Thermodynamic Chapter 6 Thermal Power Plant

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MEC 451THERMODYNAMICS

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CHAPTER 6THERMAL POWER PLANT

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PART 1


GAS TURBINE POWER PLANT

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Brayton Cycle

The Brayton cycle (a.k.a. Joule cycle) is the air-standard ideal

cycle approximation for the gas turbine engine.

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Process Description Related formula

1-2 Isentropic compression

2-3 Constant pressure heat addition

3-4 Isentropic expansion

4-1 Constant pressure heat rejection

1

2

1

1

2

2

1

k

kk

T

T

V

V

P

P

1

2

1

1

2

2

1

k

kk

T

T

V

V

P

P

Brayton Cycle Process Description

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The T-s and P-v diagrams are

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The T-s for actual cycle (including pressure drop and isentropic efficiency)

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outout

outoutoutoutout

inin

ininininin gz

VhmWQgz

VhmWQ

22

2...2...

outoutininin hmhmW...

12

.

12

.

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Energy balance: for compressor

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Isentropic efficiency for compressor:

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outout

outoutoutoutout

inin

ininininin gz

VhmWQgz

VhmWQ

22

2...2...

outoutininin hmhmQ...

23

.

23

.

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Energy balance: for boiler

2

3

10

outout

outoutoutoutout

inin

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VhmWQgz

VhmWQ

22

2...2...

outoutoutinin hmWhm...

43

.

34

.

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4

Energy balance: for turbine

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Isentropic efficiency for turbine:

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outout

outoutoutoutout

inin

ininininin gz

VhmWQgz

VhmWQ

22

2...2...

outoutoutinin Qhmhm...

14

.

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1

Energy balance: for condenser

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th Brayton net

in

out

in

W

Q

Q

Q, 1

where the pressure ratio is rp = P2/P1

th Braytonpk kr

, ( )/ 11

1

Upon derivation

Thermal efficiency of the Brayton cycle is defined as

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For fixed values of Tmin

and Tmax, the net work

of the Brayton cycle

first increases with the

pressure ratio, then

reaches a maximum at

rp = (Tmax/Tmin)k/[2(k - 1)],

and finally decreases.

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43

12

TTc

TTc

W

Wr

p

p

t

cbw

Back Work Ratio

Part of the work output from turbine is used to drive the

compressor, which in turn requires a work input. Therefore

the back work ratio can be written as

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Example GT-1

The ideal air-standard Brayton cycle operates with air

entering the compressor at 95 kPa, 22°C. The pressure

ratio rp is 6:1 and the air leaves the heat addition process at

1100 K. Determine the compressor work and the turbine

work per unit mass flow, the cycle efficiency and the back

work ratio. Assume constant properties.

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Cycle Improvement - Regenerative

Therefore, a heat exchanger can be placed between the

hot gases leaving the turbine and the cooler gases leaving

the compressor.

This heat exchanger is called a regenerator or recuperator.

For the Brayton cycle, the turbine exhaust temperature is

greater than the compressor exit temperature.

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The regenerator effectiveness regen is defined as the ratio of

the heat transferred to the compressor gases in the

regenerator to the maximum possible heat transfer to the

compressor gases.

q h h

q h h h h

q

q

h h

h h

regen act

regen

regenregen act

regen

,

, max '

,

, max

5 2

5 2 4 2

5 2

4 2

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For ideal gases using the cold-air-standard assumption

with constant specific heats, the regenerator effectiveness

becomes

5 2

4 2regen

T T

T T

Upon derivation the thermal efficiency becomes

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Example GT-2

Air enters the compressor of a regenerative gas turbine

engine at 100 kPa and 300 K and is compressed to 800 kPa.

The regenerator has an effectiveness of 65 percent, and the

air enters the turbine at 1200 K. For a compressor efficiency

of 75 percent and a turbine efficiency of 86 percent,

determine

(a) The heat transfer in the regenerator.

(b) The back work ratio.

(c) The cycle thermal efficiency.

Assume air is an ideal gas with constant specific heats.

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Cycle Improvement – Intercooling and Reheating

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The T-s diagram for this cycle is shown below.

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Example GT-3

An ideal gas turbine with two-stage compression and two-

stage expansion has an overall pressure ratio of 8. Air

enters each stage of the compressor at 300 K and each

stage of turbine at 1300 K. Determine the back work ratio

and the thermal efficiency of the cycle assuming (a) no

regenerators and (b) an ideal regenerator with 100%

effectiveness and (c) a regenerator with 60%

effectiveness. Assume constant specific heats.

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Aircraft Engine

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Alstom GT26 Gas Turbine

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PW 4000 112-INCH FAN ENGINE

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Rolls – Royce Trent

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General Electric (GE) Gas Turbine

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PART 2


STEAM POWER PLANT

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Based on the Carnot cycle the heat engine may be composed of the following components.

Carnot Cycle

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0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.00

100

200

300

400

500

600

700700

s [kJ/kg-K]

T [

C] 6000 kPa

100 kPa

Carnot Vapor Cycle Using Steam

1

23

4

th Carnot net

in

out

in

L

H

W

Q

Q

Q

T

T

,

1

1

The thermal efficiency of this cycle is given as

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Reasons why the Carnot cycle is not used:

Pumping process 1-2 requires the pumping of a mixture

of saturated liquid and vapor.

Low quality steam at the turbine exit.

The impracticalities of the Carnot cycle can be eliminated by:

Condensing the steam completely in the condenser. Superheating the steam to take advantage of a higher

temperature.

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Process Description

1-2 Isentropic compression in pump

2-3 Constant pressure heat addition in boiler

3-4 Isentropic expansion in turbine

4-1 Constant pressure heat rejection in condenser

Rankine Cycle

The simple Rankine cycle has the same component layout as the Carnot cycle. The processes in a simple Rankine cycle are:

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The pump work can be determined by:

Cycle Analysis

121

12

2211

PPm

hhmW

hmWhm

pump

pump

The energy balance for boiler:

23

3322

hhmQ

hmQhm

in

in

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The energy balance for turbine:

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4433

hhmW

hmWhm

out

out

The energy balance for condenser:

14

1144

hhmQ

hmQhm

out

out

Based on the Second Law the thermal efficiency becomes:

23

1243

hh

hhhh

q

w

in

netth

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Cycle Improvement

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Actual Cycle

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Example SPP-1

A power plant operates on a simple Rankine cycle. Steam

enters the turbine at 3 MPa and 350°C and is condensed

in the condenser at a pressure of 75 kPa. Determine (a)

the work net, (b) the heat input in the boiler, and (c) the

thermal efficiency of the cycle.

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Reheat Rankine Cycle

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Component Process First Law Analysis

Boiler Constant Pressure qin = (h3 - h2) + (h5 - h4)

Turbine Isentropic wout = (h3 - h4) + (h5 - h6)

Condenser Constant Pressure qout = (h6 - h1)

Pump Isentropic win = (h2 - h1) = v1(P2 - P1)

4523

126543

hhhh

hhhhhh

q

w

in

netth

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Example SPP-2

Consider a steam power plant operating on the ideal reheat

Rankine cycle. Steam enters the high-pressure turbine at

15 MPa and 600°C and is condensed in the condenser at a

pressure of 10 kPa. If the moisture content of the steam at

the exit of the low-pressure turbine is not to exceed 10.4

percent, determine (a) the pressure at which the steam

should be reheated, and (b) the thermal efficiency of the

cycle. Assume the steam is reheated to the inlet

temperature of the high-pressure turbine.

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Regenerative Cycle

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Regenerative Cycle with Open FWH

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Regenerative Cycle with Closed FWH

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Example SPP-3

Consider a steam power plant operating on the ideal

regenerative Rankine with one open feedwater heater. Steam

enters the turbine at 150 bar and 600°C and is condensed in

the condenser at a pressure of 0.1 bar. Some steam leaves

the turbine at a pressure of 12 bar and enters the open

feedwater heater. Determine the fraction of steam extracted

from the turbine and the thermal efficiency of the cycle.

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Plant Layout

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Plant Monitoring System

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PART 3


SPECIAL APPLICATION

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Cogeneration

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Cogeneration Stand-Alone System

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Trigeneration Plant

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Combined Cycle

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Karlsruhe Power Station (Germany)

Topping cycle = 2 x Bottoming cycle

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Sultan Ismail Power Station (Paka)

Combined Cycle (808 MW)

Gas Turbine – 137 MW x 4 Units

Steam Turbine – 130 MW x 2 Unit


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Pasir Gudang Power Station (Johor)

Combined Cycle (404 MW)

Gas Turbine – 137 MW x 2 Units

Steam Turbine – 130 MW x 1 Unit


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Connaught Bridge Power Station (Klang)

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Binary Vapor Cycle

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Education

Thermodynamic Chapter 6 Thermal Power Plant