Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics. Lecture 26. Use of Regeneration in Vapor Power Cycles

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1 Department of Mechanical Engineering ME 322 Mechanical Engineering Thermodynamics Lecture 2 Use of Regeneration in Vapor Power Cycles

2 What is Regeneration? Goal of regeneration Reduce the fuel input requirements (Q in ) Increase the temperature of the feedwater entering the boiler (increases average Th in the cycle Result of regeneration Increased thermal efficiency Energy source for regeneration High pressure steam from the turbines Regeneration equipment Feedwater heater (FWH) This is a heat exchanger that utilizes the high pressure steam extracted from the turbine to heat the boiler feedwater 2

3 Regeneration Open FWH Increased temperature into the boiler due to regenerative heating 3

4 Keeping Track of Mass Flow Splits y y 2 y Define a mass flow fraction, n m m n mass flow rate at any state n mass flow rate entering the HPT y 7 y y 5 y 3 y Determination of the flow fractions requires application of the conservation of mass throughout the cycle and the conservation of energy around the feedwater heater(s). Note: If a mass flow rate is known or can be calculated, then the flow fraction approach is not necessary!

5 Regeneration Closed FWH There are two types of closed feedwater heaters Closed FWH with Drain Pumped Forward Closed FWH with Drain Cascaded Backward 5

6 Regeneration Closed FWH Example Closed FWH with Drain Cascaded Backward y y 2 y 3 y y 5 y y 8 y 7

7 7 Regeneration Multiple FWH

8 Regeneration Example Given: A Rankine cycle is operating with one open feedwater heater. Steam enters the high pressure turbine at 500 psia, 900 F. The steam expands through the high pressure turbine to 00 psia where some of the steam is extracted and diverted to an open feedwater heater. The remaining steam expands through the low pressure turbine to the condenser pressure of psia. Saturated liquid exits the feedwater heater and the condenser. Find: (a) the boiler heat transfer per lbm of steam entering the high pressure turbine (b) the thermal efficiency of the cycle (c) the heat rate of the cycle 8

9 Regeneration Cycle P 500 psia T 900F P2 00 psia P3 psia P 00 psia P5 00 psia P P 7 P x psia 0 9

10 Known Properties P 500 psia T 900F P2 00 psia P3 psia P 00 psia P5 00 psia P P 7 P psia The next step is to build the property table 0

11 Unknown Properties P 500 psia T 900F P2 00 psia P3 psia P 00 psia P5 00 psia P P 7 P psia

12 Array Table The resulting property table... P 500 psia T 900F P2 00 psia P3 psia P 00 psia P5 00 psia P P 7 P psia Now, we can proceed with the thermodynamics! 2

13 Boiler Modeling P 500 psia T 900F The heat transfer rate at the boiler can be found by applying the First Law, Q m h h in 7 P P 7 P 00 psia P2 00 psia P5 00 psia P3 psia P psia No flow rate information is given. However, we can find the heat transferred per lbm of steam entering the HPT, Q in qin h h7 m 3

14 Turbine Modeling P 500 psia T 900F The thermal efficiency of the cycle is given by, P2 00 psia P3 psia th W Q net in W t W Q in p Wt / m W / m Q / m p in P P 7 P 00 psia P5 00 psia P psia The turbine power delivered is, W m h m h m h t W m m t h h h m m m W t wt h y2h2 y3h3 m The flow fractions need to be determined!

15 Pump Modeling There are two pumps in the cycle. Therefore, W W W p p p2 p W m h h m h h 5 7 W p m m m m m h5 h h7 h W p wp y h5 h y h7 h m Then... th W / m W / m w w Q / m q t p t p in in P P 7 P 00 psia P 500 psia T 900F P2 00 psia P5 00 psia P3 psia This is an important step in the analysis. All specific energy transfers need to be based on the same flow rate. The common value is chosen to be the inlet to the high pressure turbine (HPT). P psia 5

16 Mass Conservation P 500 psia T 900F The flow fractions must be found. The easy flow fractions are... P2 00 psia P3 psia y y y7 y y y 3 5 P P 7 P 00 psia P5 00 psia P psia Conservation of mass applied to the FWH gives, m m m 2 5 m m m m m m 2 5 y y y 2 5

17 Closing the System P 500 psia T 900F Where is the missing equation? Mass is conserved in the FWH, but so is energy. Therefore, we need to apply the First Law to the FWH, P P 7 P 00 psia P2 00 psia P5 00 psia P3 psia P psia m h m h m h m m m 2 5 h2 h5 h m m m y h y h y h The equations can be solved! The result is a new property table with a column for the mass flow fractions. 7

18 Augmented Array The updated property table... P 500 psia T 900F P2 00 psia P3 psia P 00 psia P5 00 psia P P 7 P psia From previous analysis, th w t w q in w h y h y h t p w y h h y h h p 5 7 8

19 Cycle Performance Parameters The heat rate of the cycle is, P 500 psia T 900F HR Qin Qin / m qin W W / m W / m w w net t p t p P2 00 psia P3 psia P 00 psia P5 00 psia EES Solution (Key Variables): P P 7 P psia 9