ME 00 Thermodynamics I Spring 015 Lecture 35: Vapor power systems, Rankine cycle Yong Li Shanghai Jiao Tong University Institute of Refrigeration and Cryogenics 800 Dong Chuan Road Shanghai, 0040, P. R. China Email : liyo@sjtu.edu.cn Phone: 86-1-3406056; Fax: 86-1-3406056 1.1
Lecture Contents Last Lecture: Internally reversible, steady-state flow processes» Expressions for the Heat Transfer» Expressions for the Work W m W m int rev int rev V vdp 1 v( p p1) V g( z» Work in a Polytropic Process 1 1 z ( ke pe 0 v c) ) W m W m int rev W m 1 pump vdp Q T W n ( p v p v ) n1 int 1 1 rev m m s compressor m1s 1 polytropic, n 1 W p1v1 ln( p / p1 ) polytropic, n = 1 m int rev Isothermal process» Work in a Polytropic Process of Ideal Gas W nr polytropic, n 1 W int ( T T1 ) m rev n 1 Ideal gas int RT ln( p / p1 ) m rev polytropic, n = 1 1. Isothermal process
Vapor Power Cycle Goal: To generate electricity from heat input» A---Heat converted to work-----power cycle» B---Chemical (Nuclear, solar) energy converted to heat--- Combustion (Boiler) Nuclear reaction (Reactor)» C---Heat rejection----evaporation (Cooling tower)» D---Work to electricity----magnetic induction (Generator) 1.3
Continue Vapor Power Cycle Heat Input Sources:» Fossil fuels (oil, coal, natural gas)» Nuclear fission Controlled nuclear reaction taking place in an isolated reactor building. pressurized water, a liquid metal, or a gas such as helium can be used to transfer energy released in the nuclear reaction to the working fluid in specially designed heat exchangers.» Solar radiation Use receivers for concentrating and collecting solar radiation to vaporize the working fluid.» Geothermal» Garbage/Trash Cycle Working Fluid:» Water/Steam Environmental Issues: Air pollution Thermal pollution Waste products Safety 1.4
Continue Vapor Power Cycle T H T T L Question: What is the upper limit of vapor power cycle performance? Answer: Q H = TS 3 3 1 Q L = TS 14 4 1: adiabatic, reversible compression 3: isothermal, reversible heat input 34: adiabatic, reversible expansion 41: isothermal, reversible heat rejection W net s Note : W Q Q net H L 3 Q Tds T S H H 3 1 Q Tds T S L L 41 4 1.5
Continue Carnot Power Cycle System Schematic: hot reservoir @ T H W in Q H T H = constant reversible, isothermal reversible, adiabatic reversible, isothermal W out Q L T L = constant Thermal Efficiency : cold reservoir @ T L th,c W Q net H Q Q Q H H L Q 1 Q L L 41 H T S 1 T S H 3 T 1 T L H 1.6
Continue Vapor Power Cycle th,c T 1 T L H Example 7.1: Given: T H = 500 K and T L = 300 K Find: Thermal efficiency of Carnot Cycle Solution: th,c = 0.4 Question: How to implement the Carnot Power Cycle with a real working fluid? 1.7
Carnot power cycle Answer: Put the cycle in the two-phase dome T T H 3 T L 1 4 s 1 =s s 3 =s 4 However, in real life we have to deal with deviations from Carnot: - finite temperature difference ΔT across the heat exchange - difficult to compress -phase mixture - non-isentropic compression and expansion Introduce Rankine power cycle s 1.8
Analyzing Rankine Cycle---I Assumptions: 1) Steady state. ) ΔKE=ΔPE=0. 3)stray Q=0 Turbine de dt Q W i m ihi Vi gz i e m e h e Ve gz e Condenser Pump Boiler 1.9
Analyzing Rankine Cycle---I Turbine Condenser Pump Boiler Thermal efficiency of the power cycle Concepts Back work ratio 1.10
Ideal Rankine Power Cycle System Schematic and T-s diagram: 1.11
Continue Ideal Rankine Power Cycle Define ideal Rankine power cycle:» Neglect the finite temperature difference of heat exchanger T HX = 0, T Boiler = T H and T Cond = T L» Isentropic pumping and expansion processes s 1 =s and s 3 =s 4 Difference between ideal Rankine and Carnot power cycle:» Instead of isentropic compression from 1 (Carnot), it is isentropic pumping from saturated liquid (state 1) to compressed (subcooled) liquid ( state ) (Rankine) due to the use of actual (real) working fluid (water/ steam)» Quality at turbine exit (state 4) not too low» Non-isothermal heat addition (from state to state 3) 1.1
Example: Ideal Rankine Cycle Known: 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. Find:, bwr, mass flow rate of steam, Q in, Q out Assumptions: 1) Steady state. ) ΔKE=ΔPE=0. 3) ds turbine, pump =0 ; 4) state 1, state 3 saturated. Analysis: State State 3 saturated liquid at 0.008 MPa, so h 3 =173.88 kj/kg. State 4 begin with a systematic evaluation of the specific enthalpy at each numbered state. 1.13
Example: Ideal Rankine Cycle Analysis: Water 1.14
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