Maximum Pressure in Nuclear ReactorContainment Vessel

Dayan Paez2.672 Project Laboratory

Department of Mechanical EngineeringMassachusetts Institute of Technology

Cambridge, Massachusetts 02139Email: paez@mit.edu

Michael AllshouseDepartment of Mechanical EngineeringMassachusetts Institute of Technology

Email: mra23@mit.edu

Qinyuan LiuDepartment of Mechanical EngineeringMassachusetts Institute of Technology

Email: qin87liu@mit.edu

This experiment simulates a catastrophic break in a nuclearreactor by releasing pressurized saturated steam and/or wa-ter to an attached containment vessel. A model is developedto determine the maximum pressure in the containment ves-sel. Various tests are performed to verify the model. With aworking model, the necessary design parameters for a con-tainment vessel that will successfully withstand the pressuresinvolved in a sudden release of steam and liquid are found.

NomenclatureVR Volume of reactorVCV Volume of containment vesselHR Height of reactorHCV Height of containment vesselh Height of initial water columnPsat Initial pressure of steamTsat Initial temperature of steamP0 Initial pressure of air in containment vesselPfinal Final pressure in containment vesselRs Gas constant for steamRair Gas constant for aircs Specific heat for steamcair Specific heat of air

1 IntroductionBecause a nuclear reactor tank is under high pressure,

it is important to design a safety mechanism to contain thespill should a rupture occur. For cost reasons, it is impor-tant to properly design such a containment vessel to resistthe maximum pressure caused by the sudden release of thereactor’s contents without over engineering it.

Fig. 1. Pressure vs. time for steam rupture, Psat= 110kPa, gauge

In this experiment, a nuclear reactor is simulated bybringing steam and water to a two-state phase equilibrium,and then releasing a steam or water valve connected to a con-tainment vessel. The derivation of an adequate model for thisrupture process will aid in the design recommendations.

A typical time-evolution plot of pressure in the contain-ment vessel is shown in Fig. 1. The plot shows an exponen-tial rise to a peak pressure and a subsequent steady decline.One possible explanation for this phenomenon is that me-chanical equilibrium happens significantly faster than ther-mal equilibrium, and the slow decrease in pressure in Fig. 1is evidence of this difference in time scales.

Proper modeling of the system requires a set of assump-tions which insert a discrepancy between the model and re-ality. In this report, different assumptions are examined andtheir consequent discrepancies from reality scrutinized.

Steam valve

?

Water valve6

ReactorVR, HR

SteamPsat, Tsat

Sat. H2O

Containment

Vessel

VCV, HCV

Air, P06

h

Fig. 2. Schematic of the nuclear reactor system.

2 Experimental Setup

A simulated reactor tank is connected to hot water andsteam inputs. The temperature in this tank can be continu-ously measured. A pressure gauge indicates instantaneouspressure. Water level is tracked on an external guide and canbe used to determine the qualityχ of the system.

Two valves—a low one for water, a top one for steam—can be opened to simulate a catastrophic break in the reactortank letting the discharge fill a containment vessel. This pres-sure tank is of a larger volume. The pressure in the contain-ment vessel is continuously tracked with a sensor1 connectedto a computer.

The system is shown schematically in Fig. 2.

2.1 Steam release

To simulate a steam rupture, the reactor was filled withsteam to a pre-determined pressurePsat and allowed to cometo equilibrium. With a fixed pressure in the reactor, the ini-tial pressure in the containment vesselP0 was varied to threepre-determined values by filling the vessel with air (the ves-sel was assumed to be at room temperature). The procedurewas then repeated for two other values ofPsat to get a roughsample of the dependency ofPfinal on the two parameters.

In addition, a run through several values ofPsat at con-stantP0 = atmosphere were also carried out (Fig. 4).

2.2 Water release

To simulate a reactor failure at the water end, the tankwas filled with a controlled proportion of water and steamand allowed to come to thermal equilibrium (in a two-phasestate). The water valve was then opened while the pressurein the containment vessel was recorded (see Fig. 1).

While maintaining the water levelh constant, the proce-dure described in Sec. 2.1 was carried out in order to obtaina relationship betweenPfinal and the two parametersPsat andP0. Similar to the procedure for steam breach, various runsthrough several values ofPsat at constantP0 = atmospherewere run to get a better sample of the relationship betweennuclear reactor pressure and containment vessel peak pres-sure.

Fig. 3. Pfinal vs. Psat for varying values of initial containment pres-

sure P0 (absolute pressures) in a steam breach

Fig. 4. Pfinal vs. Psat for P0 = atmand a steam breach

3 Data

3.1 Steam Breach

Fig. 3 shows howPfinal varies as a function ofPsat, thepressure of the steam in the reactor. The entire parameterspace was sampled judiciously due to time constraints, but itis nonetheless evident that the maximum pressurePfinal varieslinearly with initial reactor and containment vessel pressure.

This relationship is also evident in Fig. 4, which showshow peak pressure varies with saturated pressure for a rangeof points.

3.2 Water Breach

Fig. 5 plotsPfinal as a function ofPsat for varying val-ues ofP0–similar to Fig. 3 but this time simulating a ruptureat the water end. It is evident from the plot that the peakpressure in the containment vessel for this kind of rupture isproportional with the initial pressure in the reactor vessel. AsFig. 6 indicates, the peak pressure is generally higher for awater rupture than for a steam rupture at the same conditions.

Fig. 5. Pfinal vs. Psat for varying values of initial containment pres-

sure P0 (absolute pressures) in a water breach

Fig. 6. Pfinal vs. Psat for P0 = atmand a water breach

Fig. 7. Tsat vs. Psat during steam breach

3.3 Pure Substance ModelIn order to be sure that the pure substance model was

applicable, the temperature at thermal equilibrium beforerupture—the so-calledTsat—was measured along with thecorrespondingPsat. The resulting plot is shown in Fig. 7along with a linear fit. Here again, the results were sur-prisingly linear, giving confidence to the assumption that thesteam and the water are in a two-phase state before rupture.

1Model: Omega PX303-100G5V

(a) (b) (c)

Fig. 8. Three different models, a) Steam release, mixing; b) Water

release, no mixing; c) Water release, mixing.

4 Model4.1 Scenarios

Depending on which valve is opened the system behavesdifferently. Because of this, two general models were de-vised. In the case that both valves are opened, simulatingtwo ruptures in the nuclear reactor—one at the steam endand one at the water end—the solution can be obtained bysuperposition or considering the more dominant of the twoeffects.

In addition, opening the water valve can be further sub-divided into two different scenarios. In the first, the initialpressure difference from reactor to containment vessel is lessthan the hydrostatic pressure of the water,

Psat−P0 < ρgh. (1)

In this case, when the valve is opened, only a fraction ofthe water will move from reactor to containment vessel whilethe gases remain in their separate chambers.

In the second scenario, the pressure difference is largerso that even after all the water has rushed from reactor to con-tainment vessel, the remaining steam pressure is still higherand forces steam to flow through the valve and into the con-tainment vessel. In this slightly more complicated case, thecontainment vessel contains all three substances in mechani-cal equilibrium.

The three different scenarios are depicted pictorially inFig. 8.

4.2 Steam ReleaseIn the case of a steam break, the pressurized steam

rushes from reactor to containment vessel until the pressurein the two chambers is equalized. Although the two gasesmix, it is assumed that the temperature of each remains un-changed as there is no effective heat transfer between themin so small a time scale. Force balance around the valve isthen

PRs = PCV

s +PCVair = Pfinal, (2)

wherePRs is the partial pressure of the steam that remains in

the reactor,PCVs is the partial pressure of the steam in the

containment vessel, and similarly for the partial pressure ofair (= P0). The former two are related by conservation of

mass of the steam. Applying the ideal gas law yields

Pfinal =VR

VtotPs+

VCV

VtotP0. (3)

4.3 Water ReleaseThe possible results depending on the initial pressures

in the chambers when the water valve is opened are shown inFig. 8b, c. The first of these is a simpler case resulting in nomixing of the gases as the water moves partially from reactorto containment vessel.

4.3.1 No Mixing RuptureIn this scenario, the individual gases can be treated as

undergoing a completely isothermal expansion or compres-sion, since they remain separate from each other (heat trans-fer rate from liquid to gas—convection—is a relatively slowprocess). A force balance requires that at equilibrium, thepressure at either side of the open valve be the same,

Pseq +ρghR = Pfinal +ρghCV (4)

whereh represents the height of the water column in thechambersafterequilibrium. Conservation of mass of the wa-ter relates the initial water column height to the final ones

hHR

VR =hR

HRVR +

hCV

HCVVCV. (5)

Adiabatic expansion ((PV)1 = (PV)2) of each of thegases provides two more equations relating the initial pres-sures and volumes of each gas to their final conditions. Usingsystem parameters:

Pseq = PsatHR−hHR−hR

, (6)

Paireq = P0HCV

HCV−hCV. (7)

Eqns. 4–7 are a system of four non-linear equations infour unknowns which can be solved forPfinal.

4.3.2 Mixing RuptureMixing of the gases occurs after the system has achieved

a state as in Sec. 4.3.1 but is still not at equilibrium, with ahigher pressure in the reactor than the combined air and hy-drostatic pressure in the containment vessel. In this scenario,steam must flow into the containment vessel in an effort toequalize the system, bubbling through the water and mixingwith the air above it.

Although the steam bubbles through the water providingconsiderable opportunity for heat transfer between the two

fluids, their temperature is so close having been at thermalequilibrium just before, that the temperature of the steam canbe safely assumed to remain unchanged by the process.

The model is developed similarly as in scenario (b).Force balance at the water valve requires the pressure on bothtanks to be the same,

PRF = PCVF +ρghF

wherePCVF is the final gas pressure in the containment ves-sel: the value to design for. It is a sum of the partial pressuresof each of the gases that end up in the vessel, which again areassumed to remain at their respective temperatures. Becauseof this, the final air pressure is related to the initial air pres-sureP0 by Boyle’s Law for an ideal gasPV = const.

Mass conservation for the water relates the water col-umn height in the containment vessel to the original watercolumn height in the reactor. In addition, mass conservationarguments for the steam relate the initial steam pressure tothe final steam pressure in both tanks,

PsFVF

Rs=

PsatχVR

Rs−

PRfVR

Rs.

Solving the system of equations yields the final relation-ship forPfinal = PCVF in terms of the system parameters

Pfinal = PsatχVR−(

ρghF +P0VCV

VF

)VR

VF +VR(8)

where

hF = hHR

HCV

VCV

VR(9)

VF = χCVVCV =(

1− hF

HCV

)VCV. (10)

4.4 Subsequent ModelsThe models of Sec. 4.3 are based on the assumption that

the mechanical equilibrium condition is one of equal pres-sure on either side of the open valve.

In the initial state of the system, the steam and waterin the reactor are in a two-state phase. Because of this, itis appropriate to model the system using the pure substancemodel and not the ideal gas law, particularly during this stageof the process. Early attempts at modeling the system as-sumed the ideal gas relationship to hold for the steam in thereactor for the sake of simplicity, but data taken definitivelydemonstrates the inaccuracies associated with so simplistic amodel (see Fig. 11).

Fig. 9. Experimental and model prediction for Pfinal vs. P0 for steam

rupture (Psat= 165kPa absolute)

Fig. 10. Experimental and model prediction for Pfinal vs. P0 for

steam rupture (Psat= 185kPa absolute)

Indeed, the ideal gas assumption was only accuratewhen varying the initial pressure of the containment vessel,P0, due to air at room temperature.

This expression is equivalent to an independent, adia-batic expansion in each of the gases. Fig. 9 and Fig. 10 showthe actual and predicted dependency of maximum, final pres-sure on the initial air pressure of the containment vessel. Inthis domain, where ideal gas relationships hold, the correla-tion is pleasingly accurate.

4.4.1 Ideal Gas and1st LawTreating the steam as an ideal gas, and using the First

Law of thermodynamics around a control volume encom-passing the entire system (no net work or heat transfer acrossboundary layer), the model predicts the relationship,

PsatχVRcs

Rs+

P0VCVcair

Rair= PfinalVtot

[cs

Rs+

cair

Rair

], (11)

whereχ = 1− hHR

is the quality of the steam initially in thereactor, andVtot = χVR +VCV is the total volume availableto the gases. The peak pressure isPfinal in Eqn. 11. This is

Fig. 11. Experimental and model prediction for Pfinal vs. Psat for

steam rupture (P0 = 135kPa absolute)

evidently an incorrect model for the steam rupture as shownin Fig. 11. A similar result can be observed for scenarios (b)and (c), where there is a water rupture.

4.4.2 Isentropic ProcessThe preceding models assumed that the time scale for

the mechanical equilibrium was much quicker than for ther-mal equilibrium. Because of this, it was further assumedthat the temperature of the steam remained the same be-fore rupture and at peak pressure. Finally, in the preced-ing models, the equilibrium condition was understood to bereached when the pressures on either side of the open valvewas equal.

Michael Allshouse later noted during the data acquisi-tion that water flow in the case of water rupture continuedeven after peak pressure, possibly due to the inertia of thesystem.

A new model (see Sec. A.2) was then derived byQinyuan Liu which rejected the condition of equal pressuresand used chiefly the 1st and 2nd law of Thermodynamics andconservation of mass. Such a model inherently assumed theprocess to be isentropic, an assumption presumably conser-vative.

How appropriate a model fits the data can be seen fromthe predicted slope betweenPfinal vs. Psat. For the case ofonly steam rupture, the data and a linear fit are shown inFig. 3. The data are (surprisingly) linear2. The isentropicprocess, however, predicts a slope that is roughly one orderof magnitude less than expected. This non-conservative errorin the slope is a common theme of the various methods andmodels tried. Systematic errors in the data acquisition mayhave contributed to this global discrepancy (see Sec. 5.2.1).

4.4.3 Pure Substance ModelSome work, though admittedly not exhaustive, was done

with the pure substance model. First, it was verified that thesteam mass calculations found with the ideal gas lawPV =mRT were accurate even for a pure substance by means of

2See Sec. 5.2 for possible discrepancies.

Fig. 12. Linear fit with slope 1 between the predicted masses using

ideal gas and pure substance model

plotting one model’s prediction against the other for a givenset of saturated pressures. The result is shown in Fig. 12.

The figure demonstrates concretely that mass calcula-tions were not a source of error in the previous models. Incontrast, the energy predicted by each model varied signifi-cantly,

EPS

EIG≈ 1000, (12)

a surprising result of great consequence since it hints at theroot for the global discrepancy discussed in Sec. 4.4.2.

Because of interpolation errors in using the steam ta-bles, however, solutions using a complex set of equations(See Sec. A.3) led to and highly nonlinear fit.

4.4.4 Energy DrivenIn an effort to simplify the model to basic principles, a

rough estimate model was created taking the steam as thecontrol volume (to simulate the simpler case of only a steambreach). Application of the First Law on the steam yields

∆Esteam= W (13)

whereW is the work done on the gas by expanding it. Conse-quently, by usingW = PV and assuming all the energy of thegas goes into this expansion (worst case scenario:Efinal = 0,

Einit = msug = PfinalVfinal−PsatχVr (14)

where againχ is the quality of the steam.The result of this model is shown in Fig. 13. The gross

underestimation in slope means that at higher pressures themodel fails, a result consistent with the previous models.

Fig. 13. Energy driven model and data. Expected slope for data is

around 0.74.

5 Discussion

Of the aforementioned models, the ones that showedmost promise were rooted in the pure substance model. It isgenerally better to simplify the models in order to get morereliable results and introduce fewer parameters and sourcesof errors.

Generally speaking, the data show that a water ruptureinduces a greater pressure in the containment vessel thansteam rupture alone. It is important to note that in all casesthe peak pressure in the containment vessel was less than thesaturated pressure in the reactor, though the ratio of the twowas near unity for the case of water rupture.

This general result—that the peak pressure is on the or-der of the reactor vessel—should be a guiding principle inthe safe design of the containment vessel. Taking into ac-count the original pressure of the containment vessel,P0, thecontainment vessel should be designed to withstandP0+Psat

(see Sec. 5.1.1). Indeed, the data show that a vessel so de-signed would be grossly overengineered. Moreover, physicsdictates that in no circumstance, can the pressure gain in thelarger containment vessel exceed that of the saturated pres-sure in the reactor.

5.1 Parameter Effects

5.1.1 Sensitivity toP0

The effect of changingP0 is to increase the overall peakpressure of the containment vessel after rupture. This is truein all circumstances, regardless of water or steam rupture, asmentioned in Sec. 3.

One advantage to pressurizing the containment vessel isthat it reduces the shock of the rupturing reactor. Indeed,were the vessel to be at the same or higher pressure as thereactor, there would never be a spill! But of course, such acontainment vessel would be equally dangerous as the reac-tor it contains, rendering it useless.

Because of this, though helpful in reducing the initialimpact, or shock, of the breaking reactor, it is not advisableto pressurize the containment vessel.

5.2 Sources of ErrorsThere were various sources of errors in this experiment,

some of which were outside the control of the experimenters.These include errors due to insufficient time, resources, andinabilities to change the experimental setup.

5.2.1 Experimental SetupA significant error was systematically introduced in the

system by the experimental setup. Largely out of the exper-imenters’ control, the setup instruments read at times non-zero pressure gauges even with a flushed system. In addi-tion, having to conform to a pre-existing setup without thepossibility of measuring other quantities (e.g.T0, or the tem-perature in the containment vessel) precluded the decisivecorroboration of certain assumptions.

6 ConclusionIt is indubitable that none of the models explored for

the reactor system provide sufficient explanation or fit to thedata due to the number of assumptions that need to be madein order to make the system solvable analytically. Notwith-standing, it is possible to make safe recommendations basedon the data acquired for the design of the containment vesselbased on the initial states of the system. It is best to assume,when designing, that a water rupture will occur (and not asteam rupture) as this would result in conservative errors. Itis not obvious that further development of a model beyondan empirical nature is necessary or beneficial, as the exercisewould not yield any more understanding of the system thanof the way to manipulate equations in mathematical software.

AchnowledgementsThe authors would like to thank supervising professor

Richard Kimball, visiting faculty for the Massachusetts In-stitute of Technology Mechanical Engineering Department.In addition, the main author wishes to thank Harry Cheng,of the University of California, Davis, for producing theASME Journal Style LATEX template on which this documentis based.

A MatLAB FunctionsThe following appendix includes a list of MatLAB functions used.

A.1 v2p.m% Converts Omega PX303-100G5V output voltage to pressure in kPa

function p = v2p (voltage)

p = 20*(voltage-0.5)*101.3/14.7;

A.2 modelChen.m% Model for steam rupture% Ideated by Chen Liu, with input from MRA and DP%% [pfinal] = modelChen( PSAT, P0 )%% PSAT = Initial steam pressure in reactor% P0 = Initial air pressure in containment vessel%% pfinal = Peak pressure in containment vessel%% Assumes: Sgen = 0% T0 is constant and at room temp (300 K)% Adiabatic system (no heat transfer to surroundings)

function pfinal = modelChen(Psat, P0)

% ParametersHr = 0.32; % m height of reactorHcv = 0.7747; % m height of vesselVr = 0.0148; % mˆ3 volume of reactorVcv = 0.0645; % mˆ3 volume of vesselCs = 1560; % J/kg-K specific heat of steamCa = 1012; % J/kg-K specific heat of airRs = 461.9; % J/kg-K gas constant of steamRa = 287.05; % J/kg-K gas constant of air

% Variablesh = 0.13; % m water column height in reactorchi = 1-h/Hr; % - quality of steamVs1 = chi*Vr; % mˆ3 initial volume of steam in reactorVa1 = Vcv;Vs2 = Va1+Vs1; % mˆ3 final volume available to gasesVa2 = Va1+Vs1;Ps1 = Psat; % Pa absolute initial pressure of reactorPa1 = P0; % Pa absolute initial pressure of vesselTs1 = steam_t(Ps1);Ta1 = 300; % K room temperature

Ms = Ps1*Vs1 / (Rs*Ts1); % kg mass of steamMa = Pa1*Va1 / (Ra*Ta1); % kg mass of air

% Equation 1: 1st Law + Ideal gaseq1 = ’Ps2 + Pa2*(Rs*Va2*Ca)/(Ra*Vs2*Cs) - Ps1*(Vs1/Vs2) - Pa2*(Va1*Ca*Rs)/(Vs2*Cs*Ra)’;eq2 = ’Ms*(Cs*log((Ps2*Vs2)/(Ps1*Vs1)) + Rs*log(Vs2/Vs1)) + Ma*(Ca*log((Pa2*Va2)/(Pa1*Va1))+Ra*log(Va2/Va1))’;

% Range through Ps2, Pa2Ps2Range = (0.01e4:1e4:4e5); % Ranges for Ps2Pa2Range = Ps2Range; % Ranges for Pa2

bestPs2 = 0;bestPa2 = 0;runningMin = 1e12;

for i = 0:length(Ps2Range)-1% disp(i);for j = 0:length(Pa2Range)-1

% Evaluate eq1, eq2

Ps2 = Ps2Range(i+1);Pa2 = Pa2Range(j+1);

eqIndex = i*length(Pa2Range)+j+1;

Evaluation = [abs(eval(eq1)), abs(eval(eq2))];

if ( sum(Evaluation) < runningMin )bestPs2 = Ps2bestPa2 = Pa2runningMin = sum(Evaluation);pause

endend

end

%size(EQUATION1);%size(EQUATION);

% Find minimum values%[minVal, minIndex] = min(EQUATION);

%jOpt = mod(minIndex, length(Ps2Range))%iOpt = (minIndex-jOpt)/length(Ps2Range)

%PsOpt = Ps2Range(jOpt+1)%PaOpt = Pa2Range(iOpt+1)

Ps2Pa2pfinal = Ps2+Pa2;

A.3 modelMRA.m% Model for steam rupture% Ideated by Michael Allshouse, with input from QL and DP%% X = modelMRA( )%% Assumes: Sgen = 0% T0 is constant and at room temp (300 K)% Adiabatic system (no heat transfer to surroundings)

function X = modelMRA()

Psat = 150e3;P0 = 100e3;

% ParametersHr = 0.25; % m height of reactorHcv = 0.7747; % m height of vesselVr = 0.0148; % mˆ3 volume of reactorVcv = 0.0645; % mˆ3 volume of vesselCs = 2080; % J/kg-K specific heat of steamCa = 1012; % J/kg-K specific heat of airRs = 461.9; % J/kg-K gas constant of steamRa = 287.05; % J/kg-K gas constant of air

% Variablesh = 0.13; % m water column height in reactorchi = 1-h/Hr; % - quality of steamVs1 = chi*Vr; % mˆ3 initial volume of steam in reactorVa1 = Vcv;Vs2 = Va1+Vs1; % mˆ3 final volume available to gasesVa2 = Va1+Vs1;Ps1 = Psat; % Pa absolute initial pressure of reactorPa1 = P0; % Pa absolute initial pressure of vesselTs1 = steam_t(Ps1)Ta1 = 300; % K room temperature

Ms = Ps1*Vs1 / (Rs*Ts1) % kg mass of steamMa = Pa1*Va1 / (Ra*Ta1) % kg mass of air

global alpha beta gamma;

alpha = Cs*Vs2 / Rs;beta = Ca*Va2 / Ra;gamma = Psat*Ca*Vs1/Rs + P0*Ca*Va1/Ra;

Pa2 = fzero(@paFun, 1e5)Ps2 = gamma/alpha - Pa2*beta/alpha

X = Pa2 + Ps2;

function y = paFun(x)

Psat = 150e3;P0 = 100e3;

% ParametersHr = 0.32; % m height of reactorHcv = 0.7747; % m height of vesselVr = 0.0148; % mˆ3 volume of reactorVcv = 0.0645; % mˆ3 volume of vesselCs = 2080; % J/kg-K specific heat of steamCa = 1012; % J/kg-K specific heat of airRs = 461.9; % J/kg-K gas constant of steamRa = 287.05; % J/kg-K gas constant of air

% Variablesh = 0.13; % m water column height in reactorchi = 1-h/Hr; % - quality of steamVs1 = chi*Vr; % mˆ3 initial volume of steam in reactorVa1 = Vcv;Vs2 = Va1+Vs1; % mˆ3 final volume available to gasesVa2 = Va1+Vs1;Ps1 = Psat; % Pa absolute initial pressure of reactorPa1 = P0; % Pa absolute initial pressure of vesselTs1 = steam_t(Ps1);Ta1 = 300; % K room temperature

Ms = Ps1*Vs1 / (Rs*Ts1); % kg mass of steamMa = Pa1*Va1 / (Ra*Ta1); % kg mass of air

alpha = Cs*Vs2 / Rs;beta = Ca*Va2 / Ra;gamma = Psat*Ca*Vs1/Rs + P0*Ca*Va1/Ra;

Pa2 = [0.9e5:100:2e5]; % Independent variable in "Eq. 9"

y = real(Ms*Cs*log( (gamma/alpha - x*beta/alpha)*Vs2/Psat/Vs1) + Ms*Rs*log( Vs2 / Vs1) + Ma*Ca*log( x*Va2/P0/Va1 ) + Ma*Ra*log( Va2 / Va1));

A.4 modelDP.m% iterative run through steam tables%% Solves: Pmax Vtot - Psat Vinit = ms ug1 - ms ug2%% DP

function runningPmax = modelDP(Psat)

Vinit = 0.0089; % mˆ3, chi*VrVtot = 0.0734; % mˆ3, chi*Vr + VcvRs = 462; % J/kg-KTsat = steam_t(Psat); % K

ms = Psat*Vinit / (Rs*Tsat); % mass of steam

PmaxRange = [1e4:1e3:3e5];SolnRange = 0*PmaxRange;

% RUNNING INDEX FOR OPTIMIZATIONrunningMin = 1e12;runningPmax = 0;

for i = 1:length(PmaxRange)Pmax = PmaxRange(i);

[T1, Uf1, Ufg1] = steam_t(Psat);Ug1 = Uf1+Ufg1;

[T2, Uf2, Ufg2] = steam_t(Pmax);Ug2 = Uf2+Ufg2;

soln = abs((Pmax*Vtot - Psat*Vinit) - ms*(Ug1-Ug2));SolnRange(i) = soln;

if ( soln < runningMin )runningMin = soln;runningPmax = Pmax;

endend

%plot(PmaxRange, SolnRange);%hold on;% plot(PmaxRange, 0);