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.