It is about turbo machinary

1. STEAM TURBINE
BY
DR. K. S. RAMBHAD
1

2. Module-V
Syllabus
• Steam Turbine-
• Basic of steam turbine, Classification, compounding of turbine,
Impulse turbine –velocity diagram, Condition for max efficiency
Reaction turbine, Numerical on Simple Impulse turbine (De-Laval
turbine) of single stage only. Degree of reaction, Parson's turbine,
Condition for maximum efficiency, Numerical on Parson’s turbine
only.
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 A steam turbine is rotary machine which is designed to convert the
energy of high pressure and high temperature steam into
mechanical power.
 The operation of steam turbine wholly depends upon the dynamic
action of the steam.
 In this, the steam is first expanded in a set of nozzles where in the
pressure energy of the steam is converted into kinetic energy.
 The nozzles are fixed to the casting. If the resultant high velocity
steam is passed over the curved vanes, or blades, the steam
changes its direction and it would leave in the direction.

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 Due to this there is a change in momentum and it will exerts a
resultant force on the blades. If these blades are attached to a disc
on a rotor or shaft which is free to rotate, the resultant force would
cause the rotor to rotate. Thus the motive power is developed. The
principle of operation of steam turbine is shown in figure.
 A pair of ring of nozzle (fixed blades) fixed to the casing and a ring
of moving blades fixed to the turbine rotor is called a stage or a
turbine pair. Both the fixed and moving blades are so designed that
the steam jet shall not strike the blades but it should glide over in
the direction of blade surfaces.

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There are mainly of two types:
1. Impulse turbine:
 These turbines use the principle of impulse in which the kinetic
energy of steam obtained after passing over a ring of fixed nozzle is
used to exert a force on a ring of moving blades.
 The pressure of steam while passing over the moving blades remain
constant (neglecting losses) and its kinetic energy is converted into
mechanical work. The examples of impulse turbine, are De-Laval ,
Curtis and Rateau etc.
2. Reaction turbine:
 In case of a reaction turbines, there is a continuous pressure drop of
steam while passing over the rings of fixed and moving blades.
Accordingly, the moving blade passages are suitably designed for
steam to expand, therefore these blades also act as nozzles.
Type of Steam Turbine

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 The reactive force along with that due to change in
momentum of the steam provides the motive force for the
turbine to develop power. It follows that these turbines are
basically the impulse reaction turbine but in practice these are
called reaction turbines. Example of such a turbine is parson’s
reaction turbine.
 Mostly the steam turbines are axial flow type in which the
steam flows over the blades in direction parallel to the axis of
turbine rotor.
 The only important radial flow turbine is Lungstorm reaction
turbine is which the steam enters at the blades tip nearest to
the axis of the wheel and flows towards the circumference.

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Advantages of steam turbines
1.It is a rotary high speed machine.
2.It is compact and it has low weight to power ratio.
3.It has perfect balance and runs vibration free.
4.It needs less floor area.
5.It has low initial and maintenance cost.
6.It needs no internal lubrications, therefore its condensate
is not contaminated and it is suitable as feed water.
7.It is suitable for electrical generators.

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Classification of steam turbine
Classification of steam turbines may be done as
following:
1. According to action of steam
(a) Impulse turbine
(b) Reaction turbine
(c) Combination of both
2. According to direction of flow:
(a) Axial flow turbine
(b) Radial flow turbine

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3. According to number of stages
(a)Single stage turbine
(b)Multi stage turbine
4 .According to steam pressure at inlet of Turbine:
(a) Low pressure turbine
(b) Medium pressure turbine.
(c) High pressure turbine
(d) Super critical pressure turbine.

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5. According to method of governing:
(a) Throttle governing turbine.
(b) Nozzle governing turbine.
(c) By pass governing turbine.
6. According to usage in industry:
(a) Stationary turbine with constant speed.
(b) Stationary turbine with variable speed.
(c) Non stationary turbines.

12. Simple impulse Turbine
• An impulse turbine as the name indicates, is a
turbine which runs by the impulse of steam jet.
• In this turbine, the steam is first made to flow
through a nozzle. Then the steam jet impinges on
the turbine blade (which are curved like buckets)
and are mounted on the circumference of the
wheel.
• The steam jet after impinges glides over the
concave surface of the blades and finally leaves
the turbine.
• In the impulse turbine, the steam is expanded
within the nozzle and there is no any change in
the steam pressure as it passes over the blades.
12

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De-Laval Impulse Turbine
A De-Laval turbine is the simplest type of
impulse steam turbine, and is commonly
used. It has the following main
components.
1. Nozzle: It is a circular guide mechanism,
which guides the steam to flow at the
designed direction and velocity. It also
regulates the flow of steam. The nozzle is
kept very close to the blades, in order to
minimize the losses due to windage.

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2. Runner and blades: The runner of a De-
Laval impulse turbine essentially consists of
a circular disc fixed to a horizontal shaft. On
the periphery of the runner, a number of
blades are fixed uniformly. The steam jet
impinges on the buckets, which move in the
direction of the jet. This movement of the
blades makes the runner to rotate.
The surface of blades is made very smooth
to minimize the frictional losses. The blades
are generally made of special steel alloys.

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In most of the cases, the blades are bolted to
the runner disc. But sometimes the blades and
disc are cast as a single unit.
It has been experienced that all the blades do
not wear out equally with the time. A few of
them get worn out and damaged early and
need replacement. This can be done only if
the blades are bolted to the disc.
3. Casing: It is an air tight metallic case, which
contains the turbine runner and blades. It
controls the movement of steam from the
blades to the condenser, and does not permit
it to move into the space. Moreover, it is
essential to safeguard the runner against any
accident

16. Pressure and velocity of steam in an impulse
turbine
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 The pressure of steam jet is reduced in the nozzle and remains
constant while passing through the moving blade. The
velocity of steam is increased in the nozzle, and is reduced
while passing through the moving blades.
 Figure 22.2 shows the pressure and velocity graphs of the
steam in a simple impulse turbine while it flows in the nozzle
and blades. The pressure graph 1-2-3-4 represents steam
pressure at entrance of the nozzle, exit of the nozzle, entrance
of the blades and exit of the blades respectively. Similarly,
velocity graph 5-6-7-8 represents the velocity of steam at
entrance of the nozzle, exit of the nozzle, entrance of the
blades and exit of the blades respectively.

18. Compounding in Steam Turbine
• The compounding is the way of reducing the wheel or rotor
speed of the turbine to optimum value. It may be defined as
the process of arranging the expansion of steam or the
utilization of kinetic energy or both in several rings.
• There are several methods of reducing the speed of rotor to
lower value. All these methods utilize a multiple system of
rotors in series keyed on a common shaft, and the steam
pressure or jet velocity is absorbed in stages as the steam
flow over the blades.
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19. Different methods of compounding are:
• 1. Velocity Compounding
• 2. Pressure Compounding
• 3. Pressure Velocity Compounding
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20. Velocity Compounding of Impulse turbine
• There are number of moving blades separated by rings of fixed
blades. All the moving blades are keyed on a common shaft.
When the steam passed through the nozzles where it is
expanded to condenser pressure. Its Velocity becomes very
high. This high velocity steam then passes through a series of
moving and fixed blades.
• When the steam passes over the moving blades it's velocity
decreases. The function of the fixed blades is to re-direct the
steam flow without altering it's velocity to the following next
row moving blades where a work is done on them and steam
leaves the turbine with allow velocity as shown in diagram.
20

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22. Pressure Compounding of Impulse turbine
• There are the rings of moving blades which are keyed on a same shaft
in series, are separated by the rings of fixed nozzles.
• The steam at boiler pressure enters the first set of nozzles and
expanded partially. The kinetic energy of the steam thus obtained is
absorbed by moving blades.
• The steam is then expanded partially in second set of nozzles where
it's pressure again falls and the velocity increase the kinetic energy so
obtained is absorbed by second ring of moving blades.
• This process repeats again and again and at last, steam leaves the
turbine at low velocity and pressure. During entire process, the
pressure decrease continuously but the velocity fluctuate as shown in
diagram.
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24. Pressure velocity compounding of Impulse turbine
• This method of compounding is the combination of two
previously discussed methods. The total drop in steam pressure
is divided into stages and the velocity obtained in each stage is
also compounded. The rings of nozzles are fixed at the
beginning of each stage and pressure remains constant during
each stage as shown in figure.
• The turbine employing this method of compounding may be
said to combine many of the advantages of both pressure and
velocity staging By allowing a bigger pressure drop in each
stage, less number stages are necessary and hence a shorter
turbine will be obtained for a given pressure drop.
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26. Velocity triangle for moving blade of an impulse
turbine
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 We have already discussed that in
an impulse turbine, the steam jet
after leaving the nozzle impinges
on one end of the blade.
 The jet then glides over the inside
surface of the blade and finally
leaves from the other edge.
 It may be noted that the jet enters
and leaves the blades tangentially
for shockless entry and exit.
 Consider a steam jet entering a
curved blade after leaving the
nozzle at C.

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 Not let the jet glides over the
inside surface and leaves the
blade at D.
 Now let us draw the velocity
triangle at inlet and outlet tips of
the moving blade.
 Vb = Linear velocity of the
moving blade (AB)
 V = Absolute velocity of steam
entering the moving blade (AC)
 Vr = Relative velocity of jet to the
moving blade (BC). It is the
vectorial difference between Vb
and V

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 Vf = Velocity of flow at entrance of the
moving blade. It is the vertical component of
V.
 Vw = Velocity of whirl at entrance of the
moving blade. It is the horizontal component
of V.
  = Angle made by inlet relative velocity (Vr)
with the direction of motion of blade (Vb), it
is known as blade angle at inlet.
  = Angle with the direction of motion of the
blade at which the jet enters the blade, it is
called as inlet jet angle or nozzle angle.
 V1, Vr1, Vf1, Vw1, , , = Corresponding
values at exit of the moving blade.

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 It may be seen from the above, that the
original notation (i.e. V, Vr, Vf and Vw)
stand the inlet triangle.
 The notation with suffis1 (i.e. V1, Vr1,
Vf1 and Vw1) stand for the outlet
triangle.
 It may be noted that as the steam jet
enters and leaves the blades without
any shock i.e. tangentially therefore
shape of the blades will be such that Vr
and Vr1 will be along the tangents to the
blade at inlet and outlet respectively.
 The angle  is called the blade angle
inlet and angle  is the blade angle at
exit of the moving blade.

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 In figure, PC is the axis of the nozzle, which delivers
the steam jet with a high velocity (V) at an angle 
with the direction of motion of the blade.
 The jet impinges on a series of turbine blades
mounted on the runner disc.
 The axial component of V i.e. EC which does no
work on the blade, is known as velocity of flow
(Vf).
 It causes the steam to flow through the turbine
and also an axial thrust on the rotor.
 The tangential component of V i.e. AE is known as
velocity of whirl at inlet (Vw).
 The linear velocity or mean velocity of the blade
i.e. Vb is represented by AB in magnitude and
direction.
 The length AC represents the relative velocity Vr of
the steam jet with respect to the blade.

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 The jet now glides over and leaves the blade
with a relative velocity Vr1, which is
represented by DA.
 The absolute velocity of jet V1 as it leaves
the blade, is represented by DB inclined at
an angle  with the direction of the blade
motion.
 The tangential component of V1 ie BF is
known as velocity of whirl at exit (Vw).
 The axial component of V1 i.e. DF is known
as velocity of flow at exit (Vf1).
 Note:
 1. The inlet of velocity triangle represented
by AEC, Whereas the outlet triangle by AFD.
 2. The relations between inlet and outlet
velocity triangle (until and unless given) is Vr
= Vr1

33. Combined velocity triangle for moving blade
33
 In the last article, we have discussed the inlet
and outlet velocity triangles separately.
 For the sake of simplification, a combined
velocity triangle for the moving blade is
drawn, for solving problems on steam
turbines.
 1. first of all, draw a horizontal line and cut
off AB equal to velocity of blade (Vb) to some
suitable scale.
 2. now at B, draw a line BC at an angle  with
AB. Cut off BC equal to V (i.e. velocity of
steam jet at inlet of the blade) to the scale.
 3. join AC which represents the relative velocity at inlet (Vb). Now at A
draw a line AD at an angle  with AB

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 4. Now with A as centre and radius equal
to AC, draw an arc meeting the line
through A at D, such that AD = AC or Vr1
= Vr.
 5. Join BD, which represents velocity of
jet at exit (V1) to the scale.
 6. From C and D draw perpendiculars
meeting the line AB produced at E and F
respectively.
 7. Now EB and CE represents the velocity
of whirl and velocity of flow at inlt (Vw
and Vf) to the scale.
 Similarly, BF and DF represents the
velocity of whirl and velocity of flow at
outlet (Vw1 and Vf1) to the scale

35. Power produced by impulse turbine
• Consider an impulse turbine working under the action of a steam jet. Let us
draw a combined velocity triangle for the impulse turbines.
• Let,
• m = mass of steam flowing through the turbine in kg/s
• (Vw+Vw1)= change in the velocity of whirl in m/s
• We know that according to the Newton’s second law of motion, force in the
direction of motion of the blades.
• Fx = mass of steam flowing per second x change in the velocity of whirl
• Fx = m[Vw - (-Vw1)]….N
• Fx = m[Vw + Vw1)]….N……(i)
• Fx = m x EF…N
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36. • Work done in the direction of motion of blade
• WD = Force x Distance
• WD= m[Vw + Vw1)]Vb….N-m/s…….(ii)
• WD= m x EF x AB….N-m/s
• Therefore, power produced by the turbine
• P = m x EF x AB….Watt
• P = m[Vw + Vw1)]Vb…. Watt
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37. • Similarly, we can find out the axial thrust on the wheel which is due to
the difference of velocities of flow at inlet and outlet.
• Mathematically axial thrust on the wheel is given by
• Fy = Mass of steam flowing per second x change in the velocity of flow
• Fy = m(Vf-Vf1) = m(CE-DF) ….N……..(iii)
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38. • Note
• 1. in equation (i), the value of Vw1 is taken as negative because of
the opposite direction of Vw with respect to the blade motion. In
other words, when when point F in the velocity diagram lies on the
right of the point B then Vw1 is negative.
[Vw - (-Vw1)] = [Vw + Vw1)]
• If Vw1 is in the same direction with respect to the blade motion, then
Vw1 is taken as positive. In other words , when point F in the velocity
diagram lies on the left of point B then Vw1 positive, then change in
velocity of whirl will be given by
• [Vw - (+Vw1)] = [Vw - Vw1)]
38

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EX 1: In a De-Laval turbine, the steam enters the wheel through a nozzle with a velocity of
500 m/s and at an angle of 20o to the direction of motion of the blade. The blade speed is
200 m/s and the exit angle of the moving blade is 25o. Find the inlet angle of the moving
blade, exit velocity of steam and its direction and work done per kg of steam.
Solution: Given V= 500 m/s, = 20o
Vb = 200 m/s,  = 25o
Find , V1, 

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1. First of all, draw a horizontal line and cut of AB equal to 200 m/s, to a
some suitable scale, representing the blade speed Vb.
2. Now at B, draw a line BC at an angle of 20o (nozzle angle, ) and cut off BC
equal to 500 m/s to the scale to represents the velocity of steam jet
entering the blade (V).
3. Join AC, which represents relative velocity at inlet (Vr).
4. At A draw a line AD at angle of 25o (exit angle of the mobbing blade ).
Now with A as centre, and radius equal to AC, draw an arc meeting the line
through AD.
5. Join BD which represents the velocity of steam jet at outlet (V1).
6. From C and D draw perpendicular meeting the line AB produced at E and F
respectively.
7. CE and EF represents the velocity of flow at inlet (Vf) and outlet (Vf1)
respectively.

41. Graphical Method
• Inlet angle of moving blade,  = 32o
• Direction of the exit steam,  = 59o
• Exit velocity of Steam, V1 = BD = 165 m/s
• Vw = BE = 470 m/s
• Vw1 = BF = 90 m/s
• Work done per kg of steam
WD = m[Vw + Vw1)]Vb = 1x[470+90]x200 = 112000 N-m
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42. Analytical Method
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Ex 2: Steam at 5 bar 200oC is first made to pass through nozzles. It is then supplied to an
impulse turbine at the rate of 30 kg/minute. The steam is finally exhausted to a condenser at
0.2 bar. The blade speed is 300 m/s. the nozzles are inclined at 25o with the direction of motion
of the blade and the outlet blade angle is 35o. Neglection friction, find the theoretical power
developed by the turbine.
Solution: Given
P1 = 5 Bar, T1 = 200oC, m = 30 kg/min = 0.5 kg/s, P2 = 0.2 Bar, Vb = 300 m/s, =25o, =35o
First of all, let us draw Mollier diagram for the
flow of steam through the nozzle. From this
diagram, we find that heat drop during flow
Velocity of steam at inlet of the blade (outlet of the nozzle)
𝑉 = 44.72 ℎ𝑑
𝑉 = 44.72 510
V = 1010 m/s

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Now draw the combined velocity triangle:
1. First of all, draw a horizontal line and cut off
AB equal to 300 m/s, to some suitable scale,
to represent the blade speed (Vb).
2. Now draw inlet velocity triangle ABC on the
base AB with  = 25o and V = 1010 m/s. from
the velocity triangle we find that Vr = 850 m/s
to the scale.
3. Similarly, draw the outlet velocity triangle
ABD on the same base AB with  = 35o and
Vr1 = Vr = 850 m/s to the scale.
4. From C and D draw perpendicular to meet
the line AB produced at E and F.
By measurement from the velocity diagram, we
find that
EF = Vw+Vw1 = 1400 m/s

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46. Effect of friction on combined velocity triangle
• In the last article we have discussed that the relative velocity of
steam jet is the same at the inlet and outlet tips of the blade.
• In other words, we have assumed that the inner side of the curved
blade offers no resistance to the steam jet.
• But in actual practice, some resistance is always offered by the blade
surface to the gliding steam jet, whose effect is to reduce the relative
velocity of the jet, i.e. to make Vr1 less than Vr.
• The ratio of Vr1 to Vr is known as blade velocity coefficient or
coefficient of velocity or friction factor.
• Usually it is denoted by k.
• Mathematically blade velocity coefficient
• K=
𝑉𝑟1
𝑉𝑟 46

47. • It may be noted that the effect of friction on the combined velocity
triangle will be to reduce the relative velocity at outlet (Vr1) as shown
in figure.
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Ex 3: The velocity of steam, leaving the nozzles of an impulse turbine, is 1200 m/s and the
nozzle angle is 20o. The blade velocity is 375 m/s and the blade velocity coefficient is 0.75.
Assuming no loss due to shock at inlet, calculate for a mass flow of 0.5 kg/s and symmetrical
blading: (a) blade inlet angle; (b) driving force on the wheel; (c) axial thrust on the wheel; and
(d) power developed by the turbine.
Solution: Given
V = 1200 m/s,  = 20o, Vb = 375 m/s, K=Vr1/Vr = 0.75, m = 0.5 kg/s ,  =  for symmetrical
bladding
1. First of all, draw a horizontal line, and cut off
AB equal to 375 m/s to some suitable scale
representing the velocity of blade (Vb).
2. Now at B, Draw a line BC at an angle of 20o
(nozzle angle ) and cut off BC equal to
1200 m/s to the scale to represent the
velocity of steam jet entering the blade (V)

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3. Join CA, which represents the relative
velocity at inlet (Vr). By measurement,
we find that CA = Vr = 860 m/s.
Now cut off AX equal to 860 x 0.75 = 645 m/s
to the scale to represents the relative velocity
at exit (Vr1)
4. AT A, draw a line AD at an angle  equal to the angle , for symmetrical blading.
Now with A as centre and radius equal to AX, draw an arc meeting the line through A at
D, such that AD = Vr1.
5. Join BD, Which represents the velocity of steam jet at outlet (V1),
6. From C and D, draw perpendiculars meeting the line AB produced at E and F
respectively CE and DF represents velocity of flow at inlet (Vf) and outlet (Vf1)
respectively.

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Following values are measured from the velocity diagram
 (a) Blade angle at inlet,  = 29o,
 Whirl velocity at inlet, Vw= BE =1130m/s,
 Whirl velocity at inlet, Vw1 = BF = 190 m/s,
 Flow velocity at inlet, Vf = CE = 410 m/s and
 Flow velocity at outlet, Vf1 = DF = 310 m/s
(b) Driving force on the wheel
Fx = m (Vw + Vw1) = 0.5 (1130+190) = 660 N
(c) Axial thrust on the wheel
Fy = m (Vf + Vf1) = 0.5 (410+310) = 50 N
(d) Power developed by the turbine
P = m (Vw + Vw1)Vb = 0.5 (1130+190)375 = 247500 W

51. Reaction Turbine
• In a reaction turbine, steam enters
the rotor under pressure and flows
over the blades. While gliding,
steam propels the blades and makes
them to move. The rotor is rotated
by reactive forces of steam jets. The
motion of blades is similar to recoil
of a gun. Pure reaction turbine is not
possible in actual practice and all
turbines employ both impulse and
reaction principles. The driving
force is partly impulsive and partly
reactive.
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52. 52
PARSON'S REACTION TURBINE
A Parson's reaction turbine is the simplest type of reaction turbine and is
commonly used. The main components of it are:
1. Casing.
2. Guide mechanism.
3. Runner.
4. Draft tube.
Casing: The casing is an air tight metallic case in which steam from boiler under
high pressure is distributed around the fixed blades which are positioned at the
entrance. The casing is so designed that steam enters the fixed blades with
uniform velocity.

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Guide Mechanism: The guide mechanism consists of fixed or guide blades.
They allow the steam to enter the rotor without shock and they allow required
quantity of steam to enter the turbine. The guide blades may be opened or
closed by a regulating shaft which allows steam to flow according to the need.
Runner: The runner consists of moving blades. These blades are designed
properly to allow steam to enter and leave the blades without shock. The
steam after passing through the rotor flows to condenser through a draft tube.
It minimizes losses due to eddies.
Draft tube: The steam after passing through the runner flows into the
condenser through a tube called draft tube. It may be noted that if this tube is
not provided in the turbine, then the steam will move freely and will cause
steam eddies.

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VELOCITY DIAGRAM OF A REACTION TURBINE
 Not let the jet glides over the inside
surface and leaves the blade at D.
 Now let us draw the velocity triangle
at inlet and outlet tips of the moving
blade.
 Vb = Linear velocity of the moving
blade (AB)
 V = Absolute velocity of steam
entering the moving blade (AC)
 Vr = Relative velocity of jet to the
moving blade (BC). It is the vectorial
difference between Vb and V

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 Vf = Velocity of flow at entrance of the
moving blade. It is the vertical
component of V.
 Vw = Velocity of whirl at entrance of
the moving blade. It is the horizontal
component of V.
  = angle which the relative velocity of
jet to the moving blade (Vr) makes with
the direction of moving of the blade.
  = Angle with the direction of motion
of the blade at which the jet enters the
blade.
 V1, Vr1, Vf1, Vw1, ,, =
Corresponding values at exit of the
moving blade.

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 It may be seen from the above, that the
original notation (i.e. V, Vr, Vf and Vw)
stand the inlet triangle.
 The notation with suffis1 (i.e. V1, Vr1,
Vf1 and Vw1) stand for the outlet
triangle.
 It may be noted that as the steam jet
enters and leaves the blades without
any shock i.e. tangentially therefore
shape of the blades will be such that Vr
and Vr1 will be along the tangents to the
blade at inlet and outlet respectively.
 The angle  is called the blade angle
inlet and angle  is the blade angle at
exit of the moving blade.

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Combined velocity triangle for moving blade
1. First of all, draw a horizontal line
and cut off AB equal to the velocity
of blade (Vb), to some suitable
scale.
2. Now at B, draw a line BC at
angle  with AB. Similarly at A,
draw a line AC at angle  with EA
meeting the first line at C. now CA
and CB represents the relative
velocity (Vr) and absolute velocity
(V) of steam at inlet to the scale.
3. At A, draw a line AD at an angle  (such that =) with AB. Similarly at B draw a line BD at
an angle  (such that =) with AB meeting the first line at D. now DA and DB represents the
relative velocity (Vr1) and absolute velocity (V1) of steam at outlet, to the scale.

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4. From C and D draw perpendiculars
meeting the line AB produced at E and F.
5. Now EB and CE represents the velocith
of whirl and velocity of flow at inlet (Vw
and Vf) to the scale. Similarly BF and DF
represent the velocity of whirl and
velocity of flow at outlet (Vw1 and Vf1)
to the scale.
Note: A careful study of the combined velocity diagram of Parson’s reaction
turbine will reveal that it is symmetrical about the central line. Therefore
following relations exist in the combined velocity diagram.
Vf=Vf1; V=Vr1; Vr = V1; EA = BF

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Power produced by a reaction turbine
Consider a reaction turbine working under the action of steam pressure. Let
us draw a combined velocity triangle for the reaction turbine as shown in the
figure.
Let, m= mass of the steam flowing through the turbine in kg/s and
(Vw+Vw1)= change in velocity of whirl in m/s
We know that according to the Newton’s second law of motion, force in the
direction of motion of the blades,
Fx = mass of steam flowing per second X change in the velocity of whirl
Fx = m[Vw – (-Vw1)] = m[Vw + Vw1] ……….N ……..(i)
(Note: Vw1 is taken negative due to the opposite direction of Vw with respect
to the blade motion)

60. 60
And work done in the direction of motion of the blades
WD = Force X Displacement
WD = m[Vw + Vw1] Vb …………Nm/s
Power produced by the turbine
P = m[Vw + Vw1] Vb …………Watts ( 1 Nm/s = 1 Watt)
Similarly we can find out the axial thrust on the wheel, which is due to
difference of velocities of flow at inlet and outlet. Mathematically, axial
thrust,
Fy = mass of steam flowing per second x change in the velocity of flow
Fy = m(Vf – Vf1) ……………………N ………. (ii)

61. 61
Degree of Reaction
Definition: The ratio of the enthalpy drop in the moving blades to the total
enthalpy drop in the stage is known as degree of reaction.
Mathematically,
𝐷𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 =
Enthalpy drop in the moving blades
Total enthalpy drop in the stage
=
ℎ2−ℎ3
ℎ1−ℎ3
The enthalpy drop in the fixed blades per kg of
steam is given by
ℎ1 − ℎ2 =
𝑉2−𝑉1
2
2000
𝑘𝐽/𝑘𝑔
Enthalpy drop in the moving blades,
ℎ2 − ℎ3 =
𝑉𝑟1
2−𝑉𝑟
2
2000
𝑘𝐽/𝑘𝑔

62. 62
Total enthalpy drop in the stage
ℎ1 − ℎ3 = (ℎ1 − ℎ2) + (ℎ2 − ℎ3)
ℎ1 − ℎ3 =
𝑉2−𝑉1
2
2000
+
𝑉𝑟1
2−𝑉𝑟
2
2000
=
2(𝑉𝑟1
2
−𝑉𝑟
2)
2000
= 2(ℎ2 − ℎ3) kJ/kg
(Note for Parson’s turbine, V = Vr1 and V1= Vr)
Hence,
Degree of reaction =
ℎ2 − ℎ3
ℎ1 − ℎ3
=
ℎ2 − ℎ3
2(ℎ2 − ℎ3)
=
1
2
= 0.5 𝑜𝑟 50%
Thus we see that a Parson’s reaction turbine is a 50 percent reaction
turbine.

63. 63
Ex1: In one stage of a reaction steam turbine, both the fixed and moving blades have inlet
and outlet blade tip angles of 35o and 20o respectively. The mean blade speed is 80 m/s
and the steam consumption is 22500 kg per hour. Determine the power developed in the
pair, if the isentropic heat drop of the pair is 23.5 kJ/kg
Solution: Given  =  = 35o,  =  = 20o,Vb = 80 m/s, m = 22500 kg/h = 6.25 kg/s,
hd= 23.5 kJ/kg
1. First of all, draw a horizontal line and cut off AB equal to 80 m/s (Vb) to some suitable
scale.
2. Now at B, draw a line BC at an angle  = 20o, with AB, similarly, at A draw a line AC at
an angle  = 35o with BA meeting the first line at C.
3. At A, draw a line AD at angle  = 20o
(because  =  ) with AB. Similarly, at B
draw a line BD at an angle  = 35o (because
 =  ) with AB meeting the first line at D.

64. 64
4. From C and D draw perpendiculars meeting the line AB produced at E and
F.
By measurement, we find that the change in the velocity of whirl,
(Vw+Vw1) = 235 m/s
We know that power developed in the pair,
P = m (Vw + Vw1) Vb = 6.25 X 235 = 117500 W = 1117.5 kW

65. 65
Ex2: A Parson’s reaction turbine, while running at 400 rpm consumes 30 tonnes of steam
per hour. The steam at a certain stage is at 1.6 bar with dryness fraction of 0.9 and the
stage develops 10 kW. The axial velocity of flow is constant and equal to 0.75 of the blade
velocity. Find mean diameter of the drum and the volume of steam flowing per second.
Take blade tip angle at inlet and exit as 35o and 20o respectively.
Solution: Given N = 400 rpm, m = 30 T/h = 8.33 kg/s, p = 1.6 bar, x = 0.9, P= 10 kW
=10000 kW, Vf = 0.75 Vb,  =  = 35o,  =  = 20o,
1. First of all, draw a horizontal line and
cut off AB equal to 25mm to represent
the blade velocity (which is required to
be found out)
2. Now at B, draw a line BC at an angle 
=20o with AB. Similarly at A, draw a
line AC at an angle = 35o meeting the
first line at C

66. 66
Let, D = Mean diameter of the drum
Now let us draw the combined velocity triangle,
3. At A, draw a line AD at an angle  = 20o with AB. Similarly at B, Draw a
line BD at an angle  = 35o with AB meeting the first line a D.
4. From C and D draw perpendiculars meeting the line AB produced at E
and F.
By measurement we find that the change in the velocity of whirl
(Vw+Vw1)=73.5 mm
(𝑉
𝑤 + 𝑣𝑤1)
𝑉𝑏
=
73.5
25
= 2.94
(Vw+Vw1)=2.94Vb

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We know that the power developed (P)
10 x 103 = m(Vw+Vw1)Vb =8.33 x 2.94 Vb x Vb = 24.49 (Vb)2
Vb = 20.2 m/s
We know that the blade velocity (Vb)
20.2 =
𝜋𝐷𝑁
60
20.2 =
𝜋𝐷400
60
= 20.94 𝐷
D = 0.965 m = 965 mm
Volume of steam flowing per second
From steam tables, corresponding to a pressure of 1.6 bar, we find that
specific volume of steam
Vg= 1.091 m3/kg
Volume of steam flowing per second = m xVg
= 8.33 x 0.9 x 1.091 = 8.18 m3/s