = 95,650
Specific Gravity = .95
Pump efficiency = .83
Motor efficiency = .94
I substituted the second equation into the first equation and solved for the Pump Head (in ft).
My answer was 54,715.41 (would the units be ft??)
Does this answer seem right?
Should this number be divided by 4???
Sorry for all the questions, but I keep getting bad grades on these homework assignments and I need to learn this stuff.
By the way I attend USC (Columbia) and the class is Intro to Nuclear Reactor systems.

Take it all with a grain of salt.  I had to unlearn truths when I started working in a real plant.

[Fermi2]

you need to divide total flow by 4.  then you need the KW of the individual pump involved in one loop.  In a Westinghouse plant  4 loop 1200 MW Electric gross power output,  this should be about 650 - 700 KW.  When you multiply all of this out, you should have:

Pump Head = 3,960* (.83) * (.94) * 650 KW
                          13,678 GPM * (.95) * (.7457)

                = 2,008,234.8       
                          9,690.3

                = 207.24 Feet (Water column)        Note 2.26 Feet of Water Column = 1 PSI

    Thus     207.24 Feet/2.26 Feet/PSI  =  91.69 PSI Differential head across the pump


    Each plant has a different KW number for the pump based on normal load conditions, temperature and pressure.  But this is very realistic for the 4 loop Westinghouse design.  I have worked on the 1275 MW 4 loop plants and pump KW loads were as high as 800 KW depending on coolant temp and core fuel status of life.

Where did you get those BS Numbers? IN a 1200 MWE WEstinghouse 4 Loop REactor the RCPs are 7000 HP apiece which equates to roughly 5000 KW Cold and 6000 KW Hot (Which goes well with the TR of HP times .73 = KW.)
The to make it easier the developed Head of an RCP in a WEstinghouse 4 Loop Rector is 277 Feet. These numbers come straight out of the Calorimetric and the FSAR. Take 277 times .44 and you'll get your pressure drop.

Mike

[S3GLMS]

Just like I said in my post you can put any numbers you want in the equations based on how a plant system runs specifically loop flow, HP or KW of the motor and you will get the differntial pressure number that defines the loop.  There are Many commercial 4 loop reactors and not all are the same especially around the world.  So, put in the specific plant data. I would expect to get about 70 different sets of data from the plants in the US based on different loop configurations, flow rates and pump designs.  Remember your plant is not the one the original poster is asking about, the person was asking about the one in the "Classroom textbook table exercise" Which has specific variables.  So, relax and feel good about knowing your plant parameters. 

[wlrun3@aol.com]

   ...your response was a pleasure to read...

[Fermi2]

All Westinghouse 1200 MWE 4 Loop Reactors use the same RCPs. So no you don't get different KW ratings for the pumps. In fact Westinghouse gives a number to use in the calorimetric that is used in all Westinghouse 4 loop reactors.

[S3GLMS]

Okay, lets use real numbers from a real Westinghouse US four loop Plant.

WBN-1, May 96
Power 1246 MWe
Flow   144,000,000 LBM/Hour = 288,461.5 GPM/ 4 Loops = 72,115 GPM/Loop
Motor KW = 5850
Pump Eff  = .73
Motor Eff  = .91
.7457 = conversion KW to HP
3960 = conversion of FT- LB/Sec through HP and seconds to Minutes
Pump curve point is determined from initial design curve (with original S/G's and pump seals installed)

Pump Head = 3,960* (.73) * (.91) * 5850 KW      (Reference Camerons hydraulics Chapter 1-18th Edition)
                          72,115 GPM * (.95) * (.7457)

                = 15,389,173.8       
                     51,087.6

                = 301.23 Feet (Water column)        Note 2.26 Feet of Water Column = 1 PSI

    Thus     301.23 Feet/2.26 Feet/PSI  =  133.2 PSI Differential head across the pump (Calculated Point)

Reference (Displayed as 125 on westinghouse P2500 computer).

This exercise that the student was asking about is not an NRC test about the FSAR, it is about how do you get the numbers that are in the design background for the plant and what do they mean.  Any one can do this same one point steady state approach at their plant using their known equipment ratings, flows and KW of the system. 

This is how it works, the why, behind the numbers that are read in the control room and in the technical manuals.  Granted this takes some basic liberties by ignoring transient analysis, but for a 1 point assessment of an steady state operating condition these are the basics of how you approach the solution.  That was what was asked initially and answered to stay on topic.  (By the way I have the numbers from this time frame because I downloaded them from the P2500 and verified them on the Eagle 21 system during full unit surveillances during initial startup and full power testing).

[Fermi2]

But originally you used a made up number and pretended it was a fact. You never looked for the real number until challenged.

Mike

[S3GLMS]

     You have made an assumption about the information provided for the student.  Only data was given and stated as "realistic".  I wanted the student to get their own information on paper and get an answer.  The student never actually did provide all of the data they were using specifically the KW of their system from their table in the textbook they had.  Without their specific numbers I could not and would not do their homerwork problem for them. I was not going to do their calculations and let them run to their classroom with the homework done.  As it turns out the student never even posted their answer when Beercourt had asked them to let us know how it turned out for their exercise.  "Give a person a fish and they eat for a day, teach them to fish and they will feed themselves".  The first rule of getting someone to do their own work is teach them to "Fish" aka solve, and then they can solve for themselves.