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lecture05 - LECTURE 5 THERMODYNAMICS An Engineering...

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Unformatted text preview: LECTURE 5 THERMODYNAMICS An Engineering Approach VAPOR POWER SYSTEMS GAS POWER SYSTEMS Dr. MinJun Kim Department of Mechanical Engineering & Mechanics Drexel University Rankine Cycle 4 Qin Boiler 1 2 2 V − V2 D D D 0 = QCV − Wt + m h1 − h2 + 1 + g ( z1 − z2 ) 2 Work Pump Turbine Work Condenser 3 Qout cooling water 2 D Wt = h1 − h2 D m D QOUT = h2 − h3 D m D Wp = h4 − h3 D m η= DDDD Wt / m − W p / m (h1 − h2 ) − (h4 − h3 ) = D D QIN / m h1 − h4 D QIN = h1 − h4 D m 1 Ideal Rankine Cycle Process 1-2: Isentropic expansion. T 1’ 1 a Process 2-3: Heat transfer from the working fluid as it flows at constant pressure. Process 3-4: Isentropic compression Process 4-1: Heat transfer to the working fluid as it flows at constant pressure. D D QIN / m 4 DD Wt / m 3 c 2 2’ b 4 Wp = ∫ vdp m int − rev 3 DD Wp / m s D D QOUT / m Wp ≈ v3 ( p4 − p3 ) m int − rev Example (I) Steam is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine at 8.0 MPa and saturated liquid exits the condenser at a pressure of 0.008 MPa. The net power output of the cycle is 100 MW. Determine for the cycle (a) the thermal efficiency, (b) the mass flow rate of the steam, in kg/h, (c) the rate of heat transfer into the working fluid as it passes through the boiler, in MW, (d) the mass flow rate of the condenser cooling water, in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C. 4 Qin Boiler P = 8.0MPa 1 T 8.0 MPa 1 Saturated vapor Work Pump Turbine Work 4 0.008 MPa Condenser 3 Saturated liquid Qout cooling water 2 3 2 P = 0.008MPa s 2 Effects of Boiler & Condenser Pressure Since the ideal Rankine cycle consists entirely of internally reversible processes, an expression for thermal efficiency can be obtained in terms of average temperatures during the heat interaction processes. T 1’ 1 a C QIN m C 1 = ∫ Tds = area1 − b − c − 4 − a − 1 int − rev 4 C QIN m C = Tin ( s1 − s4 ) int − rev 4 3 c 2 2’ b Qout m s = Tout ( s2 − s3 ) = area 2 − b − c − 3 − 2 int − rev > > (Q / m)int − rev = 1 − Tout ηideal = 1 − >out > (Qin / m)int − rev Tin Principle Irreversibilities and Losses The principal irreversibility experienced by the working fluid is associated with the expansion through the turbine Heat transfer from the turbine to the surroundings represents a loss. T 1 4 4s >> (W / m) = (h1 − h2 ) ηt = > t > (Wt / m)s h1 − h2 s The work input to the pump required to overcome frictional effects also reduces the net power output of the plant. 3 2s 2 ηp s >> (W / m) = (h = >> (W / m) (h p s p 4s − h3 ) 4 − h3 ) The pump work for the isentropic process appears in the numerator. The actual pump work, being the larger quantity, is the denominator. Because the pump work is so much less than the turbine work, irreversibilities in the pump have a much smaller impact on the net work of the cycle than do irreversibilities in the turbine. 3 Example (II) Reconsider the vapor power cycle of Example (I), but include in the analysis that the turbine and the pump each have an isentropic efficiency of 85%. Determine for the modified cycle (a) the thermal efficiency, (b) the mass flow rate of steam, in kg/h, for a net power output of 100MW, (c) the rate of heat transfer into the working fluid as it passes through the boiler, in MW, (d) the rate of heat transfer from the condensing stream as it passes through the condenser, in MW, (e) the mass flow rate of the condenser cooling water, in kg/h, if cooling water enters the condenser at 15°C and exits at 35°C. T 8.0 MPa 1 4 4s 0.008 MPa 3 2s 2 s Superheat and Reheat As we are not limited to having saturated vapor at the turbine inlet, further energy can be added by heat transfer to the steam, bringing it to a superheated vapor condition at the turbine inlet. This is accomplished in a separate heat exchanger called a superheater. The combination of boiler and superheater is referred as a steam generator. Low-pressure turbine Reheat section 3 Qin 6 Work Pump Turbine High-pressure turbine 2 Steam generator 1 T 1 T1 3 T3 Work 4 2 5 Condenser Qout cooling water 6 5 4’ 4 s 4 Example (III) Steam is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first-stage turbine at 8.0 MPa, 480°C, and expands to 0.7 MPa. It is then reheated to 440°C before entering the second-stage turbine, where it expands to the condenser pressure of 0.008 MPa. The net power output is 100 MW. Determine (a) the thermal efficiency of the cycle, (b) the mass flow rate of steam, in kg/h, (c) the rate of heat transfer from the condensing steam as it passes through the condenser, in MW. Reheat section T3 = 440 °C 3 Qin 6 Work Pump 2 Steam generator T1 = 480 °C P1 =8.0 MPa Low-pressure turbine P2 =0.7 MPa 1 Turbine Work 4 High-pressure turbine 5 Pcond =0.008 MPa Condenser Qout cooling water Regenerative Vapor Power Cycle . Qin (1) 1 Steam generator 2 (y) 7 (1) Open feedback heater 5 Pump2 6 . Wp2 (1- y) 3 Condenser . Qout 4 . Wturbine (1- y) Pump1 . Wp1 5 Preliminaries for Gas Power Systems The vapor power systems studied in lecture use a working fluid that is alternately vaporized and condensed. The present lecture is concerned with power plants utilizing a working fluid that is always a gas. Gas turbines in the power plants Internal combustion engines Air-Standard Analysis (1) A fixed amount of air modeled as an ideal gas is the working fluid. (2) The combustion process is replaced by a heat transfer from an external source. (3) There are no exhaust and intake processes as in an actual engine. (4) All processes are internally reversible. Air-Standard Otto Cycle p 3 Process 1-2: An isentropic compression of the air. Process 2-3: A constant-volume heat transfer to the air from an external source. s=c 2 s=c b T 3 2 v=c 4 1 v The air-standard Otto cycle consists of two processes in which there is work but no heat transfer, Processes 1-2 and 3-4, and two processes in which there is heat transfer but no work, Processes 2-3 and 4-1. Expressions for these energy transfers are obtained by reducing the closed system energy balance assuming that changes in kinetic and potential energy can be ignored. Process 3-4: An isentropic expansion (Power stroke). Process 4-1: A cycle completed by the constantvolume in which heat is rejected from the air. v=c 1 b 4 a s 6 Air-Standard Otto Cycle W12 = u2 − u1 m W34 = u3 − u 4 m Q23 = u3 − u2 m The net work of the cycle is expressed as Q41 = u 4 − u1 m WCYCLE W34 W12 = − = (u 3 − u 4 ) − (u 2 − u1 ) m m m Alternatively, the net work of the cycle can be evaluated as the net heat added WCYCLE Q23 Q41 = − = (u 3 − u 2 ) − (u 4 − u1 ) m m m The thermal efficiency is the ratio of the net work of the cycle to the heat added η= (u 3 − u 2 ) − (u 4 − u1 ) u − u1 = 1− 4 (u 3 − u 2 ) u3 − u 2 Analysis for the Isentropic Processes For the isentropic process 1-2 and 3-4, V v v r 2 = v r1 2 = r 1 V r 1 V v r 4 = v r 3 4 = rv r 3 V 3 Where r denotes the compression ratio. Note that since V3 = V2 and V4 = V1, r = V1/V2 = V4/V3. T2 V1 = T1 V2 k −1 = r k −1 T4 V3 = T3 V4 T k −1 = 1 r k −1 3’ Where k is the specific heat ratio, k = cp/cv. Effect of Compression Ratio on Performance: 2’ v=c 2 v=c 1 b a c (T − T ) η =1− v 4 1 cv (T3 − T2 ) T η = 1− 1 T2 T4 / T1 − 1 T /T −1 3 2 3 4 T 1 η = 1 − 1 = 1 − k −1 T2 r s 7 Air-Standard Diesel Cycle The air-standard Diesel cycle is an ideal cycle that assumes the heat addition occurs during a constantpressure process that starts with the piston at top dead center. Process 1-2: An isentropic compression of the air. 4 s=c 1 a T p=c 2 4 v=c 1 b a s b 3 v Process 2-3: A constant-pressurea heat transfer to the air from an external source (Power stroke). Process 3-4: An isentropic expansion. Process 4-1: A cycle completed by the constantvolume in which heat is rejected from the air. p 2 3 s=c On the T-s diagram, area 2-3-a-b-2 represents the heat added per unit of mass and area 1-4-a-b-1 is the heat rejected per unit of mass. On the p-v diagram, area 1-2-a-b-1 is the work input per unit of mass during the compression process. Area 2-3-4-b-a-2 is the work done per unit of mass as the piston moves from top dead center to bottom dead center. Air-Standard Diesel Cycle In the Diesel cycle the heat addition takes place at constant pressure. Accordingly, Process 2-3 involves both work and heat. The work is given by The heat added in Process 2-3 can be found by applying the closed system energy balance W2 3 3 = ∫ pdv = p 2 (v3 − v2 ) m 2 m(u 3 − u 2 ) = Q23 − W23 Q23 = (u 3 − u 2 ) + p(v3 − v2 ) = (u 3 + pv3 ) − (u 2 − pv2 ) = h3 − h2 m As in the Otto cycle, the heat rejected in Process 4-1 is given by Q41 = u 4 − u1 m The thermal efficiency is the ratio of the net work of the cycle to the heat added η= WCYCLE / m Q /m u − u1 = 1 − 41 = 1− 4 Q23 / m Q23 / m h3 − h2 8 Air-Standard Dual Cycle Process 1-2: An isentropic compression. p 3 4 s=c 5 s=c 1 v Process 2-3: A constant-volume heat addition. Process 3-4: A constant-pressure heat addition (power stroke). Process 4-5: An isentropic expansion (the remainder of the power stroke). Process 5-1: A cycle completed by a constantvolume heat rejection process. 2 T p=c 2 v=c 3 4 W12 = u2 − u1 m Q34 = h4 − h3 m W34 Q23 = p (v4 − v3 ) = u3 − u 2 m m W45 = u 4 − u5 m Q51 = u5 − u1 m 5 1 v=c The thermal efficiency is the ratio of the net work of the cycle to the heat added η= s WCYCLE / m Q51 / m = 1− (Q23 / m + Q34 / m) (Q23 / m − Q34 / m) u5 − u1 (u3 − u2 ) + (h4 − h3 ) η = 1− Example (I) At the beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is 300K and the pressure 0.1 MPa. The pressure ratio for the constant-volume part of the heating process is 1.5:1. The volume ratio for the constant pressure part of the heating process is 1.2:1. Determine the thermal efficiency. p 3 4 p3 = 1.5 p2 s=c T 4 p=c 2 v=c 3 5 1 v=c 2 V4 = 1.2 V3 5 p1=0.1MPa 1 v T1=300K V1 = 18 V2 s=c s 9 Gas Turbine Power Plants Air-Standard Analysis (1) The working fluid is air, which behaves as an ideal gas. (2) The temperature rise that would be brought about by combustion is accomplished by a heat transfer from an external source. Air-Standard Brayton Cycle Regenerative Gas Turbines Regenerative Gas Turbine with Reheat and Intercooling Gas Turbines for Aircraft Propulsion fuel combustion chamber QIN Heat exchanger compressor turbine Net work out compressor turbine Net work out air products Heat exchanger QOUT Air-Standard Brayton Cycle QIN Wt = h3 − h4 m 3 Wc = h2 − h1 m Qout = h4 − h1 m 2 Heat exchanger Qin = h3 − h2 m Net work out compressor turbine Heat exchanger η= 4 Wt / m − Wc / m (h3 − h4 ) − ( h2 − h1 ) = h3 − h2 Qin / m 1 QOUT The back work ration for the cycle is bwr = DD Wc / m h2 − h1 = DD Wt / m h3 − h4 Typical back work ratios of gas turbines range from 40% to 80%. In comparison, the bwrs of vapor power plants are normally only 1 or 2% 10 Ideal Air-Standard Brayton Cycle p a 2 3 Process 1-2: An isentropic compression. Process 2-3: A constant-pressure heat addition. Process 3-4: An isentropic expansion. s= Process 4-1: A constant-pressure cooling. c s=c On the T-s diagram, 4 v 3 Area 2-3-a-b-2: the heat added per unit of mass Area 1-4-a-b-1: the heat rejected per unit of mass On the P-v diagram, Area 1-2-a-b-1: the compressor work input per unit of mass Area 3-4-b-a-3: the turbine work output per unit of mass 4 b 1 T p=c 2 p r 2 = p r1 1 a p=c b s p2 p1 ( k −1) / k pr 4 = pr 3 p p4 = pr3 1 p2 p3 ( k −1) / k p T2 = T1 2 p 1 p T4 = T3 4 p 3 p = T3 1 p 2 ( k −1) / k Effect of Pressure Ratio on Performance Assuming that the specific heat cp is constant, η= c p (T3 − T4 ) − c p (T2 − T1 ) c p (T3 − T2 ) = 1− (T4 − T1 ) (T3 − T2 ) On further rearrangement, η = 1− T1 T2 T4 / T1 − 1 T /T −1 3 2 η = 1− η = 1− T1 T2 1 ( k −1) / k p2 p 1 11 Example (II) Air enters the compressor of an ideal air-standard Brayton cycle at 100kPa, 300K, with volumetric flow rate of 5 m3/s. The compressor pressure ration is 10. The turbine inlet temperature is 1400K. Determine (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW. p2/p1=10 2 QIN Heat exchanger T3=1400K 3 T T3=1400K p = 1000kPa 3 compressor turbine Net work out 2 p = 100kPa 4 Heat exchanger 1 QOUT 4 1 T1=300K s p1=100kPa, T1=300K Principal Irreversibilities and Losses T 2 p=c 3 T p=c 2s 2 4 3 4s p=c 1 s 4 1 p=c s ηt = DD (W / m) = h − h DD (W / m) h − h t 3 t s 3 4 ηc = 4s DD (W / m ) = h DD (W / m ) h c c c 2s 2 − h1 − h1 12 Example (III) Reconsider Example (II), but include in the analysis that the turbine and compressor each have an efficiency of 80%. Determine for the modified cycle (a) the thermal efficiency of the cycle, (b) the back work ratio, (c) the net power developed, in kW. T T3=1400K p = 1000kPa 2s 2 3 4s p = 100kPa 1 T1=300K 4 s Reheat & Intercooling 10 regenerator Compressor 1 Compressor 2 Combustor 1 Combustor 2 5 4 6 7 8 9 D QIN ,1 2 1 Intercooler Turbine 1 D QIN , 2 6 Turbine 2 3 T 5 4s 4 2s 2 3 1 10 7s D WCYCLE 8 7 9s 9 D QOUT s 13 Gas Turbine for Aircraft Propulsion Turbine Combustor Compressor 5 4 2 Air in a 1 diffuser 3 gas generator T 2 Product, Gases out nozzle 3 4 5 1 a s Preparations Next Lecture: Refrigeration and Heat Pump Systems Thermodynamic Property Relations Homework#5: Will be posted in WebCT 14 ...
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