Theory and Applica>on of Gas Turbine Systems

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1 Theory and Applica>on of Gas Turbine Systems Part II: Actual Sha. Power Cycles Munich Summer School at University of Applied Sciences Prof. Kim A. Shollenberger

2 Actual Power Cycles: Compressor and Turbine Performance Due to irreversibili>es, entropy must increases across turbine(s) and compressor(s); use isentropic efficiencies to quan>fy High fluid veloci>es in turbo-machinery some>mes results in significant changes in kine>c energy between inlet and outlet; use stagna>on proper>es in isentropic efficiencies to quan>fy

3 Effect of Irreversibili>es NOTE: entropy must increase across any real compressor or turbine.

4 Kine>c Energy Changes using Stagna>on Proper>es Stagna&on (or Total) Proper&es (T 0, p 0, ρ 0, and h 0 ) proper>es a fluid has if brought to rest isentropically; for an ideal gas: T 0 T " k 1% =1+ $ 'Ma 2 1 # 2 & ρ 0 ( " k 1% + = * 1+ $ 'Ma 2 - ρ ) # 2 &, ( ) 1/ k 1 1 p 0 p ( " k 1% + = * 1+ $ 'Ma 2 - ) # 2 &, ( ) k/ k 1 1 h 0 = h +V 2 2 Ma = V, a = k R T is the speed of sound for ideal gas a NOTE: For Ma < 0.3 (incompressible) then T 0 T

5 T-s Diagram for Stagna>on Proper>es T p 0 p T 0 T s = 0 s Kine>c energy increases the temperature and pressure of the gas at stagna>on condi>ons.

Measuring Stagna>on Proper>es Pitot-tube typically used to measure p 0 Thermocouple in tube can be used to measure T 0 Approximately 98% of T 0 compared to 60-70% of T 0 for thermocouple in direct

6 Measuring Stagna>on Proper>es Pitot-tube typically used to measure p 0 Thermocouple in tube can be used to measure T 0 Approximately 98% of T 0 compared to 60-70% of T 0 for thermocouple in direct gas stream Large hole faces upstream and small hole with area less than 5% of inlet provides ven>la>on For significant radia>on must use a radia>on shield (such as a polished stainless steel sheet twisted into a helix as shown)

7 1 st Law of Thermodynamics with Kine>c Energy Changes For control volume (CV) with inlet at (1) and outlet at (2): de cv dt = Q! cv W! " cv +!m $ h 1 h 2 # ( ) + V 1 2 V 2 2 For steady state, adiaba>c, and negligible changes in PE:!W cv =!m ( h 1 h 2 ) + V V 2 2 For an ideal gas with constant specific heats: 2 =!m ( h 01 h 02 ) ( ) + g z 1 z 2 % ' &!W cv =!m c p ( T 1 T 2 ) + V V 2 2 =!m c p ( T 01 T 02 )

8 Isentropic Efficiencies for Compressor and Turbine Assuming adiaba>c and negligible KE and PE changes: η c =!W c!m ( ) s!w c!m h 2s h 1 h 2 h 1 η t =!W t!m h h 3 4 (!W t!m ) h 3 h 4s s Accoun>ng for KE changes and for an ideal gas with constant specific heats: η c = h 02s h 01 h 02 h 01 T 02s T 01 T 02 T 01 η t = h 03 h 04 h 03 h 04s T 03 T 04 T 03 T 04s

9 Calcula>ng Compressor Stagna>on Temperature from Isentropic Efficiency Solving for stagna>on temperature: η c = T 02s T 02 T 01 1 T 01 1 p 02s p 01! = T 02s # " T 01 $ & % k ( k 1) T 02 =1+ 1 (! * p # 02 T 01 η c ) *" p 01 $ & % ( k 1) k + 1-, - NOTE: p 02 = p 02s T 02 T 02s Can assume ambient condi>ons at inlet: T 01 T a and p 01 p a - p in (accounts for inlet losses due to fric>on)

10 T-s Diagram for Compressor Use diagram to show how stagna>on proper>es change for compressor

11 Calcula>ng Turbine Stagna>on Temperature from Isentropic Efficiency Again, Solving for stagna>on temperature: η t = 1 T T p s 1 T 04s T 03 p 03! = T 04s # " T 03 $ & % k ( k 1) ( T 04 " =1 η t * 1 p 04 $ T 03 ) * # p 03 % ' & ( k 1) k + -, - NOTE: p 04 = p 04s T 04 T 04s Can assume p 04 p a to account for kine>c energy loss at the turbine exit and losses due to fric>on.

12 T-s Diagram for Turbine Use diagram to show how stagna>on proper>es change for turbine

13 Polytropic Efficiency Notes η c tends to decrease as pressure ra>o increases above design condi>ons because the increase in temperature due to fric>on requires more work for compression (can be called a preheat effect ) η t tends to increase as pressure ra>o increases above design condi>ons because the increase in temperature allows more power to be extracted (can be called a rehea>ng effect ) Polytropic efficiency (that does not change with pressure ra>o) is defined to account for this

14 Overall Compressor Efficiency using For the compressor Polytropic Efficiency η c = dt s dt = ln p 02 p 01 ln T 02 T 01 For the turbine ( ( ) k 1 ) k ( ) η c = ( p 02 p 01 ) k 1 ( p 02 p 01 ) k 1 ( ) k 1 ( ) ( η c k) 1 η t = dt dt s = ln( T 03 T 04 ) ( ) k 1 ln p 03 p 04 ( ) k ( ) η t( k 1) k ( ( ) k 1 ) k η t = 1 p 04 p 03 1 p 04 p 03

15 Calcula>ng Stagna>on Temperatures from Polytropic Efficiencies Combine equa>ons to get rela>onships similar to those for a polytropic process: T 02! = p 02 # T 01 " p 01 T 04! = p 04 # T 03 " p 03 $ & % $ & % ( k 1) η c k ( ) ( ) k η t k 1 NOTE: Use the inlet condi>ons given earlier.

16 Isentropic Efficiency versus Pressure Ra>o for Turbine and Compressor Isentropic Efficiency 100% 95% 90% 85% 80% 75% Turbine Compressor η = 85% Pressure Ratio

17 Actual Power Cycles: Pressure Losses Fluid fric>on results in pressure losses in inlet and exhaust ducts, combus>on chambers, and possibly heat exchangers: Intake and exhaust duc>ng losses are included in isentropic efficiencies for compressor and turbine Combus>on chamber losses ( p b ) are due to: Aerodynamic drag of flame-stabilizing and mixing devices Momentum changes for exothermic reac>on For added heat exchangers there are fric>onal pressure losses on both the air-side ( p ha ) and gas-side ( p hg ) NOTE: Decreases turbine pressure ra>o rela>ve to compressor pressure ra>o and can significantly reduces net work output

18 T-s Diagram for Pressure Losses and Heat Exchanger

19 Accoun>ng for Pressure Losses To determine the turbine pressure ra>o adjust the inlet and outlet pressures using: # p 03 = p 02 1 Δp b Δp ha % $ p 02 p 02 & " ( p 04 = p a 1+ Δp % hg $ ' ' # & p a NOTE: Pressure drops may change with pressure ra>o because fric>onal losses are approximately propor>onal to flow velocity squared for incompressible

20 Actual Power Cycles: Heat Exchangers For finite sized counter-flow heat exchangers, temperature differences at outlets must exist Typical designs include counter-flow or cross-flow: Recuperators - hot and cold streams exchange heat through a separa>ng wall Regenerators streams are brought cyclically into contact with a matrix that alternately absorbs and rejects heat

21 1 st Law of Thermodynamics for Heat Exchanger Using mass flowrates from turbine and compressor!q HX = c p46!m t ( T 04 T 06 ) = c p25!m c ( T 05 T 02 ) Heat Exchanger Effec&veness - actual energy received by the cold air to maximum possible (when T 05 = T 04 ) ε = c! p25 m c T 05 T 02 c p24!m c T 04 T 02 ( ) ( ) T T T 04 T 02 NOTE: Asympto>c increase with volume/surface area

22 Counter-Flow Heat Exchanger: Overall Heat Transfer Coefficient Define Overall Heat Transfer Coefficient for heat exchanger: 1 ( U A) = 1 h A ( ) 25 + R f,25 + R wall + R f, h A ( ) 46 R f is fouling resitance and R wall is wall conduction resitance From energy balance across each flow can show:!q HX = ( U A) ΔT lm ΔT lm = ΔT 1 ΔT 2 ln( ΔT 1 ΔT 2 ) ΔT 1 = T 6 T 2 ΔT 2 = T 4 T 5

23 Closed System with Counter-Flow Heat Exchanger: NTU Method Define number of transfer units: Using heat exchanger effec>veness, ε, combine to get:!q HX = ε!m c c p46 ( T 04 T 02 ) & ( ε = ' ( ( ) C * = 1 exp" # NTU( 1 C * ) $ % 1 C * exp" # NTU( 1 C * ) $ % for C * <1 NTU ( 1+ NTU) for C * =1 (!m c ) ( p min!m c ) p max ( )!! NTU = U A ( m c ) p min

24 Heat Exchanger Selec>on Typical heat exchangers have 90% effec>veness Maximum turbine exit temperature (due to material limita>ons) is approximately 900 K which limits the inlet temperature to the heat exchanger Thermal stresses during start-up can be significant and also limit maximum temperature and cycling frequency Not always used for power produc>on because they do not always offer a significant advantage over simple high pressure ra>o cycles or combined cycle plants Microturbines typically do require heat-exchangers to be commercially viable

25 Actual Power Cycles: Mechanical Fric>on Bearing fric>on and windage (or aerodynamic fric>on) during power transmission from the turbine to the compressor is approximately 1% of power necessary to drive the compressor, thus!w c =!m c c p12 ( T 02 T 01 ) η m where η m = 99% Addi>onal losses due to mechanical fric>on in ancillary equipment (such as from fuel and oil pumps that can be significant for small low power turbines); accounted for by subtrac>ng from net output of unit

26 Actual Power Cycles: Property Varia>ons Working fluid proper>es vary due to changes in opera>ng condi>ons; for real gases for normal ranges, assume c p and k are only a func>on of temperature fuel/air ratio Cp c p [ ]1.2 (kj/kg K) 1.1 Equilibrium at p = 1 bar 1.3 ky Temperature/K (K)

27 Actual Power Cycles: Property Varia>ons, cont. Working fluid proper>es also vary due to combus>on Open-cycle plants use: Air in the compressor Mixture of air and fuel (typically kerosene; can be approximated as C n H 2n ) in the combus>on chamber Products of combus>on in the turbine Average proper>es for c p and k are calculated using composi>on and temperature; for increasing fuel/air ra>o, c p increases and k decreases Average molecular weight is typically the same as that for air, thus can use R/M = R air = 287 J/kg K

28 Actual Power Cycles: Property Varia>ons, cont. Products of combus>on analysis with dissocia>on is much more difficult: Occurs at temperatures above approximately 1500 K Pressure does now have a significant effect, in addi>on to temperature, on c p and k Required for detailed analysis and design of components For preliminary design calcula>ons can use the following for the air and combus>on gasses: c p,air =1, 005 J/kg K, k air =1.40 c p,cg =1,148 J/kg K, k cg =1.333

29 Actual Open Power Cycles with Internal Combus>on Performance of actual cycles is defined in terms of Specific Fuel Consump&on (SFC) - fuel mass flowrate per net power output SFC =! m f!w cycle = f!w cycle!m air Need to calculate Fuel/Air Ra&o ( f =!m f!m air ) required to transform a unit mass of air at T 02 and f kg of fuel at T f to (1 + f ) kg of products at T 03 (maximum cycle temperature)

30 Actual Open Power Cycles with Internal Combus>on To define cycle efficiency, specify the following: Compressor inlet temperature Turbine inlet temperature Fuel composi>on Calculate ideal fuel/air ra&o, f, and combus&on efficiency, η b, for incomplete combus>on Calculate Specific Fuel Consump&on, SFC Calculate efficiency in terms of the SFC and hea>ng value for the fuel

31 Combus>on Fundamentals Rapid oxida>on of fuel (or chemical reac>on that breaks molecular bonds of combus>ble elements) Produces large thermal energy release and products reactants (fuel + oxygen) -> products Fuel is typically carbon, hydrogen, and/or sulfur Air is typically used for oxygen: 0.79 moles N 2 (mostly inert) per 0.21 moles O 2 or 0.79/0.21 = 3.76 moles of N 2 for each mole of O 2

32 Combus>on Fundamentals, cont. Products for complete combus>on are carbon dioxide, CO 2, water, H 2 O, and/or sulfur dioxide, SO 2 Mass is conserved, but number of moles is not Stoichiometric coefficients precede the chemical symbols to give equal amounts by mass of each element on both sides of a balanced equa>on Mixtures for combus>on are called Stoichiometric for just enough oxygen Lean for not enough fuel Rich for excess fuel

33 Balanced Equa>ons for Combus>on Stoichiometric mixture of methane burning in for air: CH ( O N 2 ) CO H 2 O N 2 For 200% theore>cal air (or a lean mixture): CH ( O N 2 ) CO H 2 O N O 2 NOTE: Excess oxygen now appears in the products

34 Example #5 Propane (C 3 H 8 ) is burned with dry air. For the following cases obtain the balanced reac>on equa>on and fuel-air ra>o: a. stoichiometric mixture and b. lean mixture with 120% theore>cal air (or 20% excess air).

35 1 st Law of Thermodynamics Analysis for Combus>on For steady control volume with negligible KE and PE: 0 = Q! CV W! CV + (!m h) R (!m h) P Enthalpy at state T and p given by: h( T, p) = h! f + h( T, p) h( T ref, p ) ref h! f + c ( p T T ) ref! where enthalpy of forma&on ( h f ) - energy released or absorbed when a compound is formed from its elements (typically for T ref = 25 C and p ref = 1 atm)

36 Enthalpy of Combus>on or Hea>ng Values Enthalpy of Reac&on or Combus&on h RP =!m ( P h " f + Δh)!m ( P R h " f + Δh) R! h RP = (! "m h ) f ( "!m h ) P f R for all components at T ref Higher Hea&ng Value (HHV) for liquid water in products (usually measured in calorimeters where products are at low temperature and high pressure) Lower Hea&ng Value (LHV) for water vapor in products (typical for actual systems where products are at high temperatures)

37 1 st Law of Thermodynamics Analysis for Combus>on For adiaba>c combus>on chamber (or adiaba&c flame temperature), steady flow, and negligible KE and PE: (!m h) R = (!m h) P h air ( T o2 ) + f h ( f T ) f =!m i!m air h i ( T 03 ) c p,air ( T 02 T ) ref + f c ( p, f T f T ) ref = ( 1+ f ) c ( p,g T 03 T )! ref + f h RP where c p,g is the average specific heat of the products over the temperature range of T ref to T 03 and the LHV should be used for the enthalpy of reac>on.

38 Calcula>ons assume: Complete combus>on Hydrocarbon fuel: 13.92% H and 86.08% C by mass, f = for stoichiometric, h RP = -43,100 kj/kg Combus:on Temperature Rise T 03 T 02 (K) Theore:cal Fuel/Air Ra:o 150

39 Theore:cal Fuel/Air Ra:o 950 Combus:on Temperature Rise T 03 T 02 (K) Calcula>ons assume: Complete combus>on Hydrocarbon fuel: 13.92% H and 86.08% C by mass, f = for stoichiometric, h RP = -43,100 kj/kg

40 Fuel-Air Ra>o for Incomplete Combus>on Define Combus&on Efficiency as η b = f for theortical complete combustion for given ΔT f for actual incomplete combustion for given ΔT NOTE: Not the same as the ra>o of actual energy released to theore>cal maximum, but typically combus>on is 98-99% complete and this defini>on is sufficiently accurate.

41 Thermal Efficiency for Open Power Cycles with Combus>on Define thermal efficiency based on air mass flowrate:!m air W η th =! cycle!q in!m air For combus>on heat transfer into system is given by the lower hea>ng value for the fuel η th = f SFC f h RP = 1 SFC h RP NOTE: Must be careful with units because SFC is typically in [kw h/kg] and h RP is in [kj/kg]

42 Actual Combus>on Generally incomplete combus>on due to kine>cs, thus fuel ends up in products Other products of combus>on such as NO, NO 2, NO X, CO, etc. exist for high temperatures Fuels are generally mixtures, not just methane, kerosene, propane, butane, etc. Water vapor in incoming air and products

43 Actual Open Power Cycles with Internal Combus>on, cont. Fuel injected into the combustor will increase the mass flowrate through the turbine Compressed air bled off for turbine blade cooling may reduce the mass flowrate through the turbine For some cases, net effect is that mass flowrate through the compressor and turbine can s>ll be modeled as constant For high turbine inlet temperatures (above 1350 K) with significant cooling (called an air-cooled turbine), mass flowrate changes are significant

44 Bleed (or Diverted) Air Cooling Flows Extracted air can be used to cool the turbine disk (part of the central hub), stator (or sta>onary) blades, and rotor (or moving) blades Diverted flow can be up to 15% or more of the compressor flow in advanced engines and must be accounted for in accurate calcula>ons Overall air cooling systems are complex, but simplified models can be used to es>mate the overall impact on gas turbine performance

45 Example of Simple Model for Cooling System For an example consider single-stage turbine with cooling of three components: disk (mass frac>on of compressor flow of β D ) stator blades (mass frac>on of compressor flow of β S ) rotor blades (mass frac>on of compressor flow of β R )

46 Bleed Flow Example Disk bleed flow prevents hot gas flow down the face of turbine disk, but it does s>ll pass through rotor and produces work output Stator bleed also s>ll passes through rotor and produces work output Rotor bleed does not pass through the rotor, thus it does not contribute to the work output; also results in an increase in temperature drop and pressure ra>o for a specified power output

47 Bleed Flow Example, Cont. Air mass flowrate available to rotor is given by:!m R = ( 1 β R )!m a +!m f Air mass flowrate available for combus>on is given by:!m a, comb = ( 1 β D β S β R )!m a Fuel mass flowrate is given by:!m f = f ( 1 β D β S β R )!m a

48 Notes on Bleed Flows Cooling flows from stator and disc (at compressor delivery temperature) cause a reduc>on in rotary entry temperature; can be es>mated using 1 st Law of Thermo. analysis assuming perfect mixing Addi>onal small drop in efficiency due to mixing of bleeds with the main stream

49 Example #6 For an open gas turbine power cycle with regenera>on with the specifica>ons given in Table 1 complete the following: a. sketch a process diagram for the cycle, b. sketch the T-s diagram for the cycle, c. determine the specific work output for the cycle, d. determine the specific fuel consump>on, and e. determine the thermal efficiency. List your assump>ons for your calcula>ons for parts c, d, and e.

50 Table 1. Specifica:ons for Example #6 Compressor pressure ra>o 4.0 Turbine inlet temperature (K) 1100 Isentropic efficiency of compressor 0.85 Isentropic efficiency of turbine 0.87 Mechanical transmission efficiency 0.99 Combus>on efficiency 0.98 Heat exchanger effec>veness 0.80 Pressure loss, combus>on chamber 2% of p 02 Pressure loss, heat-exch. air side 3% of p 02 Pressure loss, heat-exch. gas side (kpa) 4.0 Ambient temperature (K) 288 Ambient pressure (kpa) 100

Free Power Turbine with Twin-Sha@ For flexible opera>on (such as with a variable speed load in marine applica>ons) a mechanically independent (or free) power turbine is desirable The high-pressure

51 Free Power Turbine with For flexible opera>on (such as with a variable speed load in marine applica>ons) a mechanically independent (or free) power turbine is desirable The high-pressure turbine drives the compressor and the combina>on acts as a gas generator for the lowpressure turbine that generates power Easier to start (just the gas generator ini>ally), but requires more complicated control J Gas generator.j Power turbine

52 Example #7 For an open gas turbine opera>ng with a simple power cycle and a free power turbine with the specifica>ons given in Table 2 complete the following: a. sketch a process diagram for the cycle, b. sketch the T-s diagram for the cycle, c. determine the specific work output for the cycle, d. determine the specific fuel consump>on, and e. determine the thermal efficiency. List your assump>ons for your calcula>ons for parts c, d, and e.

53 Table 2. Specifica:ons for Example #7 Compressor pressure ra>o 12.0 Turbine inlet temperature (K) 1350 Isentropic efficiency of compressor 0.86 Isentropic efficiency of turbines 0.89 Mechanical transmission efficiency 0.99 Combus>on efficiency 0.98 Pressure loss, combus>on chamber 6% of p 02 Pressure loss, 2 nd turbine exhaust (kpa) 3.0 Ambient temperature (K) 288 Ambient pressure (kpa) 100

54 Example #8 For an open gas turbine opera>ng with ideal reheat power cycle with the specifica>ons given in Table 3 complete the following: a. sketch a process diagram for the cycle, b. sketch the T-s diagram for the cycle, c. determine the mass flowrate of the air, d. determine the specific fuel consump>on, and e. determine the thermal efficiency. f. How would adding regenera>on effect your results and impact using this as a topping cycle? List your assump>ons for your calcula>ons for parts c and d.

55 Table 3. Specifica:ons for Example #8 Power produc>on (MW) 240 Compressor pressure ra>o 30 Turbines 1 and 2 inlet temperature (K) 1525 Isentropic efficiency of compressor 0.89 Isentropic efficiency of turbine 0.89 Mechanical turbine efficiency 0.99 Combus>on efficiency 0.99 Pressure loss, combus>on chamber 1 2% of p 02 Pressure loss, combus>on chamber 2 4% of p 04 Ambient temperature (K) 288 Ambient pressure (kpa) 101

56 Effect of Reheat Pressure on Example #8 Turbine Exit Gas Temperature (K) Thermal Efficiency Specific Cycle Work (kj/kg) Reheat Pressure (bar)

57 Actual Simple Gas Turbine Cycle: Thermal Efficiency versus Pressure Ra>o Efficiency depends on maximum cycle T Efficiency has peak value due to increasing work to drive compressor Due to materials limita>ons, typically use lowest pressure ra>o that s>ll has a reasonable efficiency Thermal Efficiency Ta 288 K Compressor Pressure Ra:o

58 Actual Simple Gas Turbine Cycle: Specific Work versus Pressure Ra>o Increase in specific work with maximum cycle temperature is significant Consequent reduc>on in size of plant for a given power output is an important design considera>on Specific Work (kj/kg) Compressor Pressure Ra:o

59 Actual Simple Gas Turbine Cycle: Other Variables for Op>mum Condi>ons (T 03 = 1500 K, r p near maximum η th ) Increasing polytropic efficiency by 5% of either compressor or turbine increases thermal efficiency by 4% and increases specific work output by 65 kj/kg Reducing combus>on chamber pressure loss by 5% increases cycle efficiency by 1.5% and increases specific work output by 12 kj/kg Increasing ambient temperature from 15 to 40 C reduces efficiency by 25% and specific work output by 62 kj/kg which is very significant!

60 Actual Regenera>ve Cycle Specific work output has only a slight reduc>on Significant change in thermal efficiency Reduces op>mum pressure ra>o Thermal Efficiency Compressor Pressure Ra:o

61 Actual Regenera>ve Cycle with Reheat: Thermal Efficiency versus Pressure Ra>o As shown for ideal cycles, reheat best with regenera>on Typically only used when there are other reasons for using two turbines Thermal Efficiency Ta 288 K Effectiveness 0.75 Compressor Pressure Ra:o

62 Actual Regenera>ve Cycle with Reheat: Specific Work versus Pressure Ra>o As shown for ideal cycles, significant increase in work output Op>mum is to much higher pressure ra>os Specific Work (kj/kg) Compressor Pressure Ra:o

Theory and Applica>on of Gas Turbine Systems

Theory and Applica>on of Gas Turbine Systems Part I: Ideal Sha- Power Cycles Munich Summer School at University of Applied Sciences Prof. Kim A. Shollenberger Outline for Theory of Gas Turbine Systems