Week 2 Chapter 2 Shaft Cycles Part 1_2016

7/21/2019 Week 2 Chapter 2 Shaft Cycles Part 1_2016

http://slidepdf.com/reader/full/week-2-chapter-2-shaft-cycles-part-12016 1/78

Lecturer: Dr CR Bester

Office: B3 Lab 207

Tel (W): 011-559-4184

EMail: [email protected]

[email protected]

[email protected]

Assistant: Mr D Tyczynski

[email protected]

Thermomachines TRM4A11

1st semester 2016

mailto:[email protected]










• Gas turbines

• Reciprocating Internal Combustion Engines(“RICEs”)

• Turbomachinery

Thermomachines TRM4A11

Curriculum



Textbook for Gas Turbines

The lectures presented here are based on theprescribed textbook, i.e. “Gas Turbine Theory,” bySaravanamuttoo, H.I.H., Rogers, G.F.C., Cohen, H., andStraznicky, P.V., 6th Edition, published by Pearson

Prentice Hall, England, in 2009.

In the lecture notes the textbook will be referred to bythe acronym “CRSS”.



• Two gas turbine arrangements are of importance inthis course, i.e. shaft power cycles (land based) andcycles for aircraft propulsion (airborne and on ground).

• Shaft power cycles are discussed first. This chaptercovers land-based shaft cycles for power generation.

• Ideal cycles, assuming no component inefficiencies,

pressure- or other losses, are introduced, followed bymore realistic, or “real” cycles.

• Cycle states, processes, heat input, efficiency, work,important parameters and optimisation are discussed.

Gas Turbine Arrangements



States are pressure, temperature, entropy, specific

volume

Processes are compression (win), combustion (q

in),

expansion (wout), exhaust (qout)

Efficiency is a measure (ito fraction or %) of heat

supplied by fuel (qin) that is converted to work (wnet)

Work w is a useable form of energy, e.g. to drive a

generator

Chapter 2: Shaft Power Cycles



Ideal cycles

(CRSS p. 46)



Ideal cycles

(continued on CRSS p. 47)



Notes

• The thermodynamic cycle of a gas turbine is a so-called air-standard cycle

(like the Otto-, Diesel-, Dual- and Stirling cycles)

• Air is the working fluid, or substance

• The gas-turbine cycle is also known as the Joule- orBrayton cycle

• A multitude of gas turbine configurations exists

• The cycle of the most simple configuration isappropriately known as the simple cycle

• A schematic and the T-s- and P-v diagrammes of asimple gas turbine cycle are shown in Figure 2.1.1



Fig. 2.1.1: Simple gas turbine cycle

https://en.wikipedia.org/wiki/Brayton_cycle





Note

Copy or draw the P-v- and T-s diagrammes and keep

them next to you so that you can see where you are on

the cycle during the derivations and calculations



In Figure 2.1.1, and equations that follow, heat, entropy

and volume are expressed per unit mass flow, i.e.

• specific heat input q,

• specific volume v

• and specific entropy s

denoted by symbols in small letters

The same applies to specific work w, which will be

addressed later

Note



1-2: Compressor

• Work input to the air by the compressor (or to other

working fluid, “substance” - more about that to follow)• Isentropic compression

• Air pressure and temperature increase

• Air specific volume decreases

• Air specific entropy is constant



2-3: Combustion chamber, “burner”

• Heat added to the air by the burning fuel

• Heat input at constant pressure• Heat input at high temperature

• Air temperature increases

• Air specific volume increases

• Air specific entropy increases



3-4: Turbine

• Work extracted from the air by the turbine

• Isentropic expansion

• Air pressure and temperature decrease

• Air specific volume increases

• Air specific entropy is constant



4-1: Exhaust

• Heat transferred (rejected) from the air to theatmosphere by the exhaust

• Heat rejection at constant pressure• Heat rejection at low temperature

• Air temperature decreases

• Air specific volume decreases

• Air entropy decreases



Thermal efficiency and work of an

ideal, simple gas turbine cycle

Starting from first principles, from the First Law of

Thermodynamics:

(1)

where

q is specific heat inputu is specific internal energy

w is specific work output

d is the differential operator

dwdudq



Integration of equation (1) over a cycle:

(2)

Consider the terms in equation (2) one-by-one:

I ) The term on the left hand side of equation (2) is:

(3)

i.e. the sum of heat input and -output (or heat rejected)

dwdudq

outin qqdq





Note

Sign convention for heat input and -output:

(4a)

(4b)

0in q

0out q

Th l ffi i d k f



II ) The first term on the right hand side of equation (2)

is:

(5)

where C v is specific heat at constant volume

In equation (5) the end temperature T end

of the cycle

equals the start temperature T start

of the cycle, which

equals the start temperature of the next cycle

0startend T T C dT C dT C du vvv



Th l ffi i d k f



III ) The second term on the right hand side of

equation (2) is:

(6)

where wnet

is the net specific work done by the

cycle

A closer look at it later

netwdw





Note

Sign convention for work input and –output:

(7a)

(7b)

0out w

0in w



Textbook uses W (capital W ) for specific work

Use small w with subscripts here to distinguish

•compressor specific work wC (= win)• turbine specific work wT (= w

out)

•net specific work wnet

The same applies to specific heat

• textbook uses Q for specific heat

•use qin

and qout

here to distinguish between specific

heat in- and outputs

Note

Th l ffi i d k f



Substitution of equations (3), (5) and (6) into equation

(2) gives the First Law of Thermodynamics for the air-

standard Brayton- / Joule cycle

(8)netoutin

wqq



Th l ffi i d k f



Consider the terms in equation (8) one-by-one:

Start by writing the heat input and -output in terms of thecycle temperatures:

I ) First term on the left hand side of equation (8)gives the heat input per unit mass flow of air:

(9)

where C P is specific heat at constant pressure

2323in

T T C m

Qqq

P

a

in





(10)0in23 qT T

Note

Th l ffi i d k f



II ) Second term on the left hand side of equation (8):

(11)

Equation (11) gives the heat rejected per unit

mass flow of air

41

out

41out T T C

m

Qqq

P

a





(12)0out41 qT T

Note

Th l ffi i d k f



III ) Term on right hand side of equation (8):

Substitution of equations (9) and (11) into equation

(8) gives:

(13)

4123outinnet

T T T T C qqw P



Eq ation for thermal efficienc



Thermal efficiency equals net work per unit heat input:

(14)

Substitution of equations (9) and (13) into equation (14)

gives:

(15)

in

net

q

wh

23

4123

23

4123

in

outin

T T

T T T T

T T C

T T T T C

q

qq

P

P

h

Equation for thermal efficiency

of an ideal, simple gas turbine cycle



Note

(16)*

and wnet

is positive

An efficiency of smaller than unity is expected

* The slope of a constant pressure line for heat inputon a T-s diagramme (high pressure) is larger than for

heat rejection (low pressure), so that

outin1423 qqT T T T

1243 T T T T




A little mathematical manipulation follows on equation

(15):

(17)

1

1

111

2

3

2

1

41

23

14

23

41

T

T T

T

T T

T T

T T

T T

T T h






In order to proceed, cycle pressures are required.

In the ideal cycle the combustion chamber pressure is

constant:

(18)

Similarly, in the ideal cycle the exhaust pressure isconstant:

(19)

23 P P

14 P P





The textbook uses r for pressure ratio

(of the compressor or turbine)

Use r P here instead of r

r T for temperature ratio

r V for volume ratio (later, for reciprocating internalcombustion engines)

to clarify which ratio is used

Note




From equations (18) and (19), the pressure ratios of the

compressor and turbine are equal:

(20)1

2

4

3

P

P

P

P r P






By application of the relationship between pressure ratio

and temperature ratio for isentropic compression and

expansion processes, the temperature ratios of the

turbine and compressor are obtained as:

(21)

i.e. the temperature ratio of the turbine is equal to that of

the compressor

1

2

1

1

2

11

4

3

4

3

T

T

P

P r

P

P

T

T r P T






It follows from equation (21) that:

(22)

Substitution of equation (22) into equation (17) gives:

(23)

2

3

1

4

T

T

T

T

122

1 111

T T T

T h






Subsequent substitution of equation (21) into equation

(23) gives the thermal efficiency of an ideal, simple gas

turbine cycle as follows in terms of the temperature ratio:

(24)

A graph of the thermal efficiency h vs temperature ratio

r T is shown in Figure 2.1.2

T r

11h





This graph is not shown in the textbook

It is important to note that both r T and h are

dimensionless

The graph was generated and plotted in Matlab, by

writing a Matlab .m-file “gammaT.m”, and printed to a“jpeg”-file “gammaT.jpg”

Notes



Matlab .m-file gammaT.m% gammaT.m % % This .m-file generates and plots the graph of the

% thermal efficiency of a simple gas turbine vs% temperature ratio r_T = T2/T1, or T3/T4 % % Generate temperature ratio- (rT) vector, % from 1 to 10 in steps of 0,1 % rT=[1:.1:10];% % Calculate eta-vector using eta = 1 - (1/rT) and

% then multiply with 100 to get eta in percentage % eta=1-1./rT;eta=100*eta;% % Plot eta against temperature ratio and set axis, % switch grid and zoom function on, add x- & y-labels % and graph title, change background colour to white %

plot(rT,eta,'b')axis([0 10 0 100])grid on;zoom onxlabel('Temperature ratio r_T')ylabel('Efficiency eta (%)')title('EFFICIENCY OF SIMPLE GAS TURBINE vs TEMPERATURE RATIO')whitebg(figure(1),'w')% % Finally, print the graph to a jpeg-file

% for inclusion into an MSWord-file % print gammaT.jpg -djpeg

Fi 2 1 2 Th l ffi i f



Figure 2.1.2: Thermal efficiency of

an ideal gas turbine cycle vs r T



• Thermal efficiency increases with temperature ratio• As is to be expected thermal efficiency is zero at a

temperature ratio of 1 (i.e. no compression)

• Some other points on the graphh equals (exactly)

– 50% at an r T ratio of 2

– 80% at 5 – 90% at 10

Discussion of Figure 2.1.2

Ideal simple gas turbine cycle



Substitution of equation (21) into equation (24) gives thethermal efficiency of an ideal, simple gas turbine cycleas follows in terms of the pressure ratio:

(25)

(CRSS p. 48 equation 2.1)

h

1

11

P r

Ideal, simple gas turbine cycle

efficiency in terms of pressure ratio



The thermal efficiency h not only depends on r P but alsoon the ratio of specific heats

h is plotted against pressure ratio for two values of i.e.

• 7/5 (=1,4) and

• 5/3 (=1,667)

in Figure 2.1.3 (CRSS p. 48 Figure 2.2a)

Notes

Fi 2 1 3 f i l



Figure 2.1.3: of a simple gas

turbine vs r P

for values of 7/5 & 5/3



• Efficiency h increases with r P

• h also increases with

• As with the dependance on r T , h is zero at an r P of 1

(no compression)

Discussion of Figure 2.1.3



• Diatomic gases like oxygen and nitrogen have -

values of (very close to) 7/5 (=1,4)

• Monatomic gases like helium and argon have -

values of approximately 5/3 (1,667)

Why for -values of 7/5 and 5/3?



Composition of Air

• 78,09% per vol N2

• 20,94% O2

• 0,93% Ar

• 1770 ppm CH4

• 500 ppm H2• 385 ppm CO2

• 86 ppm Xe

• 18 ppm Ne

• 5,2 ppm He• 1,1 ppm Kr

• 6.0e-11 ppm Rn

at 25°C and 101 kPa

The use of monatomic gases as



Advantage of using monatomic gases, like helium, as a

working fluid

• High thermal efficiency (see Figure 2.1.3) due to high

Disadvantage of using helium as a working fluid

• It is difficult to prevent it from leaking, especially at high

pressures

For the rest of this section on ideal cycles, air with a -

value of 1,4 will be considered as working fluid

The use of monatomic gases as

working fluids





Simple gas turbine work






Equations for compressor- and turbine specific work:

(26)

(27)

0 2112 T T C

m

W ww P

a

C

C

0 4334 T T C

m

W ww P

a

T T




From equations (26) and (27):

(28)

From equation (13):

(29)

4321 T T T T C ww P T C

4123outin

T T T T C qq P




It can be seen from equations (13), (28) and (29) that:

(30)

(which could also have been obtained by direct

application of the First Law)

netoutin wqqww

T C




As , from equations (4a) & (4b), and, from equation (16) - it follows from equation

(30) that

(31)

(32)

0;0 outin qq

outin qq

0net w

C T ww


S



Therefore the net specific work is positive

It has two advantages:

• The turbine work is sufficient to drive the compressor

• Any excess (net) work may be used to drive additional

shafts, e.g. to a propeller, rotor, fan or generator


N t



• The maximum cycle temperature is at the combustion

chamber outlet, which equals that at the turbine entry,

i.e. T 3

(“ TET ” ), also called “turbine inlet temperature”

• Work output increases with temperature T 3

• Ideally, it is desirable to increase T 3

as much as

possible

• However, material properties of gas turbine

components, especially turbine blades, limit T 3

• A means of obtaining net specific work in terms of T 3

is

required

Notes



Net specific work in terms of T 3

Recall Equation (13):

(13)

Define the maximum-to-minimum temperature ratio as:

(33)

4123outinnet

T T T T C qqw P

1

3

T

T

t



Division of equation (13) by C P T 1 gives net specific work

as follows:

(34)

Consider the terms on the right hand side of equation(34) one-by-one

1

4

1

2

1

3

1

net

1 T

T

T

T

T

T

T C

w

P

Net specific work in terms of t



I) From equation (33)

II) From equation (21)

III) From Equation (22)

(35)

13 T T t

1

12 /

P r T T

2

1

1

3

2

3

1

4

T

T

T

T

T

T

T

T




It follows from Equation (21) that:

(36)

Substitution of equations (33) and (36) into equation (35)

gives:

(37)

1

2

1 1

P r T

T

1

1

4

P r

t

T

T




Finally, substitution of equations (21), (33) and (37) into

equation (34) gives:

(38)

(CRSS p. 48 equation 2.2)

1111 1

11

1

1

net

P

P P

P

P

r r

t r

t r t

T C

w


Class example



Class example

Ideal simple gas turbine

The following data apply to an ideal, simple gas turbine

operating on air:

Ambient pressure 101 325 Pa Ambient temperature 15 °C

Pressure ratio 8,0:1

Turbine inlet temperature 1 400 °C

Class example



Calculate:

Cycle states P 1

to P 4

and T 1

to T 4

Combustion chamber temperature difference

Exhaust temperature differenceHeat input into cycle

Heat rejection from cycle

Compressor specific work

Turbine specific workNet specific work

Cycle thermal efficiency

Maximum-to-minimum temperature ratio

Class example


Class example



The air has the following thermodynamic properties:

Gas constant:

Ratio of specific heats:

J/kgK 1,287 R

4,1

Class example


Class example



Solution:

C P is going to be needed for the heat- and workcalculations, therefore first calculate C P :

1

1

R

C C RC C RC V V V V P

J/kgK 9,100414,1

1,287.4,1

1

RC

P

Class example


Class example



Class example


State 1:

State 2:

Pa1013251 P

K 15,28815,273151

T

kPa)6,810(Pa810600101325.812

P r P P

K 97,521815,288

4,114,11

12

P r T T

Class example



State 3:

State 4:

kPa)6,810(Pa81060023

P P

K 15,167315,27314003

T

kPa)(101,325Pa10132514 P P

K 65,923

8

115,1673

1 4,114,1

134

P r

T T

Class example


Class example



Combustion chamber- (“burner”-) and exhaust

temperature differences:

Compressor- and turbine temperature differences:

K 18,115197,52115,167323

T T T B

K 50,63565,92315,28841 T T T

E

K 82,23315,28897,52112 T T T C

K 50,74915,167365,92334 T T T T

Class example


Class example



Heat input into and rejected from cycle:

Compressor- and turbine specific work:

B P P

T C T T C q kJ/kg77,115618,1151.9,100423in

E P P

T C T T C q kJ/kg6,63850,635.9,100441out

C P P

T C T T C w kJ/kg95,23482,233.9,100421in

T P P T C T T C w kJ/kg13,75350,749.9,100443out

Class example


Class example



Net work:

Thermal efficiency:

Maximum-to-minimum temperature ratio:

kJ/kg18,51810.6,63810.77,1156 33

outinnet qqw

448,010.77,1156

10.18,5183

3

in

net q

wh

807,515,288

15,1673

1

3 T

T t

Class example


Notes



• Read blade cooling CRSS pp. 366 – 376 Section 7.6“The cooled turbine”

• You may just get a question on turbine blade cooling

in a test or exam

• Extracts of the section – see CRSS pp. 375 & 376

Figures 7.32 & 7.33

Notes

CRSS p 375 Figure 7 32



CRSS p. 375 Figure 7.32

CRSS p 376 Figure 7 33



CRSS p. 376 Figure 7.33

Rolls-Royce RB-211 & Trent Engines



The Rolls-Royce RB-211 engines are used in a numberof aircraft, e.g.

•Boeing 747

•Boeing 757

•Boeing 767

•Tupolev Tu-204

Modern Rolls Royce RB-211 Trent engines are used inthe following aircraft, to name but a few:

•Trent 900: Airbus A380

•Trent 1000: Boeing 787

Rolls-Royce RB-211 & Trent Engines

Homework problem for 2016 02 22



Plot CRSS Figure 2.2b on p. 48, i.e.

vs r P , ranging from 1 to 16,

for t - values from 2 to 6, in increments of 1

See Figure 2.1.4, next slide

1T C w P net

Homework problem for 2016-02-22

Figure 2.1.4: w net /C P T 1 of a simple



gas turbine vs r P for t from 2 to 6




Use thermodynamic equations to show that Equation(16) of this weeks lectures applies to an ideal simple

gas turbine cycle


Tutorial 1 problem for 2016-02-19



Tutorial 1 problem for 2016 02 19

Ideal, simple gas turbine shaft cycle

Venue: DLES 307, from 9:40 to 10:25

The total specific work done by the turbine of an ideal, simple shaft cyclegas turbine is 829125 J/kg. Calculate all the cycle temperatures and

pressures if the thermal efficiency is 50%. Also calculate the specific values(i.e. per kg of working fluid) of the compressor work, net work, heat input,exhaust heat rejected to atmosphere and maximum-to-minimumtemperature ratio. Inlet temperature and -pressure are 300 K and 85 kParespectively.

Specific heat capacity at constant pressure and ratio of specific heats are1005 J/kgK and 1,4 respectively.

Draw the pressure-temperature- (P T) diagramme of the cycle What is

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Week 2 Chapter 2 Shaft Cycles Part 1_2016