First Three Lectures - Turbine Desine

Assoc. Prof. A. Abd El-Hamied

Final

Exam

(Hrs)

Mark DistributionHours / week

3

TotalFinal

Exam

OralSemester

work

TotalLabTutorialLecture

1509030306123

Design of Steam and Gas

Turbines

Prof. T. Sabry



Course Content

• Analysis Turbine losses (Additional losses) (twolectures)

a) Internal losses

b) External losses

• Design of single-stage turbine (two lectures)

• Design of velocity-stage turbine (two lectures)

• Design of multi-stages turbine (two lectures)

a) First stage,

b) Second stage,

c) Last stage,

d) Intermediate stages


• Design of Turbine stage with Long blades(two lectures).

• Cooling of gas turbine blades (two lectures)

• Start-up and shut-down of turbine (one lecture)

• Matching of gas turbine components. (one lecture)


REFERENCES

• S. M. Yaha “Turbines, Compressors and Fans”,

Tata McGraw Hill, NewDelhi, 1989.

• Z. Husain”Steam Turbine, theory and design”,

Tata McGraw Hill, NewDelhi, 1984.

• A. Kostyuk and V. Frolov” Steam and Gas

Turbines” Mir Publishers Moscow, 1985.

• S. L. Dixon” Fluid Mechanics, Thermodynamics

of Turbomachinery” 3 rd ed, pergamon press,

Oxford, 1996.

Assessments


• Quizzes: to assess understanding of a particular section

• Mid-Term Examination to assess the progress of course delivering

• Practical (Oral) Examination to assess the ability of performing

practical tasks.

• Final Examination to assess the overall understanding and

achievements

• Assessment schedule

• Assessment 1: Quizzes Two quizzes (4th,12th)

• Assessment 2: Mid-term Examination 8th week

• Assessment 3: Practical Examination 14th week

• Assessment 4: Final- Examination 15th week

• Weighting of assessments

• Quizzes 10 % = 15

• Mid-Term Exam 10 % = 15

• Practical or Oral Exam 20 % = 30.0

• Final-Term Exam 60 % = 90.0

• Total 100 % = 150.0


Assoc.Prof Abd El-Hamied

Fixed blades

Moving blades

Shaft

CasingbearingControl stage

7


Fixed row

moving row

Casing

Shaft

10


Blade Geometry

Leading

edge

Trailing

edge

Chord

Blade height

Pressure

surface

Suction

surface

Camber

line

Exit

blade

angle

Inlet

blade

angle

11


ENERGY LOSSES IN

STEAM TURBINES

Internal losses External losses

1.losses in regulating valves 1. mechanical losses

2.losses in nozzles 2. losses due to steam

3.losses in moving blades leakage through end seals

4.losses due to disc friction and windage

5.losses due to wetness of steam

6.carry-over losses

7.losses due to axial and radial clearances (leakage loss)


o o1

1th

1

ht

ho

hi

Δh

hn

hb

hfr

hleak

hwethe

S

h

po

p\o


Losses in regulating valve

• The magnitude of this loss due to

throttling when regulating valves are

fully open may be as much as 5% of

the fresh steam pressure Po

• for design purposes this pressure loss

p=(0.03 - 0.05) Po is recommended.


Losses in nozzles

Losses in nozzles can be divided into three groups:

• Profile losses arise from growth ofboundary layer and turbulence in wake.

• Secondary losses due to frictionalresistance at blade surfaces and root andperiphery of blades.

• Shock losses occur at nearly subsonic andsupersonic velocities.


0 50 100 150 200 250

l, mm

0.93

0.94

0.95

0.96

0.97

0.98

0.99

kn

Fig. 2 Velocity coefficient for convergent nozzle as

a function of blade height


Losses in moving blades

Losses in moving blade are caused due to

various factors some of them are

• impingement losses,

• frictional losses,

• turning losses

• and wake losses.


These losses depend upon several

factors such as:

• velocity of steam,

• height of blades,

• pitch of blades and

• degree of reaction


0 50 100 150

l, mm

0.82

0.84

0.86

0.88

0.90

0.92

k b27/24

33/28

36/30

40/36

Fig. 3 Velocity coefficient kb for moving blades of an impulse turbine

for various heights and blade angles


Losses Due to Disc Friction

• The friction stresses (fr ) on the surfaces of a rotating disc in turbulent motion are proportional to the square of flow velocity and to the steam density in the disc chambers i.e

fr = k u2/v ,

• where u is the blade velocity of the disc at a radius r and v is the specific volume of steam in the disc chamber.


Scheme of steam flow in the chamber of a turbine disc.


• The amount of friction forces relative to the

rotor axis can be found by integrating the

moments appearing on elementary

surface dA of the disc.

if rsh = 0.0

)2(2.22

rdrrv

ukdArM

r

r

frfr

sh

nv

rkuM fr

5

602

23


The friction power of the disc at a sufficiently low

rsh, will be determined by the relationship

The coefficient kfr in this formula depends on

• Reynolds number, Re= .r.u/,

• roughness of disc friction and

• axial clearance between the disc and

stationary chamber.

3 2

fr fr fr

u dN M ω k

2v


2 1/10 1/5

fr

3

fr

k 2.5x10 (s / r) Re

or

k (0.45 0.8)x10


• The ratio of the friction power of a disc to

the available power of the stage is equal to

the relative energy loss due to friction of

the disc

where mo v =A1 c1th

and A1 = d l1 sin 1 and 2ho = c21th

3 2

fr frfr o

o o

N k u dξ

N 2vm h

3

frfr

1 1th 1

k u dξ

πεsin α c l

• Steam leakages from the main

stream flow beyond the rotating

blade shrouds and their roots

as well as between the shaft

and diaphragm ID is the main

source (75-81%) of turbine

performance deterioration due

to:

• Missing energy of steam

leaving or bypassing the main

flow.


Leakage Losses


• Disturbances in main flow (wakes,

eddies, etc.) from collisions of

leaked steam with the main flow.

• Therefore, in order to minimize and

contain these leakages, the

modern impulse turbine stage

incorporates three types of seals:

tip seals, root seals and shaft.

• Tip Seals

• Tip seals are used to prevent steam leakage

into the space above the rotating blades.

• This leak is the largest source of efficiency

loss due to the largest leak area and the

highest reaction (i.e. pressure drop) in this

location.

• Tip seal improves the stage efficiency by

approximately 2.5% compared to the axial

rigid seal.


• Root Seals

• Root seals perform two functions.

• First, they prevent shaft leakage from

entering into the main stream flow

resulting in an increase of stage reliability

and efficiency.

• Second, they prevent leakage from the

main stream flow into the space between

the diaphragm and disk faces and into the

disk equalizing holes, maintaining high

efficiency.Assoc. Prof. A. Abd El-Hamied

• Shaft Seals

• These seals prevent steam leakage

between the shaft and diaphragm ID. The

main engineering efforts were to create a

labyrinth seal with minimal radial

clearances between its fins and rotor.

• Labyrinth seal must prevent rubbing and

wear which mostly takes place during

start-up and shut-down regimes when the

rotor goes through the 1st critical speed.



Calculation of Leakage through

turbine stage

mg

mh

mr

mbmo

Fixed

moving


Leakages in a turbine stage are flows of

steam through clearances in glands

between:

• A diaphragm and shaft ( mg)

• The moving blades and casing (mb)

• A diaphragm and disc at the roots of

moving blades (mr)

• Through discharge holes (mh)


h-s diagram of variations of the state of

steam in stepped labyrinth glands

p\\\1=p1

po p\1 p\\

1ho =const.

h\o

h\\\0

h\\o

po

housing

Turbine shaft

p1


• The flow rate of steam through slits of a

straight through gland is substantially

higher than in a stepped gland, for this

case can be written as follows:

where

kg is a correction factor (1 - 2.4)

z is the number of slits

r is the pressure ratio r = p1/po

z

r

v

pdkm

o

ggggg

1)( 0

0


• The leakage loss at the periphery of a stage canbe determined by the following formula

where

dt is the tip diameter

A1 is the cross sectional area at exit from nozzle

is the degree of reaction at the averagediameter of the stage

l is the height of blades

d is the average diameter of the stage

eq is the equivalent clearance

o

b t eq

b b bo

1

(πd δ )m lξ η ρ 1.8 η

A dm


where

a = 0.5

r = 0.7

eq

2 2

a a r r

1δ

1 1

(μ δ ) (μ δ )

δr

δa

δr

eq rδ 0.75 δ


The flow of wet steam in a turbine stage can

involve the following phenomena:

• Expansion of steam from the superheated

state near the saturation line (x = 1.0) can

cause the phenomenon of steam super-

cooling.

• At a certain ultimate degree of super-cooling,

the steam passes over from the metastable

super-cooled state to an equilibrium state

with partial condensation and formation of

finely dispersed moisture.

Losses due to wetness of steam


• Moisture droplets can deposit on the

surfaces of blades and end walls of blade

passages in blade cascades and form a

liquid film which can increase the energy

losses in the flow due to interaction with

the boundary layer of steam flow.

• Moisture droplets may grow in size when

moving in cascade passages, owing to

condensation of surrounding steam on

them or may be broken by aerodynamic

forces in the flow, evaporate, and

coagulate.


• The paths described by moisture droplets in

blade passages depend on their size. Fine

droplets of a size d < 5 µm move along flow

lines of the vapor phase. Large droplets may

deviate from vapor- phase flow lines,

especially larger ones.

The energy loss in a turbine stage due to moisture

consists of the following components:

1. Loss due to impingement of moisture droplets

on the back side of moving blades as a result

of decelerating effect of moisture particles on

the revolving rotor.


2. Loss due to steam super-cooling,

3. Loss due to acceleration of moisture

droplets by steam flow;

4. Loss in the boundary layer associated

with the formation of liquid film on the

turbine surfaces; and

5. Loss due to increases size of trailing

edge trace caused by disintegration of

liquid film into droplets as it breaks off

the trailing edge of blades.


• In case of condensing steam turbines, the

last stages usually operate under wet-

steam region and have low efficiency

because of loss of energy due to wetness.

The heat loss caused by wet steam is

given by the formula

Where x = dryness fraction of steam

hdry = enthalpy drop in the stage

which is determined after taking into

account all heat losses, except that due to

wetness.

wet dryh (1 x)h


• In practical calculations, use the following

approximate formula:

• The coefficient a in this formula may vary within

a wide range (0.4-1.4) depending on design

parameters and operating conditions.

• For rough calculations a is taken equal to 0.8 –

0.9

o 2wet

y yξ a

2


Losses associated with partial

admission of steam

• The windage power for

the inactive portion of

moving blades is equal

to

• Windage mass flow rate

o

ww wN m h

o

w 2

2

um (1 ε)πdl

v

Diagram of windage

currents in a partial

admission turbine stage


• The windage work hw of 1 kg/s of steam is

proportional to u2 ( Δhw = k u2)

• Thus the power spent for windage in a partial

admission turbine stage is determined by the

relationship

• The formula for relative energy loss due to

windage in a single row is

• Where kw = 0.065

3

w 2

2

uN k(1 ε)πdl

v

3

w ww

o 1 1th

N k 1 ε uξ

N sinα ε c


• The formula for relative energy loss

due to windage in a multi row is

• Where m is the number of moving

rows

3

w ww

o 1 1th

N k 1 ε uξ m

N sinα ε c


Appearance of segmental loss of

energy

c1

c2

α2

c1

u

w1

co

fixed

moving


• The following formula is most popular for

calculating of segment losses

Where B2 and l2 are the width and height of

moving blades

• For a two row turbine stage the product

B2l2 is replaced by the sum of products of

width and the heights of the first and

second row of blades B2l2 + 0.6 B4l4

• The sum of windage loss ξw and segment

loss ξseg constitutes the energy loss due to

partial admission

ξp=ξw+ξseg

• In order to diminish steam leakage into

meridional clearance in a partial admission

turbine stage, the design degree of

reaction is chosen at a low level (ρ = 0.03

– 0.06)



Optimal Degree of Partiality

opt 1ε (0.5 0.7) εl

1εεopt

ξ n+ ξm

ξw

ξseg

ξ

opt 1ε (0.29 0.34) εl

For a single row

For a two rows

Where l1 is measured in cm

Example• An intermediate stage of an impulse turbine has the following initial data:

• Steam flow rate 150 kg/s,

• Steam pressure before the stage 65 bar;

• Steam temperature before the stage 470 0c;

• Steam velocity at the entry to the stage 50 m/s;

• Steam pressure behind the stage 55 bar;

• Rotational speed 3000 rpm;

• Average diameter of the stage 0.9 m;

• Diameter of diaphragm gland 0.4 m,

• Clearance in diaphragm gland 0.6 mm;

• Equivalent clearance in banding gland 0.6 mm;

• Nozzle angle 150;

• Discharge coefficients 0.97 and velocity coefficients 0.96 and 0.94 respectively.

Calculate the different losses coefficients .Assoc. Prof. A. Abd El-Hamied

Assoc. Prof. Abd El-Hamied

Steam Turbine Design

Single Stage Turbine Design


Input Data

A turbine stage is calculated for the followinginitial data

• The flow rate of steam through the stage(mo).

• The steam parameters before the stage,co, Po,to.

• The pressure behind the stage, p2

• Addition data, approximate values of x,average stage diameter, and reactiondegree.


Dimensions of Blade Cascade

for Single Stage

The calculation of a turbine stage

consists of solving two interrelated

problems:

• Determining the principal dimensions

of nozzle and moving blades

blades height l1, l2

exit angles 1,2


• Choosing a proper type of blade

profile

• adjustment angle (αad)

• chord length, b

• blade pitch, t

• number of blades, Z1,Z2

• clearance and overlaps in the stage


• Determining the blade and internal

efficiencies of the stage ήb,ήi stage

power and forces acting on the

moving blades.

The solution of these problems should

obey the requirements of high

reliability and efficiency of the stage

with due allowance for the cost of

manufacture.

Blade Dimensions

Single row turbine stages(Cylindrical moving blades)

Geometrical characteristics of a moving blade cascade

Geometrical characteristics of a nozzle blade cascade


Step -1

• C1t theoretical velocity at nozzle exit

2

1 44.72000

ot o

cc h

S

h

c2o/2

ho

v1thp1

po


Step – 2

Exit Area of a Nozzle

• V1t specific volume in isentropic expansion in

the nozzle cascade

• The discharge coefficient of blade cascades

depends on the geometrical characteristics of

cascade and flow regime parameters.

• For wet steam, the discharge coefficient µw are

higher than that of superheated steam µsh.

11

1 1

o

t

th

mvA

c

Discharge Coefficient

Discharge coefficients for

superheated steam in nozzle and

moving blade cascades depending on

relative blade height l/b and turning

angle Δβ = 180 –(β1 – β2)

Effect of wetness fraction of steam at

cascade exit on the discharge

coefficient


Step -3

Stage Diameter

• Assume x = cos 1/ 2 for impulse stage

• n: number of revolutions per minute

• C1t: theoretical velocity at nozzle exit

160 tx cd

n


Step – 4

Blade Height l1

• 1 : 11 – 20o

• 1 : 12 – 16o for blades of moderate height

1 : 16 – 20 for long blade

• If ε1 l1 < 12 mm l1 = 12 -14 mm

• ε1 = 0.8 : 0.9, b = 30 : 100 mm

11

1sin

Al

d


Step – 5

Construct the Inlet Velocity Diagram

• C1 = kn c1t

• M1t: Mach number

• K = 1.3 Superheated steam,

• k = 1.135 Dry saturated steam

• K = 1.035 + 0.1x Wet steam with dryness fraction

x

• P1 and v1t pressure and specific volume

respectevely

Chose the blade profile (Fixed

11

1 1

tht

t

cM

kp v

c1

w1

u

α1

Blades Profiles [Fixed and moving]



C – 90 – 12 A

C – 90 – 12 б

Inlet angle αo

Outlet angle α1

subsonic

Inlet angle αo

Fixed blade

Fixed blade

Outlet angle α1

sonic


C – 90 – 12 P

Fixed blade

Inlet angle αo

Outlet angle α1

Supersonic


Step – 6

Height of Moving Blades• L2 = l1 + (∆1 + ∆2)

• ∆1 and ∆2 are called the root and tip

overlap of a stage

• ∆1 = 1.0 mm

• ∆2 = 1.5 : 2.0 mm, l1 < 50 mm

• ∆1 = 1.5 mm

• ∆2 = 2.5 : 4.5 mm, 50 mm < l1 < 150 mm


• Relative velocity

• Exit area of moving blade (assume µ2 = 0.96)

• Relative exit angle

• Relative exit velocity

2

12 44.7

2000t om

ww h

22

2 2

o

t

t

mvA

w

1 22

2

sinA

d l

2 2b tw k w

w2

u

c2

β2


• Chord of moving blades

• The chord of moving blades is taken within the

range b2 = 20 -80 mm

Chose the blade profile

(moving blade)

22

2 2

twM

kp v


P – 23 – 14 A

P – 23 – 14 б

Inlet angle β1

Outlet angle β2

subsonic

Inlet angleβ1

Moving blade

Outlet angle β2

sonicMoving blade


P – 23 – 14 P

Inlet angle β1

Outlet angle β2

SupersonicMoving blade

Number of Blades

• Number of blades

Where d is the mean diameter of the stage

t= b*t\ is the pitch

b is blade chord

t\ is the relative pitch

Ε is the partial admission degree

dZ

t

Coefficient of Energy Losses

• The coefficient of energy losses for the

selected profile is found by the formula:

ξ=k1k2k3ξ\

Where the coefficients k1, k2 and k3from the

curves that describe the characteristics of

fixed and moving blades

Internal Efficiency of the

Turbine Stage

Notes

• Some of the losses mentioned may be

absent in a particular turbine stage .

• For example, with the flow of superheated

steam there is naturally no energy loss

due to wetness of the steam.

• Loss due to partial admission does not

take place in turbine stage with the degree

of partiality ε =1


ExampleDesign of an intermediate stage of an impulse

turbine for the following initial data:

• Steam flow rate

147 kg/s

• Steam pressure before the stage

6.25 MPa

• Steam temperature before the stage 470o C

• Steam velocity at the entry of the stage 58

m/s

• Steam pressure behind the stage 5.5

MPa

Assoc. Prof. Abd El-hamied

Velocity Stage

Two-Rows Turbine Stage

Input data

• Mass flow rate (kg/s)

• Inlet steam conditions Po, to

• Available heat drop in the stage Ho kJ/kg


Required

• Stage Dimensions

• Blade profile

• Stage losses

• Stage efficiency

• Draw velocity diagram

• Draw sketch for the stage


Notes

• A slight reaction degree increases the

efficiency of a two row turbine stage and at

the same time increases the optimal

velocity ratio from 0.23 for a purely impulse

to 0.3 for a stage with ρt = 12 – 15 %

• The reaction degrees of a two row turbine

stages are usually not high (0.02 – 0.06)


• Low reaction degrees are chosen in

converging flow in the passages of moving

and guide blades and thus to diminish

energy losses.

• The total degree of reaction for two rows

of a two row stage usually does not

exceed 10 – 12 %.


Adopted degree of reaction

• ρm1 = 0.02

• ρg = 0.04

• ρm2 = 0.02


Available Heat Drop

• Hon = {1-(ρm1 +ρg +ρm2 )}Ho

• Hom1= ρm1 Ho

• Hog= ρg Ho

• Hom2= ρm2 Ho

S

h

Ho

Po

to

Hon

Hom1

Hom2

Hog

v1t

v4t


• Pressure behind the exit

• Theoretical Velocities

• The specific volume v1th is determined from h-s chart

• Theoretical Mach number is calculated

• Assume α1(12-14o)

• Calculate the speed ratio

• m: number of moving rows

1 44.7th onc H

mx

2

)cos( 1

11

1 1

tht

t

cM

kp v

• Calculate the blade speed (u)

u = x c1th

• Calculate the mean diameter (d)

• Assume discharge coefficient μ

• Calculate the exit area from fixed blades A1

11

1 1

t

th

mvA

c

160 tx cd

n

11

1sin

Al

d

•Optimum degree of partiality

ε= (0.29 : 0.34)√εl1Where εl1 cm

Discharge coefficients


Effect of wetness fraction of steam

y1 = 1-x1 at cascade exit on the

discharge coefficient

Discharge coefficients for superheated

steam in nozzle and moving blade

cascades depending on relative blade

height l/b and turning angle Δβ = 180

–(β1 – β2)

• Calculate the height of fixed blades

l1=εl1/ε

• Choose the profile chord (b1)

• Select the blade profile using M1th and α1

• Then the number of fixed blades is

\

1

1tb

dz

• The velocity coefficient of the nozzle

(kn) is determined from chart or

assume.

C1=kn.C1th

• Draw inlet velocity diagram

• Drive the inlet relative velocity w1

• The energy loss in the nozzle (Δhn) is

calculated


Velocity coefficients


2

2 1 144.7 / 2000th omw H w

The theoretical relative velocity at the exit

from the first moving blades is

Mach number M2t

Area of the first row of moving blade

22

2 2

tht

t

wM

kp v

th

th

o

bw

vmA

2

21

Exit height of first moving blades

l2 = l1+Δ

Δ: overlap (3:5 mm)

The angle of flow exit is

2

21

2 sinld

Ab

For the angle β2 and M2th select the blade

profile and chord b2, and relative pitch t\

• The number of blades in the first row moving

blade is\

2

2tb

dz

• Select the velocity coefficient kb1 from chart and

then calculate w2

• Construct the exit velocity diagram and then

determine c2 and α2

• Calculate the energy loss

kgkJkw

h bb /

2000

)1( 2

1

2

11


The theoretical velocity at the exit from the

guide blades is

2

3 244.7 / 2000th ogc H c

The theoretical relative velocity at the exit

from second moving blades is

2

4 2 344.7 / 2000th omw H w


• Mach numbers3

3

3 3

tht

t

cM

kp v

44

4 4

tht

t

wM

kp v

S

h

Ho

Po

to

Hon

Hom1

Hom2

Hog

v1t

v4t

p1

p4


Exit area from guide blades

th

th

o

gc

vmA

3

3

Exit area from second moving row

th

th

o

bw

vmA

4

42

Exit Areas

3

1

3 sinld

Ag

4

21

4 sinld

Ab

Calculation of number of blades

• Number of blades (guide blades)

• Number of blades (second moving blades)

\

tbg

dgz

\

4

4tb

dz

Exit height of blades

• Exit height of guide blades

l3 = l2+Δ

• Exit height of second moving blades

l4 = l3+Δ

Δ overlap (2.0:5.0 mm)

Internal efficiency

• Blade efficiency

• heat loss in nozzles

• heat loss in the first moving blades

o

ebgbno

b

wIIwIb

H

hhhhhH

or

c

ccu

21

2

1

)(2

kgkJkc

h nthn /

2000

)1( 22

1

kgkJkw

h bb /

2000

)1( 2

1

2

11


• heat loss in guide blades

• heat loss in the second moving blades

hc k

g

g

2

2 21

2000

( )kJ/kg

kJ/kghw k

b

b

2

3

2

2

21

2000

( )

• Energy loss due to disc friction

• Leakage losses

3

frfr

1 1th 1

k u dξ

πεsin α c l

o

b t eq

b b bo

1

(πd δ )m lξ η ρ 1.8 η

A dm

b

gggg

gzA

dk

1

)(

3

w ww

o 1 1th

N k 1 ε uξ m

N sinα ε c

Components of energy losses due to

partial admission

The relative internal efficiency of the stage

plfrbi

• Effective heat drop of the stage

Hi=Hoηi

• The internal power of the stage

Ni=moHi

Example


It is required to design a two-row governing stage for

the following initial data:

Steam flow rate 60 kg/s

Steam pressure before the stage 12.0 MPa

Steam temperature before the stag 450o C

Available heat drop of the stage 170 kJ/kg

Average diameter of the stage 0.95 m

First Three Lectures - Turbine Desine

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