Help | Contact Us
NukeWorker.com
NukeWorker Menu Pressure Increase = How much heat??

Author Topic: Pressure Increase = How much heat??  (Read 8896 times)

0 Members and 1 Guest are viewing this topic.

Adnoh97

  • Guest
Pressure Increase = How much heat??
« on: Feb 23, 2009, 02:49 »
Whats up everybody, I've got another homework question that I was looking to get some help with.  The question goes as follows:
For a 1200 MW Plant

a) The amount of heat needed to provide a 100 psi increase in the pressurizer
b) The amount of water from inlet to make 100 psi down in pressurizer
c) 10 cubic feet of water inserted into pressurizer, how much did the pressure go up.
Design Pressure    2500 psi (17 MPa)
Design Temp        670 F (354 C)
Steam Volume      1200 cubic ft (34 cubic meters)
Electric heater capacity    1745 kW
Shell minimum thickness   5.9375 in (15.1 cm)
Shell outside diameter    119.875 in (3.04 m)
Overall height       41.5 ft (12.65 m)

I realize that I can get the volume for the cylinder using pi*r^2*h and then subtract the volume of steam to get the volume of water inside. But I can't use the natural gas laws because I'm am dealing with boiling water in liquid form. What equation can I use for this relation? Should I go to my thermo table and get the volume of water vapor at this pressure and temp? Any help will be greatly appreciated. Thanks for putting up with me and my LONG questions. By the way, That last question you guys answered helped my entire class. THANKS!!

S3GLMS

  • Guest
Re: Pressure Increase = How much heat??
« Reply #1 on: Feb 23, 2009, 12:07 »
To answer this question the actual conditions in the pressurizer at the starting and ending conditions need to be stated or assumed to determine the method of reaching the final conditions.  Looking at the data you have provided, a simple set of conditions can be assumed to simplify the method of solution.  As an exercise example this is the method of solution, any actual numbers can be inserted to abtain the results.

Looking at each question as its own separate problem, take a.) first:

What is the heat needed to make a 100 PSI increase in pressure?

Assumptions:
No Spray activations occur during the pressurization.
Only heaters provide heat into the pressureizer Volume.
No overall change in Steam volume occurs.
No changes in ambient heat losses during the pressurization.
Steady rate of heat input from the heaters during the pressurization.
Initial conditions of pressurizer are that the steam and water are saturated.
Final Steam conditions are slightly superheated when heaters de-energize.
Pressurizer is indeed a cylinder without hemispherical heads.

Initial conditions        To = 642 F, Po = 2100 PSIA, Steam specific Volume Vo = .17501 Cubic Feet/LBM
                                       Water specific Volume Vo = .02615 Cubic Feet/LBM ( Ref. Saturated Steam Tables)
                                       Enthalpy: H steam = 1130.5 BTU/LBM, H Water = 683.8

Final Conditions         Tf = 665 F, Pf = 2200 PSIA, Steam Volume Vo = Vf (Reference Superheated Steam Tables)
                                       Enthalpy: H steam = 1152.1 BTU/LBM, H Water = 695.5

Pressurizer Steam Volume = 1200 Cubic FT, Total Volume = 3,205 Cubic FT, Water Volume = 2005.3 Cubic feet 
(This was computed using an inside diameter of 9'-11" given above - 6" Wall thickness X 41.5' Wall height Cylinder no hemispherical heads)

Steam Volume Calculations
Find LBM Steam to be heated:    1200 Cubic Feet X 1LBM/.17501Cubic feet = 6,856.8 LBM Steam
Find Steam Enthalpy Difference: H Steam Final  - Hsteam initial = 1152.1-1130.5 = 21.6 BTU/LBM
Find Steam Heat required:       6,856.8 LBM X 21.6 BTU/LBM = 148,105.8 BTU
Convert to KW:                   148,105.8/ 3413 Btu/KW-hour = 43.39 KW-Hour


Water Volume Calculations
Find LBM Water to be heated:    2005.3 Cubic Feet X 1LBM/.02615Cubic feet = 76,684.5 LBM Water
Find Steam Enthalpy Difference: H water Final  - H water initial = 695.5-683.8 = 11.7 BTU/LBM
Find Steam Heat required:       76,684.5 LBM X 11.7 BTU/LBM = 897,074.6 BTU
Convert to KW:                   897,074.6/ 3413 Btu/KW-hour = 262.8 KW-Hour


Add Steam and Water heat numbers together for total volume heating incurred:
262.8 KW + 43.39 KW = 306.2 KW    Solution. 

There are a number of significant transients that can occur that is why it is more difficult to solve when you include the movement of fluid and volumes as well as varying heat rates into the transient of pressure increase.  But as presented this is a simplistic method to determine full volume ehaeting while ignoring some of the other more detailed issues.  Hope this helps.  If I get time I will answer B and C Later.

S3GLMS

  • Guest
Re: Pressure Increase = How much heat??
« Reply #2 on: Feb 23, 2009, 02:12 »
The Answer to the Second Question, B) requires some different assumptions about the transient change in the pressurizer to create a plan to solve it.

What is the amount of water removed from the Pressurizer to drop pressure 100 PSI?

Assumptions:

No spray actuation occurrs during the event.
Heaters are constant and no change in energy level.
No overall temperature change occurs during the time of event.
All fluid that leaves is liquid through the submerged outlet.
Ignore height diferences as small.

Initial Conditions:  Po = 2100 PSIA,  Vo = 1200 Cubic Feet

Final Conditions:   Pf = 2000 PSIA, Vf = Unknown

The relationship of the pressure of the Steam volume for a steady state system with these assumptions can be viewed as:

 PV = nRT.  An Ideal Gas relationship. Where P = Pressure, V = Volume, n = molculer ratio, R = Gas Constant

For temperatures between 65 F and 1400 F values of R and Specific Heat remain relatively stable for Water. ( ref Engineering Thermodynamics 6th Ed - Shapiro Table A-21E).

The only work or energy change of the steam is due to volumetric change.  Thus, the two states can be set equal to each other. The n and R terms are considered equal and cancel.

Po X Vo = Pf X Vf    Re-arrange this to solve for the final Volume (Vf),

Vf = Po X Vo/Pf   =  2100 PSIA X 1200 Cu Ft/2000 PSIA = 1260 Cu Ft

The difference in Volume id Vf - Vo = 1260 Cu Ft - 1200 Cu Ft = 60 Cu FT    Solution ( This is Equivelant to approximately 448 Gallons of Water.


A similar approach can be used to solve the third Question C.)

If 10 Cubic Feet of Water are added to the Pressurizer, what will be the pressure increase?

By using mostly the same assumptions as above, the P-V relationship can again be used.

No spray actuation occurrs during the event.
Heaters are constant and no change in energy level.
No overall temperature change occurs during the time of event.
All fluid that enters is liquid through the submerged surge line.
Ignore height diferences as small.

In this case though the difference in Volume is given as 10 Cu Ft Which Makes Vf = 1190 Cu FT, so the unknown value is Pf.

so set:

 Pf = Po X Vo/Vf   =  2100 PSIA X 1200 Cu Ft/ 1190 Cu FT  = 2117.6 PSIA 

Take the difference between the Vf and Vo = 2117.6 PSIA -2100 PSIA = 17.6 PSIA  Solution.

This is the simplified method of steam volume problems, there are a host of other issues that can be considered when you allow temperature and heat addition to vary during the process transient, but the results are similar in scale.

 


NukeWorker ™ is a registered trademark of NukeWorker.com ™, LLC © 1996-2024 All rights reserved.
All material on this Web Site, including text, photographs, graphics, code and/or software, are protected by international copyright/trademark laws and treaties. Unauthorized use is not permitted. You may not modify, copy, reproduce, republish, upload, post, transmit or distribute, in any manner, the material on this web site or any portion of it. Doing so will result in severe civil and criminal penalties, and will be prosecuted to the maximum extent possible under the law.
Privacy Statement | Terms of Use | Code of Conduct | Spam Policy | Advertising Info | Contact Us | Forum Rules | Password Problem?