Chapter 7
Steam Turbines
7.1 ❐ PRINCIPLE OF OPERATION OF STEAM TURBINES
A steam turbine works on the dynamic action of steam. Steam is caused to fall in pressure in a passage or nozzle. Due to this fall in pressure, certain amount of heat energy is converted into kinetic energy to give it high velocity. The high velocity steam impinges on the moving blades of the turbine and changes the direction of motion and thus gives rise to a change in momentum, therefore, a force. Essentially, the nozzles direct the steam so that it flows in a well-formed high velocity jet. Moving buckets convert this high-velocity jet to mechanical work in a rotating shaft.
In impulse turbines, there is a drop of steam pressure and consequent development of kinetic energy takes place solely in the stationary nozzles, and the work is obtained by the conversion of this kinetic energy into work on moving blades. In the reaction turbine, only a part of the kinetic energy conversion occurs in the stationary nozzles, whereas the remainder of the kinetic energy conversion is accomplished by a pressure drop in steam as it passes through the moving blades.
7.2 ❐ CLASSIFICATION OF STEAM TURBINES
On the basis of principles of operation, steam turbines are classified as follows:
- Impulse turbine
- Reaction turbine
- Impulse-reaction turbine
- Impulse Turbine: This type of turbine is called a simple impulse turbine because the expansion of steam takes place only in one set of nozzles, as shown in Fig. 7.1. The pressure of steam falls from that in the steam nozzle to that existing in the condenser, whereas the steam flows through it. Hence, the pressure in the wheel chamber is practically equal to the condenser pressure.
The high-velocity jet coming out from the nozzle impinges on the blades, so that the whole kinetic energy is converted into mechanical energy. In practice, this type of turbine is used for small power ratings. The rotor diameter is kept small and consequently, the rotational speed becomes very high. This is known as De-Laval turbine.
- Reaction Turbine: In a reaction turbine, there are guide blades instead of nozzles which convert the pressure energy into kinetic energy. The steam passing over the moving blades has the difference of pressure at the inlet tip and exit tip; hence, there is a drop of pressure in steam while passing over the moving blades. The fixed blade serves the purpose of nozzles which changes the direction of steam and at the same time, allows it to expand to a higher velocity. The pressure of steam falls as it passes over the moving blades. The diameter of each stage of a reaction turbine must increase after each group of blade rings in order to accommodate the increased volume of steam at lower pressure.
Figure 7.2 Impulse-reaction turbine
- Impulse-Reaction Turbine: If the pressure of steam at outlet from the moving blades of the turbine is less than that at the inlet side of the blades, the drop in pressure suffered by steam during its flow through the moving blades causes a further generation of kinetic energy within the blades and adds to the preparing force which is applied to the turbine shaft; such a turbine is called impulse-reaction turbine.
This type of turbine is shown in Fig. 7.2. In this type of turbine, there is application of both principles, namely, impulse and reaction. There are several rows of moving blades fixed to the shaft and equal number of fixed blades attached to the casing. The fixed blades correspond to the nozzles of an impulse turbine. Steam is admitted for the entire circumference and therefore, there is all-round admission. In passing through the first set of fixed blades, the steam undergoes a small pressure drop and the velocity is increased. It enters the first set of moving blades and suffers the change of direction and therefore momentum. This gives an impulse to the blades. However, here, the passage to the blades is so designed that there is also a small drop in pressure in the moving blades, giving rise to an increase in kinetic energy. This drop in pressure gives rise to reactions in the direction opposite to that of added velocity. Thus, the driving force is the vector sum of impulse and reaction turbines. Normally, the turbine is known as reaction turbine. This is also called Parson’s reaction turbine.
7.3 ❐ COMPARISON OF IMPULSE AND REACTION TURBINES
The comparison of impulse and reaction turbines is given in Table 7.1.
Table 7.1 Comparison of impulse and reaction turbines
7.4 ❐ COMPOUNDING OF IMPULSE TURBINES
If the entire pressure drop from boiler pressure to condenser pressure is carried out in a single stage nozzle, the velocity of steam entering the turbine blades will be very high and consequently, the turbine speed will be very high. Such high speed turbine rotors are not useful for practical purposes and a reduction gearing is necessary between the turbine and the generator driven by the turbine. There is also a danger of structural failure of the blades due to excessive centrifugal stresses. Further, the velocity of steam at the exit of the turbine is also high when a single stage of blades is used. This gives rise to considerable loss of kinetic energy, reducing the efficiency of the unit. This loss of kinetic energy is termed as ‘carry over losses’ or ‘leaving losses’.
Compounding is a method employed for reducing the rotational speed of the impulse turbine to practical limits by using more than one stage. There are three methods of compounding impulse turbines as follows:
- Velocity compounding
- Pressure compounding
- Pressure and velocity compounding
- Velocity Compounding: This system consists of a nozzle or a set of nozzles and a wheel fitted with two or more rows of blades as shown in Fig. 7.3. It has two rings of moving blades and two rows of fixed or guide blades placed. Steam entering the nozzle expands from the initial pressure to the exhaust pressure. On passing through the first row of moving blades, the steam gives only a part of kinetic energy and issues from this row of blades with a fairly high velocity. It then enters the guide blade and the moving blade. There is a slight drop in velocity in the guide blade due to friction. While passing through the second row of moving blades, the steam suffers a change in momentum and gives a part of kinetic energy to rotor. The steam leaving from the second row of moving blades is redirected to the second row of guide blades. By doing work on the third row of moving blades, the steam finally leaves the wheel in more or less axial direction with a certain residual velocity, about 2% of the initial velocity of steam at nozzle exit. Thus, the whole of velocity is not utilised on one row of moving blades but in a number of stages, as in a Curtis turbine.
Figure 7.3 Velocity-compounded impulse turbine
- Pressure Compounding: This arrangement consists of allowing the expansion of steam in a number of steps by having a number of simple impulse turbines in the series on the same shaft, as shown in Fig. 7.4. The exhaust steam from one turbine enters the nozzles of succeeding turbines. Each of the simple impulse turbines is called a ‘stage’ of the turbine comprising its sets of nozzles and blades. This is equivalent to splitting up the whole pressure drop into a series of smaller pressure drops, hence, the term ‘pressure compounding’.
The pressure compounding causes a smaller transformation of heat energy into kinetic energy to take place in each stage than in a simple impulse turbine. Hence, the steam velocity is much lower, with the result that the blade velocity and rotational speed may be lowered. The leaving loss is only 1–2% of the total available energy. Rateau turbine is a pressure compounded turbine.
- Pressure and Velocity Compounding: It is a combination of pressure compounding and velocity compounding. The total pressure drop of steam is divided into stages and the velocity of each stage is compounded. This allows a bigger pressure drop in each stage and hence, less stages are necessary which require a shorter turbine for a given pressure drop. Such a turbine is shown in Fig. 7.5. The diameter of such a turbine increases in each stage in order to accommodate for a larger volume of steam at the lower pressure. The pressure is constant in each stage.
Figure 7.4 Pressure compounded impulse turbine
7.5 ❐ VELOCITY DIAGRAMS FOR IMPULSE STEAM TURBINE
Let u = circumferential or tangential linear velocity of blades
v_{a}_{1} _{}= absolute velocity of steam at inlet of moving blade
v_{a}_{2} = absolute velocity of steam at outlet of moving blade
v_{w}_{1 } = velocity of whirl at the entry of moving blade
= v_{a}_{1 }cos α_{1}
v_{w}_{2 } = velocity of whirl at the exit of moving blade
= v_{a}_{2 }cos α_{2}
v_{f} _{1 } = velocity of flow at inlet of moving blade
= v_{a}_{1} sin α_{1}
v_{f} _{2 } = velocity of flow at outlet of moving blade
= v_{a}_{2} sin α_{2}
v_{r}_{1 } = relative velocity of steam at entrance to moving blade
v_{r}_{2 } = relative velocity of steam at exit of moving blade
ṁ = mass rate of flow of steam, kg/s.
D = diameter of blade drum.
h = height of blade.
α_{1 }= angle which the absolute velocity of steam at inlet makes with the plane of moving blades, or nozzle angle or outlet angle of fixed blades.
α_{2 }= angle which the absolute velocity at outlet makes with the plane of moving blades or inlet angle of fixed blade.
β_{1 }= inlet angle of moving blade.
β_{2 }= exit angle of moving blade.
The jet of steam impinges on the moving blades at angle α_{1 } to the tangent of wheel with a velocity v_{a}_{1}. This velocity v_{a}_{1} has the following components.
- Tangential or whirl component v_{w}_{1} and since it is in the same direction as the motion of blades, it is the actual component which does work on the blade.
- Axial or flow component v_{f} _{1} and since it is perpendicular to the direction of motion of blade, it does no work. However, this component is responsible for the flow of steam through the turbine. This component also causes axial thrust on the rotor.
The velocity diagrams at inlet and outlet of a moving blade are shown in Fig. 7.6(a). Figure 7.6(b) shows the combined velocity diagrams.
Figure 7.6 Velocity diagrams for impulse turbine: (a) Velocity diagrams at inlet and exit, (b) Combined velocity diagrams
For blades with a smooth surface, it can be assumed that friction loss is very less or zero. However, there is always a certain loss of velocity during the flow of steam over the blade and this loss is taken into account by introducing a factor called blade velocity coefficient, K. It is given by
Note that K accounts for the loss in relative velocity due to friction of blades.
- Power Developed by the Turbine:
Work done per kg of steam,
w = Force in the direction of blade motion × Distance travelled in the direction of force
= Rate of change of momentum × Distance travelled
= [v_{w1} − (− v_{w2})] × u
= (v_{w1} + v_{w2}) u
Power developed by the turbine for a mass rate of flow of ṁ kg/s of steam,
- Diagram or Blade Efficiency, η_{d} or η_{b}:
For a single blade stage,
- Gross or Stage Efficiency, η_{s}:
Stage efficiency takes into account the losses in the nozzle.
- Axial Thrust: The axial thrust on the wheel is generated due to the difference between the velocity of flow at inlet and outlet.
Axial thrust, F_{a} = Mass rate of flow of steam × Change in axial velocity
- Energy converted into heat due to blade friction:
= Loss of kinetic energy during flow of steam over the blades
7.5.1 Condition for Maximum Blade Efficiency
Blade efficiency,
From velocity diagrams given in Fig. 7.6(b), we have
where
Now BE = AE − AB = v_{a}_{1} cos α_{1} − u
Let speed ratio,
For η_{b} to be maximum,
∴ cos α_{1} − 2ρ = 0
For symmetrical blades, β_{1} = β_{2, }so that C = 1 and with no friction over the blades, K = 1
7.5.2 Maximum Work Done
Now, w = (v_{w}_{1} ^{}+ v_{w}_{2})u
Equations (7.11) and (7.12) represent parabola. The blade efficiency has been plotted against speed ratio in Fig. 7.7, without losses and with losses for α_{1} = 20°, β_{1} = β_{2} and K = 0.85.
Figure 7.7 Blade efficiency v’s speed ratio
7.5.3 Velocity Diagrams for Velocity Compounded Impulse Turbine
The velocity diagrams for the first and second stage moving blades of a velocity-compounded impulse turbine are shown in Fig. 7.8(a) and Fig 7.8(b), respectively. Consider that the final absolute velocity of steam leaving the second row is axial. The velocity u of the blades for both the rows is the same as they are mounted on the same shaft and are of equal height.
Work done in the first row of moving blades,
If there is no friction loss, v_{r}_{1} = v_{r}_{2} and for symmetrical blades, β_{1} = β_{2}.
∴ w_{1} = 2u v_{r}_{1} cos β_{1} = 2u(v_{r}_{1} cos α_{1} − u)
Now, v_{a}_{3} = v_{a}_{2}
Work done in the second row of moving blades,
Figure 7.8 Velocity diagrams for velocity compounded impulse turbine: (a) First stage, (b) Second stage
For no friction, v_{r}_{3} = v_{r}_{4}, and for symmetrical blades, β_{3} = β_{4}.
∴ w_{2} = 2uv_{r}_{3} cos β_{3} = 2u (v_{a}_{3} cos α_{3} − u)
For α_{3} = α_{2}
∴ w_{2} = 2u [(v_{a1} cos α_{1} − 2u) − u] = 2u (v_{a1} cos α_{1} − 3u)
Total work done, w_{t} = w_{1} + w_{2}
Blade efficiency, η_{b} =
For η_{b} to be maximum,
For
where n = number of rotating blade rows in series.
7.5.4 Effect of Blade Friction on Velocity Diagrams
In an impulse turbine, the relative velocity at the outlet will be the same as the relative velocity at the inlet, if friction is neglected. In practice, there is a frictional resistance to the flow of steam jet over the blade, the effect of which is to cause a slowing down of the relative velocity. Usually, there is a loss of 10−15% in the relative velocity due to friction. Owing to frictional resistance of the blades, it will be found that v_{r}_{2} = Kv_{r}_{1}, where K is a coefficient, which takes the blade loss due to friction into account. The velocity diagrams considering blade friction are shown in Fig. 7.9. Here, the inlet diagram is first drawn and the line BC, of an unknown length, is drawn at the correct angle β_{2}. Mark off on line BD = v_{r}_{1}, the friction loss of relative velocity DD′, then BD′ = Kv_{r}_{1}. With B as centre, draw an arc of radius BD′ to cut BC at C. Then BC = v_{r}_{2} = Kv_{r}_{1}.
By joining A and C, the line AC representing v_{a}_{2} is obtained. This completes the outlet velocity diagram.
Figure 7.9 Velocity diagrams for impulse turbine considering blade friction
7.5.5 Impulse Turbine with Several Blade Rings
The stage of an impulse turbine, which is compounded for velocity, consists of alternate rings of nozzles, and moving and fixed blades. As the blade velocity u is constant for all the moving blade rings of the stage, the velocity diagrams for all the blade rings can be superimposed on the same base, represented by u. Figure 7.10 shows the velocity diagram for a stage consisting of two moving and one fixed blade rings. Let us assume that the following data is known:
- Blade velocity, u.
- Nozzle angle, α_{1}.
- The moving blade angles, β_{1} and β_{2}, which are assumed to be same for both blade rings.
- The velocity of steam v_{a}_{1} discharged from nozzle.
- Blade friction loss, 10%, i.e., K = 0.9.
The following steps may be followed for drawing the velocity diagrams:
- Draw AB = u to any convenient scale. Draw AD inclined at angle α_{1} to AB so that AD = v_{a}_{1} to the same scale. Join BD. Then BD = v_{r}_{1} for the first moving blade ring.
- Mark off DD′ = 0.1 × BD so that v_{r}_{2} = 0.9 v_{r}_{1} = BD′. With B as centre, draw an arc of radius BD′ to cut BC at C where the line BC is drawn at angle of β_{2} to BA. Join AC. Then AC = v_{a}_{2} for the first moving blade ring.
- The steam now flows over the fixed blade ring and will lose 10% of its velocity during the passage. Hence, mark off CC′ = 0.1 × AC. With A as centre, draw on arc of radius AC′ to cut BD at G. Then AG = v_{a}_{3} representing the steam velocity when entering the second moving blade ring. The velocity diagram for the second blade ring is triangle AGB. BG = v_{r}_{3}, the relative velocity.
Figure 7.10 Velocity diagrams for impulse turbine considering friction with several blade rings
- The steam now flows over the second moving blade and loses one-tenth of its relative velocity due to friction., Hence, mark off GG′ = 0.1 × BG so that BG′ = 0.9 BG. With B as centre and radius BG′, draw an arc to cut BC at H, then AH = v_{a}_{4} and BH = v_{r}_{4}.
α_{2} = angle of discharge from first moving blade
α_{4} = angle of discharge from second moving blade
α_{3} = outlet angle of fixed blade.
Work done per kg of steam for first moving blade ring,
w_{1} = (v_{w1} + v_{w2})uWork done per kg of steam for second moving blade ring.
w_{2} = (v_{w3} + v_{w4})uTotal Work done, w_{t} = w_{1} + w_{2}
= [(v_{w1} + v_{w2}) + (v_{w3} + v_{w4})]uPower developed per stage =
Blade efficiency, η_{b} =
Stage efficiency, η_{s} =
Total axial thrust =
7.6 ❐ ADVANTAGES AND LIMITATIONS OF VELOCITY COMPOUNDING
7.6.1 Advantages
- Fewer number of stages are required and therefore, the initial cost is less.
- The space required is less.
- The pressure in the housing is considerably less which requires cheaper and thin casing for the turbine.
7.6.2 Limitations
- The friction losses are larger due to the high velocity of steam.
- The maximum blade efficiency and efficiency range decrease with the increase in the number of stages.
- The power developed in each successive blade row decreases with increase in number of rows.
- All the stages are not used with equal economy.
7.7 ❐ VELOCITY DIAGRAMS FOR IMPULSE-REACTION TURBINE
The steam passes to the fixed blades with velocity v_{a}_{1}. Fixed blades also act as nozzles with pressure drop occurring, while steam passes through them so that there is gain in kinetic energy and steam leaves the fixed blades with velocity v_{a}_{2}. The expansion of steam and the enthalpy drop occur in fixed and moving blades. The velocity diagrams for an impulse-reaction turbine are shown in Fig. 7.11.
- Degree of Reaction: The degree of reaction R_{d} of a reaction turbine is defined as the ratio of enthalpy drop over moving blades to the total enthalpy drop in the stage. Thus
Enthalpy drop through fixed blades,
Enthalpy drop along the moving blades, ∆h_{m} =
Kinetic energy supplied to moving blades, K.E.
Neglecting friction of blade surface,
Axial thrust on rotor = (v_{f }_{1} − v_{f }_{2}) + (∆p)_{m} × area of blade disc
Now total enthalpy drop in the stage = Work done by steam in the stage
Δh_{f} + Δh_{m} = u (v_{w1} + v_{w2})Figure 7.11 Velocity diagrams for impulse-reaction turbine
From Fig. 7.11, we have
v_{r2} = v_{f2} cosec β_{2} and v_{r1} = v_{f1} cosec β_{1}and v_{w}_{1} + v_{w}_{2} = v_{f }_{1} cot β_{1} + v_{f }_{2} cot β_{2}
The velocity of flow generally remains constant through the blades.
∴ v_{f }_{1} = v_{f }_{2 } = _{}v_{f}
If R_{d} = 0.5, then
From Fig. 7.11, we have
u = v_{f} (cot β_{2} − cot α_{2}) = v_{f} (cot α_{1} − cot β_{1})This means that the moving and fixed blades must have the same shape if the degree of reaction is 50%. This condition gives symmetrical velocity diagrams. This type of turbine is known as Parson's Reaction Turbine.
- Efficiency: We assume that the degree of reaction is 50%, that is, ∆h_{f} = ∆h_{m}, the moving and fixed blades are of the same shape, and the velocity of steam at exit from the preceding stage is the same as the velocity of steam at the entrance to the succeeding stage.
The work done per kg of steam, w = u(v_{w}_{1} + v_{w}_{2}) = u [v_{a}_{1} cos α_{1} + (v_{r}_{2} cos β_{2} − u)]
Now, β_{2} = α_{1} and v_{r}_{2} = v_{a}_{1}
where is the speed ratio.
K.E. supplied to fixed blade =
K.E. supplied to moving blade
Total energy supplied to stage,
For symmetrical blades, v_{r}_{2} = v_{a}_{1}
From Fig. 7.11, we have
For η_{b} to be maximum,
(1 + 2ρ cos α_{1} − ρ^{2}) (4 cos α_{1} − 4ρ) − 2ρ (2 cos α_{1} − ρ) (2 cos α_{1} − ρ) = 04 (cos α_{1} − ρ) (1 + 2ρ cos α_{1} − ρ^{2}) − 4ρ (cos α_{1} − ρ) (2 cos α_{1} − ρ) = 0(cos α_{1} − ρ) [(1 + 2ρ cos α_{1} − ρ^{2}) − ρ (2 cos α_{1} − ρ)] = 0∴ cos α_{1} − ρ^{2} = 0
- Height of Turbine Blading:
The height and thickness of blading are shown in Fig. 7.12.
Volume flow = Area × flow velocity
where d = mean diameter of turbine wheel
h_{1}_{,} h_{2} = blade heights at inlet and outlet respectively
t_{1}_{,} t_{2} = thickness of blades at inlet and outlet respectively.
n = number of blades
For most of the turbines,
v_{f1} = v_{f2} = v_{f}, h_{1} = h_{2 }= h and t_{1} = t_{2} = tEquation (7.31) gives height of blades.
The pitch of blades, p is given by:
ṁv_{s} = [n (p - t)h] v_{f}Generally t << p,
∴ ṁv_{s} = n ph v_{f} = π dhv_{f}
7.8 ❐ REHEAT FACTOR
The expansion of steam through a number of stages of turbine is shown in Fig. 7.13. A_{1}B_{1} represents isentropic expansion in the first stage. The actual state of steam with frictional reheating is shown by point A_{2}. Therefore, the actual heat drop in the first stage is A_{1}C_{1}. Similarly, the isentropic and actual heat drops in the succeeding stages are shown by A_{2}B_{2}, A_{3}B_{3}, ….., A_{2}C_{2}, A_{3}C_{3}, ….., and so on. A_{1}D represents the isentropic heat drop between supply and condenser state of steam.
Stage efficiency,
For first stage,
For second stage, and so on.
Turbine internal efficiency
If η_{s}_{1} = η_{s}_{2} = ….. = η_{s′} then
From Eqs. (7.34) and (7.35), we get
Combining Eqs (7.33) and (7.36), we get
Reheat factor is always greater than one. Normally R_{f} lies between 1.02 to 1.05. The line passing through A_{1}, A_{2}, … is called the condition curve.
7.9 ❐ LOSSES IN STEAM TURBINES
The various losses in steam turbines and their causes are as follows:
- Residual Velocity Loss: This loss occurs due to the absolute exit velocity of steam and is equivalent to where v_{a}_{2} is the absolute exit velocity of steam. In a single stage turbine, it may be about 10−12% and can be reduced by using multi-stages.
- Frictional and Turbulence Loss: Friction loss mainly occurs in nozzles and turbine blades. The nozzle efficiency is used to account the friction loss. The loss due to friction and turbulence is about 10%.
- Leakage Loss: It occurs at the following points:
- Between the turbine shaft and bearings
- Between the shaft and stationary diaphragms carrying nozzles, and blade tips
- Through the labyrinth glands
The total leakage loss is about 1−2%.
- Mechanical Friction Loss: Friction between shaft, bearing, and regulating valves accounts for this loss. It can be reduced by proper lubrication.
- Wet Steam Loss: The velocity of water particles is less than that of steam; therefore, water particles have to be dragged with the steam causing loss of kinetic energy.
- Radiation Loss: This is due to much higher temperature of turbine as compared to the surroundings. This loss can be reduced by proper insulation.
7.10 ❐ TURBINE EFFICIENCIES
- Blade or diagram efficiency, η_{b}
- Stage efficiency,
- Internal efficiency,
- Overall or turbine efficiency,
- Net efficiency or efficiency ratio,
7.11 ❐ GOVERNING OF STEAM TURBINES
The purpose of governing of steam turbines is to maintain the speed of a turbine sensibly constant, irrespective of the load. The various methods of governing are as follows:
- Throttle governing
- Nozzle control governing
- By-pass governing
- Combination of throttle governing and nozzle control governing
- Combination of throttle governing and by-pass governing
- Throttle governing: The line diagram of throttle governing is shown in Fig. 7.14. The steam flow to the turbine is throttled by a balanced throttle valve actuated by a centrifugal governor. An oil differential relay is incorporated to magnify the small force produced by the governor for a small change of speed to actuate the throttle valve. The throttle valve is moved by a relay piston. A floating differential lever is fixed to its one end and a piston valve is fixed at some intermediate point. The pilot piston valve consists of two piston valves covering ports without any overlap. The piston valves are also operated by lubricating oil supplied by a pump at about 3 to 4 bar. The oil from this chamber is returned by the oil drain.
Operation: Let the turbine work at full-rated load at the rated constant speed. If the load is reduced, the energy supplied to the turbine will be in excess and the turbine rotor will accelerate. Thus, the governor sleeve will lift. As the throttle valve position is assumed to be the same momentarily, the pilot piston valve spindle will get lifted, opening the upper port to oil pressure and lower port to oil return. The relay piston will thus close the throttle valve partially. The lowering of the throttle valve spindle will lower the pilot piston spindle and close the ports. As soon as the ports are closed, the relay piston gets stabilised in one position corresponding to the reduced load. The change in the available enthalpy for throttle governing is shown in Fig. 7.15.
- Nozzle control governing: The principle of nozzle control governing is accomplished by uncovering as many steam passages as are necessary to meet the load by poppet valves. An arrangement is shown in Fig. 7.16. The nozzles are divided into groups N_{1}, N_{2}_{,} and N_{3} under the control valves V_{1}, V_{2}_{,} and V_{3}, respectively. The number of nozzle groups may vary from three to five or more.
- By-pass governing: The principle of by-pass governing is shown in Fig. 7.17. Steam entering the turbine passes through the main throttle valve which is under the control of the speed governor and enters the nozzle box or the steam chest. For loads greater than the economical load, a by-pass valve is opened, allowing steam to pass from the first stage, nozzle box into the steam chest, and so into the nozzles of the fourth stage. The by-pass valve is not opened until the lift of the throttle valve exceeds a certain amount. As the load diminishes, the by-pass valve closes first. The by-pass valve is under the control of speed governor for all loads.
Figure 7.15 T-s diagram for throttling
Figure 7.16 Nozzle governing
7.12 ❐ LABYRINTH PACKING
Labyrinth seals are characterised as controlled clearance seals without rubbing contact with the moving parts and with some tolerable leakage. The fluid throttling is achieved in steps, using a series of small chambers, where a sudden irreversible acceleration with subsequent deceleration of the leaking fluid takes place. Every step down in this process of pressure dissipation is accompanied by a loss in pressure. Due to the lack of direct rubbing contact with the moving shaft, the labyrinth seal is well suited for sealing shafts operating at high rotational speeds, as in centrifugal compressors and steam turbines. It requires neither lubrication nor maintenance. The simplest design of a labyrinth packing is with a straight shaft and a straight housing is shown in Fig. 7.18(a). The modified labyrinth designs are of the staggered, stepped, and interference configurations (Figs. 7.18(b)–(e)).
Figure 7.18 Labyrinth seal designs: (a) Straight shaft and straight housing, (b) Staggered, (c) Staggered and stepped, (d) Stepped, (e) Interference
7.13 ❐ BACK PRESSURE TURBINE
There are several industries, such as paper making, textile, chemical, dyeing, sugar refining, and so on, in which there is a dual demand for power and steam for heating and process work. Producing steam separately for power and heating is wasteful. If the turbine is operated with normal exhaust pressure and the temperature of the exhaust steam is too low to be of any use for heating purposes, by suitable modification of the initial and exhaust pressures, it would be possible to generate the required power and still have available for process work a large quantity of heat in the exhaust steam.
Figure 7.19 Back pressure turbine plant
The back pressure turbine may be used in cases where the power, which may be generated by expanding steam from an economical initial pressure down to the heating pressure, is equal to, or greater than the power requirements. The layout of such a plant is shown in Fig. 7.19.
Steam is generated in the boiler at a suitable working pressure and admitted to the turbine. The exhaust steam from the turbine will normally be superheated and in most cases, is not suitable for process work. A de-superheater is used to make it suitable for process work by spraying a jet of water, thermostatically controlled, on the entering steam. The steam is cooled and the spray water evaporated. The new saturated steam enters the heaters and is entirely condensed. The condensed steam may or may not be returned to the boiler. The thermodynamics of back pressure turbine is shown in Fig. 7.20.
7.14 ❐ PASS OUT OR EXTRACTION TURBINE
In many cases, the power available from a back pressure turbine through which the whole of the heating steam flows is appreciably less than that required in the factory. In such cases, it would be possible to install a back pressure turbine to generate additional power. However, generally, the functions of both machines are combined in a single turbine. Such a turbine is shown in Fig. 7.21. The steam enters the turbine and expands through the high pressure stages before the extraction branch. Here, a certain quantity of steam is continuously being extracted for heating purposes, the remainder passes through the pressure control valve into the low-pressure part of the turbine.
7.15 CO-GENERATION
The exhaust steam from a steam power plant is often rejected to the atmosphere. When the heat of exhaust steam is used for heating of buildings, and heating required by many industrial processes, the functions of heating and power production can often be combined effectively. This combination is often called co-generation (Fig. 7.22). The process heater replaces the condenser of an ordinary Rankine cycle. The pressure at the exhaust from the turbine is the saturation pressure corresponding to the temperature desired in the process heater. Such a turbine is called back pressure turbine.
Figure 7.22 Co-generation
Example 7.1
Steam at 4.9 bar and 160°C is supplied to a single stage impulse turbine at the rate of 60 kg/min. From there, it is exhausted to a condenser at a pressure of 0.196 bar. The blade speed is 300 m/s. The nozzles are inclined at 25° to the plane of the wheel and outlet blade angle is 35°. Neglect friction losses and estimate (a) the theoretical power developed by the turbine, (b) the diagram efficiency, and (c) the stage efficiency. [IES, 1990]
Solution
Given that p_{1} = 4.9 bar, t_{1} = 160°C, ṁ = 60 kg/min, p_{2} = 0.196 bar, u = 300 m/s, a_{1} = 25°, β_{2} = 35°, K = 1
From Mollier diagram, at 4.9 bar and 160°C, h_{1} = 2700 kJ/kg
Assuming isentropic expansion, at 0.196 bar, h_{2} = 2200 kJ/kg
Inlet velocity of steam to turbine = outlet velocity of steam from nozzle
The velocity diagrams are shown in Fig. 7.23.
or v_{r}_{1 } = 739 m/s
Example 7.2
The first stage of an impulse turbine is compounded for velocity and has two rings of moving blades and one ring of fixed blades. The nozzle angle is 20° and the leaving angles of the blades are respectively: first moving 20°, fixed 25° and second moving 30°. The velocity of steam leaving the nozzle is 600 m/s, the blade speed is 125 m/s, and the steam velocity relative to the blades is reduced by 10% during the passage through each ring. Find the diagram efficiency under these conditions and the power developed for a steam of 4 kg/s.
Solution
Given that u = 125 m/s, v_{a}_{1} = 600 m/s, K = 0.9, α_{1} = 20°, β_{2} = 20°, β_{3} = 25°, β_{4} = 30°, ṁ = 4 kg/s
Choose a scale of 1 cm = 50 m/s. The velocity diagrams are drawn in Fig. 7.24.
Join bd. By measurement, bd = 9.7 cm, bd′ = 9.7 × 0.9 = 8.7 cm
Join ac. By measurement, ac = 6.4 cm, ac′ = 0.9 × 6.4 = 5.7 cm
Figure 7.24 Velocity diagrams for a two stage impulse turbine
Join ah
Blade efficiency, η_{b} =
Power developed =
Example 7.3
In an impulse turbine, the mean diameter of the blades is 1.05 m and the speed is 3000 rpm. The nozzle angle is 18°, speed ratio is 0.42, and the friction factor is 0.84. The outlet blade angle is to be made 3° less than the inlet angle. The steam flow is 10 kg/s. Draw the velocity diagrams for the blades and calculate (a) the tangential thrust, (b) the axial thrust, (c) the resultant thrust, (d) the power developed, and (e) the blading efficiency.
Solution
Given that d_{m} = 1.05 m, N = 3000 rpm, α_{1} = 18°, ρ = 0.42, K = 0.84, β_{2} = β_{1} − 3°, ṁ = 10 kg/s
Blade speed, u =
Draw the velocity diagrams as shown in Fig. 7.25.
Figure 7.25 Velocity diagrams for an impulse turbine
or β_{1} = 30.2°
- Tangential thrust ṁ = (v_{w}_{1} + v_{w}_{2}) = 10 (373.48 + 15.33) = 3888 N
- Axial thrust = ṁ (v_{f} _{1} − v_{f} _{2}) = 10 (121.35 − 92.64) = 287.1 N
- Resultant thrust = [(3888)^{2} + (287.1)^{2}]^{1}/^{2} = 3898.6 N
- Power developed
- Blade efficiency, η_{b} =
Example 7.4
One stage of impulse turbine consists of a converging nozzle ring and 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 nozzle is 660 m/s, find the blade speed so that the steam passes on without shock. Find the diagram efficiency neglecting losses if the blades are run at this speed.
[IES, 1992]
Solution
Given that α_{2} = 22°, β_{1} = β_{2} = 35°, v_{a}_{1} = 660 m/s
The velocity diagrams are shown in Fig. 7.26.
v_{f} _{1} = v_{a}_{1} sin 22° = v_{r}_{1} sin 35°
or (431.05)^{2} = u^{2} + (660)^{2} − 2u × 660 cos 22°
or 185804 = u^{2} + 435600 − 1223.88 u
or u^{2} − 1223.88 u + 249796 = 0
For K = 1, v_{r}_{2} = v_{r}_{1} = 431.05 m/s
Example 7.5
A single-stage steam turbine is supplied with steam at 5 bar, 200°C at the rate of 50 kg/min. It expands into a condenser at a pressure of 0.2 bar. The blade speed is 400 m/s. The nozzles are inclined at an angle of 20° to the plane of the wheel and the outlet blade angle is 30°. Neglecting friction losses, determine the power developed, blade efficiency, and stage efficiency.
[IES, 1994]
Solution
Given that p_{1} = 5 bar, t_{1} = 200^{o}C, p_{2} = 0.2 bar, ṁ = 50 kg/min, u = 400 m/s, α_{1} = 20°, β_{2} = 30°, v_{r}_{1} = v_{r}_{2} as K = 1
At 5 bar, 200°C, h_{1} = 2855.4 kJ/kg, s_{1} = 7.0592 kJ/kg.K
At 0.2 bar, s_{f} _{2} = 0.8319 kJ/kg. K, s_{fg}_{2} = 7.0766 kJ/kg.K
or 7.0592 = 0.8319 + 7.0766 x_{2}
or x_{2} = 0.88
Enthalpy drop, ∆h = h_{1} − h_{2} = 2855.4 − 2326.68 = 528.7 kJ/kg
or β_{1} = 31.84°
Power developed,
Blade efficiency, η_{b } = = 0.8652 or 86.52%
Example 7.6
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 nozzle is 600 m/s. The turbine uses 3600 kg/h of steam. The absolute velocity at exit is along the axis of the turbine. Determine: (a) the inlet and exit angles of blades, (b) the power output of the turbine, (c) the diagram efficiency, and (d) the axial thrust (per kg steam per second). Assume inlet and outlet angles to be equal.
[IES, 1998]
Solution
Given that u = 200 m/s, α_{1} = 20°, v_{a1} = 600 m/s, ṁ = 3600 kg/h, v_{a2} = v_{f}_{2}
The velocity diagrams are shown in Fig 7.27.
- v_{f}_{1} = v_{a1} sin α_{1} = 600 sin 20° = 205.2 m/s
v_{w}_{1} = v_{a1} cos α_{1} = 600 cos 20° = 563.8 m/s
or b_{1} = 29.42°
β_{2} = β_{1} = 29.42°v_{w2} = 0 - Power output of turbine =
- Diagram efficiency, = = 0.6264 or 62.64%
- Axial thrust, F_{a} = ṁ(v_{f1} - v_{f2})
v_{f }_{2} = u tan β_{2} = 200 tan 29.42° = 112.79 m/sF_{a} = 1 × (205.2 − 112.79) = 92.41 N
Example 7.7
Steam at 2.94 bar absolute and dry saturated comes out of rotor of a reaction turbine having identical bladings. The velocity of steam entering into the turbine is 100 m/s. The mean blade height is 4 cm and the exit angle of the moving blade is 20°. The blade velocity is 4/3 times of axial flow velocity at the mean radius. If the steam flow rate is 2.5 kg/s, find (a) the rotor speed, (b) the power developed, (c) the diagram efficiency, (d) the percentage increase in relative velocity in the moving blades and (e) the enthalpy drop of steam in blade passage.
[IES, 1999]
Solution
The velocity triangles are shown in Fig. 7.28.
Given that p = 2.94 bar, h = 4 cm, v_{a}_{1} = 100 m/s, ṁ = 2.5 kg/s, α_{1} = β_{2} = 20°, β_{1} = α_{2}, u = (4/3)v_{f}
From velocity triangles, we have (Fig. 7.28)
Specific volume of steam at 2.94 bar (from steam tables), v_{s} = 0.6173 m^{3}/kg
Example 7.8
In a multi-stage Parson’s reaction turbine, at one of the stages, the rotor diameter is 125 cm and the speed ratio 0.72. The speed of the rotor is 3000 rpm. Determine: (a) the blade inlet angle if the blade outlet angle is 22°, (b) diagram efficiency, and (c) percentage increase in diagram efficiency and rotor speed if the turbine is designed to run at the best theoretical speed.
[IES, 2000]
Solution
Given that d = 1.25m, N = 3000 rpm, = 0.72, α_{1} = β_{2} = 22°
- m/s
In Parson’s reaction turbine, (see Fig. 7.29)
v_{r2} = v_{a1}, v_{r1} = v_{a2} and a_{2} = b_{1}v_{f1} = v_{a1} sin α_{1} = 272.7 sin 22° = 102.15 m/sv_{w1} = v_{a1} cos α_{1} = 272.7 cos 22° = 252.84 m/sor β_{1} = 61°
v_{r2} = 272.7 m/sFigure 7.29 Velocity diagrams for Parson’s reaction turbine
- v_{w2} = v_{r2} cos β_{2} - u = 272.7 cos 22° − 196.35 = 56.5 m/s
Diagram efficiency,
- Power developed,kW
At best theoretical rotor speed, u = v_{a1} cos α_{1} = 272.6 cos 22° = 252.84 m/s
Percentage increase in rotor speed = = 0.2877 or 28.77%
Diagram efficiency =
Percentage increase in diagram efficiency =
Example 7.9
A De-Laval turbine has a mean blade velocity of 180 m/s. The nozzles are inclined at 17° to the tangent. The steam flow velocity through the nozzles is 550 m/s. For a mass flow of 3300 kg/hr and for axial exit conditions, find (a) the inlet and outlet angles of the blade system, (b) the power output, and (c) the diagram efficiency.
[IES, 2001]
Solution
Given that u = 180 m/s, v_{a1} = 550m/s, ṁ = 3300 kg/h, α_{1} = 17°, α_{2} = 90°
The velocity diagrams are shown in Fig. 7.30.
- v_{f1} = v_{a2} (see Fig. 7.30)
v_{f1} = v_{a1} sin α_{1} = 550 sin 17° = 160.8 m/s
tan β_{2} = v_{f1}/u = 160.8/180 = 0.89336
or β_{2} = 41.78°
tan β_{1} = v_{f1}/(v_{a1} cos α_{1} − u)= 160.8/(550 cos 17° − 180) = 0.46478or β_{1} = 24.93°
- Power output, P = ṁuv_{w1} = (3300/3600) × (180 × 550 cos 17°/10^{3}) = 85.785 kW
- Diagram efficiency = = 2 × 180 × 550 cos 17°/550^{2} = 0.626 or 62.6%
Example 7.10
A steam turbine is governed by throttling. The full load output of a steam turbine, measured at the coupling is 5 MW and the losses due to bearing friction, the governor and oil pump drive, etc., are 200 kW. The steam is supplied at a pressure of 20 bar and 300°C. The exhaust pressure of steam is 0.07 bar. The internal efficiency ratio at full load is 0.75. Calculate the coupling power of turbine when the steam flow through the turbine is 20% of that at full load. Assume that the external losses are the same as that at full load, exhaust pressure is the same, and the internal efficiency ratio is reduced to 70% at this load.
[IES, 2005]
Solution
The h–s (Mollier) diagram is shown in Fig 7.31.
At 20 bar, 300°C, h_{1} = 3020 kJ/kg
At 0.07 bar, h_{2} = 2100 kJ/kg
At full load:
Internal efficiency,
or
or h_{2'} = 2330 kJ/kg
Internal power developed,
or 5000 = ṁ × 690
or ṁ = 7.246 kg/s
Nozzle box pressure at part load = steam mass flow ratio × p_{1}
Coupling power = 679.71 – 200 = 479.71 kW
Example 7.11
The turbine rotor has a mean blade ring diameter of 500 mm and the blade angles are equal. The nozzle angle is 20° and the steam leaves the nozzles with a velocity of 900 m/s. Assuming a blade friction factor of 0.85, determine the (a) the best blade angles, (b) the turbine speed in rpm, (c) the steam consumption rate in kg/h if the turbine generates 10 kW, and (d) the maximum blade efficiency.
[IES, 2006]
Solution
Given that α_{1} = 20°, d_{m} = 0.5 m, v_{a}_{1} = 900 m/s, K = 0.85, β = β_{1} = β_{2} so that C = 1, P = 10 kW
The velocity diagrams are shown in Fig. 7.32.
For maximum blade efficiency, the speed ratio,
Power developed by turbine, P =
Mass flow rate, = 0.03023 kg/s or 108.83 kg/h
Maximum blade efficiency,
or v_{r}_{1} = 523.03 m/s
or β = 36°
7.16 ❐ EROSION OF STEAM TURBINE BLADES
The steam turbine blades are subjected to high pressure and temperature. In the intermediate pressure stages, the steam is wet. Therefore, the material of blades should be able to withstand corrosion and erosion due to the presence of water particles. In addition to corrosion and erosion, blades are subjected to high centrifugal stresses. When the speed is high and the dryness fraction is less than 0.9, the effect of moisture is the most prominent. The most affected position is the back of the inlet edge of the blade, where either grooves are formed or even some portion breaks away. Due to centrifugal force, the water particles tend to concentrate in the outer annulus and their tip speed is greater than the root speed. Hence, erosion effect is the most prominent on the tips, as shown in Fig. 7.33.
The following methods may be adopted to prevent erosion:
- By raising the temperature of steam at inlet so that at the exit of the turbine, the dryness fraction does not fall below 0.9.
- By adopting reheat cycle so that the dryness fraction at exit remains within limits.
- By providing drainage belts on the turbine so that the water droplets which are on outer periphery, due to centrifugal force, are drained.
- By providing a shield of a hard material on the leading edge of the turbine.
The most satisfactory solution to prolong the blade life is providing tungsten shield.
Example 7.12
In a De-Laval turbine, steam enters the wheel through a nozzle with a velocity of 500 m/s and at an angle of 20° to the direction of motion of the blade. The blade speed is 200 m/s and the exit angle of the moving blade is 25°. Find the inlet angle of the moving blade, exit velocity of steam and its direction and work done per kg of steam.
Solution
Given that v_{a}_{1} = 500 m/s; α_{1} = 20°; u = 200 m/s; β_{2} = 20°, v_{r}_{2} = v_{r}_{1}_{.}
Now let us draw the combined velocity triangles, as shown in Fig. 7.34, as explained below:
- First, draw a horizontal line and cut off AB equal to 200 m/s, to some suitable scale, representing the blade speed, u.
- Now at B, draw a line BC at an angle of 20° (nozzle angle, α_{1}) and cut off BC equal to 500 m/s to the same scale to represent the velocity of steam jet entering the blade (v_{a}_{1}).
- Join AC, which represents the relative velocity at inlet (v_{r}_{1}).
- At A, draw a line AD at an angle of 25° (exit angle of the moving blade, β_{2}). Now with A as centre, and radius equal to AC, draw an arc meeting the line through A at D.
- Join BD, which represent the velocity of steam jet at outlet (v_{a}_{2}).
- From C and D, draw perpendiculars meeting the line AB produced at E and F respectively. CE and DF represent the velocity of flow at inlet (v_{f} _{1}) and outlet (v_{f} _{2}) respectively.
The following values are measured from the velocity diagram:
By measurement from the velocity diagram, the inlet angle of the moving blade, β_{1} = 32°
By measurement from the velocity diagram, the exit velocity of steam, v_{a}_{2} = 165 m/s
By measurement from the velocity diagram, the direction of the exit steam, α_{2} = 59°
Work done per kg of steam = ṁ (v_{w}_{1} + v_{w}_{2})
Example 7.13
The velocity of steam leaving the nozzles of an impulse turbine is 1200 m/s and the nozzle angle is 20°. 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 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 that v_{a}_{1} = 1200 m/s; α_{1} = 20°; u = 375 m/s; K = v_{r}_{2}/v_{r}_{1} = 0.75; ṁ = 0.5 kg/s; β_{1} = β_{2}, for symmetrical blading.
Now draw the combined velocity triangle, as shown in Fig. 7.35, as explained below:
- First, draw a horizontal line, and cut off AB equal to 375 m/s to some suitable scale representing the velocity of blade(u).
- Now at B, draw a line BC at an angle of 20° (Nozzle angle, α_{1}) and cut off BC equal to 1200 m/s to the scale to represent the velocity of steam jet entering the blade (v_{a}_{1}).
- Join CA, which represents the relative velocity at inlet (v_{r}_{1}). By measurement, we find that CA = v_{r}_{1} = 860 m/s. Now cut off AX equal to v_{r}_{2} = v_{r}_{1} × K = 860 × 0.75 = 645 m/s to the scale to represent the relative velocity at exit (v_{r}_{2}).
- At A, draw a line AD at an angle β_{2} equal to the angle β_{1}, 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 = v_{r}_{2}.
- Join BD, which represents the velocity of steam jet at outlet (v_{a}_{2}).
- From C and D, draw perpendiculars meeting the line AB produced at E and F respectively. CE and DF represents the velocity of flow at inlet (v_{f} _{1}) and outlet (v_{f} _{2}), respectively.
The following values are measured from the velocity diagram:
- By measurement from the velocity diagram, the blade angle at inlet, β_{1} = 29°
- Driving force on the wheel, F_{x} = m (v_{w}_{1} + v_{w}_{2}) = 0.5 (1130 + 190) = 660 N
- Axial thrust on the wheel, F_{y} = m (v_{f} _{1} − v_{f} _{2}) = 0.5 (410 − 310) = 50 N
- Power developed by the turbine,
Example 7.14
A single row impulse turbine receives 3 kg/s steam with a velocity of 425 m/s. The ratio of blade speed to jet speed is 0.4 and the stage output is 170 kW. If the internal losses due to disc friction etc., amount to 15 kW, determine the blade efficiency and the blade velocity coefficient. The nozzle angle is 16° and the blade exit angle is 17°.
Solution
Given that ṁ = 3 kg/s; v_{a}_{1} = 425 m/s; u/v_{a}_{1} = 0.4; stage output = 170 kW; internal losses = 15 kW; α_{1} = 16°; β_{2} = 17°.
- Blade speed, u = v_{a}_{1} × 0.4 = 425 × 0.4 = 170 m/s
Total power developed, P = Stage output + Internal losses
= 170 + 15 = 185 kWLet v_{w}_{1} + v_{w}_{2} = Change in the velocity of whirl
Power developed, P = ṁ (v_{w}_{1} + v_{w}_{2}) u
185 × 10^{3} = 3 (v_{w1} + v_{w2}) 170∴ v_{w}_{1} + v_{w}_{2} = 363 m/s
Now, draw the combined velocity triangle, as shown in Fig. 7.36, and explained below:
- First, draw a horizontal line and cut off AB equal to 170 m/s, to some suitable scale, to represent the blade speed (u).
- Now, draw the inlet velocity triangle ABC on the base AB with α_{1} = 16° and v_{a}_{1} = 425 m/s, to the scale, chosen.
- Similarly, draw the outlet velocity triangle ABD on the same base AB with β_{2} = 17° and (v_{w}_{1} + v_{w}_{2}) = 363 m/s to the scale.
- From C and D, draw perpendiculars to meet the line AB at E and F. From the geometry of the figure, we find that v_{w}_{2} is in the opposite direction of v_{w}_{1}. Therefore, (v_{w}_{1} − v_{w}_{2}) = 363 m/s.
By measurement from the velocity diagram, we find that
Relative velocity at inlet, v_{r}_{1} = 265 m/s
and relative velocity at outlet, v_{r}_{2} = 130 m/s
Blading efficiency,
- Blade velocity coefficient,
Example 7.15
In one stage of a reaction steam turbine, both the fixed and moving blades have inlet and outlet blade tip angles of 35° and 20°, 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 for the pair is 23.5 kJ per kg.
Solution
Given that β_{1} = α_{2} = 35°; β_{2} = α_{1} = 20°; u = 80 m/s; ṁ = 22500 kg/h = 6.25 kg/s; Δh = 23.5 kJ/kg.
Now, let us draw the combined velocity triangle, as shown in Fig. 7.37 and explained below:
- First, draw a horizontal line and cut off AB equal to 80 m/s (u) to some suitable scale.
- Now at B, draw in line BC at an angle α_{1} = 20° with AB. Similarly, at A draw a line AC at an angle β_{1} = 35° with BA meeting the first line at C.
- At A, draw a line AD at angle β_{2} = 20° (because β_{2} = α_{1}) with AB. Similarly, at B draw a line BD at an angle α_{2} = 35° (because α_{2} = β_{1}) with AB meeting the first line at D.
- 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,
Power developed in the pair,
Example 7.16
A 50% reaction turbine with a mean blade diameter of 1 m runs at a speed of 50 rps. The blades are designed with exit angles of 50° and inlet angles of 30°. If the turbine is supplied with steam at the rate of 20 kg/s and gross efficiency is 85%, determine the following: (a) power output of the stage; (b) specific enthalpy drop in the stage; and (c) percentage increase in relative velocity in the moving blades due to steam expansion.
Solution
Given that d_{m} = 1 m; N = 50 rps; β_{1} = α_{2} = 50°; β_{2} = α_{1} = 30°; m = 20 kg/s; η_{s} = 85%
- Blade velocity, u = π d_{m} N = π × 1 × 50 = 157 m/s
Now let us draw the combined velocity triangle, as shown in Fig. 7.38, and explained below:
- First, draw a horizontal line and cut off AB equal to 157 m/s, to some suitable scale, to represent the blade velocity u.
- Now, draw the inlet velocity triangle ABC on the base AB with α_{1} = 30° and β_{1} = 50°.
- Similarly, draw the outlet velocity triangle ABD on the same base AB with β_{2} = 30° and α_{2} = 50°.
- From C and D, draw perpendiculars to meet the line AB produced at E and F.
By measurement from the velocity triangle, we find that change in the velocity of whirl.
v_{w1} + v_{w2} = EF = 450 m/sRelative velocity at inlet, v_{r}_{1} = CA = 230 m/s
and relative velocity at outlet, v_{r}_{2} = DA = 350 m/s
Power output of the stage,
P = ṁ (v_{w1} + v_{w2}) u = 20 × 450 × 157 = 1413000 W = 1413 kW. - Let Δh = specific enthalpy drop in the stage.
We know that stage efficiency (η_{s});
- Increase in the relative velocity in the moving blades due to steam expansion.
Example 7.17
At a stage of a 50% reaction turbine, the rotor diameter is 1.4 m and speed ratio 0.7. If the blade outlet angle is 20° and the rotor speed 3000 rpm, find the blade inlet angle and diagram efficiency.
Find the percentage increase in diagram efficiency and rotor speed, if the turbine is designed to run at the best theoretical speed.
Solution
Given that D = 1.4 m; ρ = u/v_{a}_{1} = 0.7; α_{1} = β_{2} = 20°; N = 3000 rpm.
Blade velocity,
and velocity of steam at inlet to the blade.
Now draw the combined velocity triangle, as shown in Fig. 7.39, as explained below:
- First, draw a horizontal line and cut off AB equal to 220 m/s, to some suitable scale, to represent the blade velocity (u).
- Now, draw the inlet velocity triangle ABD on the same base AB with α_{1} = 20° and v_{a}_{1} = 314.3 m/s, to the scale.
- Similarly, draw the outlet velocity triangle ABD on the same base AB with β_{2} = 20° and v_{r}_{2} = 314.3 m/s to the scale.
- From C and D, draw perpendiculars to meet the line AB produced at E and F.
By measurement from velocity diagram, we find that the blade inlet angle.
By measurement from velocity diagram, we find that velocity of steam at outlet,
Diagram efficiency,
Maximum efficiency of the turbine,
∴ Percentage increase in diagram efficiency = = 0.131 or 13.1%
Let N_{1} = Maximum rotor speed.
For best theoretical speed (or in other words, for maximum efficiency), the blade velocity,
Blade velocity (u);
∴ Percentage increase in rotor speed = = 0.348 or 34.8%
Example 7.18
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 shall pass on without shock. Find the diagram efficiency neglecting losses if the blades are run at this speed.
[IES, 1992]
Solution
Given that α_{1} = 22°, β_{1} = β_{2} = 35°, v_{a}_{1} = 660 m/s
The velocity triangles for impulse turbine are shown in Fig. 7.40.
Maximum blade efficiency,
For β_{1} = β_{2}, C = 1 and for no friction on the blades, K = 1.
∴ (η_{b})_{max} = cos^{2} α_{1} = cos^{2} 22° = 0.8597 or 85.97%
Speed ratio, ρ =
Blade speed, u = 660 × = 305.97 m/s.
Figure 7.40 Velocity triangles for one-stage impulse steam turbine
Example 7.19
Steam expands in a turbine from 40 bar, 450°C to 0.1 bar isentropically. Assuming ideal conditions, determine the mean diameter of the wheel if the turbine were of (a) single impulse stage, (b) single 50% reaction stage, (c) four pressure (or Rateau) stages, (d) one two-row Curtis stage, and (e) four 50% reaction stages. Take the nozzle angle as 15° and speed 3000 rpm.
Solution
Given that p_{1} = 40 bar, t_{1} = 450°C, p_{2} = 0.9 bar, α_{1} = 15°, N = 3000 rpm
At 40 bar, 450°C, from steam tables, we have
At 0.1 bar: s_{f} _{2} = 0.6492 kJ/kg.K, s_{fg}_{2} = 7.5010 kJ/kg.K,
Now for isentropic flow, s_{1} = s_{2} = s_{f} _{2} + x_{2} s_{fg}_{2}
or 6.9362 = 0.6492 + x_{2} × 7.5010
or x_{2} = 0.838
Example 7.20
Steam at 20 bar, 500°C expands in a steam turbine to 0.01 bar. There are four stages in the turbine and the total enthalpy drop is divided equally among the stages. The stage efficiency is 75% and it is the same in all the stages. Calculate the interstage pressures, the reheat factor, and the turbine internal efficiency.
Solution
Given that p_{1} = 20 bar, t_{1} = 500°C, p_{2} = 0.09 bar, η_{s} = 0.75
At 20 bar, 500°C, h_{1} = 3467.6 kJ/kg, s_{1} = 7.4316 kJ/kg.K
At 0.1 bar, h_{f}_{6} = 191.81 kJ/kg, h_{fg}_{6} = 2392.8 kJ/kg, s_{f}_{6} = 0.6492 kJ/kg.K, s_{fg}_{6} = 7.5010 kJ/kg.K
Now s_{1} = s_{6} = s_{f}_{6} + x_{6} s_{fg}_{6}
or 7.4316 = 0.6491 + x_{6} × 7.5010
or x_{6} = 0.9042
The Mollier diagram is shown in Fig. 7.41.
The inter stage pressures read from Mollier chart shown in Fig. 7.41 are:
Reheat factor =
Turbine internal efficiency, η_{it} =
Example 7.21
Find the maximum blade efficiency and corresponding blade angles for a single row impulse steam turbine assuming equiangular blades, when the nozzle angle α_{1} = 20° and blade velocity coefficient K = 0.9.
If the blade efficiency is 85% of the maximum value, what are the possible blade speed ratios for the same nozzle angle α_{1}, blade velocity coefficient, K and equiangular blades? Find the corresponding blade angles.
[IAS, 2003]
Solution
Given that α_{1} = 20°, K = 0.9
Maximum blade efficiency, (η_{b})_{max} = (1 + KC)
where K = and C =
or β_{1} = 36°
Let the blades be symmetrical so that β_{2} = β_{1} = 36° and C = 1
The velocity diagrams at inlet and outlet of a moving blade are shown in Fig. 7.42.
Example 7.22
Steam expands in a steam turbine isentropically from inlet to exhaust having an enthalpy drop = 12000 kJ/kg. Assuming ideal conditions, determine the mean diameter of the wheel if the turbine were of:
- Single impulse stage
- Single 50% reaction stage
- One two-row Curtis stage
- Ten 50% reaction stages
Take the nozzle angle as 18° and blade speed as 4000 rpm
[IAS, 2002]
Solution
Given that α_{1} = 18°, N = 4000 rpm, ∆h = 12000 kJ/kg,
Example 7.23
The velocity of steam entering a simple impulse turbine is 1000 m/s and the nozzle angle is 20°. The mean peripheral velocity of blades is 400 m/s. The blades are asymmetrical. If the steam is to enter the blades without shock, what will be the blade angles?
Neglecting the friction effects on the blades, calculate the tangential force on the blades and the diagram power for a mass flow of 0.75 kg/s. Calculate the axial thrust and diagram efficiency.
[IAS, 2001]
Given that v_{a}_{1} = 1000 m/s, α_{1} = 20°, u = 400 m/s, v_{r}_{1} = v_{r}_{2}, ṁ = 0.75 kg/s. For symmetrical blades,
Steam is to enter the blades without shock.
The velocity diagrams are shown in Fig. 7.43.
or v_{r}_{1} = 638.94 m/s
or β_{1} = 32.36°
∴ α_{2} = 32.36°
Tangential force on the blades, F_{t} = ṁ (v_{w}_{1} + v_{w}_{2}) = 0.75 (939.7 + 200.4) = 855.07 N
Diagram power = = 342.03 kW
Figure 7.43 Velocity diagrams for simple impulse turbine
Diagram efficiency = 0.9121 or 91.21%
Axial thrust, F_{a} = ṁ (v_{f} _{1} − v_{f} _{2}) = 0.75 (342.02 − 218.53) = 92.6 N
Summary for Quick Revision
- Steam turbines are of three types: impulse turbines, impulse reaction turbines, and reaction turbines.
- Compounding of steam turbines is a method employed for reducing the rotational speed of the impulse turbine to practical limits by using more than one stage.
- Impulse steam turbines can be compounded by velocity compounding, pressure compounding, and mixed velocity and pressure compounding.
- Impulse turbine:
- Power developed, P = kW
- Diagram or blade efficiency, η_{b} =
- Gross or stage efficiency, η_{s} =
- Nozzle efficiency, η_{n} =
- Axial thrust, F_{a} =
- Energy converted into heat by blade friction
- For maximum blade efficiency: speed ratio,
where
where β_{1} = inlet angle of moving blade, β_{2} = exit angle of moving blade, α_{1} = outlet angle of fixed blade
For symmetrical blades, β_{1} = β_{2} and C = 1
(η_{b})_{max} = cos^{2} α_{1}Work done, w = (v_{a}_{1}cos α_{1} − u) (1 + KC)u
w_{max} = 2u^{2} for K = 1 and C = 1
- Velocity compounded impulse turbine:
- For symmetric blades and no friction loss, i.e. β_{1} = β_{2} and v_{r}_{1} = v_{r}_{2}
Total work done, w_{t} = 4u(v_{a}_{1}cos α_{1} − 2u)
- Blade efficiency, η_{b} = 8ρ(cos α_{1} − 2ρ)
For maximum blade efficiency,
and (η_{b})_{max} = cos^{2} α_{1}
(w_{t})_{max} = 8u^{2} - If n = number of rotating blade rows in series, then
and work done in the last row =
- For symmetric blades and no friction loss, i.e. β_{1} = β_{2} and v_{r}_{1} = v_{r}_{2}
- Reaction turbine:
- Degree of reaction, R_{d} =
- For R_{d} = 0.5,
u = v_{f} (cot β_{2} − cot α_{2}) = v_{f} (cot α_{1} − cot β_{1})
∴ β_{1} = α_{2} and β_{2} = α_{1}
Also v_{r}_{2} = v_{a}_{1}
- Work done per kg of steam, w = [2ρcos α_{1} − ρ^{2}]
- Blade efficiency, η_{b} =
- For maximum blade efficiency, ρ = cos α_{1}
- If n = number of blades, and d = mean diameter of turbine wheel then pitch of blades, p =
- Volume flow rate, ṁ v_{s} = (πd − nt_{1})h_{1}v_{f} _{1} = (πd − nt_{2}) h_{2}v_{f} _{2}
For v_{f} _{1} = v_{f} _{2} = v_{f }, t_{1} = t_{2} = t and h_{1} = h_{2} = h
ṁ v_{s} = (πd − nt) hv_{f}Height of blades, h =
- Reheat factor, R_{f} =
- Turbine internal efficiency, η_{i} = η_{s} × R_{f}
where η_{s} = stage efficiency
- Degree of reaction, R_{d} =
- Turbine efficiencies:
- Governing of steam turbines:
- Throttle governing
- Nozzle control governing, and
- By-pass governing.
Multiple-choice Questions
- In steam turbine terminology, diaphragm refers to
- The separating wall between rotors carrying nozzles
- The ring of guide blades between rotors
- A partition between low and high pressure dies
- The flange connecting the turbine exit to the condenser
- A three-stage Rateau turbine is designed in such a manner that the first two stages develop equal power with identical velocity diagram, whereas the third one develops more power with higher blade speed. In such a multistage turbine, the blade ring diameter
- is the same for all the three stages
- gradually increases from the first to the third stage
- of the third stage is greater than that of the first two stages
- of the third stage is less than that of the first two stages
- In a De Laval nozzle expanding superheated steam from 10 bar to 0.1 bar, the pressure at the minimum cross-section will be
- 3.3 bar
- 5.46 bar
- 8.2 bar
- 9.9 bar
- A single-stage impulse turbine with a diameter of 120 cm runs at 300 rpm. If the blade speed ratio is 0.42, then, the inlet velocity of steam will be
- 79 m/s
- 188 m/s
- 450 m/s
- 900 m/s
- In an ideal impulse turbine, the
- absolute velocity at the inlet of moving is equal to that at the outlet
- relative velocity at the inlet of the moving blade is equal to that at the outlet
- axial velocity at the inlet is equal to that at the outlet
- whirl velocity at the inlet is equal to the outlet
- For a Parson’s reaction turbine, if α_{1} and α_{2} are fixed angles at inlet and exit, respectively, and β_{1} and β_{2} are the moving blade angles at entrance and exit, respectively, then
- α_{1} = α_{2} and β_{1} = β_{2}
- α_{1} = β_{1} and α_{2} = β_{2}
- α_{1} < β_{1} and α_{2} > β_{2}
- α_{1} < β_{2} and β_{1} > α_{2}
- The isentropic enthalpy drop in moving blade is two-thirds of the isentropic enthalpy drop in pin fixed blades of a turbine. The degree of reaction will be
- 0.4
- 0.6
- 0.66
- 1.66
- The efficiency of the nozzle-governed turbines is affected mainly by losses due to
- partial admission
- throttling
- inter-stage pressure drop
- condensation in last stages
- The main aim of compounding steam turbine is to
- improve efficiency
- reduce steam consumption
- reduce motor speed
- reduce turbine size
- A throttle-governed steam develops 20 kW with 281 kg/h of steam and 50 kW with 521 kg/h of steam. The steam consumption in kg/hr when developing 15 kW will be nearly
- Governing of steam turbines can be done by the following:
- Nozzle control
- Throttle control
- Providing additional value and passage
The correct answer will be
- 1, 2, and 3
- 1 and 2 only
- 2 and 3 only
- 1 and 3 only
- What is the cause of reheat factor in a steam turbine?
- Reheating
- Superheating
- Supersaturation
- Blade friction
- In a steam power plant, feed water heater is a heat exchanger to preheat feed water by
- live steam from steam generator
- hot flue gases coming out of the boiler furnace
- hot air from air preheater
- extracting steam from turbine
- Match List I (turbines) with List II (classification) and select correct answer using the codes given below the List:
List I List II A. Parson’s 1. Pressure compounded B. De- Laval 2. Reaction C. Rateau 3. Simple impulse D. Curtis 4. Velocity compounded Codes:
A B C D
- 3 2 1 4
- 2 3 4 1
- 2 3 1 4
- 3 2 4 1
- The outward radial flow turbine in which there are two rotors rotating in opposite directions is known an
- 50% reaction radial turbine
- Cantilever turbine
- Ljungstrom turbine
- Pass-out turbine
- In reaction turbines, with reduction of inlet pressure
- the blade heights increase as the specific volume of steam decreases
- the blade heights increase as the specific volume of steam increases
- the blade heights decrease as the specific volume of steam increases
- the blade heights decrease as the specific volume of steam decreases
- Which of the following statements are correct?
- Impulse turbine rotor blades are thick at the centre
- Rateau turbine is more efficient than Curtis turbine
- Blade velocity coefficient for an impulse turbine is the order of 60%
Select the correct answer using the codes given below:
- 1, 2, and 3
- 1 and 2
- 1 and 3
- 2 and 3
- Given that α_{1} = nozzle angle, n = number of rows of moving blades in a velocity compounded, impulse turbine, the optimum blade speed ratio is
- 2 cos α_{1}n
- In a Parson’s turbine stage, blade velocity is 320 m/s at the mean radius and the rotor blade exit angle is 30°. For minimum kinetic energy of the steam leaving the stage, the steam velocity at the exit of the stator will be
- 640 m/s
- For a free vortex design of blade in the rotor of a reaction axial turbine, the specific work along the blade height is
- higher at the blade hub and lower at the blade tip
- constant from hub to tip
- lower at the hub and tip but different from the mean section
- the same at the hub and tip but different from the mean section
- Which of the following sketches represents an impulse turbine blade?
- The graph shown in Fig. 7.44 represents the variation of absolute velocity of steam along the length of a steam turbine.
Figure 7.44
The turbine in question is
- Curtis turbine
- De-Laval turbine
- Radial turbine
- Parson’s turbine
- In a simple impulse turbine, the nozzle angle at the entrance is 30°. What will be the blade-speed ratio for maximum diagram efficiency?
- 0.433
- 0.25
- 0.5
- 0.75
- A reaction turbine stage has angles α_{1}, β_{1}, β_{2} as nozzle angle, inlet blade and outlet blade angle, respectively. The expression for maximum efficiency of the turbine is given by
- The correct sequence of the given steam turbines in the ascending order of efficiency at their design point is
- Rateau, De- Laval, Parson’s, Curtis
- Curtis, De- Laval, Rateau, Parson’s
- De -Laval, Curtis, Rateau, Parson’s
- Parson’s, Curtis, Rateau, De- Laval
- Consider the following statements regarding the nozzle governing of steam turbines:
- Working nozzles receive steam at full pressure
- High efficiency is maintained at all loads
- Stage efficiency suffers due to partial admission
- In practice, each nozzle of the first stage is governed individually
Of these statements,
- 1, 2, and 3 are correct
- 2, 3, and 4 are correct
- 1, 3, and 4 correct
- 1, 2, and 4 are correct
- Among other things, the poor part-load performance of De Laval turbines is due to the
- formation of shock waves in the nozzle
- formation of expansion waves at the nozzle
- turbulent mixing at the nozzle exit
- increase in profile losses in the rotor
- List I gives the various velocities in the velocity diagrams of a two-stage impulse turbine. List II gives the blade angles. Match the velocity from List I with the angle in List II and select the correct answer using the codes given below the lists:
List I List II A. Relative velocity of steam an inlet tip of blade 1. Nozzle angle B. Absolute velocity of steam at inlet tip of blade 2. Moving blade leading edge angle C. Relative velocity of steam at outlet tip of blade 3. Moving blade trailing edge angle D. Absolute velocity of steam at outlet tip of blade 4. Fixed blade leading edge angle Codes:
A B C D
- 1 2 4 3
- 2 1 4 3
- 2 1 3 4
- 1 2 3 4
- Which of the following relationship between angles of fixed blades and moving blades corresponds to that of Parson’s turbine?
- α_{1} = α_{2}
- α_{1} = β_{2}
- α_{2} = β_{2}
- β_{1} = β_{2}
- The following data refer to an axial flow turbine stage:
Relative velocity of steam at inlet to the rotor = 79.0 m/s., Relative velocity at the rotor exit = 152 m/s. What is the approximate degree of reaction?
- 0.9
- 0.8
- 0.7
- 0.6
- The clearance flow between the blade tips and casing of a steam turbine is
- greater in the reaction turbine than in the impulse type
- greater in the impulse turbine than in the reaction type
- independent of type of the turbine
- independent of the size of the turbine
- Running speeds of steam turbine can be brought down to practical limits
- By using heavy flywheel
- By using a quick response governor
- By compounding
- By reducing fuel feed to the furnace
- Only 3
- 1, 2, 3, and 4
- 1, 2, and 4
- 2 and 3
- The net result of pressure-velocity compounding of steam turbine is
- less number of stages
- large turbine for a given pressure drop
- shorter turbine for a given pressure drop
- lower friction loss
- Given, v_{b} = Blade speed
v_{a}_{1} = Absolute velocity of steam entering the blade, α_{1}= Nozzle angle
The efficiency of an impulse turbine is maximum when
- v_{b} = 0.5 v_{a}_{1} cos α_{1}
- v_{b} = v_{a}_{1} cos α_{1}
- v_{b} = 0.5 v^{2}_{a}_{1} cos α_{1}
- v_{b} = v^{2}_{a}_{1} cos α_{1}
- An impulse turbine produces 50 kW of power when the blade mean speed is 400m/s. What is the rate of change of momentum tangential to the rotor?
- 200 N
- 175 N
- 150 N
- 125 N
- At a particular section of a reaction turbine, the diameter of the blade is 1.8 m, the velocity of flow of steam is 49 m/s and the quantity of steam flow is 5.4 m^{3}/s. The blade height at this section will be approximately:
- 4 cm
- 2 cm
- 1 cm
- 0.5 cm
- Consider the following statements:
If steam is reheated during the expansion through turbine stages,
- Erosion of blade will decrease
- The overall pressure ratio will increase
- The total heat drop will increase
Of these statements,
- 1, 2, and 3 are correct
- 1 and 2 are correct
- 2 and 3 are correct
- 1 and 3 are correct
- In an impulse-reaction turbine stage, the heat drop in fixed and moving blades are 15 kJ/kg and 30 kJ/kg respectively. The degree of reaction for this stage will be
- 1/3
- 1/2
- 2/3
- 3/4
- If ‘D’ is the diameter of the turbine wheel and ‘U’ is its peripheral velocity, then the disc friction loss will be proportion to
- (DU)^{3}
- D^{2}U^{3}
- D^{3}U^{2}
- DU^{4}
- In a two-row Curtis stage with symmetrical blading.
- work done by both rows of moving blades are equal
- work done by the first row of moving blades is double of the work done by second row of moving blades
- work done by the first row of moving blades is three times the work done by second row of moving blades
- work done by the first row of moving blades is four times the wok done by the second row of moving blades
- The compounding of steam turbines is done to
- The expression for the maximum efficiency of a Parson’s turbine is (α_{1} is the angle made by absolute velocity an inlet)
- Consider the following statements regarding effects of heating of steam in a steam turbine:
- It increases the specific output of the turbine
- It decreases the cycle efficiency
- It increases blade erosion
- It improves the quality of exit steam.
Which of these statements are correct?
- 1, 2, and 3
- 2 and 3
- 3 and 4
- 1 and 4
- Match List I (Different turbine stages) with List II (Turbines) and select the correct answer using the codes given below the lists:
List I List II A. 50% reaction stage 1. Rateau B. Two-stage velocity compounded turbine 2. Parson C. Single-stage impulse 3. Curtis D. Two-stage pressure compounded turbine 4. De-Laval 5. Hero Codes:
A B C D
- 5 1 2 3
- 5 3 2 1
- 2 3 4 1
- 3 1 4 2
- Partial admission steams turbine refers to the situation where the
- steam is admitted partially into the blades through nozzles
- nozzle occupy the complete circumference leading into the blade annulus
- nozzle do not occupy the complete circumference leading into the blade annulus
- steam is admitted partially into the blades directly
- Consider the following statements regarding a 100% reaction turbine:
- Change in absolute velocity of steam across the moving blades is zero.
- Change in absolute velocity of steam across the moving blades is negative.
- Enthalpy drop in fixed blades is zero.
Which of these statements is/are correct?
- Only 1
- Only 2
- 2 and 3
- 1 and 3
- Which one of the following pairs is not correctly matched?
- Internal efficiency of steam turbine: Product of stage efficiency and reheat factor
- Stage efficiency of a turbine: Ratio of adiabatic heat drop to the isentropic heat drop per stage
- Dryness fraction of steam within a stage: Decreases due to reheating
- Steam condensation during expansion through the turbine: Enhances blade erosion
- Consider the following characteristics:
- High steam and blade velocities
- Low steam and blade velocities
- Low speeds of rotation
- High carry-over loss
Which of these characteristics are possessed by a simple impulse turbine?
- 1 and 2
- 2 and 3
- 1 and 4
- 3 and 4
- Velocity triangle for a reaction turbine stage is shown in the given Fig. 7.45 (AB = υ_{a}_{1} = absolute velocity at rotor blade inlet; CB = v_{r}_{1} = relative velocity at rotor blade inlet; CE = v_{r}_{2} = relative velocity at rotor blade exit and CD = CB)
Figure 7.45
The ratio of reaction force to impulse force is
- CE/CB
- CD/CE
- DE/BD
- AE/AB
- Consider the following statements:
- Throttle governing improves quality of steam in the last few stages
- Internal efficiency of steam is not seriously affected by throttle governing
- Throttle governing is better than nozzle governing
Which of these statements are correct?
- 1, 2, and 3
- 1 and 3
- 2 and 3
- 1 and 2
- Which one of the following statements is correct?
- Reheat factor is zero if efficiency of the turbine is close to unity
- Lower the efficiency, higher will be the reheat factor
- Reheat factor is independent of steam conditions at turbine inlet
- Availability of reheat is higher at low pressure end
- For maximum blade efficiency of a single-stage impulse turbine, the blade speed ratio, (α_{1} is, the angle made by absolute velocity at inlet) should be
- cos 2α_{1}
- Figure 7.46 shows the variation of certain steam parameter in case of a simple impulse turbine. The curve A-B-C represents the variation of
Figure 7.46
- pressure in nozzle and blades
- velocity in nozzle and blades
- temperature in nozzle and blades
- enthalpy in nozzle and blades
- For a reaction turbine with degree of reaction equal to 50%, (υ_{a}_{1} is the absolute steam velocity an inlet and α_{1} is the angle made by it to the tangent on the wheel) the efficiency is maximum when the blade speed is equal to
- 2υ_{a}_{1} cos α_{1}
- υ_{a}_{1}cos^{2}α_{1}
- υ_{a}_{1} cos α_{1}
- Which one of the following is the correct statement? The degree of reaction of an impulse turbine
- is less than zero
- is greater than zero
- is equal to zero
- increases with steam velocity at the inlet
- Why is compounding of steam turbines done?
- To improve efficiency
- To reduce the speed of rotor
- To reduce exit losses
- To increase the turbine output
- A four-row velocity compounded steam turbine develops 6400 kW. What is the power developed by the last row?
- 200 kW
- 400 kW
- 800 kW
- 1600 kW
- Which of the following is used to bring down the speed of an impulse steam turbine to practical limits?
- A centrifugal governor
- Compounding of the turbine
- A large flywheel
- A gear box
- Consider the following for a steam turbine power plant:
- Reduction in blade erosion
- Increase in turbine speed
- Increase in specific output
- Increase in cycle efficiency
Which of the above occur/occurs due to reheating of steam?
- Only 1
- 1 and 2
- 1, 3, and 4
- 2 and 3
- Ratio of enthalpy drop in moving blades to the total enthalpy drop in the fixed and moving blades is called
- reheat factor
- blade efficiency
- degree of reaction
- internal efficiency
- Employing superheated steam in turbines leads to
- increase in erosion of blading
- decrease in erosion of blading
- no erosion in blading
- no change in erosion of blading
- Steam enters a De laval steam turbine with an inlet velocity of 30 m/s and leaves with an outlet velocity of 10 m/s. The work done by 1 kg of steam is
- 400 Nm
- 600 Nm
- 800 Nm
- 1200 Nm
- In a 50% reaction stage, absolute velocity angle at inlet is 45°, mean peripheral speed is 75 m/s and the absolute velocity at the exit is axial. The stage specific work is
- 2500 m^{2}/s^{2}
- 3270 m^{2}/s^{2}
- 4375 m^{2}/s^{2}
- 5625 m^{2}/s^{2}
- Consider the following statements:
Steam turbines are suitable for use as prime movers for large steam power plants because
- a single steam turbine can be designed for a capacity of 1000 MW or more
- much higher speed may be possible as compared to a reciprocating engine
- they are more compact when compared to a gas turbine power plant
- the maintenance cost and running cost may not increase with years of service
Which of these statements are correct?
- 1, 2 and 3
- 2, 3 and 4
- 1, 2 and 4
- 1, 3 and 4
- Expansion line EFG of a 2-stage steam turbine is shown in Fig. 7.47 (h-s diagram). The reheat factor for this turbine is
Figure 7.47
- 1.08
- 0.7
- 0.648
- 1.43
- Which of the following is the features of pressure-compounding (Rateau staging)?
- Low efficiency at low rotational speed
- High efficiency with low fluid velocities
- High efficiency with high fluid velocities
- Low efficiency at high rotational speed
- In Parson’s reaction turbines, the velocity diagram triangles at the inlet and outlet are
- asymmetrical
- isosceles
- right-angled
- congruent
- In Parson’s turbine if α_{1} is nozzle angle, then what is the maximum efficiency of the turbine?
- What is the value of the reheat factor in multistage turbine?
- 1.03 to 1.04
- 1.10 to 1.20
- 0.90 to 1.00
- 1.20 to 1.25
- In which one of the following steam turbines, is steam taken from various points along the turbine, solely for feed-water heating?
- Extraction turbine
- Bleeder turbine
- Regenerative turbine
- Reheat turbine
- Velocity diagram shown in Fig. 7.48 is for an impulse turbine stage.
Figure 7.48
What are the tangential force and axial thrust per kg/s of steam, respectively?
- 450 N, 8 N
- 560 N, 8 N
- 680 N, 4 N
- 910 N, 4 N
- Consider the following statements in respect of impulse steam turbines:
- Blade passages are of constant cross-sectional area.
- Partial admission of steam is permissible.
- Axial thrust is only due to change in flow velocity of steam an inlet and outlet of moving blade.
Which of the statements given above are correct?
- 1, 2, and 3
- Only 1 and 2
- Only 2 and 3
- Only 1 and 3
- In an axial flow impulse turbine, energy transfer takes place due to
- change in relative kinetic energy
- change in absolute kinetic energy
- change in pressure energy
- change in energy because of centrifugal force
- In a reaction turbine the enthalpy drop in a stage is 60 kJ/kg. The enthalpy drop in the moving blades is 32 kJ/kg. What is the degree of reaction?
- 0.533
- 0.284
- 0.466
- 1.875
- An emergency governor of a steam turbine trips the turbine when
- Shaft exceeds 100% of its rated speed
- Condenser becomes hot due to inadequate cooling water circulation
- Lubrication system fails
- Balancing of turbine is not proper
Select the correct answer form the codes given below:
Codes:
- 1, 2, and 3
- 2, 3 and 4
- 3, 4, and 1
- 4, 1, and 2
- Steam enters the rotor of a reaction turbine with an absolute velocity of 236 m/s and relative velocity of 132 m/s. It leaves the rotor with a relative velocity of 2.32 m/s and absolute velocity of 136 m/s. The specific work output is
- 35.8 kW
- 40.1 kW
- 43.8 kW
- 47.4 kW
- The work output from a reaction turbine stage is 280 kW per kg/s of steam. If the nozzle efficiency is 0.92 and rotor efficiency is 0.90, the isentropic static enthalpy drop will be
- 352 kW
- 347 kW
- 338 kW
- 332 kW
- Shock waves in nozzles would occur while turbines are operating
- at overload conditions
- at part load conditions
- above critical pressure ratio
- at all off-design conditions
- Consider the following statements:
The erosion of steam turbine blades increases with the increase of
- moisture in the steam
- blade speed
Which of these statement(s) is/are correct?
- 1 alone
- alone
- 1 and 2
- Neither 1 nor 2
- Consider the following statements regarding an impulse turbine:
- Relative velocity at the inlet and exit of the rotor blades are the same.
- Absolute velocity at the inlet and exit of the rotor blades are the same.
- Static pressure within the rotor blade channel is constant
- Total pressure within the rotor blade channel is constant
Of these statements,
- 1 and 4 are correct
- 2 and 3 are correct
- 1 and 3 are correct
- 2 and 4 are correct
- If in an impulse turbine designed for free vortex flow, the tangential velocity of steam at the root radius of 250 mm is 430 m/s and the blade height is 100 mm, then the tangential velocity of steam at the tip will be
- 602 m/s
- 504 m/s
- 409 m/s
- 307 m/s
- The blade passage for the nozzle blade row of the first stage of an impulse turbine is best represented an
- If u, υ_{a}, and u_{r} represent the peripheral, absolute, and relative velocities, respectively, and suffix 1 and 2 refer to inlet and outlet, which one of the following velocity triangles could be a reaction turbine stage with reaction more than 50%?
Explanatory Notes
- 3. (b)
For superheated steam, n = 1.3
- 4. (c)
Speed ratio =
- 7. (a)
Degree of reaction,
- 10. (c) Let steam consumption, y = A + Bx
where x = kW
281 = A + 20 B521 = A + 50 B - Solving we get, A = 121, B = 8
y = 121 + 15 × 8 = 241 kg/h
- 19. (c) u = 320 m/s, b_{2} = 30°
- 23. (a) For maximum diagram efficiency,
- 30. (d) Degree of reaction,
- 35. (d) P = ṁ u (v_{w1} − v_{w2})/10^{3} kW
Rate of change of momentum =
- 36. (b)cm
- 38. (c) Degree of reaction
- 57. (b) Power developed in the last row kW
where n = number of rows
- 62. (a) Work done by 1 kg of steam N.m
- 63. (d) u = v_{w} = v_{w}_{1} + v_{w}_{2} = 75
Specific work = u (v_{w}_{1} + v_{w}_{2}) = u^{2} = 75^{2} = 5625 m^{2}/s^{2}
- 65. (a) Reheat factor
- 71. (d) Tangential force = ṁ (v_{w}_{1 } − v_{w}_{2}) = 1(910 − 0) = 910 N
Axial thrust = ṁ(v_{f} _{1 } − v_{f} _{2}) = 1 × 4 = 4N
- 74. (a)
- 76. (a) v_{a}_{1} = 236 m/s, v_{r}_{1} = 132 m/s, v_{r}_{2} = 232 m/s, v_{a}_{2} = 136 m/s
Specific work output
- 77. (c) P = 280 kW, h_{n } = 0.92, h_{b} = 0.90
- 81. (d) r_{1} = 250 mm, u_{1} = 430 m/s, h = 100 mm
In a free vortex flow,
or u_{1}r_{1} = u_{2}r_{2}
Review Questions
- Explain the principle of operation of a steam turbine.
- How do you classify steam turbines?
- Compare impulse and reaction turbines.
- What do you mean by compounding of steam turbines? What are the various methods of compounding impulse steam turbines?
- Define blade efficiency and stage efficiency.
- What is the condition for maximum blade efficiency of an impulse turbine considering blade friction?
- Define speed ratio and write the expression for a turbine with n blade rings.
- What are the advantages of velocity compounding?
- What are the limitations of velocity compounding?
- What is the necessity of compounding impulse turbine?
- Define degree of reaction.
- What is the condition for blade angles for 50% degree of reaction?
- Write the formula for maximum blade efficiency of a reaction turbine.
- Define reheat factor and turbine internal efficiency.
- List the various losses in a steam turbine.
- What are the various methods for governing of a steam turbine?
- Explain the working of a back-pressure turbine.
- What is an extraction turbine?
- Explain co-generation.
- What is the role of Labyrinth packing?
Exercises
7.1 In a single stage impulse turbine, the blade angles are equal and the nozzle angle is 20°. The velocity coefficient for the blade is 0.83. Find the maximum blade efficiency possible.
If the actual blade efficiency is 90% of maximum blade efficiency, find the possible ratio of blade speed to steam speed.
[Ans. 0.813, 0.3235]
7.2 The following data refer to a compound impulse turbine having two rows of moving blades and one row of fixed blades in between them.
Nozzle angle = 15°, Exit velocity of steam from the nozzle = 450 m/s
Moving blades tip discharge angles = 30°
Fixed blade discharge angle = 20°
Friction loss in each blade rows = 10% of the relative velocity
Calculate the blade velocity, blade efficiency, and specific steam consumption for the turbine.
[Ans. 98m/s, 78.8%, 45 kg/kWh]
7.3 In a stage of impulse-reaction turbine, steam enters with a speed of 250 m/s at an angle of 30° in the direction of blade motion. The mean blade speed is 150 m/s when the rotor is running at 3000 rpm. The blade height is 10 cm. The specific volume of steam at nozzle outlet and blade outlet are 3.5 m^{3}/kg and 4 m^{3}/kg, respectively. The turbine develops 250 kW. Assuming the efficiency of nozzle and blades together considered as 90% and carry over coefficient as 0.8, find (a) the enthalpy drop in each stage, (b) degree of reaction, and (c) stage efficiency.
[Ans. 29.3 kJ/kg, 0.39, 71%]
7.4 At a stage of reaction turbine, the mean diameter of the rotor is 1.4 m. The speed ratio is 0.7. Determine the blade inlet angle if the blade outlet angle is 20°. The rotor speed is 3000 rpm. Find the diagram efficiency, percentage increase in it, and rotor speed if the rotor is designed to run at the best theoretical speed, the exit angle being 20°.
[Ans. 53.8°, 90.5%, 3.65%, 93.8%]
7.5 In a 50% reaction turbine stage running at 50 rps, the exit angles are 30° and the inlet angles are 50°. The mean diameter is 1 m. The steam flow rate is 10^{4} kg/min and the stage efficiency is 85%. Calculate (a) the power output of the stage, (b) the specific enthalpy drop in the stage, and (c) the percentage increase in the relative velocity of steam when it flows over the moving blades.
[Ans. 11.6 kW, 82 kJ/kg, 52.2%]
7.6 A single row impulse turbine develops 130 kW at a blade speed of 180 m/s using 2 kg/s of steam. The steam leaves the nozzle at 400 m/s. The friction coefficient for blades is 0.9 and steam leaves the blades axially. Determine (a) the nozzle angle and (b) the blade angles at entry and exit, assuming no shock.
[Ans. 25.47°, 43.5°, 35.8°]
7.7 A simple impulse turbine has one ring of moving blades running at 150 m/s. The absolute velocity of steam at exit from the stage is 80 m/s at an angle of 75° from the tangential direction. The blade speed coefficient for blades is 0.85 and the rate of steam flowing through the stage is 3 kg/s. If the blades are equiangular, determine (a) the blade angles, (b) the nozzle angle, (c) the absolute velocity of steam issuing from the nozzle, and (d) the axial thrust.
[Ans. 24.35°, 14.5°, 362.4 m/s]
7.8 In a single stage impulse turbine, the nozzle angle is 20° and blade angles are equal. The velocity coefficient for blades is 0.85. Calculate the maximum blade efficiency possible. If the actual blade efficiency is 92% of the maximum blade efficiency, find the possible ratio of blade speed to steam speed.
[Ans. 81.6%, 0.336 or 0.603]
7.9 In a single stage steam turbine saturated steam at 10 bar is supplied through a convergent-divergent steam nozzle of 20° angle. The mean blade speed is 400 m/s. The steam pressure leaving the nozzle is 1 bar. Find (a) the best blade angles if they are equiangular and (b) the maximum power developed by the turbine if 5 nozzles are used of 0.6 cm^{2} area at the throat. Assume nozzle efficiency 90% and blade friction coefficient 0.87.
[Ans. 35.46°, 115.8 kW]
7.10 The first stage of an impulse turbine is compounded for velocity and has two rows of moving blades and one ring of fixed blades. The nozzle angle is 15° and leaving angles of blades are respectively, first moving 30°, fixed 20°, second moving 30°. The velocity of steam leaving the nozzle is 540 m/s. The friction loss in each blade row is 10% of the relative velocity. Steam leaves the second row of moving blades axially. Calculate (a) the blade velocity, (b) the blade efficiency, and (c) the specific steam consumption.
[Ans. 117.3 m/s, 78.6%, 31.39 kg/kWh]
7.11 The first stage of a turbine is a two-row velocity compounded impulse wheel. The steam velocity at inlet is 600 m/s and the mean blade speed is 120 m/s. The nozzle angle is 16° and the exit angles for the first row of moving blades, the fixed blades, and the second row of moving blades are 18°, 21° and 35°, respectively. Calculate (a) the blade inlet angles for each row, (b) the driving force for each row of moving blades and axial thrust on the wheel for a mass flow rate of 1 kg/s, (c) the diagram efficiency and diagram power for the wheel per kg/s steam, and (d) the maximum possible diagram efficiency. Assume blade velocity coefficient as 0.9 for all blades.
[Ans. First row: Moving blade 20°, fixed blade 24.5°, Second row: moving blade 34.5°; 875 N, 294 N; 42 N; 140.28 kW, 77.93%, 92.4%]
7.12 The following data refer to a stage of a Parson’s reaction turbine:
Speed of turbine = 1500 rpm
Mean diameter of rotor = 1 m
Stage efficiency = 80%
Blade outlet angle = 20°
Speed ratio = 0.7
Calculate the isentropic enthalpy drop in the stage.
[Ans. 13.08 kJ]
7.13 A stage of a Parson’s reaction turbine delivers dry saturated steam at 2.7 bar from the fixed blades at 90 m/s. The mean blade height is 4 cm, and the moving blade exit angle is 20°. The axial velocity of steam is 3/4 of blade velocity at the mean radius. Steam is supplied to the stage at the rate of 2.5 kg/s. The effect of blade tip thickness on the annulus area can be neglected. Calculate (a) the wheel speed, (b) the diagram power, (c) the diagram efficiency, and (d) the enthalpy drop in this stage.
[Ans. 1823 rpm, 13.14 kW, 78.7%, 5.25 kJ/kg]
7.14 In a stage of a reaction turbine, the mean rotor diameter is 1.5 m, speed ratio = 0.72, blade outlet angle = 20° and rotor speed = 3000 rpm. Calculate (a) the diagram efficiency and (b) the percentage increase in diagram efficiency and rotor speed if the rotor is designed to run at the best theoretical speed, the exit angle being 20°.
[Ans. 91%, 2.96%, 3914.7 rpm]
7.15 The nozzle angle of a single stage impulse turbine is 20°. The moving blade angles are equal and the velocity coefficient for the blade is 0.83. Find the maximum blade efficiency possible. If the actual blade efficiency is 90% of the maximum blade efficiency, find the possible ratio of blade speed to steam speed.
[ESE, 1979]
7.16 The total power developed by a 20-stage reaction turbine is 11200 kW. Steam is supplied at 15 bar abs. and 300°C. The pressure of steam leaving the turbine is 0.1 bar abs. The turbine has a stage efficiency of 75% and a reheat factor of 1.05. Determine the mass flow rate of steam through the turbine. Assume that all stages develop equal power.
[ESE, 1980]
7.17 At a certain stage of the turbine in Exercise 7.16, the pressure of steam is 1 bar and it is dry saturated. If the blade exit angle is 20° and the ratio of the blade speed to steam speed is 0.7, find the mean, diameter of the rotor of this stage, and the rotor speed. Assume the blade height as the mean drum diameter and neglect the thickness of the blades. The velocity diagrams for the blades are symmetrical.
[ESE, 1980]
7.18 An impulse stage of a steam turbine is supplied with dry and saturated steam at 15 bar. The stage has a single row of moving blades running at 3600 rpm. The mean diameter of the blade disc is 9.0 m. The nozzle angle is 15° and the axial component of the absolute velocity leaving the nozzles is 93.42 m/s. The height of nozzles at their exit is 100 mm. The nozzle efficiency is 0.9 and the blades velocity coefficient is 0.965. The exit angle of the moving blades is 2° greater than that at the inlet. Determine (a) the blade inlet and outlet angles, (b) the isentropic heat drop in the stage, (c) the stage efficiency, and (d) power developed by the stage.
[ESE, 1982]
7.19 A single row impulse turbine stage running at 3000 rpm has a blade to steam speed ratio of 0.48. The nozzles are set at angle of 15° with respect to the place of the blade disc. The mean velocity of the blades is 144 m/s. The moving blades are symmetrical about their axis parallel to the direction of rotation. The nozzle efficiency is 0.92 and the blade velocity coefficient is 0.97. The height of the nozzles at their exit section is 100 mm. Steam leaving in the nozzles is saturated and dry and its pressure is 10 bar. Determine (a) the isentropic heat drop in the stage, (b) the energy lost in the nozzles and the moving blades due to friction, (c) energy lost due to finite velocity of steam leaving the stage, (d) mass flow rate, (e) power developed by the stage, and (f) the efficiency of the stage.
Assume that steam is accelerated from rest in the nozzles.
[ESE, 1983]
7.20 An impulse stage of steam turbine has a mean diameter of 1.2 m. The speed of rotor is 3000 rpm. The mass flow rate of steam is 20 kg/s. Steam is supplied to the stage at 15 bar and 300°C, where it expands to 10 bar. Determine the efficiency ant the power output of the stage if the nozzle efficiency is 0.9 and the blade velocity coefficient is 0.92. Assume acceleration from rest for the steam expanding in the nozzle. Assume nozzle angle to be 25°.
[ESE, 1986]
7.21 Steam enters a single-stage impulse turbine at 380 m/s and the blade speed is 170 m/s. The steam flow rate is 2.2 kg/s and turbine develops 112 kW. Assume the blade velocity coefficient to be 0.8. For an axial discharge of steam, find (a) nozzle angle, (b) blade angles, and (c) diagram efficiency.
[ESE, 1987]
7.22 Steam enters in a stage of impulse-reaction turbine with a speed of 280 m/s at an angle of 22° in the direction of blade motion. The mean diameter of the rotor, which rotates at 3200 rpm, is 1.0 m. The blade height is 10 cm. The specific volume of steam at nozzle outlet and blade outlet are 3.6 m^{3}/kg and 4.1 m^{3}/kg, respectively. The turbine develops 462 kW. Find (a) the weight of steam used per second, (b) blade angle, (c) enthalpy drop in each stage, (d) degree of reaction, and (e) stage efficiency. Assume combined efficiency of nozzles and blades as 90% and carry over coefficient as 0.8.
[ESE, 1988]
7.23 The nozzles of a Lavel turbine deliver steam at the rate of 0.9 kg/s with a velocity of 730 m/s to a set of blades revolving at the rate of 30000 rpm The diameter of the wheel is 11.5 cm. The nozzles are inclined at an angle of 20° to the plane of wheel rotation. Calculate (a) the diagram efficiency, (b) power developed by the blades, (c) energy lost in the blades per second, and (d) the condition for maximum efficiency of the turbine. Assume the blade velocity coefficient as 0.72 and outlet blade angle 25°C.
[ESE, 1989]
7.24 The nozzles of a two-row velocity-compounded stage have outlet angles of 22° and utilise an isentropic enthalpy drop of 200 kJ per kg of steam. All moving and guide blades are symmetrical, and the mean blade speed is 150 m/s. Assuming an isentropic efficiency for the nozzles of 90%, find graphically all the blade angles and calculate the specific power output produced by the stage. The velocity at inlet to the stage can be neglected.
[ESE, 1991]
7.25 One stage of an impulse turbine consists of a converging nozzles ring and one ring of moving blades. The nozzles are inclined at 22° to the blade 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 shall pass on without shock. Find the diagram efficiency neglecting losses if the blades are run at this speed.
[ESE, 1992]
7.26 A single steam turbine is supplied with steam at 5 bar, 200°C at the rate of 50 kg/min. It exhausts into a condenser at a pressure of 0.2 bar. The blade speed is 400 m/s. The nozzles are inclined at an angle of 20° to the plane of the wheel and the outlet blade angle is 30°. Neglecting friction losses, determine the blade efficiency, the stage efficiency and the power developed by the turbine.
[ESE, 1994]