APPLIED THERMODYNAMICS. TUTORIAL No.3 GAS TURBINE POWER CYCLES. Revise gas expansions in turbines. Study the Joule cycle with friction.

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1 APPLIED HERMODYNAMICS UORIAL No. GAS URBINE POWER CYCLES In this tutorial you will do the following. Revise gas expansions in turbines. Revise the Joule cycle. Study the Joule cycle with friction. Extend the work to cycles with heat exchangers. Solve typical exam questions. D.J.Dunn

2 . REVISION OF EXPANSION AND COMPRESSION PROCESSES. When a gas is expanded from pressure p to pressure p adiabatically, the temperature ratio p is given by the formula p he same formula may be applied to a compression process. Always remember that when a gas is expanded it gets colder and when it is compressed it gets hotter. he temperature change is - If there is friction the isentropic efficiency ( is) is expressed as is = (ideal)/(actual) for a compression. is = (actual)/(ideal) for an expansion. An alternative way of expressing this is with POLYROPIC EFFICIENCY For a compression from () to () the temperature ratio is expressed as follows. p and for an expansion from () to () p p where p is called the polytropic efficiency. WORKED EXAMPLE No. A gas turbine expands kg/s of air from bar and 900oC to bar adiabatically with an isentropic efficiency of 87%. Calculate the exhaust temperature and the power output. =. c p = 005 J/kg K SOLUION = (/) -/. = 7 (/) = 56.8 K Ideal temperature change = = 60.5 K Actual temperature change = 87% x 60.5 = 5.5 K Exhaust temperature = = 6.85 K he steady flow energy equation states Since it is an adiabatic process = 0 so P = H/s = m c p = x 005 x (5.5) P = -.5 x 06 W (Leaving the system) P(out) =.5 MW D.J.Dunn

3 SELF ASSESSMEN EXERCISE No.. A gas turbine expands 6 kg/s of air from 8 bar and 700oC to bar isentropically. Calculate the exhaust temperature and the power output. =. c p = 005 J/kg K (Answers 57. K and.68 MW). A gas turbine expands kg/s of air from 0 bar and 90oC to bar adiabatically with an isentropic efficiency of 9%. Calculate the exhaust temperature and the power output. =. c p = 00 J/kg K (Answers 657. K and.6 MW). A gas turbine expands 7 kg/s of air from 9 bar and 850oC to bar adiabatically with an isentropic efficiency of 87%. Calculate the exhaust temperature and the power output. =. c p = 005 J/kg K (Answers K and.0 MW) D.J.Dunn

4 . HE BASIC GAS URBINE CYCLE he ideal and basic cycle is called the JOULE cycle and is also known as the constant pressure cycle because the heating and cooling processes are conducted at constant pressure. A simple layout is shown on fig.. Figure Illustrative diagram. he cycle in block diagram form is shown on fig.. Fig. Block diagram D.J.Dunn

5 here are ideal processes in the cycle. - Reversible adiabatic (isentropic) compression requiring power input. P in = H/s = m c p ( - ) - Constant pressure heating requiring heat input. in = H/s = m c p ( - ) - Reversible adiabatic (isentropic) expansion producing power output. P out = H/s = m c p ( - ) - Constant pressure cooling back to the original state requiring heat removal. out = H/s = m c p ( - ) he pressure - volume, pressure - enthalpy and temperature-entropy diagrams are shown on figs. a, b and c respectively.. EFFICIENCY Fig a p-v diagram Fig b p-h diagram Fig c -s diagram he efficiency is found by applying the first law of thermodynamics. nett in P - out P th nett in nett P out - P out in in mc mc p p ( ( - ) - ) ( ( - ) - ) It assumed that the mass and the specific heats are the same for the heater and cooler. It is easy to show that the temperature ratio for the turbine and compressor are the same. p p r p p p r p r p is the pressure compression ratio for the turbine and compressor. D.J.Dunn 5

6 D.J.Dunn 6. since ) - ( ) - ( p p th th r r his shows that the efficiency depends only on the pressure ratio which in turn affects the hottest temperature in the cycle.

7 WORKED EXAMPLE No. A gas turbine uses the Joule cycle. he pressure ratio is 6/. he inlet temperature to the compressor is 0oC. he flow rate of air is 0. kg/s. he temperature at inlet to the turbine is 950oC. Calculate the following. i. he cycle efficiency. ii. he heat transfer into the heater. iii. he net power output. =. c p =.005 kj/kg K SOLUION r P th th in nett r p 0.86 p mc P nett in p ( 6 8 x 6 - ) 0. x kw K or 0% 0. x.005 x ( - 7.) 50.8 kw SELF ASSESSMEN EXERCISE No. A gas turbine uses the Joule cycle. he inlet pressure and temperature to the compressor are respectively bar and -0oC. After constant pressure heating, the pressure and temperature are 7 bar and 700oC respectively. he flow rate of air is 0. kg/s. Calculate the following.. he cycle efficiency.. he heat transfer into the heater.. the nett power output. =. c p =.005 kj/kg K (Answers.7 %, 06.7 kw and 88.6 kw) D.J.Dunn 7

8 . HE EFFEC OF FRICION ON HE JOULE CYCLE. URBINE he isentropic efficiency for a gas turbine is given by: i = (Actual change in enthalpy)/(ideal change in enthalpy) i = (Actual change in temperature)/(ideal change in temperature). COMPRESSOR For a compressor the isentropic efficiency is inverted and becomes as follows. i = (Ideal change in enthalpy)/(actual change in enthalpy) hi = (Ideal change in temperature)/(actual change in temperature) Remember that friction always produces a smaller change in temperature than for the ideal case. his is shown on the -s diagrams (fig.a and b). Fig.a urbine expansion. Fig.b Compression process. i = ( )/( ) i = ( )/( ) he power output from the turbine is hence P(out) = m c p ( ) i he power input to the compressor is hence P(in) = m c p ( )/i. HE CYCLE WIH FRICION It can be seen that the effect of friction on the gas turbine cycle is reduced power output and increased power input with an overall reduction in nett power and thermal efficiency. Figs. 5a and 5b show the effect of friction on -s and p-h diagrams for the Joule cycle. Fig.5a emperature - Entropy Fig.5b. Pressure - Enthalpy D.J.Dunn 8

9 Note the energy balance which exists is P(in) + (in) = P(out) + (out) P(nett) = P(out) - P(in) = (nett) = (in) - (out) WORKED EXAMPLE No. A Joule Cycle uses a pressure ratio of 8. Calculate the air standard efficiency. he isentropic efficiency of the turbine and compressor are both 90%. he low pressure in the cycle is 0 kpa. he coldest and hottest temperatures in the cycle are 0oC and 00oC respectively. Calculate the cycle efficiency with friction and deduce the change. Calculate the nett power output. =. and c p =.005 kj/kg K. ake the mass flow as kg/s. SOLUION No friction th = - rp / - = 0.8 or 8.8 % With friction ' = 9 x = 5 K i = 0.9 = (5-9)/(-9) = 5 K ' = 7/ = 8.7 K i = 0.9 = (7-)/(7-8.7) = th = - (out)/(in) = - (-)/(-) th= 0.6 or 6% he change in efficiency is a reduction of 8.8% (in) = m c p ( - ) = x.005 x (7-557) = 760 kw Nett Power Output = P(nett) = th x (in) = 0.6 x 760 = 99 kw D.J.Dunn 9

10 SELF ASSESSMEN EXERCISE No. A gas turbine uses a standard Joule cycle but there is friction in the compressor and turbine. he air is drawn into the compressor at bar 5oC and is compressed with an isentropic efficiency of 9% to a pressure of 9 bar. After heating, the gas temperature is 000oC. he isentropic efficiency of the turbine is also 9%. he mass flow rate is. kg/s. Determine the following.. he net power output.. he thermal efficiency of the plant. =. and c p =.005 kj/kg K. (Answers 6 kw and 0.%) D.J.Dunn 0

11 . VARIANS OF HE BASIC CYCLE In this section we will examine how practical gas turbine engine sets vary from the basic Joule cycle.. GAS CONSANS he first point is that in reality, although air is used in the compressor, the gas going through the turbine contains products of combustion so the adiabatic index and specific heat capacity is different in the turbine and compressor.. FREE URBINES Most designs used for gas turbine sets use two turbines, one to drive the compressor and a free turbine. he free turbine drives the load and it is not connected directly to the compressor. It may also run at a different speed to the compressor. Fig.6a. shows such a layout with turbines in parallel configuration. Fig.6b shows the layout with series configuration. Fig. 6a Parallel turbines Fig.6b. Series turbines D.J.Dunn

12 . INERCOOLING his is not part of the syllabus for the power cycles but we will come across it later when we study compressors in detail. Basically, if the air is compressed in stages and cooled between each stage, then the work of compression is reduced and the efficiency increased. he layout is shown on fig. 7a.. REHEAING he reverse theory of intercooling applies. If several stages of expansion are used and the gas reheated between stages, the power output and efficiency is increased. he layout is shown on fig. 7b. Fig.7a. Intercooler Fig.7b. Reheater D.J.Dunn

13 WORKED EXAMPLE No. A gas turbine draws in air from atmosphere at bar and 0oC and compresses it to 5 bar with an isentropic efficiency of 80%. he air is heated to 00 K at constant pressure and then expanded through two stages in series back to bar. he high pressure turbine is connected to the compressor and produces just enough power to drive it. he low pressure stage is connected to an external load and produces 80 kw of power. he isentropic efficiency is 85% for both stages. Calculate the mass flow of air, the inter-stage pressure of the turbines and the thermal efficiency of the cycle. For the compressor =. and for the turbines =.. he gas constant R is 0.87 kj/kg K for both. Neglect the increase in mass due to the addition of fuel for burning. SOLUION c p R and R c p cv hence cp c v Hence c p =.005 kj/kg K for the compressor and.9 kj/kg K for the turbines. COMPRESSOR 0.86 r p 8 x 5 8. K Power input to compressor = m c p ( - ) Power output of h.p. turbine = m c p ( - ) Since these are equal it follows that.005(89.8-8)=.9(00- ) =09. K D.J.Dunn

14 HIGH PRESSURE URBINE ηi 0.85 hence 987. K p p p hence LOW PRESSURE URBINE 5.9 i hence p hence K.9 bar 88.5 K NE POWER he nett power is 80 kw hence 80 = m c p ( - 5 ) = m x.9( ) m = 0. kg/s HEA INPU (in) = m c p ( - ) = 0. x.9 ( ) = 5. kw HERMAL EFFICIENCY th = P(nett)/(in) = 80/5. = 0. or.% D.J.Dunn

15 SELF ASSESSMEN EXERCISE No. A gas turbine draws in air from atmosphere at bar and 5oC and compresses it to.5 bar with an isentropic efficiency of 8%. he air is heated to 00 K at constant pressure and then expanded through two stages in series back to bar. he high pressure turbine is connected to the compressor and produces just enough power to drive it. he low pressure stage is connected to an external load and produces 00 kw of power. he isentropic efficiency is 85% for both stages. For the compressor =. and for the turbines =.. he gas constant R is 0.87 kj/kg K for both. Neglect the increase in mass due to the addition of fuel for burning. Calculate the mass flow of air, the inter-stage pressure of the turbines and the thermal efficiency of the cycle. (Answers 0.6 kg/s and 0. %) D.J.Dunn 5

16 .5. EXHAUS HEA EXCHANGERS Because the gas leaving the turbine is hotter than the gas leaving the compressor, it is possible to heat up the air before it enters the combustion chamber by use of an exhaust gas heat exchanger. his results in less fuel being burned in order to produce the same temperature prior to the turbine and so makes the cycle more efficient. he layout of such a plant is shown on fig.8. Fig.8 Plant layout In order to solve problems associated with this cycle, it is necessary to determine the temperature prior to the combustion chamber ( ). A perfect heat exchanger would heat up the air so that is the same as 5. It would also cool down the exhaust gas so that 6 becomes. In reality this is not possible so the concept of HERMAL RAIO is used. his is defined as the ratio of the enthalpy given to the air to the maximum possible enthalpy lost by the exhaust gas. he enthalpy lost by the exhaust gas is H = m g c pg ( 5-6 ) his would be a maximum if the gas is cooled down such that 6 =. Of course in reality this does not occur and the maximum is not achieved and the gas turbine does not perform as well as predicted by this idealisation. H(maximum) = H = m g c pg ( 5-6 ) he enthalpy gained by the air is H(air) = m a c pa ( - ) Hence the thermal ratio is.r. = m a c pa ( - )/ m g c pg ( 5 - ) he suffix a refers to the air and g to the exhaust gas. Since the mass of fuel added in the combustion chamber is small compared to the air flow we often neglect the difference in mass and the equation becomes c pa. R. c pg 5 D.J.Dunn 6

17 WORKED EXAMPLE No.5 A gas turbine uses a pressure ratio of 7.5/. he inlet temperature and pressure are respectively 0oC and 05 kpa. he temperature after heating in the combustion chamber is 00 oc. he specific heat capacity c p for the exhaust gas is.5 kj/kg K. he adiabatic index is. for air and. for the gas. Assume isentropic compression and expansion. he mass flow rate is kg/s. Use the chart below to determine c p for air. Calculate the air standard efficiency if no heat exchanger is used and compare it to the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.88 is used. SOLUION Referring to the numbers used on fig.8 the solution is as follows. Air standard efficiency = - rp (-/) = = 0.8 or.8% Solution with heat exchanger (-/) 0.86 = r p = 8 (7.5) = 50.6 K (-/) = /r p = 57/(7.5) = K Use the thermal ratio to estimate with a typical value of c p =.005 kj/kg K K he chart below shows the effect of pressure and temperature on C p. Post compressor pressure is about 7.5 bar and the mean temperature of air in the heat exchanger is about 78 K. From the chart c p will be around.08 kj/kg K D.J.Dunn 7

18 .08 Recalculate In order find the thermal efficiency, it is best to solve the energy transfers. P(in)= mc pa ( - ) = x.08 (50.6-8) = 8. kw P(out) = mc pg ( - 5 ) = x.5 ( ) = 75.9 kw P(nett) = P(out) - P(in) = 77.7 kw (in)combustion chamber) = mc pg ( - ) (in)=.5( ) = kw th = P(nett)/(in) = 77.7/777.5 = 0.6 or 6.% K D.J.Dunn 8

19 SELF ASSESSMEN EXERCISE No. 5. A gas turbine uses a pressure ratio of 7/. he inlet temperature and pressure are respectively 0oC and 00 kpa. he temperature after heating in the combustion chamber is 000 oc. he specific heat capacity c p is.005 kj/kg K and the adiabatic index is. for air and gas. Assume isentropic compression and expansion. he mass flow rate is 0.7 kg/s. Calculate the net power output and the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.8 is used. (Answers kw and 57%). A gas turbine uses a pressure ratio of 6.5/. he inlet temperature and pressure are respectively 5oC and bar. he temperature after heating in the combustion chamber is 00 oc. he specific heat capacity c p for air is.005 kj/kg K and for the exhaust gas is.5 kj/kg K. he adiabatic index is. for air and. for the gas. he isentropic efficiency is 85% for both the compression and expansion process. he mass flow rate is kg/s. Calculate the thermal efficiency when an exhaust heat exchanger with a thermal ratio of 0.75 is used. (Answer 8.%) D.J.Dunn 9

20 WORKED EXAMPLE No.6 A gas turbine has a free turbine in parallel with the turbine which drives the compressor. An exhaust heat exchanger is used with a thermal ratio of 0.8. he isentropic efficiency of the compressor is 80% and for both turbines is he heat transfer rate to the combustion chamber is.8 MW. he gas leaves the combustion chamber at 00oC. he air is drawn into the compressor at bar and 5oC. he pressure after compression is 7. bar. he adiabatic index is. for air and. for the gas produced by combustion. he specific heat c p is.005 kj/kg K for air and.5 kj/kg K for the gas. Determine the following. i. he mass flow rate in each turbine. ii. he net power output. iii. he thermodynamic efficiency of the cycle. SOLUION = 98 K = 98(7.) (-/.) = 5 K = 7 K 5 = 7(/7.) (-/.) = 88.5 K COMPRESSOR i = 0.8 = (5-98)/( -98) hence = K URBINES reat as one expansion with gas taking parallel paths. i = 0.85 = (7-5 )/(7-88.5) hence 5 = 98.7 K HEA EXCHANGER hermal ratio = 0.8 =.005( )/.5( ) hence = 890. K COMBUSION CHAMBER (in)= mc p ( - ) = 80 kw 80 = m(.5)(7-890.) hence m =.665 kg/s COMPRESSOR P(in) = mc p (-) =.665(.005)( ) = kw D.J.Dunn 0

21 URBINE A P(out) = kw = m A c p (-5) = =.665(.5)(7-98.7) hence m A =.8 kg/s Hence mass flow through the free turbine is.68 kg/s P(nett) = Power from free turbine =.68(.5)(7-98.7) = 65.7 kw HERMODYNAMIC EFFICIENCY th = P(nett)/(in)= 65.7/80 = 0.9 or.8 % SELF ASSESSMEN EXERCISE No. 6. List the relative advantages of open and closed cycle gas turbine engines. Sketch the simple gas turbine cycle on a -s diagram. Explain how the efficiency can be improved by the inclusion of a heat exchanger. In an open cycle gas turbine plant, air is compressed from bar and 5oC to bar. he combustion gases enter the turbine at 800oC and after expansion pass through a heat exchanger in which the compressor delivery temperature is raised by 75% of the maximum possible rise. he exhaust gases leave the exchanger at bar. Neglecting transmission losses in the combustion chamber and heat exchanger, and differences in compressor and turbine mass flow rates, find the following. (i) he specific work output. (ii) he work ratio (iii) he cycle efficiency he compressor and turbine polytropic efficiencies are both 0.8. Compressor c p =.005 kj/kg K =. urbine c p =.8 kj/kg K =. Note for a compression and for an expansion p p p p D.J.Dunn

22 . A gas turbine for aircraft propulsion is mounted on a test bed. Air at bar and 9K enters the compressor at low velocity and is compressed through a pressure ratio of with an isentropic efficiency of 85%. he air then passes to a combustion chamber where it is heated to 75 K. he hot gas then expands through a turbine which drives the compressor and has an isentropic efficiency of 87%. he gas is then further expanded isentropically through a nozzle leaving at the speed of sound. he exit area of the nozzle is 0. m. Determine the following. (i) he pressures at the turbine and nozzle outlets. (ii) he mass flow rate. (iii) he thrust on the engine mountings. Assume the properties of air throughout. he sonic velocity of air is given by a= (R) ½. he temperature ratio before and after the nozzle is given by (in)/(out) = /(+) D.J.Dunn

23 . (A). A gas turbine plant operates with a pressure ratio of 6 and a turbine inlet temperature of 97oC. he compressor inlet temperature is 7oC. he isentropic efficiency of the compressor is 8% and of the turbine 90%. Making sensible assumptions, calculate the following. (i) he thermal efficiency of the plant. (ii) he work ratio. reat the gas as air throughout. (B). If a heat exchanger is incorporated in the plant, calculate the maximum possible efficiency which could be achieved assuming no other conditions are changed. Explain why the actual efficiency is less than that predicted. D.J.Dunn