Department of Mechanical EngineeringME 322 – Mechanical Engineering

Thermodynamics

Lecture 26

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 toheat the boiler feedwater

3

Regeneration – Open FWH

Increased temperature into the boiler due to regenerative heating

4

Keeping Track of Mass Flow Splits

1 1y

2y

3y

4y

5y6y7y

Define a mass flow fraction,

1

mass flow rate at any state mass flow rate entering the HPT

nn

m nym

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 FWHThere are two types of closed feedwater heaters

Closed FWH with Drain Pumped

Forward

Closed FWH with Drain Cascaded

Backward

6

Regeneration – Closed FWHExample – Closed FWH with Drain Cascaded Backward

1 1y

2y3y

4y

5y6y

7y8y

7

Regeneration – Multiple FWH

Regeneration Example

8

Given: A Rankine cycle is operating with one open feedwater heater. Steam enters the high pressure turbine at 1500 psia, 900°F. The steam expands through the high pressure turbine to 100 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 1 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

Regeneration Cycle

9

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

Known Properties

10

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The next step is to build the property table

Unknown Properties

11

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

Array Table

12

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The resulting property table ...

Now, we can proceed with the thermodynamics!

Boiler Modeling

13

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The heat transfer rate at the boiler can be found by applying the First Law,

1 1 7inQ m h h

No flow rate information is given. However, we can find the heat transferred per lbm of steam entering the HPT,

1 71

inin

Qq h h

m

Turbine Modeling

14

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The thermal efficiency of the cycle is given by,

t pnetth

in in

W WWQ Q

The turbine power delivered is,

1 1 2 2 3 3tW m h m h m h

2 31 2 3

1 1 1

tm mW

h h hm m m

1 2 2 3 31

ttWw h y h y hm

The flow fractions need to be determined!

1 1

1

/ //

t p

in

W m W mQ m

Pump Modeling

15

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

There are two pumps in the cycle. Therefore,

1 2p p pW W W

4 5 4 6 7 6pW m h h m h h

4 65 4 7 6

1 1 1

p m mWh h h h

m m m

4 5 4 6 7 61

pp

Ww y h h y h h

m

1 1

1

/ //

t p t pth

in in

W m W m w wQ m q

Then ...

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).

Mass Conservation

16

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The flow fractions must be found. The easy flow fractions are ...

2 5 6m m m

2 5 6

1 1 1

m m mm m m

2 5 6y y y

1 6 7 1y y y

3 4 5y y y

Conservation of mass applied to the FWH gives,

Closing the System

17

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

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,

2 2 5 5 6 6m h m h m h

2 5 62 5 6

1 1 1

m m mh h h

m m m

2 2 5 5 6 6y 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.

Augmented Array

18

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The updated property table ...

From previous analysis,t p

thin

w wq

1 2 2 3 3tw h y h y h

4 5 4 6 7 6pw y h h y h h

Cycle Performance Parameters

19

1

1

1500 psia900 F

PT

2 100 psiaP

3 1 psiaP

4

4

1 psia0

Px

5 100 psiaP 6

6

100 psia0

Px

7 1P P

The heat rate of the cycle is,

1

1 1

/HR

/ /in in in

net t p t p

Q Q m qW W m W m w w

EES Solution (Key Variables):