Chapter 7 Vapor and Ga Power Sytem In thi chapter, we will tudy the baic component of common indutrial power and refrigeration ytem. Thee ytem are eentially thermodyanmic cycle in which a working fluid i alternatively vaporized and condened a it flow through a et of four procee and return to it initial tate. We will ue the firt and econd law of thermodynamic to analyze the performance of ower and refrigeration cycle. 7. Rankine Cycle We can ue a Rankine cycle to convert a foil-fuel, nuclear, or olar power ource into net electrical power. Figure 7.-a how the component of a Rankine cycle and Figure 7.-b identifie the tate of the Rankine cycle on a T diagram. Turbine W t Fuel air Q H Boiler 4 Rankine cycle Q C Condener W p Pump Figure 7.-a The ideal Rankine cycle Figure 7.-b The path of an ideal Rankine cycle Moran, M. J. and Shapiro H. N., Fundamental of Engineering Thermodynamic, Wiley, 008, pg. 95 7-
In the ideal Rankine cycle, the working fluid undergoe four reverible procee: Proce -: Ientropic expanion of the working fluid from the turbine from aturated vapor at tate or uperheated vapor at tate to the condener preure. Proce -: Heat tranfer from the working fluid a it flow at contant preure through the condener with aturated liquid at tate. Proce -4: Ientropic compreion in the pump to tate 4 in the compreed liquid region. Proce 4-: Heat tranfer to the working fluid a it flow at contant preure through the boiler to complete the cycle. We will conider the Rankine cycle with water a the working fluid. Proce -: Auming that bulk kinetic and potential energy and heat tranfer are negligible, the power produced by the turbine i given by W t = m (h h ) (7.-) In thi equation m i the ma flow rate of the working fluid. If the team enter a a uperheated vapor, it doe not condene ignificantly in the turbine. If the team i aturated a it enter the turbine, a ignificant of liquid i formed which caue eroion and wear of the turbine blade. Proce -: The team enter the condener and exit in tate a aturated liquid water. The change of phae occur at contant preure with the heat removed from the flowing team given by Q c = m (h h ) (7.-) Proce -4: A pump raie the preure of the liquid. High-preure water exit the pump in tate 4 where the work delivered to the liquid i given by W p = m (h 4 h ) m v (p 4 p ) (7.-) In thi equation the pump work i integrated from W p m = 4 vdp v (p 4 p ) The liquid volume i aumed to be contant. Since the pecific volume of the liquid i ignificantly le than that of the vapor, the work required by the pump i much le than that produced by the turbine. Typically, a mall fraction of the power produced by the turbine i ued to compre the liquid, and the remaining power i the net power obtained by the cycle. 7-
Proce 4-: The high-preure liquid i brought back to a aturated or uperheated vapor tate in the boiler where the rate of heat tranfer to the working fluid i given by Q H = m (h h 4 ) (7.-4) The vapor exit the boiler in tate and the cycle i repeated. Example 7.-. ---------------------------------------------------------------------------------- Steam enter the turbine in a power plant at 600 o C and 0 MPa and i condened at a preure of 0. MPa. Aume the plant can be treated a an ideal Rankine cycle. Determine the power produced per kg of team and the efficiency of the cycle. Determine the efficiency of a Carnot cycle operated between thee two temperature. Solution ------------------------------------------------------------------------------------------ Table E7.4- State of the ideal Rankine cycle Specific Internal Specific Specific Temp Preure Volume Energy Enthalpy Entropy Quality Phae C MPa m/kg kj/kg kj/kg kj/kg/k 600 0 0.087 4 65 6.90 Dene Fluid (T>TC) 99.6 0..566 49 505 6.90 0.946 Liquid Vapor Mixture 99.6 0. 0.0004 47. 47.4.0 0 Saturated Liquid 4 00. 0 0.0009 47.4 47.7.0 Compreed Liquid o 600 C, 0 MPa T 4 0. MPa Figure E7.- The procee on T diagram. Table E7.- lit the tate of team in the ideal Rankine cycle with the bold value are the two propertie ued to defined the tate. The work produced by the turbine per kg of team i W t = (h h ) = 65 505 = 0 kj/kg Koretky M.D., Engineering and Chemical Thermodynamic, Wiley, 004, pg. 40 7-
The work received by the pump i W p = (h 4 h ) = 47.7 47.4 = 0. kj/kg The net work produced by the plant i W cycle = W t W p = 0 0. = 09.7 kj/kg The rate of heat tranfer to the working fluid i given by Q H = (h h 4 ) = 65 47.7 = 97. kj/kg The efficiency of the cycle i η = W cycle Q = 09.7 H 97. = 0.47 The efficiency of a Carnot cycle operated between thee two temperature i TL 99.6 + 7.5 η Carnot = = = 0.57 T H 600 + 7.5 -------------------------------------------------------------------------------------- Figure 7.- T diagram howing irreveribilitie in pump and turbine. The ientropic turbine and pump efficiencie are given by: h h η t = h h h4 h, and η p = h h 4 The actual work obtained from the turbine i le than the ientropic work and the actual work required for the pump i larger than the ientropic work. Moran, M. J. and Shapiro H. N., Fundamental of Engineering Thermodynamic, Wiley, 008, pg. 40 7-4
Example 7.- 4. ---------------------------------------------------------------------------------- Steam i the working fluid in a modified Rankine cycle where the turbine and pump each have an ientropic efficiency of 85%. Saturated vapor enter the turbine at 8.0 MPa and aturated liquid exit the condener at a preure of 0.008 MPa. The net power output of the cycle i 00 MW. Determine for the modified cycle (a) the thermal efficiency, (b) the ma flow rate of the team, in kg/h, (c) the rate of heat tranfer, Q, into the working fluid a it pae through the boiler, in MW, (d) the rate of heat tranfer, Q c, from the condening team a it pae through the condener, in MW, (e) the ma flow rate of the condener cooling water, in kg/h, if cooling water enter the condener at 5 C and exit at 5 C. Solution ------------------------------------------------------------------------------------------ H Figure E7.-a The ideal Rankine cycle Figure E7.-b The modified Rankine cycle 5 Figure E7.-a how the four tate in the ideal Rankine cycle with the team propertie lited in Table E7.-. Figure E7.-b how the modified Rankine cycle on the T diagram with the dah line repreent the irreveribilitie in the turbine and the pump. Thee irreveribilitie caue an increae in entropy acro the turbine and the pump. 4 Moran, M. J. and Shapiro H. N., Fundamental of Engineering Thermodynamic, Wiley, 008, pg. 96 5 Moran, M. J. and Shapiro H. N., Fundamental of Engineering Thermodynamic, Wiley, 008, pg. 40 7-5
Table E7.- Steam propertie at variou tate in the Rankine cycle. Specific Internal Specific Specific State Temp Preure Volume Energy Enthalpy Entropy Quality Phae C MPa m/kg kj/kg kj/kg kj/kg/k 95. 8 0.05 570 758 5.74 Saturated Vapor 4.5 0.008. 697 795 5.74 0.6745 Liquid Vapor Mixture 4.5 0.008 0.00008 7.8 7.9 0.595 0 Saturated Liquid 4 4.76 8 0.00005 7.9 8.9 0.595 Compreed Liquid The ientropic work produced by the turbine per kg of team i ( Wt / m) = (h h ) = 758 795 = 96 kj/kg The pecific enthalpy at the turbine exit, tate, can be determined uing the turbine efficiency h h η t = h h W / m t = ( W t / m ) h = h η t (h h ) = 758 0.85(758 795) = 99.5 kj/kg The pecific enthalpy at the pump exit, tate 4, can be determined uing the turbine efficiency h4 h η p = h h 4 h 4 = h + (h 4 h )/η t h 4 = 7.9 + (8.9 7.9)/0.85 = 8. kj/kg The net power produced by the cycle i W cycle = W t W p = m [(h h ) (h 4 h )] The rate of heat tranfer to the working fluid i given by Q H = m (h h 4 ) (a) The thermal efficiency i η = W Q cycle H = ( h ) ( ) h h4 h h h 4 η = ( 758 99.5) ( 8. 7.9) 758 8. = 0.4 7-6
(b) Determine the ma flow rate of the team Solving with numerical value, we obtain the net power produced by the cycle per kg/ of team Since W cycle / m = (h h ) (h 4 h ) = (758 99.5) (8. 7.9) = 809. kj/kg W cycle = 00 MW, the ma flow rate of the team i m = W /809. kj/kg = ( ) cycle 00 MW (000 kw/mw) 809. kj/kg =.59 kg/ m = (.59 kg/)(600 /h) = 4.449 0 5 kg/h (c) Determine the rate of heat tranfer to the team, Q H = m (h h 4 ) = (.59 kg/) (758 8.) kj/kg =.8 0 5 kw Q H Q H = (.8 0 5 kw)( MW/0 kw) = 8. MW (d) Determine the rate of heat tranfer to the condening water, Q c = m (h h ) = (.59 kg/) (99.5 7.9) kj/kg =.8 0 5 kw Q c = (.8 0 5 kw)( MW/0 kw) = 8. MW (e) Determine the ma flow rate of the condener cooling water, in kg/h, if cooling water enter the condener at 5 C and exit at 5 C. Q c Specific Internal Specific Specific Temp Preure Volume Energy Enthalpy Entropy Quality Phae C MPa m/kg kj/kg kj/kg kj/kg/k 5 0.00705 0.0000 6.98 6.98 0.45 0 Saturated Liquid 5 0.00568 0.00006 46.7 46.7 0.505 0 Saturated Liquid The ma flow rate of the condening cooling water i given by m cw = h Q c h cw, out cw, in = 5.8 0 kj/ (46.7-6.98) kj/kg =.6 0 kg/ m cw = (.6 0 kg/)(600 /h) =9.9 0 6 kg/h 7-7
7. Refrigeration Cycle The mot common refrigeration cycle i the vapor compreion cycle hown in Figure 7.-. In tep 4, heat i removed at the temperature T L from the ytem being refrigerated by the evaporation of a liquid under the preure P L. In tep, aturated vapor at P L i compre ientropically to P H where it become uperheated vapor. In tep, heat Q H i tranferred to the urrounding by condenation at T H. In tep 4, the cycle i cloed by throttling the liquid to the lower preure P L. T P H Q H Condener Throttle valve Evaporator 4 Compreor W Liquid Ienthalp Q 4 L a b c Figure 7.- A vapor-compreion refrigeration cycle and it T diagram. P L Vapor The heat tranfer between the ytem and the urrounding can be obtained from the T diagram. Since Q = Td, the heat effect i the area under the curve repreenting the path. In Figure 7.- the heat Q H tranferred from the refrigerator to the high temperature environment i the area --a-c, which i negative. The heat Q L removed from the low temperature ytem i the area 4--c-b and i poitive. For the cyclic proce U = Q L + W Q H = 0 W = Q H Q L (7.-) The efficiency of the refrigeration cycle, called the coefficient of performance COP, i given by COP = W Q L (7.-) The work required for the refrigeration cycle can be obtained from the T diagram W = Q H Q L = (Area --a-c) (Area 4--c-b) W = Q H Q L = (Area ---4) + (Area -4-b-c) The refrigeration cycle hown in Figure 7.- i a emi-reverible cycle ince all tep except throttling are reverible. 7-8
Example 7.- 5 ---------------------------------------------------------------------------------- A vapor compreion refrigeration proce uing NH a the working fluid i to operate between 0 and 80 o F. Determine the coefficient of performance for the emi-reverible operation. Solution ------------------------------------------------------------------------------------------ T P H h 4 h h Evaporator 4 Q L W 80 o F 0 o F Liquid Ienthalp 4 a b c P L Vapor We will firt locate the four tate of the refrigeration cycle: State : Saturated liquid at 80 o F, h = h 4 =.7 Btu/lb. State : Saturated vapor at 0 o F, h = 66.8 Btu/lb. State 4: Liquid and vapor mixture at h 4 =.7 Btu/lb and 0 o C State : Superheated vapor at = =.95 Btu/lb o R, P H = P = 5. pia Making energy balance around the evaporator yield 0 = h 4 h + Q L Q L = h h 4 = 66.8.7 = 485. Btu/lb Making energy balance around the compreor yield 0 = h h + W W = h h = 686.6 66.8 = 69.8 Btu/lb The coefficient of performance i then Q COP = L 485. = W 69. 8 = 6.95 5 Kyle, B.G., Chemical and Proce Thermodynamic, Prentice Hall, 999, pg. 656 7-9
Example 7.----------------------------------------------------------------------------------- A vapor-compreion refrigeration proce uing ammonia a the working fluid i to operate between 0 o C and 0 o C. In tep 4 heat i upplied to the fluid at 0 o C under the preure p. The aturated vapor at P i then compreed ientropically to p, where it become uperheated vapor, tate. Removal of heat from thi vapor lead to cooling at contant preure followed by condenation at 0 o C, tep. Throttling the aturated liquid at tate to the lower preure at tate 4 cloe the cycle. The four tate are given (not in any particular order) a follow T( o C) p(mpa) h(kj/kg) (kj/kg K) 0.49 44. 5. 0.5 0.668.5.005 69.8 58.49 Determine (a) The heat tranfer (kj/kg) in tep 4, (b) the heat tranfer (kj/kg) in tep, and (c) the work upplied (kj/kg) by the compreor. Solution ------------------------------------------------------------------------------------------ T P H Q H Condener Throttle valve Evaporator 4 Q L Compreor W Liquid Ienthalp 4 a b c P L Vapor Table E7.- The four tate in proper order State T( o C) p(mpa) h(kj/kg) (kj/kg K) 0 0.49 44. 5. 69.8.668 58.49 5. 0.668.5.005 4 0 0.49.5.5 (a) The heat tranfer (kj/kg) in tep 4 Q L = 44..5 = 9.8 kj/kg (b) The heat tranfer (kj/kg) in tep Q H = 58.49.5 = 59.0 kj/kg (c) The work upplied (kj/kg) by the compreor.\ W c = 58.49 44. = 9.7 kj/kg 7-0