Unit - 4
Steam nozzles & Steam Turbines
Types of steam nozzles
Convergent nozzle
Divergent nozzle
Convergent- divergent nozzle
Applications of steam nozzle
Velocity of steam flowing through a nozzle
Consider a mass flow of steam through a nozzle.
Let =Velocity of steam at the entrance of nozzle in m/s
=Velocity of steam at any section considered in m/s
=Enthalpy or heat of steam entering the nozzle in kJ/Kg and
=Enthalpy or total heat of steam at the section considered in kJ/kg.
We know that for a steady flow process in a nozzle.
Neglecting losses in a nozzle.
………. (1)
=Enthalpy or heat drop during expansion of steam in a nozzle
=
Since the entrance velocity or velocity of approach is negligible as compared to Therefore from the equation (1)
---2
In actual practice there is always a certain amount of friction present between the steam and nozzle surfaces. This reduces the heat drop by 10 to 15 percent and thus the exit velocity of steam is also reduced correspondingly. Thus the above relation may be written as :
Where K is nozzle coefficient or nozzle efficiency.
Mass Of Steam Discharged Through Nozzle :
We have already discussed that the flow of steam, through the nozzle is isentropic which is approximately represented by the general law:
We know that gain in kinetic energy
= ….. Neglecting initial velocity of steam
And Heat drop = Work done during Rankine cycle
=
Since gain in kinetic energy is equal to heat drop therefore
= ………… (1)
We know that
………… (2)
Substituting the value of in equation 1
=
Now the volume of steam flowing per second
= Cross sectional area of nozzle x Velocity of steam =
And volume of 1kg of steam i.e. specific volume of steam at pressure
=
Mass of steam discharged through nozzle per second
Substituting the value of from equation 2
=
Steam turbines may be classified into the following types:
2. According to the exhaust condition of steam
3. According to the pressure of steam
4. According to the number of stages
Impulse turbine
De-level impulse turbine
2. Runner and blades:
3. Casing:
Pressure and velocity of steam in an impulse turbine
Reaction turbine
Parson's reaction turbine
Parsons turbine is the simplest type of reaction steam turbine and is commonly used. It has the following main components:
2. Guide mechanism:
3. Turbine runner:
4. Draft tube:
Pressure and velocity of steam in reaction turbine
Comparison between impulse turbine and reaction turbine
S.no. | Impulse Turbine | Reaction Turbine |
1. | The steam flows through the nose and impinges on the moving blades | The steam flows first through guide mechanism and then through the moving blades. |
2. | The steam impinges on the buckets with kinetic energy. | The steam glides over the moving vanes with pressure and kinetic energy. |
3. | The steam may or may not be admitted over the whole circumference. | The steam must be admitted over the whole circumference. |
4. | The steam pressure remains constant during its flow through the moving blades. | The steam pressure is reduced during its flow through the moving blades. |
5. | The relative velocity of steam while gliding over the blades remain constant. | The relative velocity of steam while gliding over the moving blades increases. |
6. | The blades are symmetrical | The blades are not symmetrical. |
7. | The number of stages required are less for the same power developed. | The number of stages required are more for the same power developed. |
Compounding of Impulse Steam Turbines
The following three methods are commonly employed for reducing the rotor speed:
2. Pressure compounding of an impulse turbine
3. Pressure velocity compounding of an impulse turbine
Velocity triangles for moving blade of an impulse turbine
Let
Linear velocity of the moving blade (AB)
V=absolute velocity of steam entering the moving blade (AC)
=relative velocity of Jet to the moving blade (BC). It is the vectorial difference between and V.
velocity of flow at entrance of the moving blade.
velocity of whirl at entrance of the the moving blade
=angle which the relative velocity of Jet to the moving blade makes with the direction of motion of the blade
=Angle with the direction of motion of the blade at which the jet enters the blade.
rresponding values at exit of the moving blade
Impulse turbines
Combined velocity triangle for moving blades
Velocity diagram for two stage impulse turbine
We know that power developed by two stage impulse turbine
Where m is the mass of steam supplied in kg/s
Velocity triangles for moving blades of reaction turbine
linear velocity of the moving blade (AB)
V=absolute velocity of steam entering the moving blade (BC)
=Relative velocity of jet to the moving blade (AC).
=Velocity of flow at entrance (EC). It is the vertical component of V.
Velocity of whirl at entrance (BE). It is a horizontal component of V.
=angle with the direction of motion of the blade at which the steam enters the blade
=angle which the relative velocity of jet makes with the direction of motion of the blade
rresponding values at exit of the moving blade
Combined velocity triangle for moving blades
Steam turbine governing and control
- Throttle governing
- Nozzle governing
- Bypass governing
- Combination of 1 and 2 and 1 and 3.
(a) (b)
Where Steam consumption in kg/h at any load M
Steam consumption in kg/h at no load
Steam consumption in kg/h at full load
M= Any other load in kW
Full load in kW and
K= Constant
varies from about 0.1 to 0.14 times the full load consumption. The equation can be written as
is called the steam consumption per kWh
2. Nozzle governing
Comparison of throttle and nozzle control governing
S.No | Aspects | Throttle Control | Nozzle control |
1. | Throttling losses | Severe | No throttling losses (Actually there are a little throttling losses in nozzles valves which are partially open) |
2. | Partial admission losses | Low | High |
3. | Heat drop available | Lesser | Larger |
4. | Use | Used in impulse and reaction turbine both | Used in impulse and also in reaction (if initial stage impulse turbines) |
5. | Suitability | Small turbines | Medium and larger turbines |
3. Bypass governing
(a)
(b)
(C)
(D
Internal losses in turbines
Strictly speaking an ideal turbine (having 100 percent gross efficiency) will do the work equivalent to the isentropic enthalpy or heat drop of the steam used in the turbine. But in actual practice, the work done by a turbine is much less than isentropic heat drop of the steam used. There are several factors which effect the performance of a steam turbine. All these factors which reduce the output of the turbine are known as internal losses. Do there are many internal losses in a steam turbine, yet the following are important from the subject point of view:
Energy losses in steam turbines
The increase in heat energy required for doing mechanical work in actual practice as compared to the theoretical value, in which the process of expansion takes place strictly according to the adiabatic process, is termed as energy loss in steam turbine.
The losses which appear in an actual turbine may be divided into two following groups
1. Internal losses: Losses directly connected with steam conditions while in its flow through the turbine. They may be further classified as
(ii) Losses in regulating valves
(iii) Losses in nozzles (guide blades)
(iv) Losses in moving blades:
(a) Losses due to trailing edge wake
(b) Impingement losses
(c) Losses due to leakage of steam through the angular space
(d) Frictional losses
(e) Losses due to turning of steam jet in the blades
(f) Losses due to shrouding
(v) Leaving velocity losses (exit velocity)
(vi) Losses due to friction of disc carrying the blades and windage losses
(vii) Losses due to clearance between the rotor and guide blade discs
(viii) Losses due to wetness of steam
(ix) Losses in exhaust piping etc.
2. External losses: Losses which do not influence the steam conditions. They may be further classified as:
Reheat factor
Let = initial pressure of the steam.
=pressure of steam leaving the first stage
=pressure of steam leaving the second stage.
=final pressure of the steam
=initial temperature of the steam
=Stage efficiency for each stage of the turbine.
Now let us draw the expansion of steam in three stages on a Mollier chart as shown in figure
It will be interesting to know from the Mollier diagram that the pressure lines diverge from left to right, which shift the isentropic expansion lines at each stage slightly towards the right side in the diagram. Or in other words enthalpy or heat drop is slightly increased. The sum of increased heat drops is known as cumulative heat drop.
Now the ratio of cumulative heat drop to the isentropic heat drop is known as reheat factor
Mathematically, reheat factor
=
=
Numericals
D = 1.05 m
N= 3000 r.p.m
=18°
m= 10 kg/s
1)
2)
V= 392.7 m/s
3) Now draw combined velocity triangle as shown in figure
i) Draw horizontal line and cut off AB equal to 164.93 m/s, to suitable scale to represent blade velocity
ii) Now draw inlet velocity triangle ABC on the base AB with and BC=V=392.7 m/s
By measurement find blade angle at inlet =30° and relative velocity at inlet
iii) Draw outlet velocity triangle ABD on the same base AB with = 270
And
iv) from C and D draw perpendiculars to meet the line AB produced at E and F
4) tangential thrust on the blades
=
=10(390) = 3900 N
5) Axial thrust
=10(125-95)
=300 N
6) Power developed
=
=643.23 kW
7) Blading efficiency
= 83.20%
ii) work done per second by Jet
=2944.5 W
2. In a single stage impulse turbine the mean diameter of the blade ring is 1m & the rotational speed is 3000 rpm. The steam is issued from th nozzle at 300 m/s & the nozzle angle is 200 . The blades are equiangular. If the friction loss in the blade channel is 19% of the kinetic energy corresponding to the relative velocity at the inlet to the blades, what is the power developed in the blading when the axial thrust on the blade is 98N ?
D= 1m
N =3000 m/s
300m/s
=20°
=0.81
Axial thrust on the blades = 98N
1)
=157.08 m/s
2) Now draw combined velocity triangle as shown in figure
i) draw a horizontal line and cut off AB equal to 15 7.08 m/s, to suitable scale to represent blade velocity
ii) Now draw inlet velocity triangle ABC on the base AB with =20°and BC = V= 300 m/s. By measurement find blade angle at inlet =38° and relative velocity of steam inlet
iii) Similarly draw outlet velocity triangle ABD on the same base AB
iv) From C and D draw perpendicular to meet the line AB produced at E and F.
3) Axial thrust=
98 = m(100-85)
m= 6.53 kg/s
4) Power developed
=238.48 kW
3. The following data refer to a single stage impulse turbine. Isentropic nozzle heat drop = 251 kj/kg; Nozzle efficiency = 90 %; nozzle angle 200 ; ratio of blade speed to whirl component of steam speed = 0.5; blade velocity coefficient = 0.9; the velocity of steam entering the nozzle = 20 m/s
Determine i) The blade angles at inlet & outlet if the steam enters into the blade without shock & leaves the blade in an axial direction. ii) Blade efficiency iii) Power developed & axial thrust if the steam flow is 8 kg/s
Isentropic heat drop =251kJ/kg
Ratio of blade to whirl component of steam speed=0.5
K=0.9
m= 8 kg/s
V=20 m
Velocity of steam entering nozzle=20 m/s
1)
0.9=
Useful heat drop = 225.9 kJ/kg
Applying the energy equations to the nozzle we get
V=672.46 m/s
2) Now draw combined velocity triangle as shown in figure
i) Draw line BC =V=672.66 m/s and
ii) Draw a line through point C which is perpendicular to the horizontal line through A and it cuts at the point E.
iii) By measurement find whirl velocity at inlet
iv) Now,
Mark point A to represent distance AB (blade velocity m/s )
v) join line AC and complete inlet velocity triangle ABC
vi)By measurement find blade angle at inlet =35° and relative velocity of steam at inlet m/s
vii) draw outlet velocity triangle ABD
3) blade angle at outlet =26°
Blade efficiency
=
= ×100
=87.77%
Power developed
P=
=1587.6 kW
Axial thrust=
=8(230-155)
=600 N
4. A stage of a turbine with Parson’s blading delivers dry saturated steam 2.7 bar from the fixed blades at 90 m/s. The mean blade height is 40mm & the moving blade exit angle is 200. The axial velocity of steam is ¾ of the blade velocity at the mean radius. Steam is supplied to the stage at the rate of 9000 kg/h. The effect of the blade tip thickness on the annulus area can be neglected. Calculate
Pressure = 2.7 bar
x=1
h=40 mm=0.04m
1)
=90 sin 20 = 30.78 m/s
2)
3) Now draw combined velocity triangle as shown in figure
i)Draw a horizontal line & cutoff AB equal to 41.04m/s to suitable scale to represent blade velocity
ii)Now draw inlet velocity triangle ABC on the base AB with & BC=V=90m/s.
iii)Similarly draw outlet velocity triangle ABD on the same base AB
The mass flow steam
In this case
=0.6686
But
D=0.43m
N=1823 rpm
Diagram power=
=2.5(127)41.04
=13.03
Rate of doing work per kg/s=
=127 41.04
=5212.08 Nm/s
Energy input to the moving blade per style
=
=
=6642 Nm/s
Diagram efficiency=
=78.47%
Enthalpy drop in the stage
Enthalpy drop in moving blade =
=2592 J/kg
=2.59kJ/kg
Total enthalpy drop per stage=2 2.59
= 5.18kJ/kg
5. The outlet angle of blade of a Parson’s turbine is 200 & the axial velocity of flow of steam is 0.5 times the mean blade velocity. If the diameter of the ring is 1.25m & the rotational speed is 3000 r.p.m. Determine i) Inlet angles of blade. ii)Power developed if dry saturated steam at 5 bar passes through the blade whose height may be assumed as 6cm. Neglect the effect of blade thickness.
D=1.25m
N=3000 rpm
H=6cm=0.06m
=196 rpm
= 98m/s
Now draw the inlet velocity triangle as shown in figure
The inlet angles by measurement
=
Mass flow rate
Here at 5 bar
=
=61.57kg/sec
Power developed P=
=3861.67kw
Reference: