Steam Turbine 1
Steam Turbine 1
Steam Turbine 1
14
Steam Turbines
14.1 INTRODUCTION
Steam turbines date back to 120 B.C. when the first steam turbine was developed by Hero of Alexandria.
Subsequently number of steam turbines came up but the practically successful steam turbine appeared
at the end of nineteenth century when Gustaf De Laval designed a high speed turbine built on the
principle of reaction turbine in 1883. Before this in 1629 G. Branca developed the first impulse turbine.
Brancas impulse turbine and Heros reaction turbine are shown in Fig. 14.1.
In nineteenth century some more steam turbines were developed by Sir Charles A. Parsons and
C.G. Curtis which gave a filip to the development to the modern steam turbine. Over the period of time
the modern steam turbines evolved with capacity from few kilowatts to 350,000 kW and in speed from
1000 rpm to 40,000 rpm. Steam turbines offer the advantages over other prime movers in terms of
simplicity, reliability and low maintenance costs. Reciprocating steam engines use pressure energy of
steam while steam turbines use dynamic action of the steam. Steam turbines require less space as
compared to diesel engine or steam engine and also the absence of reciprocating parts & reciprocating
motion in steam turbine results in lesser vibrations and lighter foundation. In steam turbine the expanding steam does not come into contact with lubricant and so exhaust steam leaves uncontaminated.
Diaphragm: Diaphragm is attached to the casing containing the nozzles and performs function of
confining steam flow to nozzle passage.
Packing: Packing is provided for preventing the leakage across the annular space between the
diaphragm and shaft, casing and shaft. Packing is provided in the form of carbon rings or labyrinth
glands. Carbon rings present an effective seal against shaft leakage in small turbines and are extensively
used with labyrinth and water seals for preventing shaft leakage in larger turbines. Carbon ring consists
of a ring of carbon rectangular in cross section and usually divided into four segments. Rings snugly fit
into a recess in casing and are kept tight against shaft by means of garter spring. Labyrinth seals consist
of series of thin strips fixed to the casing or other stationary member and arranged so as to maintain the
smallest possible clearance with the shaft. The labyrinth seals have small restrictions that increase the
velocity of leaking fluid only to have it dissipated in the pockets, thereby throttling the fluid. Figure 14.3
shows the straight and stepped labyrinth seals. The tips of the strips are extremely thin so that if rubbing
occurs the tip gets worn away without damaging the shaft. These labyrinth seals do not prevent complete leakage of fluid. These are effective in only reducing leakage. Complete leakage prevention is done
in association with other seals as carbon seals described earlier.
Steam chest: This is the steam supply chamber which houses steam before being supplied to
nozzles.
The mechanism of impulse and reaction forces getting generated is discussed here. From Newtons second law we know,
F = ma
dV
F = (m)
dt
F I
H K
In case of Fig. 14.5 (a) the impulse force is available due to change in magnitude of velocity and
shall be given by the product of mass flow rate and change in velocity. In case (b) the impulse force is
generated due to change in direction of velocity and if the blade is stationary and frictionless then there
shall be no decrease in magnitude of velocity.
Reaction force is available when the tangential velocity of fluid is increased and is opposite in
reference to the direction of velocity.
In the case shown in Fig 14.5 (b), the total force exerted on blade is actually a combination of
impulse and reaction. Impulse force is available in the entrance half of the blade where jet impinges
causing a force to right.
While in the exit half, the leaving jet exerts a reactive force on the blade which also acts to the
right. Combined effect of the two forces on the impulse blade is arbitrarily called impulse force.
Reaction force available due to increase in tangential velocity of fluid can be seen in case of nozzle
due to acceleration of fluid.
Detailed discussion on the reaction turbines running on shaft work resulting from reaction forces
is given along with reaction turbines.
(b) Based on the cylinder flow arrangement: Steam turbines may be classified based upon the
flow arrangement into following types.
(i) Single flow single casing turbine
(ii) Double flow single casing turbine
(iii) Cross flow compound turbine with single flow
(iv) Cross flow compound turbine with double flow
(v) Triple cross flow compound turbine with double flow
Various arrangements in the above listed turbines are shown by line diagram in Fig 14.7 Single
cylinder turbines are the one which have all the stages enclosed in one cylinder while in multi cylinder
(c) Based on direction of flow: Steam turbines can be classified based on the direction of flow by
which steam flows through turbine blading. Steam turbines can be:
(i) Radial flow turbine
(ii) Tangential flow turbine
(iii) Axial flow turbine
Radial flow turbines were developed by B.F. Ljungstorm of Sweden as shown in Fig. 14.8. Radial
flow turbine incorporates two shafts end to end and can be of suitably small sizes. Radial flow turbines
can be started quickly and so well suited for peak load and used as stand by turbine or peak load
turbines. These are also termed as Ljungstrom turbines.
In radial flow turbines the steam is injected in middle near shaft and steam flows radially outwards through the successive moving blades placed concentrically. In radial flow turbines there are no
stationary blades so pressure drop occurs in moving blade passage. Concentric moving blades rings are
designed to move in opposite directions.
In tangential flow turbines the nozzle directs steam tangentially into buckets at the periphery of
single wheel and steam reverses back and re-enters other bucket at its periphery. This is repeated
several times as steam follows the helical path. Tangential flow turbines are very robust but less efficient.
In axial flow turbines steam flows along the axis of turbine over blades. These axial flow turbines
are well suited for large turbo generators and very commonly used presently.
(d) Based on number of stages: Steam turbines can also be classified based upon the number of
stages in steam turbines i.e. depending upon the amount of heat drop. It can be:
(i) Single stage turbine
(ii) Multi stage turbine.
Single stage turbines have the expansion occurring in single stage while in multi stage turbines the
expansion occurs in more than one stages of turbine. When expansion occurs in two stages it is called
double stage turbine and with expansion occurring in three stages it is called triple stage turbine.
Pass out turbines are those in which certain quantity of steam is continuously extracted for the
purpose of heating and allowing remaining steam to pass through pressure control valve into the low
pressure section of turbine. Pressure control valve and control gear is required so as to keep the speeds
of turbine and pressure of steam constant irrespective of variations of power and heating loads.
(f) Based on speed of turbine: Steam turbines can be classified based upon the steam turbine as
low speed, normal speed and high speed turbines as given below.
(i) Low speed steam turbine.
(ii) Normal speed steam turbine.
(iii) High speed steam turbine.
Low speed turbines are those steam turbines which run at speed below 3000 rpm. Normal speed
steam turbines are those turbines which run at speed of about 3000 rpm while high speed steam turbines
are the one which run at more than 3000 rpm.
(g) Based on pressure in steam turbines: Steam turbines can also be classified based upon the inlet
pressure of steam turbine as follows:
High pressure steam from boiler enters the nozzle through pipings and leaves nozzle at predefined
angle so as to smoothly flow over the moving blades. Steam velocity gets increased during its flow
through nozzle due to its expansion occurring in it. During the passage of steam over the moving blades
steam undergoes change in its direction while losing the velocity and thus causing rotation of moving
blade ring mounted on shaft.
Simple impulse turbine is used where small output at very high speed is required or only a small
pressure drop is available. These are not suited for applications requiring conversion of large thermal
energy into work. Speed of operation of turbine can be regulated by compounding of impulse turbine
discussed ahead. Compounding of steam turbine is required as in case of simple impulse turbine, the
single stage may offer speed of the order of 30,000 rpm which can not be directly used for any
engineering application and needs to be reduced. Also such a high speed shall induce large stresses in the
blades. Compounding is a thermodynamic means for reducing the speed of turbine where speed reduction is realized without employing a gear box.
In velocity compounded impulse turbine the high velocity steam from boiler enters the first ring
of stationary nozzles and undergoes the complete pressure drop as desired in a stage along with increase
in velocity. Low pressure and high velocity steam leaving nozzle enters the moving blade ring where a
part of velocity drop takes place while pressure drop does not occur due to symmetrical blade profile.
Figure 14.15 gives the inlet and outlet velocity diagrams at inlet edge and outlet edge of moving
blade along with the combined inlet and outlet velocity diagram for a stage of simple impulse turbine.
The notations used for denoting velocity angles and other parameters during calculations are explained
as under, (SI system of units is used here).
U = Linear velocity of blade =
pdN
, m/s where d = mean diameter of wheel in m
60
N = Speed in rpm.
C1 = Absolute velocity of steam at inlet to moving blade or velocity of steam leaving nozzle.
(Absolute velocity is the velocity of an object relative to the earth)
C2 = Absolute velocity of steam at exit of moving bade.
Fig. 14.15
C1w = Whirl velocity at inlet to moving blade or tangential component of absolute velocity at inlet
to moving blade.
C2w = Whirl velocity at exit of moving blade or tangential component of absolute velocity at exit
of moving blade.
C1a = Flow velocity at inlet to moving blade or axial component of absolute velocity at inlet to
moving blade.
C2a = Flow velocity at exit of moving blade or axial component of absolute velocity at exit of
moving blade.
V1 = Relative velocity of steam at inlet of moving blade (Blade velocity at inlet) (Relative velocity is the absolute velocity of one moving object compared with absolute velocity of other
object.)
V2 = Relative velocity of steam at exit of moving blade (Blade velocity at exit).
m = Mass of steam flowing over blade
r = Ratio of linear velocity of blade and absolute velocity at inlet of moving blade =
U
C1
V2
V1
DCw = DV w
FD = m (C2 cos a2 + C1 cos a1)
= m (V1 cos b1 + V2 cos b2)
This driving thrust can be used for getting the rate of work done on the rotor.
Rate of work done = W = FD U
W = mDCw.U
Work done per unit of steam mass flow, w = UDCw
FG
H
IJ
K
R F V I F cos b IJ UV
U) S1 + G J G
T H V K H cos b K W
= mU V1 cos b1 1 +
= mU (C1 cos a1
V2 cos b 2
V1 cos b 1
W = m U (C1 cos a1 U ) {1 + K C}
where
V2
V1
cos b 2
cos b 1
For perfectly smooth and symmetrical blade both K and C shall have unity value.
i.e. K = 1, C = 1
Therefore for simple impulse turbine stage having perfectly smooth and symmetrical blade, rate of
work done,
C = Ratio of cosines of blade angles =
W = 2 mU (C1 cos a1 U)
From velocity diagram also the rate of work done per unit of steam mass flow w, can be
estimated as below,
w = U DCw
Combined velocity diagram if drawn to the scale gives, U = AB, DCw = EF. Thus length EF and
AB shall give D Cw and U respectively.
Hence, w = AB EF, rate of work done per unit of steam mass flow
Rate of work done W = m AB EF . The suitable scale factors are to be used while using
lengths.
The velocity diagram at inlet, outlet or combined velocity diagram can be suitably drawn to the
scale and various parameters be obtained by measuring respective geometrical entity i.e. length and
angle.
The work available at rotor can also be obtained using steady flow energy equation between
section 1 and 2. Assuming no change in potential energy from inlet to exit across the moving blade and
no heat interaction across the stage, the S.F.E.E. can be given as;
FG
H
m h1 +
C12
2
IJ = m FG h + C IJ + W
K H 2K
R
F C - C IJ UV
W = m Sa h - h f + G
H 2 KW
T
2
2
2
1
2
2
In case of impulse stage the change in enthalpy from section 1 to 2 can be given by the change in
kinetic energy associated with relative velocity from 1 to 2.
FG V - V IJ
H2 2K
1
h ) = bV - V g
2
2
2
(h1 h2) =
or,
(h1
2
1
2
2
2
1
Substituting in W,
Rate of work done in a stage:
W=
lbV
m
2
2
2
g b
gq
m
2
lbC
2
1
- C22
m
{2U DCw}
2
or, W = mUDCw
Axial thrust: Axial component of velocity or flow velocity change causes creation of axial thrust.
Axial thrust due to change in momentum because of change in flow velocity.
Axial thrust = m (C1a C2a)
= m (C1 sin a1 C2 sin a2)
Axial thrust, Fa = m DCa
From velocity diagrams the change in axial velocity is given by length DG. Length, DG = EC FD
Axial thrust = m DG = m (EC FD)
Diagram efficiency or blading efficiency: Diagram efficiency or blading efficiency refers to the
ratio of work done to the energy supplied to rotor in a stage. In case of impulse turbine this energy
supplied to rotor in a stage can be given by the kinetic energy supplied at inlet of moving blade. Assum-
FG
H
ing no expansion in moving blades the energy supplied to rotor can be given by m
IJ
K
C12
. Rate of work
2
hD =
m U DCw
C2
m 1
2
2 U D Cw
C12
2 U DCw
C12
Stage efficiency or Gross efficiency: Stage efficiency refers to the ratio of rate of work done and
energy supplied to the stage. Energy supplied to the stage can be accounted by the change in enthalpy
between section 0 and 1 i.e. inlet of nozzle to exit of nozzle. Stage efficiency is thus the output of stage
divided by the available energy for the stage.
Energy supplied to stage = m (h0 h1)
hs =
Stage efficiency,
Stage efficiency =
U DCw
h0 - h1
Nozzle efficiency: Nozzle efficiency refers to the ratio of kinetic energy available and the enthalpy
change occurring across the nozzle i.e. between inlet and outlet (sections 0 and 1).
C12
2
Nozzle efficiency, hN =
m h0 - h1
m
Nozzle efficiency =
C12
2 h0 - h1
Combining the stage efficiency, diagram efficiency and nozzle efficiency, it can be given that;
Stage efficiency = Diagram efficiency Nozzle efficiency
Overall efficiency: The overall efficiency of stage can be given by the ratio of work delivered at
turbine shaft to the energy supplied to the stage.
Overall efficiency, ho =
hD =
2U DCw
C12
hD =
2U C1 cos a 1 - U (1 + KC )
C12
hD = 2
FG U IJ FG cos a
HC KH
Here
U
C1
IJ (1 + KC)
K
U
, is non-dimensional form of velocity. Let us denote by r
C1
U
, so, hD = 2r (cos a1 r) (1 + KC)
C1
Diagram efficiency can be optimized with respect to non-dimensional velocity denoted by r. This
non-dimensional velocity r is also called as blade-steam velocity ratio or blade speed-steam velocity
ratio or blade-steam speed ratio.
Differentiating hD with respect to r, we get,
r =
i.e.
dh D
= 2 (cos a1 2r) (1 + KC)
dr
For a perfectly smooth and symmetrical blade, K = 1, C = 1
dh D
= 4 (cos a1 2r)
dr
Equating first differential to zero;
cos a1 2r = 0
r=
cos a 1
2
cos 2 a 1 (1 + KC )
2
Corresponding to r =
cos a 1
, the rate of work done can be given as the maximum rate of work
2
done. Mathematically,
Maximum rate of work done can be obtained by substituting
r=
Rate of work,
cos a 1
2U
U
=
or C1 =
C1
cos a 1
2
14.6
Impulse turbine blade height is a very important parameter. Blade height is a function of total annular
area required for the flow of fluid. It depends upon the mass flow rate through the section, specific
volume of steam at that section,
area of section through which steam is passing and flow velocity at the section. In estimating the
net annular area available for flow, edge thickness and blade angles are taken into account. The nozzle
passage and two row Curtis stage are shown in Fig 14.17.
Here estimation of both nozzle height and blade height has been described ahead.
For the nozzle passage shown the area available for flow at exit of one nozzle passage can be given
as A;
A = whn, where w is the width of flow passage at exit at mean nozzle height and hn is nozzle height.
The width of flow passage at exit at mean nozzle height can be given as,
w = (p sin a t),
where p is nozzle pitch at mean nozzle height and t is edge thickness of nozzle and a is nozzle
angle.
Therefore, A = (p sin a t)hn
In the absence of information about edge thickness considerations above equation can be used to
get nozzle height.
For full peripheral admission when nozzle diaphragm is occupied by nozzles, the number of
nozzles can be obtained as,
pd
n=
p
where d is the diameter of mean nozzle ring and n is number of nozzles.
Edge thickness in nozzle calculations can be accounted by edge thickness factor (k) denoted as,
k=
FG p sin a - t IJ
H p sina K
Thus, A = k hnp sin a , can be used to get nozzle height. Considering the total number of nozzles,
the total nozzle area can be given as,
An = nA
An = pd hn k sin a
hb
C
C sin a
= 1
= 1a = 1
V1 a
V1 sin b
hn
From velocity diagram at blade inlet it can be seen that, C1 sin a = C1a = V1 sin b = V1a.
Hence hb = hn, ideally but as described earlier the height at blade entrance is slightly increased,
which is also called blade step-up or overlap. Similarly for two row Curtis stage considered here the
blade height can be related as,
hb1 =
hn C1a
V2 a
hb2 =
hn C1a
V3 a
hb3 =
hn C1a
V4 a
Step up or overlap is arbitrarily taken and varies from 0.2 cm in high pressure stages to 2 cm in
low pressure stages of large turbines. Here in this calculation no consideration is made for increase in
specific volume of fluid in blade passage due to reheat although it shall be there but may be neglected on
account of being small.
Maximum height of blade is restricted by the stresses getting generated due to bending and
centrifugal forces.
Alternatively, the blade height can be estimated by the volume flow rate considerations. Moving
blade height at exit can be given as below;
mv =
FG l IJ (p
Hp K
where, l is the length of arc covered by nozzle and pb is the distance between two consecutive blades,
tb is edge thickness of blade as shown in Fig 14.18. Here
Steam turbines are called full admission turbines when the nozzle coverage of blades is complete
i.e. in such case l = pd, where d is mean blade diameter of the rotor.
Fig. 14.19 Stage of velocity compounded impulse turbine and velocity diagram for the stage.
Here high pressure steam first enters the nozzle and steam leaving nozzle enters the row of
moving blades. Steam leaving moving blades enters the fixed blade row. Fixed blades just act as guide
hd =
hd =
Total work
Energy supplied
k2 m U a2C cosa
FG m C IJ
H 2 K
1
- 4U
fp
4U
(2C1 cos a1 4U)
C12
2
1
U
, = r, i.e. blade speed to steam velocity ratio
C1
hd = 4 r (2cos a1 4r)
Differentiating diagram efficiency with respect to r and applying conditions to get the optimum
value of r for maximum diagram efficiency;
dh d
d
=
(8r cos a1 16r2)
dr
dr
Substituting
dh d
= 0
dr
or,
r opt =
cos a 1
or 4U = C1 cos a1
4
1
th
4
of the total work from stage. It can be extrapolated from here that if there are more than two rows of
moving blades in a velocity compounded turbine stage i.e. three, four or more number of rows, then the
optimum blade speed to steam velocity ratio r can be generalised for n number of moving blade rows.
cos a 1
cos a 1
ropt =
i.e. for three rows, ropt =
2n
6
Similarly, work from the last row of moving blades shall be,
Here it is obvious that for optimum value of r i.e. ropt the work available from second row is
Wlast row =
1
of total work.
2n
1
th of the total work and from a four
8
1
th of the total work. Here it is obvious that as the number
row stage, work from last row shall be
16
of rows increases the work from last row goes on giving diminishing work. Therefore, with increase
in number of rows the work obtained from last row decreases, so the number of rows is generally
limited to two. If we look at diagram efficiency then it shows that the efficiency also diminishes with
increase in number of stages. Graphical pattern between the diagram efficiency versus blade speed to
steam velocity ratio r is shown in Fig. 14.20.
Thus from a three row stage, work from last row shall be
Fig. 14.20 Diagram efficiency vs. blade speed to steam velocity ratio for velocity
compounded impulse turbine
14.8
REACTION TURBINES
In a reaction turbine the pressure drop occurs in both stationary and moving rows contrary to the
impulse turbine where the total pressure drop occurs in stationary nozzles alone. The difference in
blading of reaction and impulse has already been described earlier. In reaction turbine the passage
Figure 14.21 shows the schematic of a reaction turbine stage having fixed blades followed by
moving blades row. Due to the varying cross section area for steam flow the pressure drop occurs in
both stationary (fixed) blades row and moving blades row. The velocity increases in stationary blades
which act as nozzles. Thus the passage formed in the stationary blades in reaction turbine are of nozzle
type although they do not have conventional nozzle shape. Steam stream leaving stationary blades
impinges upon the moving blades. This impinging stream exerts a force to the right as evident from the
velocity diagrams of reaction blading. Velocity diagram of reaction turbine is similar in principle to the
velocity diagram in impulse turbine.
Steam entering moving blades is subjected to pass through converging area passage along with
change in direction. Thus there is increase in velocity (V2 > V1) from inlet to exit in moving blade which
results in a reaction force. Change in direction of velocity is accompanied by change in momentum thus
an impulse force. It shows that the rotation of shaft is caused by the combination of impulse and
reaction forces. The magnitude of impulse force depends upon the pressure drop in fixed blades. It may
be noted that due to shaft rotation being caused by combination of impulse and reaction forces these
reaction turbines are also termed as impulse-reaction turbine. These are also called full admission turbines as the steam enters through fixed blade row over complete annulus. The enthalpy drop over the
reaction turbine stage shows that heat drop occurs in both fixed blades and moving blades rows. If the
total enthalpy drop in stage is equally divided between the stationary and moving blades then the stage is
called 50% reaction stage. A mathematical parameter called degree of reaction is used to quantify the
proportion of enthalpy drops occurring in stationary and moving blades. The degree of reaction is
defined as the ratio of enthalpy drop in moving blades row (rotor blades) to the total enthalpy drop in the
stage. Mathematically it can be given as,
Enthalpy drop in moving blades ( rotor blades )
Degree of reaction =
Total enthalpy drop in thestage
h1 - h2
Degree of reaction,
e=
h0 - h2
In case of 50% degree of reaction, e = 0.5 and h1 h2 =
1
(h h2)
2 0
1
(h h2)
2 0
Such turbines having 50% degree of reaction are called Parsons turbine. Parsons turbine has
symmetrical blades for moving and stationary blades i.e. inlet angles of stationary and moving blades are
equal and also the exit angles of stationary and moving blades are equal. Term symmetrical blading in
reaction turbine refers to the 50 per cent reaction stage.
or,
Velocity diagram for 50% reaction turbine: 50% degree of reaction turbine stage has symmetrical
stationary and moving blades. The combined velocity diagram for a moving blade having section 1 1
as inlet section and section 2 2 as exit section is shown in Fig. 14.24.
Fig. 14.24 Combined velocity diagram for a moving blade of 50% reaction turbine.
From earlier discussions it is clear that 50% degree of reaction turbine has equal enthalpy drops
occuring in stationary blade row and moving blade rows. Steady flow energy equation (S.F.E.E.) when
applied over turbine stage yields the following assuming no heat interaction, no change in potential
energy.
Let us assume that turbine has more than one symmetrical stage such that absolute velocity of
steam leaving moving blade row equals to the velocity of steam entering fixed blade. Thus, from given
velocity diagram steam enters fixed blade with velocity C2 while leaves it with velocity C1. In case of
moving blades the relative velocities are to be considered i.e. velocity increasing from V1 to V2 from inlet
to exit.
For stationary blade row the S.F.E.E. gives,
h0 h1 =
C22 - C12
2
V22 - V12
2
Total enthalpy drop in stage, h0 h2 = (h0 h1) + (h1 h2)
and for 50% degree of reaction as C1 = V2 and V1 = C2
so,
h0 h1 = h1 h2
h0 = 2h1 h2
or,
(h0 h2) = 2 (h1 h2)
For moving blade row the S.F.E.E. gives, h1 h2 =
h1 - h2
1
=
2
h0 - h2
Hence it proves that symmetrical blading results in 50% degree of reaction turbine.
Rate of work done from reaction stage can be estimated similar to that of impulse stage. Work
done = m U DCw
where
DCw= Cw2 + Cw1 = (C1 cos a1 + C2 cos a2)
For symmetrical blading,
C2 = V1, a2 = b1, so, DCw = C1 cos a1 + V1 cos b1
or,
DCw = C1 cos a1 + (C1 cos a1 U)
DCw = 2C1, cos a1 U
i.e.
W = m U (2C1 cos a1 U)
C12
+m
2
FG V
H
2
2
- V12
2
IJ
K
m
(C12 + V22 V12)
2
m
=
(C12 + C12 V12), as C1 = V2
2
V2
Ein = m (C12 1 )
2
From velocity diagram, V12 = C12 + U 2 2C1U cos a1
Substituting value of V1 in energy input expression,
Ein =
Ein
Ein
R cC + U - 2C U cosa h UV
= m SC 2
T
W
C + 2UC cosa - U U
R
= m S
VW
2
T
W
Ein
m U 2 C1 cos a 1 - U
C12 + 2 U C1 cos a 1 - U 2
2
2U 2 C1 cos a 1 - U
C12 + 2UC1 cos a 1 - U 2
b
2 r a 2 cosa
hd =
U
1 -r
, hd =
C1
1 + 2 r cos a 1 - r 2
2
1
Substituting r =
2
1
2
1
The maximum value of diagram efficiency and the optimum value of blade speed to steam velocity
ratio, r can be estimated by differentiating with respect to r and equating to zero.
d hd
= 0
dr
or,
2 cos a1 2r = 0
or,
r = cos a1
Substituting 'r' for getting maximum value of diagram efficiency, hd,max
or,
hd,max =
hd,max =
2 cos 2 a 1
1 + cos 2 a 1
Work output for the condition of maximum diagram efficiency can be obtained by substituting
r = cos a1
or,
U
= cos a1
C1
Fig. 14.25 Velocity diagram for optimum reaction turbine stage having = cos 1
Height of blade in reaction turbine can be estimated from the volume flow rate as given below.
Reaction turbine stage is full admission stage.
Taking number of blades as N, thickness of blade as t, the mass flow rate as m, average
specific volume as v it can be written that,
m v = (pd Nt) l V2 sin b2
Height of blade, l =
mv
( d N t ) V2 sin 2
It is observed that the minimum height for reaction turbine blade should not be less than 5 per cent
of mean blade ring diameter and the maximum height should not exceed beyond 20 per cent of mean
blade ring diameter. If the blade is of tapered type then the maximum height may go up to 30 per cent of
mean blade ring diameter. The lower and upper limits of blade height are to be maintained while deciding
blade height as shorter blades have increased energy losses while longer blades may have excessive
stresses in them. Generally in reaction stages of turbine the blade height is allowed to increase gradually
so as to give increasing annular area for expanding fluid.
Fig. 14.26 Ideal and actual expansion in steam turbine showing reheat effect on hs diagram
While actually steam enters at state 1 and leaves at state 2 i.e. enthalpy at state 2 is higher than
at 2. Comparison of expansion process 12 and 12 shows that due to various losses the total available
energy (h1 h2) is not completely converted into shaft work. Portion of available energy given by
{(h1 h2) (h1 h2)} remains with expanding fluid i.e. steam itself. On hs diagram the shift in state
after expansion from 2 to 2 due to non-ideal operation is understood as reheating of steam at constant
pressure. For the sake of simplicity this reheating from 2 to 2 is considered to be of constant pressure
Fig. 14.27 Ideal and actual expansion in steam turbine on T-S diagram
In case of multi stage expansion the h s diagram shown in Fig. 14.26 indicates the four stage
expansion starting from state 1. In the first stage theoretically the ideal expansion is given by the line
12 while actually it is 12. Second stage expansion shall begin from state 2 and isentropic and nonisentropic expansion processes are shown by 23 and 23 respectively. Actual state after expansion in
second stage is 3 and the subsequent stage expansion begins from this point. Expansion in third stage
results in theoretical and actual states as 4 and 4 respectively. In the last stage starting from 4 the ideal
and actual states after expansion are at 5 and 5 respectively.
Thus for multi stage expansion the actual states at the end of each stage are 2, 3, 4 and 5. For
any stage, say first stage the isentropic and non-isentropic enthalpy drop may be related to stage efficiency as below,
Stage efficiency, hs for Ist stage =
ah - h f
ah - h f
1
Here the locus of actual state points 12 3 4 5 is also called the condition line as it
approximates actual state of steam during its exit through turbine stages. If we look at the enthalpy
drops occurring in ideal and actual conditions then it is seen that the sum of isentropic enthalpy drops in
stages i.e. (Dh12 + Dh2 3 + Dh34 + Dh45) shall be different from the actual enthalpy drop (Dh12 + Dh23
+ Dh34 + Dh45).
This sum of isentropic enthalpy drop (heat drop) is generally termed as cumulative heat drop
(Dhc). Total enthalpy drop for isentropic expansion can be given by Dh1-5 and this is also termed as
Rankine heat drop (DhRankine).
Let us look actual and ideal expansion in a stage, say second stage where expansion begins at 2
and should go up to 3 while it actually ends at 3. The shift in state from 3 to 3 indicates reheating of
steam during expansion. This phenomenon is called reheating of steam during expansion in steam
turbine and is quantified by Reheat Factor. Reheat factor is defined by the ratio of cumulative heat
RF =
Reheat factor value is always greater than unity because cumulative heat drop shall always be
more than Rankine heat drop. This may be seen from hs diagram which has diverging constant pressure lines because of which the cumulative heat drop becomes more than Rankine heat drop. Reheat
factor value increases with increase in number of stages for a given pressure range. If the stage efficiency is reduced then also the value of reheat factor increases for given pressure range and number of
stages. From designers point of view the reheat factor must have value of unity under ideal conditions.
But, since unity reheat factor value can not be attained so efforts must be made to reduce the reheat
factor value to make it close to unity. Unity reheat factor indicates that actual expansion is similar to ideal
expansion and cumulative heat drop equals to Rankine heat drop.
Reheat factor may be related to internal efficiency of turbine as detailed below. Internal efficiency
is given by the ratio of sum of actual outputs of stages and the available energy in turbine. Internal
efficiency may also be given by the ratio of internal work done of turbine to the Rankine heat drop.
Mathematically, internal efficiency,
hint =
hint =
D hactual
D hRankine
D h12 + D h2 3 + D h3 4 + D h3 5
D h15
D h15
D h15
If the stage efficiency for turbine stages is given by hS and stage efficiency remains same for all
stages, then stage efficiency can be given as,
hint =
or,
hS =
Also,
D h12
D h2 3
D h3 4
D h4 5
=
=
=
D h12
D h2 3
D h3 4
D h4 5
Let us assume the available energy in each stage to be equal as, Dhstage
Dh12 = Dh23 = Dh34 = Dh45 = Dhstage
or,
Dh12 = hSDhstage = Dh23 = Dh34 = Dh45
Substituting in internal turbine efficiency, for n number of stages
Dh15 = hS Dhstage n
hint =
h S D hstage n
h S D hstage n
=
D h15
D hRankine
hc
RF
h S D hstage n RF
D hc
hint =
hint =
D h12 RF
D h12
h S n D hstage RF
D hc
For single stage turbine,
a D h f RF
aD h f
15
15
Thus it is obvious that the internal efficiency of a turbine having multiple stages is always more
than stage efficiency as Reheat factor is more than unity.
In throttle governing the steam entering is regulated by opening and closing of valve. As the valve
is closed, the throttling or constant enthalpy process occurs across the valve with an increase in entropy
and corresponding decrease in availability of energy per unit mass flow of steam. Also due to throttling
Schematic of simple throttle governing in steam turbines is shown in Fig. 14.29. Here a centrifgal
governor is used to sense the change in speed of shaft. The relay system has a pilot valve and servomotor. The displacement of servomotor piston either upward or downward decides the opening of
throttle valve C. Servomotor piston is actuated by the high pressure oil entering from pilot valve to
upper or lower half of servomotor piston D. Under normal operation the servomotor piston occupies
middle position and pilot valves keep the inlet and exit ports in closed position. When oil enters the upper
half of servomotor then servomotor piston lowers down and the throttle valve starts closing causing
reduction of steam flow rate and so the output till the speed is maintained to normal running speed.
Simultaneously, the oil from lower half of servomotor gets drained out through pilot valve port. When oil
under pressure enters lower half of servomotor then servomotor piston gets lifted up causing lift of
throttle valve.
For throttle governing of steam turbine the steam consumption rate may be plotted with load
resulting into characteristic line called Willans line as shown in Fig. 14.30.
During throttle governing the Willans line is straight line making an intercept on y-axis. Mathematically, it can be given as,
M = KL + M0
where M is steam consumption in kg/h at any load L.
M0 is steam consumption in kg/h at no load i.e. L = 0
L is any load on turbine in kW.
K is the constant and gives slope of Willans line.
Here it shows that even at no load the steam consumption shall be M0 (kg/h) which is graphically
given by intercept on y-axis.
M
M
=K+ 0
Above equation can also be written as,
L
L
M
where
is the specific steam consumption at any load, kg/kWh
L
M
and 0 is the specific steam consumption at no load, kg/kWh.
L
Throttle governing offers following disadvantages due to throttle action at inlet:
(a) Throttling increases initial superheat at inlet and the greatest variations in steam velocity
occur in the later stages.
(b) The wetness of steam gets reduced in later stages due to throttling. Due to this reduced
wetness there occurs reduction in stage efficiency at part load operation of turbine.
(ii) Nozzle Control Governing: Nozzle control governing is the one in which steam flowing
through nozzles is regulated by valves. Nozzle control governing is generally employed at
first stage of turbine due to practical limitations. The nozzle areas in remaining stages remain
constant. If some how the nozzle governing is provided for all nozzles in each and every
stage then an ideal condition of turbine flow passage areas conforming to mass flow rate
at all loads shall exist. Under such ideal conditions the pressure, velocities and nozzle and
blade efficiencies would be constant with load. For such ideal condition the Willans line
would be straight line as indicated for throttle governing of turbine.
As the valves are being regulated for actuating nozzle control governing so there occurs some
throttling of steam at each valve. However, the amount of throttling is considerably lesser and the
decrease in availability of energy to turbine is not too much. In order to avoid this occurrence of
throttling very large number (infinite) of nozzle and governing valves may be put.
(iii) By Pass Governing: In case of by-pass governing arrangement is made for by-passing
surplus quantity of steam without allowing total steam quantity to contribute in turbine
output when load reduces. Arrangement of by pass governing is shown in Fig. 14.32.
14.14
Differences between the impulse turbine and reaction turbines are as under.
Impulse turbines
(a) Impulse turbine has profile type
blades and has constant area between
two consecutive blades.
(b) Impulse turbine stage has pressure
drop occurring only in nozzles. No
pressure drop occurs in moving blade
row.
(c) Impulse turbines have incomplete
admission of steam or steam being
admitted at selected positions around
the motor.
(d) Impulse turbine diaphragm has
nozzles mounted on it. Velocity of
Reaction turbines
(a) Reaction turbine has airfoil type blades
and has converging area between two
consecutive blades.
(b) Reaction turbine stage has pressure
drop occurring in both fixed as well
as moving blades.
(c) Reaction turbines have complete
admission of steam or steam being
admitted all around the rotor through
fixed blade ring.
(d) Reaction turbine has fixed blade
ring attached to casing to serve
Cont.
(e)
(f)
(g)
(h)
EXAMPLES
1. Single stage impulse turbine has equal blade angles and nozzle angle of 15. Determine the
maximum possible blade efficiency if the blade velocity coefficient is 0.85. Determine the blade speed
to steam velocity ratio if the actual blade efficiency is 90% of maximum blade efficiency.
Solution:
Given, a1 = 15, b1 = b2, K = 0.85
Maximum blade efficiency,
hb,max =
F
H
cos 2 a 1
cos b 1
1+ K
2
cos b 2
I
K
cos 2 15
(1 + 0.85)
2
= 0.863 or 86.3%
=
Ans.
2. A single stage of simple impulse turbine produces 120 kW at blade speed of 150 m/s when steam
mass flow rate is 3 kg/s. Steam enters moving blade at 350 m/s and leaves the stage axially. Considering
velocity coefficient of 0.9 and smooth steam entry without shock into blades, determine the nozzle angle
and blade angles. Solve using velocity diagram.
V2
= 0.9, a2 = 90
V1
Fig. 14.33
Steps:
(i) Draw AB = 3 cm =
FH 150 IK
50
corresponding to U.
Ans.
3. In a single stage simple impulse turbine the steam flows at rate of 5 kg/s. It has rotor of 1.2 m
diameter running at 3000 rpm. Nozzle angle is 18, blade speed ratio is 0.4, velocity coefficient is 0.9,
outlet angle of blade is 3 less than inlet angle. Determine blade angles and power developed.
r =
V
U
= 0.4, 2 = 0.9, a1 = 18, b2 = b1 3
C1
V1
U =
2 p 3000 1.2
= 188.5 m/s C1 = 471.25 m/s
60 2
Fig. 14.34
From velocity diagram, b1 = 29, b2 = 26, DCw = 495 m/s (= length FG)
Power developed = m U DCw
= 5 188.5 495 = 466537.5 J/s
= 466.54 kW
Blade angles at inlet and exit = 29 and 26
Power developed = 466.54 kW
Ans.
4. In a single stage impulse turbine the isentropic enthalpy drop of 200 kJ/kg occurs in the nozzle
having efficiency of 96% and nozzle angle of 15. The blade velocity coefficient is 0.96 and ratio of
blade speed to steam velocity is 0.5. The steam mass flow rate is 20 kg/s and velocity of steam entering
is 50 m/s. Determine
(a) the blade angles at inlet and outlet if the steam enters blades smoothly and leaves axially.
(b) the blade efficiency
(c) the power developed in kW
(d) the axial thrust.
Solve using velocity diagram.
Solution:
Let the state at inlet to nozzle, leaving nozzle and leaving blade be denoted using subscripts
0, 1 and 2.
Using nozzle efficiency the actual enthalpy drop may be calculated as.
Actual enthalpy drop in nozzle, DhN = 200 0.96 = 192 kJ/kg
a f
Ans.
V2
= 0.98
V1
The velocity diagram is drawn using above information. Taking scale 1 cm = 100 m/s.
Fig. 14.36
22
0.80
N
7
2
N = 226.7 rps
N = 13602 rpm
Speed of rotation = 13602 rpm
Ans.
W =
=
=
1hp =
hp per kg of steam per second =
U DCw
570 1080 = 615600 W/kg
615.6 kW/kg
0.7457 kW
826.86 hp
3600
= 4.58 kg
785.52
Ans.
Work done
Energysupplied tostage
615.6
( 757.05 )
= 0.8132
Ans.
C22
2
= 51200 J/kg
51200
Percentage energy loss at exit =
100
757.05 103
= 6.76%
Percentage energy loss at exit = 6.76%
Ans.
m U D Cw
745.7
3 200 485
745.7
= 390.23 hp
HP developed = 390.23 hp
Ans.
7. In a simple impulse steam turbine stage steam enters the nozzle at 15 bar, dry saturated with
velocity of 150 m/s. Nozzle angle is 20 and steam leaves nozzle at 8 bar and enters into smooth blades.
Considering nozzle velocity coefficient of 0.90 and blades to be equiangular determine the following
for maximum diagram efficiency.
(a) the blade angles,
(b) the blading efficiency,
(c) the stage efficiency.
Solution:
From steam tables, enthalpy at inlet to nozzle, h0 = hg at 15 bar, = 2792.2 kJ/kg
entropy, s0 = sg at 15 bar = 6.4448 kJ/kg.K
Let dryness fraction at exit of nozzle be x1 at 8 bar.
For isentropic expansion, s1 = s0 = 6.4448 kJ/kgK
s1 = 6.4448 = sf at 8 bar + x1 sfg at 8 bar
6.4448 = 2.0462 + (x1 4.6166)
x1 = 0.9528
h1 = hf at 8 bar + x1 hfg at 8 bar
= 721.11 + (0.9528 2048)
h1 = 2672.44 kJ/kg
Isentropic heat drop in nozzle = (h0 h1) = 119.76 kJ/kg
Velocity at exit of nozzle C1 =
=
a f
2b119.76 10
2 h0 - h1 + C02
3
0.9 + (150 ) 2
C1 = 487.92 m/s
For maximum diagram efficiency the blade velocity can be obtained by,
r=
cos a 1
cosa 1
U
=
C1
2
2
U = 487.92
cos15
U = 229.25 m/s
2
Velocity diagram is drawn considering C1 = 487.92 m/s, U = 229.25 m/s a1 = 20, b1 = b2.
Scale 1 cm = 50 m/s.
Fig. 14.38
Ans.
Work done per kg of steam = U DCw = (229.25 455) = 104.31 103 J/kg
C12
( 487.92 ) 2
=
2
2
= 119.03 103 J/kg
Blading efficiency =
Stage efficiency =
=
104.31 10 3
= 0.8763
119.03 10 3
Work done per kgof steam
Energysupplied per kgof steam tostage
104.31 10 3
(150 ) 2
119.76 10 3 +
2
RS
T
UV = 0.7962
W
Ans.
8. Single row impulse turbine operates between 10 bar and 5 bar with expansion efficiency of
95%. 10 kg of dry saturated steam per second enters into nozzle and leaves nozzle at angle of 20 to the
axis of rotation of blades. The blade velocity coefficient is 0.90, blade speed is 200 m/s and internal
losses due to disc friction and windage losses is 0.5 kJ/kg of steam. Consider that there is no axial thrust
on the blades.
Determine,
(a) Blade angles,
(b) Stage efficiency,
(c) Stage output in hp and prepare heat balance sheet. Also reason out the error if any by
solving using velocity diagram.
2 h0 - h1 0.95
= 2 122.65 10 3
C1 = 495.28 m/s
U = 200 m/s, a1 = 20,
Given:
V2
= 0.90
V1
Fig. 14.39
angle FBD = b1 = 32
angle ABE = b2 = 39
Work done per kg of steam = U DCw
= (200 485)
= 97000 J/kg, Actual work = (97 0.5) = 96.5 kJ/kg
b97000 10
-3
- 0.5
= 0.7474 or 74.74%
129.11
( 97 - 0.5) 10
Output in hp =
= 1294.09 hp
0.7457
Energy lost in nozzle = (h0 h1) 0.05 = 6.46 kJ/kg
FG V
H
IJ FG ( 320)
K H
- V22
=
2
= 9.73 kJ/kg
2
1
- ( 288 ) 2
2
IJ
K
C22
(180 ) 2
=
= 16.2 kJ/kg
2
2
129.11
Expenditure, kJ/kg
(i) Useful work = 96.5
(ii) Nozzle friction loss = 6.4
(iii) Blade friction loss = 9.7
(iv) Windage & disc friction loss = 0.5
(v) Exit energy loss = 16.2
129.3 kJ/kg, Error = 0.19 kJ/kg
%
74.63
4.95
7.50
0.38
12.55
Error of 0.19 kJ/kg is due to cumulative error creeping in from construction and rounding off
measurements from velocity diagram.
9. Following data refer to the velocity compounded impulse turbine with two rows of moving
blades:
Steam supply: 60 bar, 500C
Speed of rotation: 3600 rpm
Mean radius of blades: 60 cm
Steam supply: 6 kg/s
Blade speed to steam velocity ratio: 0.2
Nozzle efficiency: 90%
Nozzle angle: 20
Exit angles: 25, 25 and 40 for first moving, fixed and second moving
blades respectively
Blade velocity coefficient: 0.85 for all blades
Disc friction and Windage loss: 20 hp.
Determine,
(a) the stage pressure
(b) the blading efficiency
(c) the power output
(d) the state of steam leaving stage.
60 10 - 2 2 p 3600
= 226.2 m/s
60
U
r = 0.2 =
so C1 = 1131 m/s
C1
Blade speed,
U = rw =
Dh01 =
Dh01 =
Enthalpy at state 1,
Enthalpy at inlet to nozzle,
Entropy
h1
h0
h0
s0
=
=
=
=
C12
(1131) 2
=
= 639.58 kJ/kg
2
2
D h 01
= 710.64 kJ/kg
0.9
h0 Dh01
hat 60 bar, 500C
3422.2 kJ/kg
6.8803 kJ/kgK
F
H
I
K
Fig. 14.40
Ans.
Fig. 14.41
From diagram,
C3
= 0.85, C3 = 0.85 C2 = 518.5 m/s
C2
V 3 = 335 m/s V4 = 284.75 m/s
Velocity diagram is drawn in following steps:
(i) Draw AB corresponding to U (= 226.2 m/s)
(ii) Draw AD at angle 20 corresponding to C1 (= 1131 m/s)
(iii) Join D with B and measure BD (= 9.35 cm) or V1 = 935 m/s, Estimate V2 = 0.85 V1
(iv) Draw BE corresponding to V2 = (794.75 m/s) at angle b2 (= 25)
(v) Join E with A and measure AE (= 6.1 cm) or C2 = 610 m/s. For fixed blade C3 = 0.85
C2, Estimate C3.
(vi) Draw AG at angle a3 (= 25) corresponding to C3 (= 518.5 m/s)
(vii) Join G with B and measure BG (= 3.35 cm) or V3 = 335 m/s. Estimate V4 = 0.85 V3
for IInd moving blade.
(viii) Draw BF at angle 40 corresponding to V4 (= 284.75 m/s)
(ix) Join F with A and measure AF (= 2.8 cm) or C4 = 280 m/s.
Length KH (= 15.6 cm) or DCw Ist moving = 1560 m/s
Length JI (= 5.45 cm) or DCw II moving = 545 m/s
Total Work done = (U DCw Ist moving) + (U DCw II moving)
per kg of steam
= 226.2 (1560 + 545)
= 476.15 kJ/kg
Blading efficiency =
Net power output =
476.15
= 0.7444 or 74.44%
639.58
FH 476.15 6 IK 20
0.7457
= 3811.16 hp
Energy loss at exit per kg =
FG C IJ = (280)
H2K 2
2
4
= 39.2 kJ/kg
381116
. 0.7457
= 473.66 kJ/kg
6
C
V2
V
= 4 = 3 = 0.90
C2
V1
V3
Estimate C2 =
Fig. 14.42
BAD
IBD
ABE
HAE
BAH
ABF
=
=
=
=
=
=
a1 = 21
b1 = 26
b2 = 28.5
a2 = 39
a3 = 20
b3 = 33
Blade speed =
Ist moving blade inlet and exit angle =
Fixed blade inlet and exit angle =
IInd moving blade inlet angle =
Nozzle angle =
104.16 m/s
26, 28.5
39, 20
33
21
Ans.
Let height of nozzle first moving, fixed blade and second moving blades be ln, lm1, lf, lm2 respectively
ln Ca1 = lm1 Ca2
Ca 1
lm1
182.28
=
=
= 1
Ca 2
ln
182.28
lf
ln
Ca 2
=1
Ca 2
Ans.
U
= 0.7, hstage = 0.80, b2 = 20
C1
2 p 3000 1
= 157.08 m/s
60 2
C1 = 224.4 m/s
U = rw =
or,
Fig. 14.43
Steps:
(i) Draw AB corresponding to U = 157.08 m/s.
(ii) Draw AD at angle a1 = b2 = 20 corresponding to C1 (= 224.4 m/s).
(iii) Join point D with B which gives V1 (= 95 m/s). Measure angle FBD = b1 = 56
(iv) Draw C2 along AE at angle a2 = b1 = 56. Also draw EB at angle 20.
It intersects at E. Lines AE and BE give C2 and V2 respectively.
(v) Length FG gives DCw (= 260 m/s) of 5.2 cm.
Work output = U DCw
(per kg of steam) = 157.8 260
= 40840.8 J/kg
40840.8
State efficiency hstage = 0.80 =
D hisentropic
Dhisentropic = 51051 J/kg or 51.05 kJ/kg
Enthalpy drop in stage = 51.05 kJ/kg
Ans.
12. A Parsons reaction turbine has mean diameter of blades as 1.6 m and rotor moving at 1500
rpm. The inlet and outlet angles are 80 and 20 respectively. Turbine receives steam at 12 bar, 200C
and has isentropic heat drop of 26 kJ/kg. 5% of steam supplied is lost through leakage. Determine the
following considering horse power developed in stage to be 600 hp.
(a) the stage efficiency
(b) the blade height.
Fig. 14.44
Work done
20.29
=
= 0.7804 or 78.04%
26
Heat supplied
Ans.
13. In a reaction turbine 6 kg/s steam is admitted at 15 bar dry saturated in the first stage. Turbine
has eight pairs on mean diameter of 50 cm and run at 3000 rpm with mean blade speed to steam velocity
U=
p 50 10 - 2 2 3000
= 78.54 m/s
2 60
Steam velocity,
C1=
78.54
U
=
= 98.18 m/s
r
0.8
Angles, a1 =
DCw =
=
DCw =
20, b2 = 20
2C1 cos a1 U
{(2 98.18 cos 20) 78.54)}
105.98 m/s
Actual mass of steam flowing through rows, mactual = 6 0.90 = 5.4 kg/s
Power output = U DCw mactual no. of pairs
Ans.
Theoretical heat drop in turbine can be obtained using efficiency of working steam,
Dh =
482.4 0.7454
= 70.51 kJ/kg
6 0.85
At inlet to steam turbine the enthalpy may be seen from steam table as,
h1 = hat 15 bar, dry saturated = hg at 15 bar = 2792.2 kJ/kg
s1 = sg at 15 bar = 6.448 kJ/kgK, v1 = 0.13177 m3/kg
Hence, enthalpy at exit of turbine
hexit = h1 Dh
= 2792.2 70.51
hexit = 2721.69 kJ/kg
Pressure at exit of turbine can be approximated from the mollier diagram corresponding to hexit
and sexit (= s1) vertically below state 1. Pressure at exit = 12 bar (approx.)
Pressure at exit = 12 bar
Ans.
v1 + vexit
( 0.13177 + 0.16009)
=
2
2
= 0.14593 m3/kg
p 50 h C1a
100
14. A reaction turbine has mean blade speed of 180 m/s, blade speed to steam velocity ratio of 0.8,
outlet angles of fixed and moving blades as 20 and 30, specific volume at outlet of fixed blade as
0.5 m3 and at moving blade outlet as 0.6 m3. Areas at exit of fixed blade and moving blades are same,
Consider the efficiency of blades as 90% when considered as nozzles and K2 = 0.88 where K is blade
velocity coefficient. Determine
(i) the degree of reaction,
(ii) adiabatic heat drop,
(iii) the overall stage efficiency.
Solution:
Given: a1 = 20, b2 = 30, r = 0.8, U = 180 m/s.
Steam velocity
C1 =
=
C1 =
From continuity equation, the steam mass flow rate at exit of fixed blades and moving blades shall
be same, so
C1a
vat exit of fixed blades
C 2a
vat exit of moving blades
C
76.95
= 2a
0.6
0.5
C2a = 92.34 m/s
RS C - K C
T 2h
RV -K V
Heat drop in moving blades = S
T 2 h
Heat drop in fixed blades =
2
1
2
2
2
2
2
1
UV
W
UV
W
Here nozzle efficiency is given as 0.9 for both fixed and moving blades.
C22 = V22 + U 2 2UV 2 cos b2
= (184.68)2 + (180)2 (2 180 184.68 cos 30)
C22 = 8929.18
C2 = 94.49 m/s
RS (225)
T
- ( 0.88 8929.18)
2 0.9
= 23.76 kJ/kg
RS (184.68)
T
- ( 0.88 6309.9)
2 0.9
= 15.86 kJ/kg
Therefore, degree of reaction =
15.86
( 23.76 + 15.86)
= 0.4003 or 40.03%
=
UV
W
Ans.
UV
W
Ans.
34.45
= 0.8695 or 86.95%
39.62
Ans.
15. A multi stage steam turbine has steam entering at 20 bar, 300C and leaving at 0.05 bar and
0.95 dry. Determine reheat factor, condition of steam at exit from each of five stages considering
efficiency ratio of 0.555 and all stages doing equal work.
Solution:
At inlet to turbine, the enthalpy h1 = hat 20 bar, 300C
h1 = 3023.5 kJ/kg
s1 = 6.7664 kJ/kgK
Enthalpy at exit,
h6 = hf at 0.05 bar + x hfg at 0.05 bar
= 137.82 + (0.95 2423.7)
h6 = 2440.34 kJ/kg
h1 h6 = 583.16 kJ/kg = Actual heat drop
Given: Stage efficiency or efficiency ratio = 0.55
Since all the stages do equal work, so the useful enthalpy drop in each of stage shall be equal.
Fig. 14.45
FH h - h IK = 116.63 kJ/kg
5
1
h1 - h2
h1 - h2
116.63
0.555
= 210.05 kJ/kg
For second stage, h2 h3 = 116.63 kJ/kg
h - h3
Stage efficiency = 0.555 = 2
h2 - h3
h1 h2 =
116.63
= 210.05 kJ/kg
0.555
Similarly, for third stage h3 h4 = 210.05 kJ/kg
for fourth stage, h4 h5 = 210.05 kJ/kg
for fifth stage, h5 h6 = 210.05 kJ/kg
After isentropic heat drop up to state 6, s1 = s6 = 6.7664 kJ/kgK
s6 = sf at 0.05 bar + x6 sfg at 0.05 bar
h2 h3 =
x6 =
FH 6.7664 - 0.4764 IK
7.9187
x6 = 0.7943
Enthalpy, h6 = hf at 0.05 bar + x6 hfg at 0.05 bar
= 137.82 + (0.7943 2423.7)
h6 = 2062.96 kJ/kg
F = bh - h g + c h - h h
GG + bh - h g + ch - h h + ch - h
H
Adialatic heat drop c = h - h h
I
hJJK
5 210.05
( 3023.5 - 2062.96)
RF = 1.09
Reheat factor = 1.09
Ans.
Plotting the actual heat drop shown by 1-6 on Mollier diagram state of steam leaving each stage
can be given as under.
No. of stage
1
2
3
4
5
Pressure at exit of stage, bar
6.9
2.6
0.8
0.2
0.05
Condition of steam
235C
160C
dry sat.
0.975
0.95
dryness
dryness
factor
factor
Ans.
Fig. 14.46
Solution:
From mollier diagram,
h1 h6 = 1028 kJ/kg
h1 - h6
5
= 205.6 kJ/kg
Plotting the condition line as straight line between 1 and 6, the following states are obtained at inlet
of each stage
No. of stage
1
Pressure, at inlet bar
35
State of steam
450C
2
3
14.5
5.5
342C 235C
4
1.7
125C
5
0.35
0.95
h2 - h3
205.6
=
= 0.8358
246
h2 - h3
hstage,III =
h3 - h4
205.6
=
= 0.8567
h3 - h4
240
hstage,IV =
h4 - h5
205.6
=
= 0.8567
h4 - h5
240
hstage,V =
h5 - h6
205.6
=
= 0.8749
h5 - h6
235
Ans.
Ans.
ah - h f
ah - h f
1
1028
= 0.8847
1162
Ans.
17. One stage of an impulse turbine consists of a converging nozzle ring and one ring of moving
blades. The nozzles are inclined at 22 to the blades whose tip angles are both 35. If the velocity of
steam at exit from the nozzle is 660 m/s, find the blade speed so that the steam passes without shock.
Find diagram efficiency neglecting losses if the blades are run at this speed.
[U.P.S.C., 1992]
Solution:
Diagram efficiency of impulse turbine, when losses are neglected, shall be equal to the maximum
diagram efficiency obtained for optimum blade speed-steam velocity ratio (r).
hdiag, max =
cos 2 a 1
(1 + KC)
2
Considering turbine to have smooth and symmetrical blades i.e. K = 1,C = 1 and Blade angles,
b1 = b2
Nozzle angle = a1 = 22
cos a 1
U
=
C1
2
Ans.
18. A simple impulse turbine has a mean blade speed of 200 m/s. The nozzles are inclined at 20
to the plane of rotation of the blades. The steam velocity from nozzles is 600 m/s. The turbine uses 3500
kg/hr of steam. The absolute velocity at exit is along the axis of the turbine. Determine (i) the inlet and
exit angles of the blades, (ii) the power output of the turbine, (iii) the diagram efficiency, and (iv) the
end thrust (per kg steam per second) and its direction.
[U.P.S.C., 1998]
Solution:
Given, U = 200 m/s, a1 = 20, C1 = 600 m/s, a2 = 90, m = 3500 kg/hr
scale: 1 cm = 50 m/s
Fig. 14.47
C1 sin a 1
C1 cos a 1 - U
b1 = tan1
FG 205.2 IJ
H (600.cos 20 - 200) K
b1 = 29.42
For symmetrical blade,
b1 = b2 = 29.42
Ans.
Ans.
a m U DC f
FG m C IJ
H 2K
w
2
1
2U C1 cosa 1
C12
2U cosa 1
C1
2 200 cos 20
600
= 0.6265
Ans.
U
200
=
= 229.61m/s
cos b 2
cos 29.42
Ans.
EXERCISE
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
14.9
14.10
14.11
14.12
14.13
14.14
T1 + T2
2 T2 + h s T1 - T2
The number of stages may be considered infinite and the condition line is straight line.
14.22 A steam turbine stage is supplied with steam at 50 bar, 350C and leaves turbine at pressure of
5 bar. Determine steam supply rate if the isentropic efficiency is 82% and power output is 15224
kW
[37.8 kg/s]
14.23 A De Laval steam turbine has steam leaving nozzle with velocity of 1200 m/s at angle of 20 and
mean blade speed of 400 m/s. Considering blades to be symmetrical, blade velocity coefficient of
0.8 and steam flowing at rate of 1000 kg/hr, determine the following using velocity diagram,
(a) the blade angle
(b) the power developed
(c) the relative velocity of steam entering blades
(d) the blade efficiency.
[30, 30, 145.5 kW, 830 m/s, 72.8%]
14.24 Determine the turbine output from a single stage impulse turbine having smooth blades, steam
flow rate of 20 kg/s, steam velocity of 600 m/s, blade speed of 250 m/s, nozzle angle of 20 and
blade outlet angle of 25. Also calculate axial thrust.
[3275 kW, 800N]
14.25 Determine nozzle angle, blade angles at inlet and exit for a single stage impulse steam turbine
developing 132 kW, speed of 3340 rpm, mean rotor diameter of 1 m, steam flow rate of 2 kg/s, blade
velocity coefficient of 0.9, steam leaving nozzle at 400 m/s and steam discharges axially from
turbine. Solve using velocity diagram.
[21, 36, 32]
14.26 De-Laval turbine has nozzle angle of 20, symmetrical blades and blade velocity coefficient of 0.85.
Determine maximum blade efficiency. Also find out the blade speed to steam velocity ratio if the
actual blade efficiency is 92% of maximum blade efficiency.
[81.6%, 0.6 or 0.34]
14.27 What shall be the nozzle angle for a De-Laval turbine offering maximum possible efficiency of
90%?
[18.44]
14.28 A stage of simple impulse turbine has 10 kg/s steam entering the rotor having mean diameter of
105 cm, running at 50 r.p.s.. Nozzle angle is 18, blade speed to steam velocity ratio is 0.42, blade
velocity coefficient is 0.84 and the exit angle of blade 3 less than inlet angle. Draw the velocity
diagram and obtain (a) the axial thrust on blades, (b) the tangential thrust on blades, (c) the
resultant thrust on blades (d) the power developed.
[250 N, 3900 N, 3908 N, 641.5 kW]
14.29 A singe stage impulse steam turbine has mean diameter of rotor as 1 m and runs at 3000 rpm. Steam
leaves nozzle at 20 with velocity of 300 m/s. Determine the power developed if the axial thrust on
14.30
14.31
14.32
14.33
14.34
14.35
14.36
14.37
blades is 98 N, blades are symmetrical and there occurs 19% frictional loss of kinetic energy of
relative velocity at inlet to blades.
[181.2 kW]
Single row impulse turbine has blade speed of 175 m/s and steam leaves nozzle at 400 m/s. Steam
flows at the rate of 163.2 kg/min and leaves turbine axially. Considering blade velocity coefficient
of 0.9 determine nozzle angle, blade angles at inlet and exit, axial thrust, energy loss at exit, energy
loss in blades and diagram efficiency if turbine produces 180 kW output.
[19, 32.6, 36.3, 4.35 N, 8.27 kJ/kg, 5.53 kJ/kg. 41.36%]
Following data refer to a stage of velocity compounded impulse steam turbine;
Nozzle angle: 16
Angle at exit of first moving blade: 20
Angle at exit of fixed blade: 25
Angle at exit of second moving blade: 30.
Velocity of steam leaving nozzle: 800 m/s
Blade speed: 150 m/s
Blade velocity coefficient for first moving, fixed and second moving blades: 0.8, 0.85 and 0.85.
Steam flow rate: 5 kg/s
Determine, (a) diagram efficiency, (b) the energy carried at exit, (c) the axial thrust on each moving
blade ring, (d) the power developed.
[66.2%, 3.79 kJ/kg, 20.9 N, 27 N, 1059 kW]
A two stage velocity compound impulse steam turbine has mean blade speed of 150 m/s, steam
leaves nozzle with 675 m/s velocity at angle of 20. The exit angles of first row moving blade, fixed
blade and second row moving blade are 25, 25 and 30 respectively. For the blade velocity
coefficient of 0.9 for all blades and flow rate of4.5 kg/s determine,
(a) the power output, (b) the diagram efficiency.
[80.7 kW, 78.5%]
A simple impulse turbine has nozzles at angle of 20 to the direction of motion of moving blades
and steam leaves nozzle at 375 m/s, blade velocity coefficient is 0.85, steam flow rate is 10 kg/
s and blade speed is 165 m/s. Determine,
(a) Inlet and outlet angles for blades such that there is no axial thrust.
(b) The power output.
[34.4, 41.6, 532 kW]
A Parsons reaction turbine has blade speed of 78.5 m/s, stage efficiency of 80%, blade speed
to steam velocity ratio of 0.7 and blade outlet angle of 20. Estimate the isentropic enthalpy drop
in the stage.
[12.32 kJ/kg]
In a 50% reaction turbine the inlet and exit angles are 80 and 20 respectively. Blade speed is
113 m/s and steam is admitted in turbine at 12 bar, 200C and leaves after adiabatic enthalpy drop
of 17.5 kJ/kg with only 95% of heat drop being utilized for producing work of 620 hp. Calculate
the stage efficiency and blade speed.
[78.5%, 2.29 cm]
A reaction turbine has mean blade speed of 220 m/s and blade speed to steam velocity ratio
of 0.7. Determine the angle at inlet of blade and work done per kg of steam flow if angle at exit
is 20.
Also determine the percentage increase in diagram efficiency if turbine is run for optimum blade
speed to steam velocity ratio.
[55, 81.5 kJ/kg, 93.8%]
A reaction turbine has eight pairs with mean blade speed of 68 m/s and average blade speed to
steam velocity ratio of 0.8. Steam at 15 bar, dry saturated is supplied at the rate of 300 kg/min.
14.38
14.39
14.40
14.41
Angle at exit of fixed and moving blade is 20. Determine (a) the horse power, (b) the average blade
height.
[312.6 hp, 1.35 cm]
In a four stage velocity compounded impulse turbine, steam is supplied at 35 bar, 420C and leaves
at 0.07 bar. Obtain the initial pressure, quality at inlet of each stage and internal efficiency of
turbine if the average efficiency is 70%. Also find reheat factor.
[11.3 bar, 304C, 2.9 bar, 189C, 0.55 bar, 0.99, 74.6% 1.07]
A reaction turbine stage has steam leaving fixed blade at 3 bar, 0.94 dry and velocity of 143
m/s. Ratio of axial velocity of flow to blade velocity is 0.7 at entry and 0.75 at exit from moving
blades. Angles at exit of fixed and moving blades are same. The height of blade is 1.8 cm and
steam flow is 150 kg/min. For blade velocity of 70 m/s determine the degree of reaction.
[58.2%]
A three stage turbine has steam entering at 30 bar, 350C and leaves first, second and third stages
at 7 bar, 1 bar and 0.1 bar respectively. For equal stage efficiency of 70% determine (a) the final
condition of steam, (b) the reheat factor, (c) the overall thermal efficiency.
[0.94, 1.05, 24.5%]
A four stage turbine has steam supplied at 20 bar, 350C and leaves at 0.05 bar. Considering
overall efficiency to be 80% and equal work from the stages and straight condition line determine,
(a) the steam pressure at exit of each stage, (b) the each stage efficiency, (c) the reheat factor.
[9 bar, 2.4 bar, 0.5 bar, 0.05 bar, 72.5%, 76%, 78.4%, 82.5%, 1.04]