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SYSTEM SIMULATIONSYSTEM SIMULATION

System simulation is a calculation step for finding System simulation is a calculation step for finding out the output when the input parameters are out the output when the input parameters are changed. With the results the designer could make changed. With the results the designer could make a decision on the appropriate size or design before a decision on the appropriate size or design before making a prototype or real systems. making a prototype or real systems.

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Continuous simulationContinuous simulation Discrete simulationDiscrete simulation

Procedure of the system simulation could beProcedure of the system simulation could be

Deterministic inputDeterministic input Stochastic inputStochastic input

Steady stateSteady state DynamicDynamic

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In this course, the conditions areIn this course, the conditions are

Continuous simulationContinuous simulation Deterministic inputDeterministic input Steady stateSteady state

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Steps of System SimulationSteps of System Simulation

1. Define the system.1. Define the system.

2. Set mathematical model of each component in 2. Set mathematical model of each component in the system.the system.

3. Generate connections among the models.3. Generate connections among the models.

==>> Informative flow diagram ==>> Informative flow diagram

4. Check and prescribe the constraints needed.4. Check and prescribe the constraints needed.

5. Simulate the system. 5. Simulate the system.

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ExampleExampleSteam turbine, Condenser and cooling tower in a Steam turbine, Condenser and cooling tower in a

steam power plant steam power plant

Steam - turbine Steam - turbine

WT = WT(Pc,QT)

f1(WT,Pc,QT) =0

Cooling Tower Cooling Tower

Condenser Condenser

QC = Qc(TCw ,Pc,mcw)

f2(Tcw,Qc, Pc,mcw) =0

Tcw = Tcw (Twb ,RH,CR,mcw)

f3(Twb, RH,CR,mcw,Tcw)=0

…………………………..(1)..(1)

……………………....(2)(2)

………………....(3)(3)

WT = net power of the power plant

QT = heat rate input of the power plant

Qc = heat rate at the condenser

Pc = condenser pressure

mcw = mass flow rate of the cooling tower

Tcw = cool water temperature at the cooling tower

CR = range of the cooling tower

WT = net power of the power plant

QT = heat rate input of the power plant

Qc = heat rate at the condenser

Pc = condenser pressure

mcw = mass flow rate of the cooling tower

Tcw = cool water temperature at the cooling tower

CR = range of the cooling tower

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Thus two more energy balance equations assisted :Thus two more energy balance equations assisted :

Inputs : Twb , RH , mcw , WT

Unknowns : Pc , QT , Tcw , Qc and CR

steam cycle : QT = WT + Qc

===>>> f4 (QT , WT , Qc)=0

Condenser - Cooling tower : Qc =(mcp)CR

===>> f5(mcw, Qc ,CR)=0

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ff11(W(WTT , P , Pcc ,Q ,QT T ))

ff44 (Q (QTT , W , WTT , Q , Qcc))

ff55 ( m ( mcw cw , Q, Qcc ,CR) ,CR)

ff33( T( Twbwb , RH , CR , m , RH , CR , mcwcw , T , Tcwcw ) )

ff22 ( T ( Tcw cw , Q, Qcc , P , Pcc , m , mcwcw))

QQT T CHECK CHECKQQTT TRIAL TRIAL

QQCCQQCC

CRCR

TTwbwb

RHRH

mmCWCW

mmCWCW

mmCWCW

PPcc

TcwTcw

WWTT

WWTT

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Pt.3 : WA = CA(P3 - Patm)1/2 =>> f1 (WA , P3) = 0…….(1)

Pt.4 : WB = CB(P4 - Patm)1/2 =>> f2 (WB , P4) = 0…….(2)

0 => 1 : Patm - p1 = c1W12 + hg =>> f3 (W1 , P1) = 0…...(3)

2 => 3 : P2 - p3 = c2W12 =>> f4 (W1 , P2,P3)=0...…(4)

3 => 4 : P3 - p4 = c3W22 =>> f5 (W2 , P3,P4)=0...…(5)

Pump : => f6(w1,P1,P2) = 0……(6)

Two more equations are needed which are

W1 = WA + W2 ==>> f7 (W1 ,W2,WA) = 0…..(7)

W2 = WB ==>> f8 (W2 ,WB) = 0……(8)

h, Patm are presensibled unknowns : WA , WB , W1 ,W2 , P1 , P2 , P3 ,P4 (8 unknowns)

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ff66(W(W11 , P , P11 ,P ,P2 2 )) ff33(W(W11 , P , P11 ))

ff44(W(W11 , P , P22 ,P ,P3 3 ))

ff55(W(W22 , P , P33 ,P ,P4 4 ))

ff11(W(WAA , P , P33 ))

ff22(W(WBB , P , P44 ))

ff77(W(W11 ,W ,WAA ,W ,W22 ))

ff88(W(W22 ,W ,W3 3 ))

TRIAL WTRIAL W11

TRIAL WTRIAL W22

PP33

PP44WWBB

WW22

WW11PP11WW11

WW11

WWAA

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Sequential SimulationSequential Simulation When we start with the information input of a component

in a whole system, the an output from this first component will be the input needed to calculate the output of the next component and go on to the last component.

When we start with the information input of a component in a whole system, the an output from this first component will be the input needed to calculate the output of the next component and go on to the last component.

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Simultaneous SimulationSimultaneous Simulation

The calculation is not straightforward. Two methods normally used are Successive substitution

A value or more input variables are assumed to begin to the calculation and the process is continued till the originally-assumed variable are recalculated. The iteration is finished when the convergence of the of the variable is obtained.

Newton-Raphson method.

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Example Example Cooling Tower-CondenserCooling Tower-Condenser

Q = 293.1 Q = 293.1 XX 10 1033 kW , W = 9146.6 kg/s , T kW , W = 9146.6 kg/s , Twbwb = 20 ,25 C = 20 ,25 C

Condenser :Condenser : Pc = 1.0055 x 10-5 Q + 0.34398Tcw - 5.4006 kN/m2

f1(Pc , Q , Tcw) = 0 ……………(1)

Cooling Tower :Cooling Tower : Tcw = 0.54078Twb + 0.43889CR + 0.001233W, C

Inputs : Q,W,TInputs : Q,W,Twbwb

Unknowns : PUnknowns : Pcc,T,Tcwcw,CR ,CR

one more equation is needed :one more equation is needed :

Q = mcQ = mcpp.CR ==> f.CR ==> f33 (Q,W,CR)=0 ………………..….…(3) (Q,W,CR)=0 ………………..….…(3)

f2(Tcw,Twb,CR,W) = 0 ……...…(2)

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f1(Pc , Q , Twc)

f3 (Q,W,CR)

f2(Tcw,Twb,CR,W) = 0

TcwTwb

Q

W

Q

CR

Pc

W

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Successive Substitution Successive Substitution Example Example pump 1 : P(kPa) = 810 - 25W1 -

3.75W12

pump 2 : P(kPa) = 920 - 65W2 - 30W2

2Pressure drop : PL(kPa) = 7.2W2

h = 40 m Find W1 , W2Solution :Solution :

systemsystem : PA + gZA-PL + P = PB +gZB

Since PA =PB and ZA - ZB = 40 thus

Pa/kPa1000

m)40 m/s 9.807kg/m(1000 7.2W P

232

P = 7.2W2 + 392.28 f1(P ,W) = 0 ……………(1)

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Pump Characteristics : P = 810 - 25W1 - 3.75W12 => f2(P , W1 ) = 0…….

(2)

P = 900 - 65W2 - 30W22 => f3(P , W2 ) = 0……..

(3)Unknowns : P , W1 ,W2 , W

then one more equation is needed which is

W = W1 + W2

f4 (W1 , W2 , W) = 0…..….(4)

Successive Substitution Successive Substitution

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Successive Substitution Successive Substitution

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Successive substitution is simple but sometimes the calculation step could not be iterated

Example Example Engine and CarEngine and Car

Power to drive the car : P = 4.2+0.45V + 0.0025V2.8

; V(m/s) , P(kW)

f1(P ,V) = 0…….(1)

Engine power : P = 60 + 8V - 0.16V2 f2(P ,V) = 0…….(2)

Starting with V=50 m/s

Final result

P = 112.3905 W

V = 42.25 m/s

Starting with V=42 m/s

The final result could

not be obtained

Auto

P = 4.2+0.45V + 0.0025V2.8

Engine

P = 60 + 8V - 0.16V2

Auto

P = 4.2+0.45V + 0.0025V2.8

Engine

P = 60 + 8V - 0.16V2

PP VV PPVV

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Newton - Raphson Method Newton - Raphson Method

For y = y(x) = anxn+an-1xn-1+…+ax+a0

If xc is a solution ==> y(xc) = 0

with Taylor’s expansion around the point xc

y(x) = y(xc) + y’(xc)(x-xc) + 1/2!y’’(xc)(x-xc)2+…

= 0 +y’(xc)(x-xc) ; y’=dy/dx

If we trial with xt y(xt) = y’(xc)(xt-xc)

Since xc is not known then y’(xc) is used instead.

y(xy(xtt) = y’(x) = y’(xtt)(x)(xtt-x-xcc) ) )x('y

)x(yxx

t

ttc

)x('y

)x(yxx

old

oldoldnew

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Newton - Raphson Method Newton - Raphson Method

Example ;Example ; x+2 = ex+2 = exx Find xFind x

Solution ;Solution ; y(x) = x+2 - ey(x) = x+2 - ex x andand y’(x) = 1-e y’(x) = 1-exx

Initial trial with value of x= 2

==> y(2) = 2+2-e2 = -3.389

y’(2) = 1-e2 = -6.389

1.469 6.389-

3.389- - 2 x new,1

)x('y

)x(yxx

old

oldoldnew x

y(x)

2

-3.389 -0.876 -0.132 -0.018

1.469 1.208 1.112

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For multi-variable such as For multi-variable such as y = f(xy = f(x11 , x , x22 , x , x33) ) then then

Newton - Raphson Method Newton - Raphson Method

f1(x1 , x2 , x3) = 0

f2(x1 , x2 , x3) = 0

f3(x1 , x2 , x3) = 0From single variable From single variable y’(xnew)(xold - xnew) = y(xold)

old

3

2

1

new 3old 3

new 2old 2

new 1old 1

old3

3

2

3

1

3

3

2

2

2

1

2

3

1

2

1

1

1

f

f

f

x -x

x -x

x -x

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x

f

x1new , x2new, x3new and reiterate till the old x values are close to the new ones.

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Example Example f1 = P - 7.2w2 - 392.28

f2 = P - 810 + 25w1 + 3.75w12

f3 = P - 900 + 65w2 + 30w22

f4 = w - w1 - w2

old4

3

2

1

new old

new 2old 2

new 1old 1

new old

old

4

2

4

1

44

3

2

3

1

33

2

2

2

1

22

1

2

1

1

11

f

f

f

f

w -w

w -w

w -w

P -P

w

f

w

f

w

f

P

f

w

f

w

f

w

f

P

f

w

f

w

f

w

f

P

f

w

f

w

f

w

f

P

f

22

f1 f1 1 1 0 0 0 -14.4w 0 -14.4w

f2f2 1 25+7.5w 1 25+7.5w11 0 0 0 0

f3f3 1 1 0 65+60w 0 65+60w22 0 0

f4f4 0 -1 0 -1 -1 -1 1 1

P

1w

2w

w

old4

3

2

1

new old

new 2old 2

new 1old 1

new old

old

2

1

f

f

f

f

w -w

w -w

w -w

P -P

1 1 - 1 -0

0 60w650 1

0 0 7.5w 251

14.4w0 - 0 1

Assume Assume P = 75 , wP = 75 , w1 1 = 3 , w= 3 , w22 = 1.5 , w = 5 = 1.5 , w = 5

FinalFinal ==>>==>> P = 650.49 , wP = 650.49 , w11 = 3.991 , w = 3.991 , w22 = 1.997 , w = 5.988 = 1.997 , w = 5.988