Sustainable Power Generation Applied Heat and Power Technology. Equations, diagrams and tables

Transcription

1 Sustainable Power Generation Applied Heat and Power Technology Equations, diagrams and tables 1

2 STEAM CYCLE Enthalpy of liquid water h = c p,liquid (T T ref ) T ref = 273 K (normal conditions). The specific heat capacity for liquid water is: c p = 4.18 kj/(kg K) & Heat supply (kg/s kj/kg = kw): = m& ( ) Q Boiler h 3 h 2 Turbine power (kg/s kj/kg = kw): = m& ( ) P T h 3 h 4 Isentropic efficiency turbine: η T = h h 3 3 h h 4 4s Steam cycle efficiency: η = PT ( Pp ) Q& Fuel Boiler efficiency: η B = Q& Q& Boiler Fuel Heat balance for one open heater: m& extr h2 + ( m& m& ) h5 m& extr h6 = 0 m, h6 mextr, h2 m-mextr, h5 Heat balance for one closed heater & m extr ( h2 h7 ) = m& ( h6 h5 ) 2

3 GAS TURBINE CYCLE Fuel G Air Compressor power [kg/s kj/kg = kw ] or [kg/s kj/(kg K) K = kw ]: P C = m& ( ) or P m& c ( T ) air h 2 h 1 C = air P, T1 Turbine power [kg/s kj/kg = kw ] or [kg/s kj/(kg K) K = kw ]: P T = m& ( ) or P m& c ( T ) gas h 3 h 4 T = gas P, T4 Electric net power output: P GTel = ( P η P ) η T m C G Gas turbine efficiency: η Enthalpies: h 1 = h 1,AIR h 2 = h 2,AIR h 3 = h 3,AIR + x DH t3 h 4 = h 4,AIR + x DH t4 GT P = Q& GT Fuel = m& Fuel P GT LHV Temperature increase compressor where the temperature T 1 is in [K] and η SC is the isentropic efficiency of the compressor: T 2 T 1 T1 = η SC p p 2 1 κ C 1 κ C 1 3

4 Temperature decrease turbine where the temperature T 3 is in [K] and η ST is the isentropic efficiency of the turbine: T 3 T 4 = T 3 η ST κ T 1 κ T p p4 The heat supplied in the combustion chamber: Q& FUEL = m& fuel LHV Specific fuel consumption, β, is defined as: m& fuel β = [kg m& fuel / kg air ] air Gas flow after combustion chamber: m & gas = m& + m& = ( 1+ β ) m& air fuel air Heat balance on the combustion chamber: m& h & ( β m& h air 2 + m fuel LHV = 1+ ) Gas content x = ( 1+ f ) β 1+ β air where f is the the stoichiometric air to fuel ratio (kg air/kg fuel). For light oil, f = Solving the heat balance for the specific fuel consumption: β = LHV h h h 3, AIR 2 3, AIR ( 1+ f ) DH t3 4

5 COMBINED CYCLE Stack tg4 tfw hfw Temperature tg3 Economizer Δtpp ts Evaporator hs1 (saturation) tg2 hs2 Superheater Heat transferred Gas tg1 turbine exhaust tsh hsh Economizer tg4 tg3 To the stack hfw Feed water pump hs1 Drum Evaporator hs2 hs1 tg2 Circulation pump Superheater Superheater hsh hs2 Gas turbine Exhaust tg1 Superheated Steam Pinch point temperature difference: Δt PP = t g3 - t s Approach temperature: Δt appr = t s t eco,out Heat balance HRSG: Qrecovered mgt cpg (tg1- tg4) Gas side mst (hsh- hfw) Steam side Heat balances for each component in the HRSG are as follows (without the economizer approach temperature): Economizer: m gt c pg (t g3 t g4 ) = m st (h S1 h fw ) Evaporator: m gt c pg (t g2 t g3 ) = m st (h S2 h S1 ) Superheater: m gt c pg (t g1 t g2 ) = m st (h sh h S2 ) Heat balances for each component in the HRSG with the economizer approach temperature: Economizer: m gt c pg (t g3 t g4 ) = m st (h appr h fw ) Evaporator: m gt c pg (t g2 t g3 ) = m st (h S2 h appr ) 5

6 COMBUSTION Calculation of the amount of flue gas per kg fuel Consider a fuel with the following composition on total substance (%-mass): a% C, b% O 2, c% H 2, d% S, e% N 2, f% humidity, g% ash, a + b + c + d + e + f + g = 100% The first step is to convert the content of the different elements to moles: Element Molar mass (g/mol) Analysis (g/kg fuel) Analysis (mol/kg fuel) C 12 a*10 a*10/12 O 2 32 b*10 b*10/32 H c*10 c*10/2.02 S 32 d*10 d*10/32 N 2 28 e*10 e*10/28 H 2 O f*10 f*10/18.02 Ash 1 - g*10-1 Ash is not contributing to the combustion or the flue gas and will not be further considered. Elemental combustion reactions C + O 2 CO 2 S + O 2 SO 2 H O 2 H 2 O Needed air supply according to the elemental reactions: Element Analysis (mol/kg fuel) Oxygen needed (mol/kg fuel) C a*10/12 a*10/12 O 2 b*10/32 - b*10/32 H 2 c*10/ * c*10/2.02 S d*10/32 d*10/32 N 2 e*10/28 - H 2 O f*10/

7 Oxygen needed [O 2 ] = a*10/12 b*10/ *c*10/ d*10/32 [mol/kg fuel] Since air is most often used for combustion, N 2 has to be included in the analysis: Nitrogen in air: [N 2 ] = 3.76*[O 2 ] [mol/kg fuel] The molar amount of air needed for stoichiometric combustion of 1 kg of the fuel is: l 0 = [O 2 ] + [N 2 ] [mol/kg fuel] From the set of elemental reactions, the gases produced (which constitute the flue gas) are also obtained: Element Analysis H 2 O CO 2 SO 2 N 2 O 2 (mol/kg fuel) C a*10/12 a*10/12 O 2 b*10/32 H 2 c*10/2.02 c*10/2.02 S d*10/32 d*10/32 N 2 e*10/28 e*10/28 H 2 O f*10/18.02 f*10/18.02 N 2 in air [N 2 ] The stoichiometric amount of flue gas is the sum of the compounds: g 0 = a*10/12+ c*10/ d*10/32 + e*10/28 + f*10/ [N 2 ] [mol/kg fuel] Excess air is required for the combustion process. For an air/fuel ratio m, the real amount of air becomes: l = m*l 0 [mol/kg fuel] Since air comes in excess, in the flue gas there will be oxygen and more nitrogen. The real amount of flue gas becomes: g = g 0 + l 0 *(m-1) [mol/kg fuel] To convert the flue gas amount from mol to m n 3 the equation of state pv=nrt is used, assuming the flue gas is ideal at normal conditions (1atm, 273K) so that: 1 mol of gas in normal conditions corresponds to m n 3 (V/n = RT/p) If the calorific (higher) heating value HHV is known, the lower heating value LHV can be estimated from LHV = HHV 2.44 * (8.94 * h 2 + F) [MJ/kg] h 2 is the hydrogen mass fraction in the fuel; F is the fuel humidity is a mean value for heat of vaporization in [MJ/kg H 2 O] at 25ºC 7

8 BOILER EFFICIENCY Indirect method: η p = (100- P fg -P CO -P ash -P rad ) % P fg = Flue gas losses. Can be estimated from Siegerts formula: P fg t gas t = k * [ CO ] 2 REF DRY t ref = 25ºC, t = flue gas temperature in ºC, [CO 2 ] dry is the carbon dioxide fraction in % on dry flue gas (from combustion table), k is taken from a diagram and is dependent on fuel type. P CO are losses because of CO in the flue gas. If the CO-fraction is measured in x ppm on dry flue gas, the loss is: P CO g *[ CO] LHV * LHV t DRY CO = [%] fuel If the measurement is performed in wet (total) flue gas, g t is replaced by g. LHV CO is the heat of combustion for CO in [MJ/m n 3 CO]. LHV CO = MJ/ m n 3. P ash = Losses because of unburned in the residual ash. If [A] is the mass fraction of ash in the fuel and [bb] is the mass fraction of unburned in the ash taken out from the furnace (residual ash) then the amount of residual ash is: [ A] Re sidualashamount = 1 [ bb] The loss of combustible substance in the residual ash is: [ bb] [ A]* * LHV 1 [ bb] Pash = LHV fuel C where LHV c is the lower heating value of the combustible part of the residual ash 30 MJ/kg Emissions: Assume that the measurement of a certain emission is given as x ppm in dry flue gas. The desired unit is [mg/mj fuel]. The transition of unit is: [emission] = 10 6 * (x * 10-6 * g t * ρ emission )/LHV fuel [mg/mj fuel] where 8

9 ρ emission is the density of the emission [kg/m n 3 ] which can be found in a property table. g t is the dry flue gas amount in [m n 3 /kg fuel] If the measurement is performed in wet (total) flue gas, g t is replaced by g 9

10 NUCLEAR POWER 1 [amu] = [grams] The nucleons mass in [amu] are as follows: Neutron: [amu] Proton: [amu] 1 [amu] = 931 [MeV] The activity, which means the amount of decays to occur per time unit, is proportional to the amount of isotopes: A = λ N A 0 = λ N 0 Were A 0 represents the activity at the initial time t 0. N is the amount of isotopes and λ is the decay constant The activity in matter decreases with time due to A = λ N 0 e -λ t = A 0 e -λ t The activity can be given in the units Becquerel [Bq] or Curie [Ci], 1 [Bq] = 1 [decay/second] 1 [Ci] = [decays/second] The half-life-time T 1/2 is defined as the time taken to reach half the value of the initial activity. If this time is inserted in the activity equation above, we get: A 0 /2 = A 0 e -λ T½, were T½ represents the half-life-time The half-life-time can therefore be written as: T½ = ln 2/λ Microscopic target 1. The absorption target, σ a. 2. The fission target, σ f. 3. Radioactive capturing target, σ c. 4. The spreading target, σ s. 10

11 σ s = σ e + σ i, were σ e is the elastic and σ i the inelastic target. σ a = σ c + σ f The total microscopic target becomes: σ t = σ a + σ s The size of a target is given in the unit barns: 1 [barn] = [cm 2 ] A neutron flux with Φ 0 [neutrons / cm 2, s] irradiating perpendicular to a plate or a wall: N = the number of nuclei per cm 3 in the wall x = wall thickness A = wall frontal area in cm 2 σ = microscopic target in cm 2 The neutron flux behind the wall is Φ = Φ 0 e -σ N x The definition the macroscopic target, Σ, is written: Σ = N σ where σ represents the microscopic target and N the number of nuclei per cm 3. Therefore, Σ is the total target for a certain exchange in 1 cm 3 of the substance, and the dimension is [1/cm]. The inverse of this target is the mean free path in [cm], which is a mean distance a neutron travels before a certain type of exchange occurs. The number of nuclei per cm 3 in a material can be calculated according to: N = [ρ ] / A ρ = the density in [g/cm 3 ] A = the atomic weight in [g/mole] corresponds to Avogadro s constant, [atoms/mole] If different kinds of atoms exist inside a volume, the macroscopic target is calculated according to: Σ = N 1 σ 1 + N 2 σ 2 + N 3 σ 3 + If the calculation refers to molecules the atomic weight A is replaced of the molecular weight MW. When calculating Σ, for a molecule, the following equation is used: Σ = ([ρ 6, ] / MW) (μ 1 σ 1 + μ 2 σ μ i σ i ) where μ i represents the number of atoms, of the same kind, per molecule. R = Σ Φ [exchanges/cm 3, s] The following relation is special for fission: R f = Φ Σ f [fissions/cm 3, s] Where Φ is the neutron flux [cm -2, s -1 ] And Σ f is the macroscopic target for fission 11

12 The power per cm 3 in a nuclear core, P/V, is: P / V = R f / (3, ) [W/cm 3 ] The effective multiplication factor is defined as k = ε p f η P Where ε is the fast fission factor p is the resonance escape probability f = Σ au / [Σ au + Σ ak ] = thermal utilization factor Σ ak represents the absorption in moderator, cladding etc The thermal fission factor, termed η, states the number of produced neutrons per absorbed thermal neutron. η = ν [ Σ f (235) / {Σ c (238) + Σ c (235) + Σ f (235)}] The number ν represents the average number of emitted neutrons per fission. In the case for U-235 the value of ν is approx. equal to Shorter, η can be written η = ν [ Σ f / Σ au ] P is the probability for NO leakage of neutrons from the core Enrichment e = N U235 /(N U235 +N U238 ) The reactivity ρ is defined as: ρ k eff 1 The power increase with time t in a nuclear core is given as: P = P 0 e ρ t/a Where a is the average lifetime per neutron generation The reactor period is defined as T = a/ρ 12

13 Enthalpies and heat of vaporization of saturated water and steam as function of Pressure Pressure (bar) Temperature ( C) Enthalpy Saturated Liquid Water (kj/kg) 13 Enthalpy Saturated Steam (kj/kg) Heat of vaporization (kj/kg)

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17 Enthalpies and heat of vaporization of saturated water and steam as function of Temperature Temperature ( C) Pressure (bar) Enthalpy Saturated Liquid Water (kj/kg) 17 Enthalpy Saturated Steam (kj/kg) Heat of vaporization (kj/kg)

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25 Enthalpy-entropy diagram (h-s) for steam 25

26 Ratio of specific heats (isentropic exponent) 26

27 Enthalpy for air and gases from light oil firing as function of temperature β x = ( 1 + f ) h GAS = h AIR + x Dh where x is the gas content defined as: 1 + β. f is the stoichiometric air-to-fuel ratio, f = for light oil. The specific fuel consumption, β, is defined as β= m fuel /m air For pure air: x = 0. For pure combustion gases (= no air content): x = 1. Dh is the enthalpy difference (kj/kg) between 100% gases (x=1) and 100% air (x=0). Temperature ( C) Enthalpy for air, h AIR, (kj/kg) x=0 Dh (kj/kg) , , ,1 1, ,2 2, ,2 2, ,3 3, ,3 3, ,4 4, ,5 5, ,6 5, ,7 6, ,8 7, , ,1 8, ,2 9, ,4 10, ,6 10, ,8 11, , ,2 13, ,5 14, ,7 14, , ,3 16, ,6 17, , ,3 19, , ,1 20, ,6 21, , ,5 23, , ,5 25, , ,6 27, ,2 28, ,8 29, ,4 30, ,1 31, ,8 32,2 27 Temperature ( C) Enthalpy for air, h AIR (kj/kg) x=0 Dh (kj/kg) ,5 33, ,2 34, , ,8 36, ,6 37, ,4 38, ,2 39, ,1 40, , ,9 42, ,9 43, ,8 44, ,8 45, ,9 46, ,9 47, , ,1 51, ,3 52, ,4 53, ,6 54, ,8 55, , ,2 57, ,5 59, ,7 60, , ,3 62, ,7 63, ,4 66, ,8 67, ,2 68, ,6 69, ,5 72, , ,5 74, ,5 77,3

28 Temperature ( C) Enthalpy for air, h AIR (kj/kg) x=0 Dh (kj/kg) ,1 78, ,6 79, ,2 81, ,8 82, ,4 83, ,7 86, ,3 87, ,7 90, ,4 91, , ,9 94, ,6 95, , ,1 98, ,9 99, ,7 101, ,5 102, ,4 103, ,2 105, ,1 106, ,9 108, ,8 109, ,7 110, ,6 112, ,5 113, ,4 115, ,3 116, , ,2 119, ,2 120, ,2 122, ,2 123, ,2 125, ,2 126, ,2 128, ,2 129, ,3 131, ,3 132, ,4 134, ,4 135, ,5 137, ,6 138, ,7 140, ,8 141, ,9 143, ,8 28

29 Specific heat, cp, for gases from light oil firing as function of temperature β x = ( 1+ f ) x is the gas content defined as: 1+ β. f is the stoichiometric air-to-fuel ratio, f = for light oil. The specific fuel consumption, β, is defined as β= m fuel /m air. For pure air: x = 0. For pure combustion gases (= no air content): x = Specific heat, J/kgK x = 1.0 x = 0.9 x = 0.8 x = 0.7 x = 0.6 x = 0.5 x = 0.4 x = 0.3 x = 0.2 x = 0.1 x = Temperature, grad C 29

30 Gas and air amounts (m n 3 /kg of fuel) as function of LVH for different fuels 30

31 ' CO 2 and O 2 contents in flue gas as function of air excess factor (equivalence ratio)

32 Siegerts formula: P fg t gas t = k * [ CO ] 2 REF DRY t ref = 25ºC, t = flue gas temperature in ºC, [CO 2 ] dry is the carbon dioxide fraction in % on dry flue gas (from combustion table), k is taken from a diagram and is dependent on fuel type. 32