Department of Mechanical Engineering

ME 322 – Mechanical Engineering

Thermodynamics

Lecture 26

Use of Regeneration in Vapor Power Cycles

What is Regeneration?

• Goal of regeneration – Reduce the fuel input requirements (Qin)

– 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

Regeneration – Open FWH

3

Increased temperature

into the boiler due to

regenerative heating

Keeping Track of Mass Flow Splits

4

1 1y

2y

3y

4y

5y6y7y

Define a mass flow fraction,

1

mass flow rate at any state

mass flow rate entering the HPT

n

n

m ny

m

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!

Regeneration – Closed FWH

5

There are two types of closed feedwater heaters

Closed FWH with

Drain Pumped

Forward

Closed FWH with

Drain Cascaded

Backward

Regeneration – Closed FWH

6

Example – Closed FWH with Drain Cascaded Backward

1 1y

2y3y

4y

5y6y

7y

8y

Regeneration – Multiple FWH

7

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 7

1

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 WW

Q Q

The turbine power delivered is,

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

2 3

1 2 3

1 1 1

tm mW

h h hm m m

1 2 2 3 3

1

tt

Ww h y h y h

m The flow fractions need to

be determined!

1 1

1

/ /

/

t p

in

W m W m

Q 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 6

5 4 7 6

1 1 1

pm mW

h h h hm m m

4 5 4 6 7 6

1

p

p

Ww y h h y h h

m

1 1

1

/ /

/

t p t p

th

in in

W m W m w w

Q 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 m

m 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 6

2 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

th

in

w w

q

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 q

W W m W m w w

EES Solution (Key Variables):