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CHAPTER 2

CHAPTER 2. SHAFT POWER CYCLES. There are two main types of power cycles; 1. Shaft power cycles : Marine and Land based power plants. 2. Aircraft propulsion cycles: Performance depends upon forward speed & altitude. CHAPTER 2. SHAFT POWER CYCLES I Ideal Cycles.

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CHAPTER 2

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  1. CHAPTER2 SHAFT POWERCYCLES

  2. There are two main types of power cycles; • 1. Shaft power cycles : Marine and Land based power plants. • 2. Aircraft propulsion cycles: Performance depends upon forward speed & altitude. Chapter2 Shaft Power Cycles

  3. CHAPTER2 SHAFT POWERCYCLES I Ideal Cycles

  4. The analysis is based on the : perfection of individual components w ,h depend upon r and Tmax w : Specific power output h: Cycle efficiency r : Pressure ratio Tmax : Max. cycle Temperature Ideal Cycles Chapter2 Shaft Power Cycles

  5. a)Compression and expansion processes are isentropic b) The change of Kinetic Energy of the working fluid between the inlet and outlet of each component is negligible. Ideal Cycles - Assumptions Chapter2 Shaft Power Cycles

  6. c) No pressure losses in the inlet ducting, combustion chambers, heat exchangers, intercoolers, exhaust ducting and ducts connecting the components. d) The composition of the working fluid does not change and it is a prefect gas with constant specific heats Ideal Cycles - Assumptions Chapter2 Shaft Power Cycles

  7. Ideal Cycles - Assumptions: e)The mass flowrateof gas is constant f)Heat transfer in the Heat Exchanger is complete; so in conjunction with (d)+(e),temperature rise on the cold side is equal to the temperature drop on the hot side. • (d,e) indicates that the combustion chamber is such thatit is as if heated by an external heat source. Chapter2 Shaft Power Cycles

  8. SIMPLE Gas Turbine CYCLE The ideal cycle for a simple Gas Turbine is the BRAYTON (or JOULE) cycle. Fig. 2.1 Simple Gas Turbine Chapter2 Shaft Power Cycles

  9. Steady flow energy equation: q = hII -hI + 1/2 ( vII2 - vI2) + w where : q= heat transfer per unit mass flow. w= work per unit mass flow. w12 = - (h2-h1) = - cp (T2–T1) q23= (h3-h2) = cp (T3-T2) w34 = (h3-h4) = cp (T3-T4) SIMPLEGas Turbine CYCLE Chapter2 Shaft Power Cycles

  10. SIMPLE Gas Turbine CYCLE The efficiency of the cycle is then : Chapter2 Shaft Power Cycles

  11. SIMPLE Gas Turbine CYCLE The cycle temperatures can be related to the pressure ratio rp ; rp = p2/p1= p3/p4 For isentropic compression and expansion; p/r= RT ; p/rg = const. T2/T1=rp(g-1)/gandT3/T4= rp(g-1)/g Chapter2 Shaft Power Cycles

  12. The Efficiency of the Simple Gas Turbine Cycle Chapter2 Shaft Power Cycles

  13. The Efficiency of the Simple Gas Turbine Cycle Then the cycle efficiency is; where;( 2.1 ) Chapter2 Shaft Power Cycles

  14. SIMPLE Gas Turbine CYCLE The specific work output w , depends upon the size of the plant for a givenpower. It is found to be a function of not only pressure ratio, but also of maximum cycle temperature T3. Thus; w= cp (T3-T4) - cp (T2-T1) can be expressed as; ( 2.2 ) The specific work output is a function of " t " (T3/T1) and "rp" ; w =w (t, rp). Chapter2 Shaft Power Cycles

  15. SIMPLE Gas Turbine CYCLE Fig. 2.2Efficiency and specific work output - Simple Cycle FIG. 2.2 Efficiency and specific work output - simple cycle Chapter2 Shaft Power Cycles

  16. SIMPLE Gas Turbine CYCLE T3 :Maximum Cycle Temperature, imposed by the metallurgical limit. t = T3/T1 = 3.5 -4 for long life industrial plants, t = 5-5.5 for aircraft engines with cooled turbine blades. From the T-S diagram, it is clear that when rp= 1 or rp=(T3/T1)(g/g-1) w= 0 Thus in between there is a maximum (or minimum) value for w Chapter2 Shaft Power Cycles

  17. SIMPLE Gas Turbine CYCLE For any given value of "t" (T3/T1), the optimum value of rp for maximum specific work output can be calculated by differentiating eqn. 2.2 wrt. rp(g/g-1) and equating to zero. The result is; i.e. ( 2.3 ) Since Chapter2 Shaft Power Cycles

  18. SIMPLE Gas Turbine CYCLE This is equivalent to; So, w is maximum when compressor and turbine outlet temperatures are equal. For all values of rp between 1 and ropt= [ (T3/T1) (g/2 (g-1)) ]T4 > T2 and a heat exchanger can be incorporated to reduce the heat transfer from the external source and so increase the efficiency. Chapter2 Shaft Power Cycles

  19. Heat Exchanger Cycle Fig. 2.3 Simple cycle with heat - exchange Chapter2 Shaft Power Cycles

  20. Heat Exchanger Cycle The cycle efficiency h is; with ideal HET5 = T4 with the help of isentropicrelations; ( 2.4) Chapter2 Shaft Power Cycles

  21. Heat Exchanger Cycle Fig 2.4 Efficiency of a simple cycle with heat-exchange for rp = 1 : h = 1-1/t which is Carnot efficiency. as T3 increases, t increases and thenh increases Chapter2 Shaft Power Cycles

  22. Heat Exchanger Cycle Specific work output does not change with HE thus is thesame as the simple cycle. To obtain an appreciable improvement in h by HE in ideal cycles. a) a value of rp < ropt then work output is maximum. b) It is not necessary to use a higher cycle pressureratio as Tmax of the cycle is increased. (a) is true for actual cycles whereas (b) requires modification Chapter2 Shaft Power Cycles

  23. Reheat Cycle Fig. 2.5 Reheat cycle • A substantial increase in specific work output can be obtained by splitting the expansion and reheating the gas between low-pressure and high-pressure turbines. Chapter2 Shaft Power Cycles

  24. Reheat Cycle • Since the vertical distance between any pair of constant pressure lines increase with the increasing entropy • (T3 - T4) + (T5 - T6) > (T3 - T4 ) • thus :wreheat > wsimple • w34 + w56 = Cp (T3 - T4) + Cp (T5 - T6) • = Cp (T3 - T4) + Cp (T3 - T6) • wt = Cp T3 (1-T4/ T3) + Cp T3 (1-T6/T5) Chapter2 Shaft Power Cycles

  25. Reheat Cycle since • Denoting • then P4 = P5 Chapter2 Shaft Power Cycles

  26. Reheat Cycle • To find P4 for maximum work output; • The result is; • Hence, P3/P4 = P4/P6 • for maximum work output the optimum splitting is an equal one. Chapter2 Shaft Power Cycles

  27. Reheat Cycle • Specific work output of the cycle is then; w = Cp (T3 - T4) + Cp (T5 - T6) - Cp (T2- T1) • Thus: • ( 2.5 ) • Then the efficiency; ( 2.6) Chapter2 Shaft Power Cycles

  28. Reheat Cycle • Effect of Reheat : increase in specific output and decrease in efficiency. Fig. 2.6 Work output vs. r in aReheat Cycle • EXERCISE : For a simple Reheat Cycle, prove that specific work output is maximum whenrp(g-1/g) =(T3/T1)2/3 = t2/3 . Chapter2 Shaft Power Cycles

  29. Cycle with Reheat & Heat Exchange • The reduction in efficiency due to reheat can be overcomed by adding heat exchanger. • The high exhaust gas temperature is now fully utilized in the HE and the increase in work output is no longer offset by the increase in heat supplied. Fig. 2.7 Reheat cycle with Heat - Exchange Chapter2 Shaft Power Cycles

  30. Reheat cycle with Heat - Exchange Fig. 2.8 Efficiency - reheat cycle with heat - exchange Chapter2 Shaft Power Cycles

  31. 7 HE f f 6 3 4 5 2 1 Cycle With Reheat & Heat Exchange The reduction in efficiency due to reheat can be overcomed by adding a heat exchanger. The higher exhaust gas temperature is now fully utilized in the HE and the increase in work output is no longer offset by the increase in heat supplied. 6 Chapter2 Shaft Power Cycles

  32. Cycle With Reheat& Heat Exchange T 3 5 7 6 4 HE 2 8 1 s Chapter2 Shaft Power Cycles

  33. Intercooler 9 f HE f 3 6 8 1 2 5 7 4 LPC HPC LPT HPT 10 Cycle With Reheat, Intercooling & Heat Exchange Chapter2 Shaft Power Cycles

  34. Cycle With Reheat, Intercooling & Heat Exchange T 5 7 9 6 8 HE 4 2 10 1 3 s Chapter2 Shaft Power Cycles

  35. CHAPTER2 SHAFT POWERCYCLES II Actual Cycles

  36. ACTUAL CYCLES • The performance of real cycles differ from that of ideal cycles for the following reasons : a)Change in Kinetic Energy between inlet and outlet of each component can not necessarily be ignored. • b) Compression and expansion are actually irreversible and therefore involves an increase in entropy • c) Fluid friction causes pressure losses in components and associated ducts. • d)HE can not be ideal, terminal temperature difference is inevitable Chapter2 Shaft Power Cycles

  37. ACTUAL CYCLES • e) Slightly more work than that required for the compression process will be necessary to overcome bearing andwindage friction in the transmission between compressor and turbine and to drive ancillary components such as fuel and oil pumps. (hmech) • f) Cp and g changes throughout the cycle. Cp = f(T) h= f(T)and chemical composition. • g) Combustion is not complete (hcomb) Chapter2 Shaft Power Cycles

  38. The efficiency of any machine (which absorbs or produces work), is normally expressed in terms of the ratio of actual to ideal work transfers For a compressor; For a perfect gas; h = Cp T , This relation is sufficiently accurate for real gasses under conditions encountered in a GT if a mean Cp over the relevant range of temperature is used ACTUAL CYCLES Chapter2 Shaft Power Cycles

  39. Compressor and Turbine Efficiencies • Then for compressors; ( 2.7 ) • Similarly for turbines the isentropic efficiency defined as; ( 2.8 ) Chapter2 Shaft Power Cycles

  40. For compressors: from equation 2.7 ( 2.9 ) Similarly for turbines; ( 2.10 ) ACTUAL CYCLES Chapter2 Shaft Power Cycles

  41. Compressor & Turbine Efficiencies • Since • Thus ( 2.11 ) • Since the vertical distance between a pair of constant pressure lines on the T-S diagram increases as entropy increases, SDTs’> DT’ hs >hc( for compressors ) Chapter2 Shaft Power Cycles

  42. Now consider an axial flow compressor consisting of a number of successive stages. If the blade design is similar in successive blade rows it is reasonable to assume that the isentropic efficiency of a single stage hs remains the same through the compressor. Then the overall temperature rise; Compressor and Turbine Efficiencies Chapter2 Shaft Power Cycles

  43. Compressor and Turbine Efficiencies • The difference between hcandhs will increase with the number of stages i.e. with the increase of pressure ratio. • A physical explanation is that the increase in temperature due to friction in one stage results in more work being required in the next stage. • A similar argument can be used to show that for a turbine ht > hst . Fig. 2.9 Definition of Isentropic and Small Stage Efficiencies Chapter2 Shaft Power Cycles

  44. Polytropic Efficiency • Isentropic efficiency of an elemental stage in the process such that it is constant throughout the whole process. • For a compression process η c = dT'/dT = const. • But for an isentropic process, • in differential form dT'= η  c dT • integrating between 1 & 2 (inlet & outlet) ( 2.12 ) Chapter2 Shaft Power Cycles

  45. Polytropic Efficiency • So hccan be computed from measured values of P and T at the inlet an outlet of the compressor, as; ( 2.13 ) • Finally the relation between hc & hc; ( 2.14 ) Chapter2 Shaft Power Cycles

  46. Polytropic Efficiency • Similar relations can be obtained for turbines since , • It can be shown that for an expansion between turbine inlet 3 and outlet 4; ( 2.15 ) ( 2.16 ) Chapter2 Shaft Power Cycles

  47. Polytropic Efficiency Fig. 2.10 Variation of turbine and compressor isentropic efficiency with pressure ratio for polytropic efficiency of 85 % Chapter2 Shaft Power Cycles

  48. Polytropic Efficiency • In practice, as with hc and ht , it is normal to define the polytropic efficiencies in terms of stagnation temperatures and pressures. ( 2.17 ) • where ( 2.18 ) where • Here n is the coefficient for a polytropic process. Chapter2 Shaft Power Cycles

  49. Pressure Losses Fig. 2.11 Pressure losses • Pb= Pressure loss in Combustion Chamber • Pha = Frictional pressure loss on the air side of HE • Phg= Frictional pressure loss on the gas side of HE Chapter2 Shaft Power Cycles

  50. Pressure Losses • Pressure losses cause a decrease in the available turbine pressureratio. • Po3 = Po2 - Pb - Pha • Po4 = Pa + Phg • It is better to take Phg & Pb as fixed proportions of compressor delivery pressure; Then; Chapter2 Shaft Power Cycles

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