Presentation on theme: "Thermodynamics Lecture Series Applied Sciences Education."— Presentation transcript:
Thermodynamics Lecture Series email: firstname.lastname@example.org http://www5.uitm.edu.my/faculties/fsg/drjj1. email@example.com Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA Ideal Rankine Cycle –The Practical Cycle
Example: A steam power cycle. Steam Turbine Mechanical Energy to Generator Heat Exchanger Cooling Water Pump Fuel Air Combustion Products System Boundary for Thermodynamic Analysis System Boundary for Thermodynamic Analysis Steam Power Plant
Second Law Steam Power Plant High T Res., T H Furnace q in = q H net,out Low T Res., T L Water from river An Energy-Flow diagram for a SPP q out = q L Working fluid: Water Purpose: Produce work, W out, out
Second Law – Dream Engine Carnot Cycle P - diagram for a Carnot (ideal) power plant P, kPa, m 3 /kg q out q in 2 34 1 What is the maximum performance of real engines if it can never achieve 100%??
Carnot Principles For heat engines in contact with the same hot and cold reservoir P1: 1 = 2 = 3 (Equality) P2: real < rev (Inequality) Second Law – Will a Process Happen Processes satisfying Carnot Principles obeys the Second Law of Thermodynamics Consequence
Clausius Inequality : Sum of Q/T in a cyclic process must be zero for reversible processes and negative for real processes Second Law – Will a Process Happen reversible impossible real
Increase of Entropy Principle – closed system The entropy of an isolated (closed and adiabatic) system undergoing any process, will always increase. Entropy – Quantifying Disorder Surrounding System For pure substance : and Then
Entropy Balance – for any general system Entropy – Quantifying Disorder For any system undergoing any process, Energy must be conserved (E in – E out = E sys ) Mass must be conserved (m in – m out = m sys ) Entropy will always be generated except for reversible processes S gen Entropy balance is (S in – S out + S gen = S sys )
Steam Power Plant Vapor Cycle External combustion Fuel (q H ) from nuclear reactors, natural gas, charcoal Working fluid is H 2 O Cheap, easily available & high enthalpy of vaporization h fg Cycle is closed thermodynamic cycle Alternates between liquid and gas phase Can Carnot cycle be used for representing real SPP?? Aim: To decrease ratio of T L /T H
Efficiency of a Carnot Cycle SPP Vapor Cycle – Carnot Cycle
Impracticalities of Carnot Cycle Vapor Cycle –Carnot Cycle Isothermal expansion: T H limited to only T crit for H 2 O. High moisture at turbine exit Not economical to design pump to work in 2-phase (end of Isothermal compression) No assurance can get same x for every cycle (end of Isothermal compression) s 3 = s 4 s 1 = s 2 q in = q H T, C T crit THTH TLTL q out = q L s, kJ/kg K
Impracticalities of Alternate Carnot Cycle Vapor Cycle – Alternate Carnot Cycle s 3 = s 4 s 1 = s 2 q in = q H T, C T crit THTH TLTL q out = q L s, kJ/kg K Still Problematic Isothermal expansion but at variable pressure Pump to very high pressure Can the boiler sustain the high P?
Overcoming Impracticalities of Carnot Cycle Vapor Cycle – Ideal Rankine Cycle Superheat Superheat the H 2 O at a constant pressure (isobaric expansion) Can easily achieve desired T H higher than T crit. reduces moisture content at turbine exit Remove all excess heat at condenser Phase is sat. liquid at condenser exit, hence need only a pump to increase pressure Quality is zero for every cycle at condenser exit (pump inlet)
Vapor Cycle – Ideal Rankine Cycle Pump Boiler Turbin e Condenser High T Res., T H Furnace q in = q H in out Low T Res., T L Water from river A Schematic diagram for a Steam Power Plant q out = q L Working fluid: Water q in - q out = out - in q in - q out = net,out
T- s diagram for an Ideal Rankine Cycle Vapor Cycle – Ideal Rankine Cycle T, C s, kJ/kg K 1 2 T crit THTH T L = T sat@P4 T sat@P2 s 3 = s 4 s 1 = s 2 q in = q H 4 3 PHPH PLPL in out pump q out = q L condenser turbine boiler
Vapor Cycle – Ideal Rankine Cycle Boiler In,2 Out,3 q in = q H Energy Analysis q in – q out + in – out = out – in, kJ/kg q in – 0 + 0 – 0 = h exit – h inlet, kJ/kg q in = h 3 – h 2, kJ/kg Assume ke =0, pe =0 for the moving mass, kJ/kg Q in = m(h 3 – h 2), kJ
Vapor Cycle – Ideal Rankine Cycle Condenser Out,1 In,4 q out = q L Energy Analysis q in – q out + in – out = out – in, kJ/kg 0 – q out + 0 – 0 = h exit – h inlet - q out = h 1 – h 4, So, q out = h 4 – h 1, kJ/kg Assume ke =0, pe =0 for the moving mass, kJ/kg Q out = m(h 4 – h 1), kJ
Vapor Cycle – Ideal Rankine Cycle Energy Analysis q in – q out + in – out = out – in, kJ/kg 0 – 0 + – out = h exit – h inlet, kJ/kg - out = h 4 – h 3, kJ/kg So, out = h 3 – h 4, kJ/kg out In,3 Out,4 Turbin e Assume ke =0, pe =0 for the moving mass, kJ/kg W out = m(h 3 – h 4), kJ
Vapor Cycle – Ideal Rankine Cycle Energy Analysis q in – q out + in – out = out – in, kJ/kg 0 – 0 + in – = h exit – h inlet, kJ/kg in = h 2 – h 1, kJ/kg Pump in Out,2 In,1 For reversibl e pumps where So, W in = m(h 2 – h 1), kJ
Vapor Cycle – Ideal Rankine Cycle Energy Analysis Efficiency
T- s diagram for an Ideal Rankine Cycle Vapor Cycle – Ideal Rankine Cycle T, C s, kJ/kg K 1 2 T crit THTH T L = T sat@P4 T sat@P2 s 3 = s 4 s 1 = s 2 q in = q H 4 3 PHPH PLPL in out pump q out = q L condenser turbine boiler s 1 = s f@P1 h 1 = h f@P1 s 3 = s @P3,T3 s 4 = [s f +xs fg ] @P4 = s 3 h 3 = h @P3,T3 h 4 = [h f +xh fg ] @P4 h 2 = h 1 + 2 (P 2 – P 1 ); where Note that P 1 = P 4
Vapor Cycle – Ideal Rankine Cycle Energy Analysis Increasing Efficiency Must increase net,out = q in – q out Increase area under process cycle Decrease condenser pressure; P 1 =P 4 P min > P sat@Tcooling+10 deg C Superheat T 3 limited to metullargical strength of boiler Increase boiler pressure; P 2 =P 3 Will decrease quality (an increase in moisture). Minimum x is 89.6%.
Reheating increases and reduces moisture in turbine Vapor Cycle – Reheat Rankine Cycle T L = T sat@P1 in s 5 = s 6 s 1 = s 2 T crit THTH T sat@P4 T sat@P3 s 3 = s 4 q out = h 6 -h 1 out, II P 4 = P 5 P 6 = P 1 6 1 5 4 q reheat = h 5 -h 4 q primary = h 3 -h 2 out P3P3 3 2 T, C s, kJ/kg K
Energy Analysis Vapor Cycle – Reheat Rankine Cycle q in = q primary + q reheat = h 3 - h 2 + h 5 - h 4 q out = h 6 -h 1 net,out = out,1 + out,2 - in = h 3 - h 4 + h 5 - h 6 – h 2 + h 1
Energy Analysis Vapor Cycle – Reheat Rankine Cycle where s 6 = [s f +xs fg ] @P6. Use x = 0.896 and s 5 = s 6 Knowing s 5 and T 5, P 5 needs to be estimated (usually approximately a quarter of P 3 to ensure x is around 89%. On the property table, choose P 5 so that the entropy is lower than s 5 above. Then can find h 5 = h @P5,T5. h 6 = [h f +xh fg ] @P6
Energy Analysis Vapor Cycle – Reheat Rankine Cycle s 1 = s f@P1 where s 3 = s @P3,T3 = s 4. h 1 = h f@P1 h 3 = h @P3,T3 h 2 = h 1 + 2 (P 2 – P 1 ); where From P 4 and s 4, lookup for h 4 in the table. If not found, then do interpolation. P 5 = P 4.