GAS TURBINE POWER PLANTS

GAS URBINE POWER PLANS Prof. A. Valentini

INRODUCION I have written these notes for Italian speaking students in the last year of high school for the course of Applied Mechanics and Fluid Machinery. he main aim is to give them the chance to follow one module of their mechanics course all in English. he knowledge of basic thermodynamics studied during the fourth year of school is assumed, however, revision of the main concepts concerning gas turbo machinery is included in the first two chapters. Some exercises, that I will solve with the students during the course, are included as well as all the mathematical demonstrations that they are able to understand with their present mathematical knowledge. I hope that my work will be useful for improving the students' speaking, understanding and reading abilities in both technical and everyday English. Prof. A. Valentini - Gas urbine Power Plants

CONENS FIRS LAW OF HERMODYNAMICS FOR AN OPEN SYSEM pag. 3 2 HE ISENROPIC EFFICIENCY FOR GAS URBO MACHINERY pag. 5 3 GENERALIIES ABOU GAS-URBINE POWER PLANS pag. 7 4 HE JOULE CYCLE pag. 0 5 HE REAL CYCLE pag. 2 6 HE COMBUSION CHAMBER pag. 5 7 URBINE PLANS WIH REGENERAIVE INNER HEA EXCHANGER pag. 6 8 CYCLES WIH INER COOLING AND REHEAING pag. 8 EXERCISES pag. 20 BIBLIOGRAPHY pag. 24 Prof. A. Valentini - Gas urbine Power Plants 2

FIRS LAW OF HERMODYNAMICS FOR AN OPEN SYSEM hermodynamic systems can be open, if they exchange mass with external environment, or closed if they don't exchange mass. hey usually exchange energy as mechanical energy (work) and/or thermal energy (heat), if the system doesn't exchange mass and energy it is called isolated. Gas turbine and turbo compressors are open systems so calculations are done with the first law of thermodynamics for an open system: L i + Q e = i + E k + E g () L i internal work (it is the mechanical energy exchanged between the gas and the blades of the turbine) Q e heat exchanged through control volume i change in enthalpy between the outlet and inlet of the machine E k change in kinetic energy E g change in potential energy All the terms in equation () are measured in J picture of a gas turbine. kg (energy per mass unit). Figure shows a fig. - Gas turbine We consider, by convention, the work and the heat coming into the system as positive and the work and the heat coming out as negative. Prof. A. Valentini - Gas urbine Power Plants 3

Equation () can be simplified: Since turbo machines are so fast heat transfer is insignificant so Q e 0 For gases the change in potential energy is small in comparison with the change in enthalpy so E g 0 Normally the change in kinetic energy between the inlet and outlet gas is small and so E k 0. We consider all gases to be perfect (or ideal) so the perfect gas equation (2) is always true. P gas pressure (P a ) v specific volume ( m3 kg ) P v = R (2) R individual gas constant depending on its molecular weight ( J kg K ) gas temperature (K) For perfect gases the enthalpy is the product of pressure specific heat of the gas C p ( J K kg ) and the temperature (K) so equation () becomes: L i = C p (3) For the sign convention mentioned above the work calculated with (3) is positive for a compressor and negative for a turbine. For turbines we defined the positive "internal work obtained " (L i ) obt = L i. he mechanical power produced by a turbine and that absorbed by a turbo compressor are calculated by equations (4) and (5) respectively: P = G. (L i ) obt η m (4) P C = G L i η m C (5) Where G is the mass flow rate and η m and η m C are the mechanical efficiencies of the two turbo machines. Mechanical efficiencies take into consideration the work lost by mechanical friction and they are quite high (η m 0.96 0.99). In table the main constants for some gasses are included. Prof. A. Valentini - Gas urbine Power Plants 4

C p (J/kg K) C V (J/kg K) R = C p C V (J/kg K) k = C p C V Air 003 76 287.4 O2 908 649 260.4 N2 034 737 297.4 CO2 842 653 83.29 H2 4240 000 440.4 NH3 2060 560 500.32 CH4 2253 735 58.3 tab. - able of gas constants 2 HE ISENROPIC EFFICIENCY FOR GAS URBO MACHINERY As we said before turbo machines are adiabatic, if the thermodynamic transformation of the gas in the machines was reversible it would be isentropic. Since the fluid-dynamic friction can never be neglected in fast machines, the real thermodynamic transformation is irreversible with increasing entropy. he isentropic efficiency compares a real machine with an ideal one. Il figure 2 are represented the real (irreversible) and ideal (reversible) transformations for turbines and turbo compressors in temperature-entropy diagram. P 2 P 2 2 id irreversible compression P irreversible expansion P 2 reversibile compression reversible expansion 2 id 2 S S fig. 2 hermodynamic transformations of the gas in turbo compressors and turbines Prof. A. Valentini - Gas urbine Power Plants 5

he isentropic efficiencies are defined as below: turbo compressor P 2 compression ratio β c = P 2 P η C is = L i id L i (7) P gas turbine P expansion ratio β e = P P 2 = (L i) obt id (L i ) (8) obt η is P 2 he compression and expansion ratios are usually between 5 and 40 or sometimes more. It is useful to remember the equation for isentropic transformations studied in general thermodynamics: with k = C P C V If we combine (9) with (2) we get: p v k = constant (9) k = constant (0) pk he real transformation in a turbo machine can be considered a polytropic transformation: Where the m is exponent of the polytropic. p v m = constant () Prof. A. Valentini - Gas urbine Power Plants 6

3 GENERALIIES ABOU GAS-URBINE POWER PLANS he purpose of gas-turbine power plants is to produce mechanical power from the expansion of hot gas in a turbine. In these notes we will focus on stationary plants for electric power generation, however, gas turbines are also used as jet engines in aircraft propulsion. he simplest plant is the open turbine gas cycle used to produce electrical power as shown in figure 3.. Compressor 2. Generator 3. Fuel flow control 4. Combustion chamber 5. Fuel nozzle 6. Mixing tube 7 - Electrical starter 8 - urbine fig. 3 - Open turbine gas cycle he net power available on the shaft is transformed into electrical power by the generator while the electrical starter is an electrical engine which is only used when the plant is turned on. he schematic picture of the previous plant, non including electrical starter and generator, is shown in figure 4. G b (fuel mass flow rate) 2 3 combustion chamber turbo compressor shaft turbine 4 P U (net mechanical power) G a (air mass flow rate) fig. 4 - Open turbine gas cycle Ga + G b (exhaust gases mass flow rate) Prof. A. Valentini - Gas urbine Power Plants 7

Air at room pressure and temperature is compressed to a high pressure in the turbo compressor Fuel is added in the combustion chamber where combustion takes place resulting in high-temperature combusted gases he hot gases expand in the turbine back to the atmospheric pressure producing mechanical power. he cycle is said to be open because fresh air enters the compressor continuously and exhaust is expelled, but thermodynamically it is as if the operating fluid returns to its initial state. Part of the mechanical power generated by the turbine is used to drive the compressor. An alternative plant is the so called closed turbine gas cycle, schematically shown in figure 5. G b (fuel mass flow rate) G a (air mass flow rate) Ga + G b (exhaust gas mass flow rate) 2 3 P U 4 G (operating fluid mass flow rate) G r (refrigerator mass flow rate) fig. 5 - Close turbine gas cycle In this case the operating fluid does not undergo any chemical transformations (because it is not involved in combustion processes) but only thermodynamic transformations. Combustion takes place between air and fuel in the combustion chamber, which is also a heat exchanger. Both plants have advantages and disadvantages. In the first case it is not necessary to cool down the operating fluid (this requires a huge amount of refrigerator Prof. A. Valentini - Gas urbine Power Plants 8

mass flow rates) and so this plant can be adopted on sites with low water availability. One of the advantages of the closed turbine gas cycle is that the turbine stays clean because the combustion products do not pass through it. Another advantage is that it is possible to use gases with k > k air (k air =.4), which increases the efficiency of the cycle (as we will see in the next chapter). he net power available on the shaft in both cases is: P U = P P C (2) and it can be transformed into electric power (P el ) by an electrical machine called a turbo alternator: P el = P U η el (3) where η el is the electrical efficiency of the alternator. he thermal power produced by the combustion of the fuel is: Q = η b G b H i (4) where η b and H i are the combustion efficiency and the inferior calorific value of the fuel. he thermal efficiency (how heat is transformed in work) of the plant is: η u = P U Q (5) and the total efficiency of the electrical power transformation is: η tot = P el Q = η u η el (6) We will only consider the open-turbine-gas-cycle plant in these notes since they are the most common type. Prof. A. Valentini - Gas urbine Power Plants 9

4 HE JOULE CYCLE he Joule (or Brayton-Joule) cycle, shown in figure 6, is the ideal cycle for gas-turbine engines. he ideal cycle is made up of the following four reversible processes: -2 Isentropic compression 2-3 Constant pressure combustion 3-4 Isentropic expansion 4- Constant pressure heat rejection 3 P2 = P3 2 4 P = P4 S fig. 6 - he Brayton - Joule cycle in -S chart he efficiency of the ideal Joule cycle can easily be calculated assuming that in ideal cycles just one kilo of fluid, that undergoes only thermodynamic transformations, is considered. herefore, the (5) can be written in terms of energy as below: η u = L U Q = (L ) obt L C Q (7) he terms of the fraction can be calculated with the first law of thermodynamics for open systems: (L ) obt = C P ( 3 4 ) (8) L C = C P ( 2 ) (9) Q = C P ( 3 2 ) (20) If we put (8), (9) and (20) in (7), and simplify the specific heat, we get: Prof. A. Valentini - Gas urbine Power Plants 0

η u = ( 3 4 ) ( 2 ) 3 2 (2) (2) can be rearranged as follow: η u = ( 3 2 ) ( 4 ) 3 2 = ( 4 ) 3 2 = ( 4 ) 2 ( 3 2 ) (22) but using (0) for the isentropic compression and expansion we see that 4 = 3 2 so (22) becomes: η u = = 2 (23) using (0) again we get: 2 = ( P k 2 k k ) = βc k (24) P so (22) becomes: η u = = k β C k (25) herefore, the efficiency of the ideal cycle depends on the type of gas (though the constant k) and the compression ratio and it increases if k and/or β C increase. In figure 7 the plot of the efficiency versus the compression ratio (with k fixed) can be seen. η u 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0. 0 k =. 4 0 20 40 60 80 β C 00 fig. 7 - Graph of the thermal efficiency versus the compression ratio for the ideal cycle Prof. A. Valentini - Gas urbine Power Plants

5 HE RUE CYCLE In gas turbine power plants the true cycle is quite different from the ideal Brayton-Joule, which was described in chapter 4, due to the following fluid dynamics and mechanical losses: compression and expansion are not isentropic pressure drop in the combustion chamber variation of average gas parameters (caused mainly by temperature variation) imperfect combustion bearing friction drive power of auxiliary equipment We consider the non-isentropicity in the thermodynamic transformation in the turbo machines using the isentropic efficiencies (7) and (8). o consider the pressure drop in the combustion chamber we define the pneumatic combustor efficiency as: η πb = P 3 P 2 (26) were P 2 is the turbo-compressor outlet pressure and P 3 is the combustor outlet pressure. he pneumatic combustor efficiency in smaller than one, as are all efficiencies, and so the true cycle expansion ratio is smaller than the compression ratio β c = P 2 P (27) β e = P 3 P 4 = η πb P 2 P = η πb β c (28) In these notes we do not consider the variation of gas parameters so some average values are assumed. Imperfect and incomplete combustion are considered assuming a combustion efficiency η πb, while mechanical friction (and the drive power of auxiliary equipment, if present) is considered assuming the mechanical efficiencies of the two turbo machines η m and η C m. he variations with respect to the ideal cycle are shown in figure 8. Prof. A. Valentini - Gas urbine Power Plants 2

3 3id 3 P3 P2 2id 2 4id 4 P = P4 S fig. 8 - he true Joule cycle in -S chart he mechanical power produced by the turbines in this case can be written as: P = (G a + G b ). (L i ) obt η m = (G a + G b ). C p ( 3 4 ) η m (29) While the power absorbed by the compressor is: P C = G a L i η m C = G a C p ( 2 ) η m C (30) Approximately 55 to 65 percent of the power produced by the turbine is used to drive the compressor. he plant thermal efficiency (5) can be written as follow: η u = P U = Q (G a + G b ) C p ( 3 4 ) η m G a C p ( 2 ) η m C η b G b H i (3) For the true cycle the efficiency strongly depends on the gas turbine inlet temperature 3, higher temperatures lead to higher efficiency as shown in figure 9. Prof. A. Valentini - Gas urbine Power Plants 3

η u 0.3 0.25 0.2 3 = 500 K 0.5 0. 3 = 200 K 0.05 0 3 = 800 K 3 = 000 K 0 0 20 30 40 50 60 β C 70 fig. 9 - Graph of thermal efficiency versus compression ratio and temperature 3 urbine inlet temperature is limited by the thermal conditions that can be tolerated by the metal alloy of the turbine blades. Gas temperatures at the turbine inlet can be between 200ºC and 400ºC, but some manufacturers have been able to boost inlet temperatures as high as 600ºC by engineering blade coatings and cooling systems to protect metallurgical components from thermal damage. Prof. A. Valentini - Gas urbine Power Plants 4

6 HE COMBUSION CHAMBER As we already know the combustion chamber (or combustor or burner) is the component in a gas turbine power plant where combustion takes place. he picture of a typical combustor is shown in figure 0. fig. 0 - Main parts of a combustion chamber he energy balance in a combustion chamber is: G b i b Q = η b G b H i = (G a + G b ) i 3 (G a i 2 + G b i b ) (32) Q G a 2 3 G a+ G b i 2 = C p 2 i 3 = C p 3 were ib is the fuel enthalpy. Since the fuel is usually preheated is possible to assume that ib i2 and so (32) becomes: Q = η b G b H i = (G a + G b ) (i 3 i 2 ) (33) Prof. A. Valentini - Gas urbine Power Plants 5

he ratio of the mixture between air and the fuel is defined in the following way: α = G a G b (34) for gas turbines this ratio is usually bigger than 50, much higher than the stoichiometric ratio (about 56), which means that combustion in gas-turbine power plants takes place with air excess. his is necessary to keep the 3 temperature down in order not to damage the turbine blades. 7 GAS URBINE PLANS WIH REGENERAIVE INNER HEA EXCHANGER If outlet flue-gas temperature (4) is higher than the temperature at the end of compression (2) the exhausted gas can be used for pre-heating compressed air. In figure the schematic picture of a plant with regenerative heat exchanger is shown. G a + G b 2 G b heat exchanger 5 G a 3 P U 4 G a + G b G a fig. - Gas turbine plant with a regenerative inner exchanger From figure and 2 (regenerative cycle) it can be seen that heat is supplied starting from point 5 (heat exchanger outlet) instead of point 2 (turbo compressor outlet). Prof. A. Valentini - Gas urbine Power Plants 6

3 5id = 4 2 5 5id 4 S fig. 2 - Gas turbine plant regenerative cycle So the actual thermal power supplied in the combustor is: Q = η b G b H i = (G a + G b ) (i 3 i 5 ) (35) he use of a heat exchanger can increase the efficiency of the plant significantly because of the thermal power saved. If the heat exchanger was ideal, at the exchanger outlet the air would reach temperature 5 = 4, but for a real heat exchanger the outlet temperature 5 < 4 because of the finite exchange surface. he efficiency of the heat exchanger is defined as follows: R = 5 2 4 2 (36) his is the ratio between the actual and the theorethical heat exchanged, it is approximately in the range of 0.6-0.8. Prof. A. Valentini - Gas urbine Power Plants 7

8 CYCLES WIH INER COOLING AND REHEAING o increase the cycle efficiency (5) it is possible to: Reduce the heat supplied Reduce the compressor power Increase the turbine power he reduction of heat is achieved with regenerative cycles as seen in the last chapter. It is possible to demostrate that the minimum work needed to compress a gas is spent with isothermal compression. It is very difficult to achieve such compression because, during compression, countinous heat subtraction would be necessary to keep the temperature constant. It is possible to approximate the isothermal compression with multiple stage compression with inter cooling in between. he power of a gas turbine is increased with re-heating (re-combustion) during expansion. It is possible to have further combustion, after the first expansion, because of the large amount of excess air used. In figure 3 a scheme of a plant with regeneration, inter cooling and re-heating is shown, while in figure 4 there is the corresponding thermodynamic cycle. G a + G b + G' b 2'' G b G' b 2 ' 2' '' 5 G a 3 G a + G b 4' G a + G b + G' b 3' P U 4 G a + G b + G' b G a fig. 3 - Gas turbine plant with regenerative cycle, reheating and inter cold compression Prof. A. Valentini - Gas urbine Power Plants 8

Q 3 Q 4' 3' 5 4 2'' 2' 2 '' ' fig. 4 - Cycle of plant with regenerative cycle, reheating and inter cold compression All the formulas necessary for the calculations are written below: P C = G a C p ( 2 ) η m c + G a C p ( 2 ) η m c + G a C p ( 2 ) η m c (37) S P = (G a + G b ) C p ( 3 4 ) η m + (G a + G b + G b ) C p ( 3 4 ) η m (38) Q = Q + Q (39) Q = η b G b H i = (G a + G b ) C p ( 3 5 ) (40) Q = η b G b H i = (G a + G b + G b ) C p ( 3 4 ) (4) α = G a G b (42) α = G a + G a G b (43) Prof. A. Valentini - Gas urbine Power Plants 9

EXERCISES ) For the gas-turbine power plant shown in fig. 3 the following data are given: P = bar = 5 3 = 200 Hi = 50 MJ/kg η b = 0.95 η is = η C is = 0.85 β C = 20 η m = η C m = 0.98 Gb =.6 kg s η πb = 0.95 Cp = 090 J kgk k =.4 he plant is used to produce alternating current by way of an alternator with an electrical efficiency η el = 0.9, calculate:. he air flow rates 2. he net power available on the shaft 3. he electrical power When we have a cycle the first step is always to calculate all pressure and temperature values for the points of the real cycle. 3 3id 3 P3 point P2 2id 2 4id 4 P = bar = 00000 Pa = 5 = 288 K P = P4 point 2 S From the compression ratio: P 2 = P β C = 20 bar = 2000000 Pa. o calculate 2 we can first calculate the final temperature of the isentropic compression with (0) 2is = β C k k 677.82 K. hen with the compressor isentropic efficiencies (7) it is possible to calculate the actual temperature 2. Prof. A. Valentini - Gas urbine Power Plants 20

η C is = L id i = C p ( 2is ) L i C p ( 2 ) 2 746.6 K point 3 From the pneumatic combustor efficiency: P 3 = P 2 η πb = 9 bar = 900000 Pa. While 3 is given: 3 = 200 = 473 K point 4 P 4 = P = bar = 00000 Pa. he expantion ratio: β e = η πb β c = 9. 4 can be calculated in the same way as 2. 4is = 3 ( β e ) k k 635. K η t is = (L i) obt id (L i ) = C p ( 3 4 ) obt C p ( 3 4is ) 4 760.78 K Now we can calculate the air flow rate with the energy balance of the combustion chamber (33) : η b G b H i = (G a + G b ) C p ( 3 2 ) so G a = η b G b H i C p ( 3 2 ) G b 89.33 kg s he work required by the compressor and that obtained by the turbine is: he net mechanical power on the shaft is: L c = C p ( 2 ) 499889 J kg (L ) obt = C p ( 3 4 ) 77634.35 J kg P U = (G a + G b ) (L ) obt η m G a L c η m C 2362046 W 23.6 MW he thermal power is: Q = η b G b H i 72000000 W 72 MW he thermal efficiency is: η u = P U Q 0.327 he electrical power is: P el = P U η el 2.24 MW he total efficiency is: η tot = P el = η u η el 0.2943 Q Prof. A. Valentini - Gas urbine Power Plants 2

2) For an open cycle of a gas-turbine power plant the characteristic points are given: P = bar P 2 = 25 bar P 3 = 24 bar P 4 = P = 293 k 2 = 750 k 3 = 500 k 4 = 850 k 3 2 4 Further data are known: Hi = 50 MJ/kg η b = 0.96 η m = η C m = 0.98 Cp = 090 J. Calculate the power plant efficiency 2. o improve the efficiency a regenerative exchanger with R = 0.7 is used. Calculate the new plant efficiency. S kgk he efficiency is calculated with (3) but we don't know either the air flow rate or the fuel flow rate. By dividing the numerator and the denominator by G b, (3) can be written as follows: η u = (α + ) C p ( 3 4 ) η m α C p ( 2 ) η m C η b H i where the only unknown value is α = G a G b follows: and it can be calculated by rearranging (33) as η b H i = (α + ) C p ( 3 2 ) Prof. A. Valentini - Gas urbine Power Plants 22

α = And so the efficiency is: η b H i C p ( 3 2 ) = 0.96 50000000 090 (500 750) 57.7 η u = (57.7 + ) 090 (500 850) 0.98 57.7 090 (750 293) 0.98 0.96 50000000 0.238 If the regenerative cycle is used the air and fuel flow rates ratio is different: 3 2 5 4 α = η b H i C p ( 3 5 ) S where 5 is calculated by (36) R = 5 2 4 2 : 0.7 = 5 750 850 750 5 = 820 K α = he new efficiency becomes: 0.96 50000000 090 (500 820) 64.76 η u = (64.76 + ) 090 (500 850) 0.98 64.76 090 (750 293) 0.98 0.96 50000000 0.265 the efficiency increases by about %. N.B. In the calculation we assumed that the cycle points do not change in the regenerative cycle but this is only an approximation. Prof. A. Valentini - Gas urbine Power Plants 23

BIBLIOGRAPHY [] A. Capetti, Motori ermici, UE, 967 [2] A. Dadone, Introduzione e complementi di macchine termiche ed idrauliche, CLU [3] W. W. Bathie, Fundamentals of Gas urbines, John Wiley & Sons, 996 [4] R. Kurz, Introduction to Gas urbines and Applications [5] E. A. Baskharone, Principles of urbomachinery, Cambridge University Press, 2006 Prof. A. Valentini - Gas urbine Power Plants 24