ME200 - Lecture 350 - 11/19/2010 Modified Rankine Cycle ME...

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Unformatted text preview: 11/19/2010 Modified Rankine Cycle ME 200 Thermodynamics I Lecture 35 November 19th, 2010 One granule LiBH4 (~0.5 mm) with drop of 90% H2O2 Purdue University , Dr. Tim Pourpoint – timothee@purdue.edu Outline • Last Lecture: – Ideal Rankine Cycle and comparison with Carnot Cycle • Today: – Finish Example 34.1 – Modified Rankine Cycle 2’ 3 2 1 1’ ME 200 4 2 1 11/19/2010 Example 34.1 Steam, TH = 450oF, TL = 212oF, Wnet = 1 MW States 1 and 3 are saturated • Given: • Find: – Heat transfer in – Mass flow rate of steam – Thermal efficiency T =0.8 3 4 • System sketch: • Assumptions: – SSSF – KE=0, PE=0 – THX = 0 TH = TB and TL = TC 2 1 ME 200 P =0.6 3 Solutions for Example 34.1 • Solution: – Cycle in a T‐s diagram p2 = p3 T TH TB TC TL p1 = p4 ∆THX 2s 3 2 1 4s 4 ∆THX s ME 200 4 2 11/19/2010 Solutions for Example 34.1 • Using C.O.M, 1st and 2nd law, we will fill following table of state point properties: T [oF] x 212 0.0 3 450 1.0 4 212 State P [psia] 1 H [Btu/lbm] S [Btu/lbmR] 2 • With those properties, we will be able to check: – Overall energy balance: 0 Q B QC WT WP – Overall entropy production: total P B T C ME 200 5 Solutions for Example 34.1 3 4 • Continue Solution: (1) Pump Analysis: – mass balance: m1 m 2 mst 2 1 – energy balance: dE cv v2 v2 Qcv Wcv min (h gz )in mout (h gz )out 2 2 dt out in 0 WP m1h1 m 2 h 2 WP mst (h1 h 2 ) W isentropic pump work WP P,is pump efficiency P ME 200 6 3 11/19/2010 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Pump Analysis: – Recall T ds = dh – v dp, ds = 0 dw = dh = v dp dh dp 2 1 2 WP,is mst vdp 1 – For incompressible fluid: WP,is mst v1 (p 2 p1 ) – From Steam Tables: T [oF] x 212 0.0 3 450 1.0 ME 4200 p1 = psat(T1=TL) = 14.7 psia = p4 p2 = psat(T3=TH) = 422.1 psia = p3 v1 = vf(T1)=0.01672 ft3/lbm s1 = 0.31213 Btu/lbmR h1 = 180.2 Btu/lbm 212 State P [psia] 1 H [Btu/lbm] S [Btu/lbmR] 2 7 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Pump Analysis: 2 1 WP,is w P,is v1 (p 2 p1 ) mst w P,is 0.01672 wP w P,is P lb 144in 2 ft 3 Btu Btu (422.1 14.7) m 1.26 2 2 lb m in ft 778ft lb m lb m 1.26 Btu / lb m 2.1Btu / lb m 0.6 w P h 2 h1 st 1 Law : h 2 h1 w P (180.2 2.1)Btu / lb m 182.3Btu / lb m Then, T2 T(p 2 , h 2 ) 212.8o F. T [oF] x 212 0.0 3 450 ME 4200 212 Note : T2 T1 1o F 1.0 State 1 P [psia] 2 H [Btu/lbm] S [Btu/lbmR] and liquid water at (2) 8 4 11/19/2010 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Pump Analysis: T s 2 s1 cavg ln 2 T1 2 1 T s 2 s1 cavg ln 2 , T1 Btu Btu (212.8 460)R Btu ln 1.01 0.31333 lb m R lb m R (212 460)R lb m R n dScv Q k min sin m out s out P,tot 2nd Law : dt k 1 TR ,k in out s 2 0.31213 0 Qp Tenv mst (s1 s 2 ) P , tot P,tot Btu Btu Process is possible! s 2 s1 (0.31333 0.31213) 0.0012 st m lb m R lb m R T [oF] x 212 0.0 3 450 1.0 ME 4200 212 State P [psia] 1 H [Btu/lbm] S [Btu/lbmR] 2 9 Solutions for Example 34.1 3 4 • Turbine Analysis: Basic Equations: 2 – mass balance: m1 m 2 mst – energy balance: dE cv Qcv Wcv min (h dt in 0 WT m3 h 3 m 4 h 4 W h h4 T T ,act 3 WT ,is h 3 h 4s T [oF] x 212 0.0 3 450 212 v2 v2 gz )in m out (h gz )out 2 2 out WT mst (h 3 h 4 ) Then, WT mst T (h 3 h 4s ) Wt or mst t (h 3 h 4s ) 1.0 ME 4200 1 State 1 P [psia] 2 H [Btu/lbm] S [Btu/lbmR] 10 5 11/19/2010 Solutions for Example 34.1 3 4 • Continue Turbine Analysis: – from Steam Tables: 2 1 h3 = hg (T3 = TH) = 1205.6 Btu/lbm s3 = sg (T3 = TH) = 1.4806 Btu/lbmoR s4s = s3 [s4s > sf(p4) and s4s < sg(p4)] s 4s s f (p 4 ) x 4s s g (p 4 ) s f (p 4 ) x 4s s 4s s f (p 4 ) s g (p 4 ) s f (p 4 ) – from Steam Tables: sf(p4) = 0.3121 Btu/lbmoR and: sg(p4) = 1.7567 Btu/lbmoR Then, x 4s Also, x 4s (1.4806 0.3121)Btu / lb m o R 0.809 (80.9%) (1.7567 0.3121)Btu / lb m o R h 4s h f (p 4 ) h 4s h f (p 4 ) x 4s h g (p 4 ) h f (p 4 ) h g (p 4 ) h f (p 4 ) T [oF] x 212 0.0 3 450 1.0 ME 4200 212 State P [psia] 1 H [Btu/lbm] S [Btu/lbmR] 2 11 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Turbine Analysis: – From Steam Tables: hf(p4) = h1 = 180.2 Btu/lbm hg(p4) = hg(p1=p4) = 1150.5 Btu/lbm 1150 Btu/ – h4s = 180.2 Btu/lbm + 0.809 (1150.5 – 180.2) Btu/lbm h4s = 965.2 Btu/lbm 2 1 – Then: h4 = h3 – t (h3 – h4s) h4 = 1205.6 Btu/lbm – 0.8 (1205.6 – 965.2) Btu/lbm h4 = 1013.3 Btu/lbm – Also: State 1 P [psia] wT = h3 – h4 = 192.3 Btu/lbm >> wP = 2.1 Btu/lbm T [oF] x 212 0.0 3 450 ME 4200 212 H [Btu/lbm] S [Btu/lbmR] At this point, we are only missing x4 and s4 1.0 2 12 6 11/19/2010 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Turbine Analysis: 2 h 4 h f (p 4 ) 1013.3 180.2 x4 0.8586 h g (p 4 ) h f (p 4 ) 1150.5 180.2 1 s 4 s f (p 4 ) x 4 s g (p 4 ) s f (p 4 ) s 4 0.3121 0.8586 (1.7567 0.3121) 1.5524 2nd Law : 0 QT Tenv Btu lb m R n dScv Q k min sin m out s out cv dt k 1 Tcv,k in out st Btu T,tot s 4 s3 0.0718 mst lb m R mst (s3 s 4 ) T,tot Process is possible! T [oF] x 212 0.0 3 450 1.0 ME 4200 212 State P [psia] 1 H [Btu/lbm] 2 S [Btu/lbmR] At this point, we know all the properties at all states of the process use 1st and 2nd law on remaining elements to 13 calculation performance of the cycle Solutions for Example 34.1 3 4 • Continue Solution: – Continue Turbine Analysis: 2 1 Wnet 1MW WT WP mst (w T w P ) Btu 3.413x106 Wnet 1MW hr 17,888 lb m mst Btu w T w P (192.3 2.1) 1MW hr lb m Then, Btu WT mst (h 3 h 4 ) 3.44 106 1.008 MW hr lb Btu Btu WP mst w P 17,888 m 2.1 37,564 0.01MW hr lb m hr ME 200 14 7 11/19/2010 Solutions for Example 34.1 3 4 • Continue Solution: (3) Boiler Analysis: 2 1 – mass balance: m 2 m3 mst – energy balance: dE cv v2 v2 Qcv Wcv min (h gz )in m out (h gz ) out 2 2 dt in out 0 Q B m 2 h 2 m3 h 3 lb Btu Q B mst (h 2 h 3 ) 17,888 m (1205.6 182.3) hr lb m Btu Q B 18.21 106 hr ME 200 15 Solutions for Example 34.1 3 4 • Continue Solution: – Continue Boiler Analysis: 2 dScv Q 2nd Law : k min sin m out s out tot dt k 1 TR ,k in out Q Q 0 B mst (s 2 s3 ) B,tot B,tot mst (s3 s 2 ) B TH TH 1 n B,tot 17,888 B,tot 2851 ME 200 lb m Btu 18.21 106 Btu / hr (1.4806 0.3133) hr lb m R (550 460)R Btu hr R 16 8 11/19/2010 Solutions for Example 34.1 3 4 • Continue Solution: (4) Condenser Analysis: 2 1 1st Law : 0 Q C m 4 h 4 m1h1 Btu QC mst (h1 h 4 ) 14.9 106 hr Q 2nd Law : 0 C mst (s 4 s1 ) C,tot TL Q C,tot mst (s1 s 4 ) c Tenv C,tot 17888 C,tot 3503 lb m Btu 14.9 106 Btu / hr (0.3121 1.5524) hr lb m R (120 460)R Btu hr R ME 200 17 Solutions for Example 34.1 3 4 • Continue Solution: (7) Thermal efficiency: 2 1 th,act W W WP (3.44 106 0.04 106 )Btu / hr net T 0.186 18.21 106 Btu / hr QB QB th,act 18.6% th , C 1 ME 200 th,act th,C TL (212 460)o R 1 0.26 26% TH (450 460)o R 18.6 0.715 26 18 9 11/19/2010 Solutions for Example 34.1 3 4 • Finally: Check for overall energy balance: 2 1 dE CM QCM WCM dt 0 Q B QC WT WP Btu Btu Btu Btu 14.9 106 3.44 106 0.04 106 hr hr hr hr 0 (okay!) 0 18.31 106 Overall entropy production: total P B T C total 17,888 total 7773 lb m lb Btu Btu Btu Btu 0.0012 2851 17,888 m 0.0781 3503 hr lb m R hr R hr lb m R hr R Btu hr R ME 200 19 Solutions for Example 34.1 • Continue Solution: – Table of state point properties x h Btu/lbm s Btu/lbmR s Btu/lbmR 212 0.0 180.2 0.312 12 0.001 422.1 212.8 liquid 182.3 0.313 23 0.159 3 422.1 450 1.0 1205.6 1.481 34 0.072 4 14.7 212 0.859 1013.3 1.552 41 0.196 State p psia oF 1 14.7 2 ME 200 T 20 10 11/19/2010 Rankine Cycle Improvements • Basic Rankine Cycle: ME 200 21 Increasing Rankine cycle efficiency • Average temperature should be as high as possible during heat addition and as low as possible during heat rejection – Carnot theory • How to achieve this? – Lowering condenser pressure – Superheating steam to high temperatures – Increasing boiler pressure • Keep in mind area enclosed by cycle on T‐s or P‐v diagram represents net work/heat for internally reversible cycles ME 200 22 11 11/19/2010 Continue Rankine Cycle Performance Trends • Effect of condensing pressure on performance: • This lowers steam temperature during heat rejection • More work input to pump and heat input, but more work output from turbine • Overall efficiency increases due to lowering temperature during heat rejection • Limited by saturation pressure for temperature of cooling medium; watch for air leakage; higher moisture content in turbine (typically want x4 > 0.9 to avoid liquid drops corroding turbine blades) ME 200 23 Rankine Cycle Performance Trends Effect of superheating steam to high temperatures: • Increase in net work and heat input • Overall effect is increase in efficiency due to higher T during heat addition • Helps by decreasing moisture content in turbine • Limited by turbine blade material (620C or 1150F); ceramics ME 200 24 12 11/19/2010 Continue Rankine Cycle Performance Trends • Continue effect of superheat on performance: – Consequence of superheating (using example 34.1): Tsup (oF) 0 100 500 0.186 0.190 0.217 x4 0.858 0.913 >1.0 • Note: – Increasing Tsup has not a very big effect on thermal efficiency, but is needed to increase x4 to a “practical” value! ME 200 25 Continue Rankine Cycle Performance Trends • Effect of boiler pressure on performance: • Increases average T during heat addition • Moisture content increases (reheat) • Maximum pressure dictated by temperature of heat source and materials • Optimal pressure: trade‐off between 1st costs and operating costs • 2.7MPa (400psia) in 1922 to over 30MPa (4500psia) today producing >1000MW ME 200 26 13 11/19/2010 Continue Rankine Cycle Improvements • Two‐stage Cycle with Reheat: ME 200 27 Continue Rankine Cycle Improvements • Continue two‐stage cycle with reheat: – High quality (or superheated) vapor exiting the turbine without large superheat – Takes advantage of increased efficiency offered by higher pressures yet avoids low quality steam at the turbine exhaust • Overall thermal efficiency: WT1 WT2 WP W th net Qin QB QR ME 200 SEE EXAMPLE 8.3 IN YOUR TEXTBOOK 28 14 11/19/2010 Summary and Conclusions • In this lecture, we talked about: – methods to increase thermal efficiency of the ideal Rankine cycle – modifying the ideal Rankine cycle with superheat and reheat process • Next Lecture: Vapor Compression Cycle TRAIN YOURSELF: • Describe an ideal vapor power cycle with reheat. • What are the advantages and disadvantages of reheating process in a vapor power cycle? ME 200 29 15 ...
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This note was uploaded on 12/11/2011 for the course ME 200 taught by Professor Gal during the Fall '08 term at Purdue University-West Lafayette.

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