Relazione ORC
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Transcript of Relazione ORC
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Bottoming ORC Power Plant
Davide Occello
Introduction
This report is about the construction of a model and the design of an OrganicRankine Cycle (ORC) Power Plant, bottoming an Internal Combustion Engine,which produces a hot mixture of water and glycol. The ORC has to be designed
in order to maximize the economic benefit of the plant, and the following aimshave to be reached:
1. Construction of the plants model and determination of the degrees offreedom
2. Choice of the organic working fluid
3. Imposition of some thermodynamic constraints
4. Optimization of the cycle
5. Calculation of the efficiencies
6. Analysis of the effect of a different Water-Glycol mixture
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1 ORC Power Plant Scheme
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2 Construction of the plants model
The idea used in the construction of the model is explained in the flow chartin figure 2. The coloured tiles express particular stages of the process such asinput parameters (purple) , constrained variables (yellow) and the optimizationfunction (green).
As can be seen the different input variables have an effect only on some ofthe calculations, leaving the other untouched, this permits to run a more directoptimization.
Now an detailed analysis of each peace of the plant will be done.
Evaporation and Condensation pressuresUsing the tables extracted from SOLKANE for a given working fluid, an inter-polating function was written, so that any thermodynamic property could bedetermined using two variables in the superheated vapour zone and only onein the wet zone. In this way the calculation of the working pressures is shrunk to:
Pevap = P(Tevap,wet)
Pcond= P(Tcond,wet)
Turbine
Known the plants working pressures and the TIT (Turbine inlet temperature),
which is the evaporation temperature (input parameter), the specific work canbe calculated as follows:
Lturb = h2 h1 = iso(h2s h1)
Obviously the enthalpyh1 = h, the saturated vapour enthalpy at the evap-
oration temperature, while the h2s is calculated this way:
h1 = h(Tevap, = 1)
s1 = s(Tevap, = 1)
h2s= h(s1, Pcond)
The TOT (Turbine outlet temperature) can be now calculated:
TOT =T(h2, Pcond)
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Figure 1: ORC Model Flow diagram
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Pump
Similar calculations can be done also for the pump (isoentropic):
Lpump= v1(Pevap Pcond)
Where :
v1= v(Tcond, = 0)
The POT (Pump outlet temperature) can be now calculated:
POT =Tcond+LpumpCp
Heat exchanger notation
In order to simplify the notation of the heat exchanger formulas the graph infigure 2 will be taken as a model.
Figure 2: Heat Exchanger Scheme
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Evaporator
The following procedure was used for the calculations in the evaporator:
On the hot side (Water Glycol Mixture):
Th2 = Th1 ( mhv)
mwg Cpwg
Where hv is the latent heat of evaporation a the given evaporation tempera-ture (input parameter), m is the ORC mass flow (input parameter) and w gsubscript indicates the Water-glycol mixture parameters.
On the cold side (Evaporating organic fluid):
Tc2 = Tc1
So the graph in figure 2 can be plotted and the heat exchanger characteris-tics can be calculated:
evap =Th1 Th2Th1 Tc1
NTUevap = log(1 evap)
Uevap = 2500 [W/m2K]
;
Aevap =NTUevapCpwg mwg
Uevap
Condenser
The condenser is composed of two sections, one in which the fluid is outside thewet zone, and one in which the fluid condenses, for this reason the calculationis a little bit more complex.
Condensation zone
On the cold side (Cold water):
Tc2 = Tc1+ ( mhv)mw Cpw
Where hv is the latent heat of evaporation a the given condensation temper-ature (input parameter), m is the ORC mass flow (input parameter) and wsubscript indicates the Cold Water parameters. Tc1 = 28 [C]
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Figure 3: Evaporator T-x diagram
On the hot side (Condensing organic fluid):
Th3 = Th2 = Tcond
Cooling zone
For the cooling zone we have different border conditions, such as:
Tc3 = Tcond 2
Tc2 andTh2 are known, so:
Th1 = Th2+mwCpwmCp
(Tc3 Tc2)
Constraints The value ofTh1 calculated this way has to be lower than theTurbines outlet temperature. This constraint was applied because the modeldoesnt guarantee it and it is thermodynamically impossible to have the oppositecondition in a real plant.
So the graph in figure 5 can be plotted and the heat exchanger characteristicscan be calculated:
cond = Tc1 Tc2Th1 Tc1
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Figure 4: Condenser Model
precond=Th1 Th2
Th1 Tc2
NOTE In the calculation of the efficacy of the heat exchangers, the formulais written in a way that permits to simplify the heat capacity.
NTUcond = log(1 cond)
precond(1 eNTUprecond(1Cr)
(1 Cr eNTUprecond(1Cr) = 0 (Implicit calculation)
Ucond= 2500 [W/m2K] Uprecond= 50 [W/m
2K]
Acond=NTUcondCpw mw
UcondAprecond=
NTUprecond(Cp m)minUprecond
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Figure 5: Condenser T-Q diagram
Regenerator
The regenerator was approached as every other heat exchanger, with the simpli-fication of the face that the mass flow is the same on both sides of the exchanger.
The border conditions were the following:
Th1 = TOT(Turbine outlet temperature)
Tc1 = POT (Pump outlet temperature)
Th2 = P re CIT (Pre-Condenser inlet temperature)
And so with a simple energy balance the Tc2 can be calculated:
Tc2 = Tc1+ (CphCpc
)(Th1 Th2)
And the heat exchanger characteristics are:
reg (1 e
NTUreg(1
Cr)
(1 Cr eNTUreg(1Cr) = 0 (Implicit calculation)
Ureg = 50 [W/m2K] Areg =
NTUreg(Cp m)minUreg
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Figure 6: Regenerator T-x Diagram
Pre-Heater
The pre-heater works between the following border conditions:
Th1 = EOTh (Evaporator Outlet Temperature - Hot side)
Tc1 = ROTc (Regenerator Outlet Temperature - Cold Side)
Tc2 = Tevap (Evaporation Temperature)
And so with a simple energy balance the Th2 can be calculated:
Th2 = Th1 ( mCp
mwCpw)(Tc2 Tc1)
And the heat exchanger characteristics are:
PH(1 eNTUPH(1Cr)
(1 Cr eNT UPH(1Cr) = 0 (Implicit calculation)
UPH= 600 [W/m2K] Areg = NTU
PH(Cp m)minUPH
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Figure 7: Preheater T-x Diagram
3 Choice of the organic fluid
In order to maximize the economic benefit of the plant, the choice of the work-ing fluid is crucial. The choice was run over a wide range of candidates fromthe SOLKANE database, and a comparison of the various characteristics of thedifferent fluids was analysed.The fluid has to comply to the following characteristics:
In the wet zone atTevap andTcond
High ratio (High turbine specific work)
Possibly small evaporation pressure (< 20 bar) for mechanical reasons
Possibly condensation pressure around 1 bar
The first condition is the most important because without is the cycle cannotwork, so a first selection of fluids was done on that criteria.
Then a comparison of the remaining fluids was done on the basis of the secondcriteria, and was chosen the fluid which had the maximum , the R365mfc. Infact, choosing two mock evaporation and condensation temperatures within themaximum range (93 - 28) the pressure ratio = Pevap/Pcondwas maximum forR365mfc.
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Moreover the fluid complied to the other conditions also, and so it was chosenas the best candidate to test.
4 Thermodynamic Constraints
Because of the fact that the model has 4 degrees of freedom, it could go alsoin places where there is no thermodynamic meaning, and so in order to keep acertain degree of realism in the simulation the following constraints were intro-duced:
All the [0;1]
All Areas < 100m2
Pre-CIT TOT (as said before)
Mass flows under 25 [kg/s]
5 Optimization
The optimization was conducted manually, aiming to maximize the economicbenefit of the plant. The optimization function (Back time) was calculated asfollows:
Pnet= m(Lturbgeninv mec Lpump
pumpelp,gen) Paux
Pturb = m(Lturbgeninvmec)
Ppump= m( Lpump
pumpelp,gen )
Data
Capital cost Se=200000 [euro]
Electricity Selling price Sen= 0.28 [euro/kWh]
Electricity Buying price Sst = 0.11 [euro/kWh]
Water price Cw=1.7 [euro/h]
Working time tw=7500 [h/year]
Fixed costs Oemc=3500 [euro/year]
Variable costs Oemv=0.03 [euro/KWh]
Paux= 6 [kW]
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Economic Balance
IncomeI=Sen Pturb tw [euro/year]
CostsOem = Oemv Pnet tw [euro/anno]Oaux= (Paux+ Ppump) tw Sst [euro/anno]Ow = tw Cw
Annual EarningsR= I (Oemc +Oemv+ Oaux+Ow)
Backtimetritorno = Se/R
Free variables
The free variables chosen for the model are :
Tevap Evaporation Temperature
Tcond Condensation Temperature
mMass flow of the main cycle
mw Mass flow of the cooling water
Temperature range In order to optimize the cycle the Temperature rangewas extended as possible, until the area of the heat exchangers began to be toohigh. This because the higher the temperature range is, the higher the powerand the efficiency of the cycle could be.
Main cycle mass flow The mass flow of the main cycle acted directly tochange the power of the plant, so increasing it was wanted, but the problemwas that with too high mass flow the other constraints were not met, mainlythe 3rd one (Pre-CIT TOT), but also the areas exploded.
Cooling Water Mass flow This mass flow was one of the only parametersthat could permit to increase the plant power, by permitting to increase themain cycle mass flow, and so it was brought to the maximum.
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The final values
Tevap = 89 [C]
Tcond = 32 [C]
m= 1.05 [kg/s]
mw = 25 [kg/s]
Results
Pnet= 14.7 [kW]
tritorno = 10.24 [years]
Areas and efficiencies
Evaporator Aevap = 35.6 [m2] evap = 0.83
Condenser Acond= 27.1 [m2] cond = 0.48
Pre-Condenser Aprecond= 43.0 [m2] precond= 0.77
Regenerator Areg = 49.5 [m2] reg = 0.75
Pre-Heater APH= 10.9 [m2] PH= 0.98
6 Calculation of the efficiencies
Carnot Efficiency
carnot= 1(TcondTevap
) = 1305.15
362.15= 15.7%
Limit Efficiency
lim=a Qa+ b Qb+ c Qc
Qa+ Qb+ Qc= 14.2%
Where:
i = 1 i TcondTevap
i=
i
i
i = T
m
Tmin
i = T
m
Tmax
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Figure 8: ORC Cycle T-s Diagram
Qi are the absorbed heats
As can be seen in the cycle in figure 8 the following simple relations:
B = 1 TcondTevap
A= 1
A= (T2+T1)/2
Tevap
B = (T
5+T5)/2Tcond
B = 1
And for the heat part, the following ones:
QA= h2 h1
QB =hBC h2
QC=h3 hBC
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Real Efficiency at shaft
re = 1 QoutQin
= 13, 7%
Where:
Qin= h3 h1
Qout= h5 h6
Real Electrical Efficiency
el=LnetQin
= 7, 02%
Where:
Qin= 209, 87500 = 1.57 [GWh/anno] is the inlet heat
Lnet = 14, 7 7500 = 0.11 [GWh/anno] is the work after the efficiencychain
Combined Plant Efficiency
Because of the fact that this is a bottoming ORC cycle, the overall efficiencyof the plant will be defined by the following equation, given a certain efficiency
of the topping cycle:
comb = LtotQin,ICE
= LICE+ LORC
LICE+ Qcool+ Qrad+ Qth= 43, 6%
WhereICE= 42, 5%,LICE= 5827500,LORC= 14, 77500,Qth= 4967500,Qcool= 2357500 , Qrad = 567500
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7 Analysis of the effect of the water-glycol mix-
tureThe glycol mass fraction in the hot source had an effect on the cycle overall.This was because the glycol is fairly less dense and has also a lower cp thanwater. So leaving the MATLAB routine unchanged, increasing the glycol massfraction, the overall heat capacity of the hot flow lowered imposing a moreefficient heat exchanger to compensate. This led to an increase in the area ofboth the Evaporator and the Pre-Heater, as showed in figure 9.
Figure 9: Effect of the Glycol mass fraction in the hot source water
On the other hand
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