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. 1. What are units Quantities? Discuss their importance.Ans. A turbine operates most efficiently at its design point i.e. at a particular combination of head, discharge and power output. In practice these variables seldom remains constant. Recourse is then made to the concept of unit and specific quantities and model relationships when it is desired to:

(i) Predict the behavior of turbine working under different conditions.

(ii) Comparison between performance of turbines of same and different types.

(iii) Correlation and use of experimental data.

Q. 2. Define specific speed of pump and write its expression.

Ans. The specific speed of a centrifugal pump is defined as the speed of a geometrically similar pump which would deliver one cubic meter of liquid per second against a head of one meter. It is denoted by .

Expression:

Where = Specific speed

N = Speed in r.p.m.

H = Manometric head, m

Q = Discharge,

Q. 3. Define cavitations. What are the effects of cavitations?

Ans. Cavitations is defined as the phenomenon of formation of vapor bubbles of a flowing liquid in a region where the pressure of the liquid falls below its vapour pressure and the sudden collapsing of these vapour bubbles in a region of higher pressure. When the vapour bubbles collapse, a very high pressure is created. The metallic surfaces, above which these vapour bubbles collapse, is subjected to these high pressure, which cause pitting action on the surface. Thus cavities are formed also considerable noise and vibrations are produced.

Effects of cavitations:

(i) The metallic surfaces are damaged and cavities are formed on the surfaces.

(ii) Due to sudden collapse of vapour bubble, considerable noise and vibrations are produced.

(iii) The efficiency of a turbine decreases due to cavitations.

Q. 4. What do you mean by Thomas’s cavitations factor?

Ans Thomas’s cavitations factor a (sigma) can be used for determining the region where cavitations takes place in reaction turbines. The mathematical expression is given by:

Where

= Barometric pressure head in m of water H

= Atmospheric pressure head in m of water

= Vapour pressure head in m of water

= Suction pressure heat at outlet of reaction turbine in m of water

H = Net head on the turbine in m.

Q. 5. Explain the various methods of preventing cavitation.

Ans. Following methods can be used to prevent cavitation.

1. Design consideration: Design the system so that there is no separation of flow avoids sharp corners or curvatures of flow in the system that produces vertices or eddier.

2. Model testing: Model of the prototype must be mode and tested in laboratory.

3. Installation: The chance of cavitation reduces if the location of turbine above the tail race decreases.

4. Choice of material: Materials which are cavitation resistant can be used. e.g. Satellite, Bronze, Nickel etc.

5. Surface finishing: The surface finishing should be of high degree to reduce the occurrence of cavitation as compared to the low surface finishing.

6. Admission of air: By the infection of high pressure air into the flowing liquid at low pressure zones in a hydraulic machine, the cavitation can be reduced or prevented.

Q. 6. Define NPSH of a centrifugal pump

Ans. Net positive suction head (NPSH) is defined as the absolute pressure head at the inlet to the pump, minus the vopour pressure head (in absolute units) plus the velocity head.

NPSH = Absolute pressure head at inlet of pump

- vapour pressure head (absolute units)

+ Velocity head

Q. 7. State the significance of similarity parameters in pump?

Ans. The key performance parameters of pumps are capacity, head, BHP, BEP and specific speed. This value provides the operating window within which these parameters can be varied for satisfactory pump operations. So these parameters help a lot while designing, comparing the pumps for their performance.

Q. 8. Derive the expression for specific speed for a pump.

Ans. Q = Area x Velocity of flow

Or … (1)

D = Diameter of the impeller of the pump and

B = Width of the impeller.

We know that B D

From equation (i) we have

… (2)

We also know that tangential velocity is given by

u DN … (3)

Now tangential velocity (u) and velocity of flow are related to the manometric head

As … (4)

Substituting the value of u in equation (iii) we get

Substituting the value of D in equation (ii)

Q

O=k … (5)

Where K is constant of proportionality

If 1m and Q =1 N becomes Ns;

Substituting these values in equation (v), we get

Q= or

Q. 9. What is meant by scale effect? Ans. Scale effect: When a model is made, it is very difficult to achieve geometrical similarity with same surface roughness. The change in roughness causes change in the effective head lost. So the efficiency of a prototype is different from its model. The effect is known as scale effect.

Problem 1. A homologous model of a centrifugal pump runs at 600 rpm against a head of 8 m, the power required being 5 kW. If the prototype 5 times the model size is to develop a head of 40 m, determine its speed, discharge and power. The overall efficiency of the model is 0.8 while that of prototype is 0.85.

Solution: , h=8m, H=40m, N=600rpm

We know

n= = 600 x 5 = 600 x 5

= 1341.6 rpm.

Specific speed of model and prototype must be same

So

=0.133

=0.017kW

Power developed by the model is given by

5= x0.8

q=

Problem 2. It is proposed to design a homologous model for a centrifugal pump. The prototype pump is to run at 600 r.p.m. and develop 30 m head the flow rate being 1m3/s. The model of 1/4 scale is to run at 1450 r.p.m. Determine the head developed, discharge and power required for the model. Take overall efficiency as 80%.

Solution. Given

So

(a) Head relation between model and prototype is given by:

=3.309

= 10.95m

(b) Discharge relation between model and prototype is given by

× = 0.03776

(c) Power developed = 507 kW

Problem 3. A model 1/5 of prototype is tested in a laboratory at 1200 r.p.m. It is observed during testing that it developed l0m head when power input was 40 kW. If the prototype has to work against the head of 40 m, find out the speed and power required to run the pump and ratio of flow rates of model and prototype.

Solution. The given data is

= 1200 r.p.m.

=40kW, =40m

We can use the following equation for finding

× 1200 = 480 r.p.m.


So x 40 = 8000 kW


Problem 4. Find the number of pumps required to take water from a deep well under a total head of 89 m. All the pumps are identical and are running at 800 r.p.m. The specific speed of each pump is given as 25 while the rated capacity of each pump is 0.16 .

Solution. Given Total head = 89 m, Speed = 800 r.p.m.

Specific speed, = 25, Q = 0.16

Let = Head developed by each pump

Using equation,

= 12.8

=29.94 m

Number of pumps required =

=3 Ans.

Problem 5. A pump operates at a maximum efficiency of 82% and delivers 2.25 under a head of 18 m while running at 3600 r.p.m. speed. Computer the specific speed of the pump. Also determine the discharge head and power input to pump at a shaft speed of 2400 r.p.m. cite the assumption’ made, if any.

Solution.

Q =2.25 H= 18 m; =82% = 0.82;

N = 3600 r.p.m.

Power, P =

= 484518 W = 484.518 W

The specific speed is given by

Q =2.25 =2.25x 1/s

= 19541

However, if discharge Q is taken in

= 618

Since we have

= 1.5

X18 =8m

x 484.518 = 143.56 kW.

Q. 1. Discuss briefly the guide mechanism in reaction turbines.

Ans. It consists of a stationary circular wheel all round the runner of the turbine. The stationary guide vanes are fixed on guide mechanism The guide vanes allow the water t” strike the vanes fixed on the runner without shock at inlet. The width between two adjacent vanes of guide mechanism can be altered so that the amount of water striking the runner can be varied.

Q. 2. List the advantages of Kaplan Turbine over Francis Turbine. Ans. Advantages of Kaplan turbine over Francis turbine

(i) Runner vanes are adjustable in Kaplan turbine while in Francis turbine run vanes are not adjustable.

(ii) There is less resistance offered as the number of vanes are fewer in Kaplan turbine (in) Specific speed range 250-850 m Kaplan turbine In Francis turbine specific speed range is 5o—250.

Q. 3. Draw velocity triangles at inlet and outlet of typical Francis turbine vane.

Ans. There are three types of velocity triangles for. inlet and outlet in Francis turbine. Triangles are made for slow runner, medium runner and fast runner.

Fig. Slow runner

Fig. Medium Runner

Fig. Fast Runner

Q. 4. Define degree of reaction and Euler’s Head.

Ans. The degree of reaction (R) is defined as a ratio of change of pressure energy in the runner to the change of total energy in the runner per kg of water.

Euler’s Head: It is defined as energy transfer per unit weight.

Q.5. Why is the efficiency of Kaplan turbine nearly constant irrespective of speed variation under load?

Ans. Kaplan turbines has the concept of adjusting the runner vanes in the face of changing load conditions on the turbine, with proper adjustment of blades during its running the Kaplan turbine is capable of giving a constant and high efficiency for a wide range of load conditions. The pitch of the blades is also automatically adjusted by the governor through the action of a servo meter.

Q. 6. Define specific speed of a turbine and write down its expression.

Ans. The specific speed of a turbine may be defined as the speed of an imaginary turbine, identical with the given turbine which will develop a unit power under a unit head.

It is given by

N = Speed of the runner in r.p.m.

H =Head of water

P = Power produced.

Q. 7. Sketch different types of draft tubes.

Ans. Following are the important types of draft tubes which are commonly used.

1. Conical draft tubes

2. Simple elbow tubes

3. Moody spreading tubes

4. Elbow with circular inlet and rectangular outlet.

Fig. Types of draft tubes

Q. 8. List the various functions of surge tanks.

Ans. Surge tanks have the following functions:

1. To control the pressure variations, due to rapid changes in the pipeline flow, thus eliminating water hammer possibilities.

2. To regulate the flow of water to the turbines.

3. To reduce the distance between the free water surface and turbine, thereby reducing the water hammer effect on penstock.

4. It protects up stream tuner from high pressure rises.

Q. 9. Explain (i) Hydraulic efficiency (ii) Mechanical efficiency (iii) Overall efficiency of turbines.

Ans. (i) Hydraulic Efficiency-It is the ratio of work done on the wheel to the head of rater (or energy) actually supplied to the turbine i.e.

(ii) Mechanical Efficiency—it is the ratio of actual work available at the turbine to e energy imparted to the wheel.

(iii) Overall Efficiency—it is a measure f the performance of a turbine and is the 120 of power produced by the turbine to the energy actually supplied to the turbine.

Q. 10. Differentiate between Francis and Kaplan turbine.

Ans.

Q. 11. List the unit quantities as applied to turbo-machines.

Ans. (i) Unit power (ii) Unit speed (iii) Unit discharge.

Unit Power-The powered developed by a turbine working under a head of 1 meter, is known as unit power:

P =Power developed,

H =Head of water

Unit Speed-The speed of turbine, working under a head of 1 meter, is known as unit speed

N = speed of turbine,

H = Head of water

Unit Discharge-The discharge of a turbine, working under a head of 1 meter, is known as unit discharge.

Q = Discharge,

= Head of water

Q. 12. List the factors/criteria to choose a turbine.

Or

How to decide whether Kalpan, Francis or a pelton type tupe turbine would be used in a hydro project?

Ans. The selection of turbines depend on the following considerations.

1. Operating Head— Pelton turbine - Greater than 400 m

Francis turbine - 50-400 m

Kaplan turbine - Less than 50 m

2. Specific speed-Turbine having high specific speed is selected. High speed means a smaller size of the turbine. Francis turbines run at higher speeds (50—250) than those of pelton wheels (8—50), Kaplan turbine have the greatest specific speed (250—1000).

3. Cavitation- Cavitation occurs when the pressure at the runner outlet equals vapour pressure. Francis turbines can not be used for very high heads

because of cavitation. Pelton turbines are free from cavitation because the pressure at runner outlet is the atmospheric.

4. Performance characteristics—Turbines should be selected in such a way that their efficiencies do not fall appreciably when operating under part load. Francis turbines operate efficiently between half and full load. Kaplan turbines are more efficient at low heads.

5. Overall cost—The plant should be designed for the minimum cost as cost is the prime consideration in designing a plant

6. Number of units—It is better to go in for a larger unit as far as possible, but there must be at least two units at any particular site so that one unit is always available.

Q. 13. What is the importance of a draft tube in a Francis turbine . Discuss different types of draft tubes.

Ans. It is a pipe, which connects the turbine and outlet or tail race, through which the water exhausted from the runner, flows to the outlet channel.

It also act as a water conduit.

Draft tube has the following important function:

1. It makes the installation possible above the tail race level without the loss of head.

2. Water velocity at runner outlet is very, high. By using draft tube the velocity can be lowered. Loss of kinetic energy is converted into pressure energy.

3. Draft tube prevents the splashing of water coming out of the runner.

Different types of draft tubes used are:

(1) Conical draft tubes

(2) Simple elbow tubes

(3) Moody spreading tubes

(4) Elbow with circular inlet and rectangular outlet.

Fig. Types of draft tubes

(1) Conical Draft Tubes—This is known as tapered draft tube and used in all reaction turbines where conditions permit. It is preferred for low specific speed and Francis turbine. The maximum cone angle is 8° (a = 40°). The hydraulic efficiency is 90%.

(2) Simple Elbow Tubes-The elbow type draft tube is often preferred in most of the power plants. If the tube is large in diameter; ‘it may be necessary to make the horizontal portion of some other section. A common form of section used is over or rectangular. It has low efficiency around 60%.

(3) Moody Spreading Tubes-This tube is used to reduce the whirling action of discharge water when turbine runs at high speed under low head conditions. The draft tube has efficiency around 85%.

(4) Elbow with circular inlet and rectangular outlet—This tube has circular cross- section at inlet and rectangular section at outlet. The change from circular section to rectangular section take place in the bend from vertical leg to the horizontal leg. The efficiency is about 85%.

Q. 14. Derive the expression for specific speed of turbine. What is the range of specific speed for reaction turbine?

Ans. Power available at turbine shaft

Since and w are constant: … (1)

The tangential velocity u, the flow velocity the absolute velocity v and the head H on the turbine are related as

Now

Also

Substituting this value in expression (1)

… (2)

Where k is constant of proportionality

Now taking H =1, P= 1, then (specific speed)

Expression (ii) may be written as

Specific speed,

Specific speed for Francis turbine = 50 — 250.

Specific speed for Kaplan turbine = 250 — 850.

Q. 15. Show that in a given turbine v

H = available head, u tangential velocity, Q = discharge, P power developed.

Ans. (i) We know that

Absolute velocity v

… (1)

Also tangential velocity, … (2)

So from (1) and (2)

(ii) Q = Area of flow x Velocity

So

(iii)

so (hence proved).

Q. 16. Define draft tube efficiency. Give mathematical expression.

Ans. The efficiency of the draft tube is defined as the ratio of actual conversion of kinetic head into pressure head in the draft tube to the kinetic head at the inlet of the draft tube.

Mathematically, = =

Q .17. Why the draft tube is not used for Pelton turbine?

Ans. In case of pelton turbine all the K. E. is lost and draft tube is not used because the pressure value is just the atmospheric so there is no requirement of draft tube.

Q .18. What is the function of scross casing in reaction turbines?

Ans. Scroll casing provides the limited area around the runner to maintain the constant velocity of water flow around the runner The material of scroll casing may be cost steel, cast iron, concrete or concrete and steel.

Q.19. Explain with neat sketch the operation of Kaplan turbine, governing of Kaplan turbines and their performance characteristics.

Ans. Kaplan Turbine The figure shows a schematic diagram of Kaplan turbine The function of the guide vane is same as in case of Francis turbine

Between the guide vanes and the runner, the fluid in a propeller turbine turns through a right-angle into the axial direction and then passes through the runner. The runner usually has four or six blades and closely resembles a ship’s propeller Neglecting the frictional effects, the flow approaching the runner blades can be considered to be a free vortex with whirl velocity being inversely proportional to radius, while on the other hand, the blade velocity is directly proportional to the radius The take care of this different relationship of the fluid velocity and the blade velocity with the changes in radius, the blades are twisted. The angle with axis is greater at the tip that at the root.

Performance Characteristics of Reaction Turbine:It is not always possible in practice, although desirable, to run a machine at its maximum efficiency due to changes in operating parameters. Therefore, it becomes important to know the performance of the machine under conditions for which the efficiency is less than the maximum It is more useful to plot the basic dimensionless performance parameters (Fig 1) as derived earlier from the similarity principles of fluid machines Thus one set of curves, as shown in Fig 1, is applicable not just to the conditions of the test, but to any machine in the same homologous series under any altered conditions.

Fig.I: Performance characteristics of a reaction turbine in dimensionless parameters)

Figure 2 is one of the typical plots where variation in efficiency of different reaction turbines with the rated power is shown.

Fig. 2 Variation of efficiency with load

Governing of Reaction Turbines- Governing of reaction turbines is usually done by altering the position of the guide vanes and thus controlling the flow rate by changing the gate openings to the runner. The guide blades of a reaction turbine are pivoted and connected by levers and links to the regulating ring. Two long regulating rods, being attached to the regulating ring at their one ends, are connected to a regulating lever at their other ends. The regulating lever is keyed to a regulating shaft which is turned by a servomotor piston of the oil.

Q. 20. Write note on Surge tanks.

Ans. A surge tank is a storage reservoir fitted at some opening made on a long penstock to receive the rejected flow when the penstock is suddenly closed by a value fitted at its steed end. Surge tanks, relieves the pipe line of excessive pressure produced due to closing of the penstock, thus eliminating positive water hammer effect by admitting in it a large mass of water which would have flown out of the pipe line.

It is also used in a large pumping plant to control variations resulting from rapid changes in the flow.

Functions of surge tanks:

(1) To control the pressure variations by reliving the line of excessive pressure.

(2) Regulation of flow in power plants and pumping plants.

(3) Regulation of turbine speed.

Location of surge tank: Theoretically it should be located close to a power or pumping plant. It is enerally located at the junction of pressure tunnel and penstock or on the side of the mountain.

Types of surge tanks:(1) Single surge tanks

(2) Restricted orifice type(3) Differential type.

Q. 21. Write short note on design of runner for reaction turbine.

Ans. Suppose, H = Head

N = Running Speed

P = Power Output

The design Procedure is given as follows.

1. Assume probable values of

Hydraulic efficiency.

Overall efficiency

n, Ratio of width to diameter

Flow ratio

2. Find Discharge by using

Shaft Power

3. Area through which water enters

Where and are entrance diameter and width.

is effect for the vanes.

4. Find tangential velocity

5. Find Flow Velocity,

6. Obtain and by using

Assume

Use continuity equation

(1) Net Head, H =

(2) Hydraulic efficiency,

(3) Discharge through Kaplan turbine:

Problem 1. A Francis turbine works under a head of 25 m producing 3675 kW at 150 r.p.m. Determine the (a) Unit power and unit speed of the turbine (b) Specific speed of the turbine and (c) Power developed by this turbine if the speed is reduced to 100 r.p.m.

Solution. P= 3675 kW

H=25m

N = 150 r.p.m.

Unit power and unit speed

Unit power:

Unit speed:

Specific speed of the turbine

= 162.66

=163 r.p.m.

Power developed if the speed reduced to 100 r.p.m.

We know that

Also

Problem. 2. A Kaplan turbine runner is to be designed to develop 7357.5 kW shaft power. The net available head is 5.50 m. Assume that the speed ratio is 2.09 and flow ratio is 0.68 and the overall

efficiency is 60%. The diameter of the boss is rd of the diameter of the runner. Find the diameter of the runner, its speed and its specific speed.

Solution: Given:

Shaft power P = 7357.5 kW

Head H = 5.50m

Speed ratio

Flow ratio

Overall efficiency, = 60% = 0.60

Diameter of boss,

Using relation

0.60 =

We have

6.788m

And 6.788 = 2.262 m

Using

= 61.08 r.p.m.

Specific speed is given by

= 622 r.p.m.

Problem. 3. The following data pertains to an inward flow reaction turbine Net head = 60 m, speed = 650 r.p.m., Brake power = 275 kW Ratio of wheel width to wheel diameter at inlet = 0.10 Ratio of inner

diameter to outer diameter = 0.5 Flow ratio = 0.17, = 0.95

and = 0.85. The flow velocity remains constant and the discharge

is radial. Neglecting area blockage by blades, work out the main dimensions and blade angles of the turbine.

Solution:

Flow velocity = =5.83m/s

=5.83m/s

Power available from the turbine shaft = w Q H x

275 x = (9810 x Q x 60) x 0.85

Discharge through the turbine,

= 0.55 /s

Also

0.55 = x 0.1 d x 5.83

Diameter of wheel at inlet, = 0.5486m = 54.86cm

Width of wheel at inlet, = 0.1 x 54.86 = 5.486 cm.

Diameter of wheel at outlet, = 0.5 d = 0.5 x 54.86 = 27.43 cm

Since the discharge of water at inlet and outlet tips is same,

Width of wheel at outlet, = 0.1097 m = 10.97 cm

Angles at inlet:

Peripheral velocity at inlet, = 18.66 m/s

Hydraulic efficiency,

0.9 5

Angles at outlet:

0.6248,

Problem 4. A Francis turbine with an overall efficiency of 75% is required to produce 14825 kW power. It is working under a head of

7.62 m. The peripheral velocity = 0.26 and the radial velocity

of flow at inlet are 0.96 . The wheel runs at 150 r.p.m. and the hydraulic losses in the turbine are 22% of the available energy. Assume Radial discharge, determine

(i) The guide blade angle

(ii) The wheel vane angle at inlet

(iii) Diameter of the wheel at inlet, and

(iv) Width of the wheel at inlet.

Solution: Overall efficiency, =

Power produced = 148.25 kW

Head = 7.62 m

Peripheral velocity,

= 3.179 m/s

Velocity of flow at inlet,

= 11.738 m/s

Speed, N = 150 r.p.m.

Hydraulic losses = 22% of available energy

Discharge at outlet = Radial

Hydraulic efficiency is given as

=0.78

=0.78

= 18.34 m/s

1. The guide blade angle, a

=0.64

0.64 = 32.619°

2. The wheel vane angle at inlet,

=0.774

0.774 = 37.74

3. Diameter of wheel at inlet

=0.4047m

4. Width of the wheel at inlet

w.P.

= 2.644

Using

2.644 = x 0.4047 x x 11.738

= 0.177m

Problem 5. A hydro-turbine is required to give 25 mW at 50 m heat and 90 r.p.m. runner speed. The laboratory facilities available permit testing of 20 kW model at 5m head. What should be the model runner speed and model prototype scale ratio?

Solution: = 25 mW =20 kW

= 90 r.p.m. =5 m

=50m

Scale ratio = =6.29

=90x6.29x =179r.p.m.

Problem 6. In an inward flow reaction turbine having vertical shaft, water enters the runner from the guide blades at an angle of 155° with the runner blade angle at entry being 100°. Both these angles

are measure from the tangent at runner periphery drawn in the direction of runner rotation. The flow velocity through the runner is constant, water enters the draft tube from tile runner without whirl and the discharge from the draft tube into the tail race takes place with a velocity of 2.5 m/s. The runner has the dimensions of 40 cm external diameter and 3.8 cm inlet width. The turbine works with a net head of 35m and the loss of head in the turbine due to fluid n is 4m of water. Draw vector diagrams and calculate:

1. Speed of the runner

2. Runner blade angle at a point on the outlet edge where the radius of rotation is 9 cm.

3. Power generated by the turbine and its specific speed.

4. Inlet diameter of the draft tube.

Solution. Velocities at inlet and exit are related by the expression:

From the inlet velocity triangle

= (180—155) =25°

= (180 — 100) = 800

Since the discharge is in radial direction,

Work done = = 0.43

From the energy balance,

Head supplied

= (work done) + (kinetic heat at exit) + (losses in the runner)

3.5 + +4

= 8.45m/s

= 1.968 = 1.968 x 8.45 = l6.63m/s

1. ;1663=

N = = 794 r.p.m.

2. From outlet velocity triangle:

=8.45m/s

Peripheral velocity of the outer edge at 9 cm radius

= 16.63 x = 7.48 m/s

= 1.13 ; vane angle at outlet, =

Discharge through the turbine, Q x 0.4 x 0.038 x 8.45

= 0.4035

3. Power developed by the turbine,

= 9810 x 0.4035 x 0.43 121.5 x W = 121.5 kW

Assume a mechanical efficiency of 98%

4. Power available at turbine shaft = 121. 5 x 0.98 = 119.07

Specific speed of the turbine, Ns = 101.77

5. Inlet area of draft tube = = = 0.04775

If d is the inlet diameter of the tube,

=0.04775

d= =0.246m

Problem.7. Francis turbine develops 365 kW at an overall efficiency of 80%. When working under a static head of 5 m, the draft tube being cylindrical and of diameter 2.5 m. What increase in power and efficiency of the turbine would you expect if a tapered draft tube having an inlet diameter of 4m and efficiency of conversion of 90% is substituted for the cylindrical one? It maybe presumed that head, speed and discharge remain constant.

Solution Power available = wQH×

365 x = (9810 x Q x 5) x 0.8

Q=

When the draft tube is tapered one velocity of water at inlet to draft tube

=1.89m/s

velocity of water at outlet of draft tube

=0.74m/s

Heat gained = xO.9 = 0.14 m

Increase in efficiency = = 0.028 = 2.8%

Increase in power = increase in efficiency x original power

= 0.028 x (9810 x 9.30 x 5)

=12773W=12.77kW

Problem.8. An inward flow reaction turbine discharges radially and the velocity of flow is constant and equal to the velocity of discharge from the turbine. Show that the hydraulic efficiency can be expressed by

Where a and are respectively the guide vane angle and wheel vane angle at Intel.

Solution. From the inlet velocity triangle

For radical discharge at outlet

Thus

Or

Substituting the value of , we get

Also substituting the value of u from above, we get

Now =

Or =

Or

Problem.9. The velocity of whirl at inlet to the runner of an inward

flow reaction turbine is (3.15 ) m/s and the velocity of flow at inlet

is (1.05 ) m/s. The velocity of whirl at exit is (0.22 ) m/s in the

same direction as at inlet and the velocity of flow at exit is (0.83 ) where H is the head in meters. The inner diameter of the runner is

0.6 times the outer diameter. Assuming hydraulic efficiency of 80%, compute the angles of the runner vanes at inlet and exit.

Solution.

From inlet velocity triangle, we have

1.9091

= 62°21’

From outlet velocity triangle, we have

=0.6194

=31°46’

Problem.10. The inlet and the outlet runner blade angles of a propeller turbine are and 25° respectively to the tangential direction of the runner. The inlet guide vane angle is 30°. The speed of the turbine 30 rpm. The mean diameter of the runner blades is 3.6 m and the area of flow is 30 . Assuming that the velocity of flow is constant throughout, determine (1) Discharge (ii) Power developed (iii) Hydraulic efficiency (iv) Specific speed.

Solution.

=3.6m

N =30 r.p.m.

= 90°

= 25°

a = 30°

Flow area, a =

Runner blade angle at inlet is radial

As velocity of flow is constant so

= 5.65 m/s

Also

= 5.65 m/s

From inlet velocity triangle

=5.65xtan30°

=3.262m /s

=5.65m/s

From outlet velocity triangle,

tan 25°

+ 5.65 = =7 m/s

= 7 - 5.65 = 1.35 m/s

=3.529 m/s

We have,

[5.65 x 5.65 — 1.35 x 5.65]

H = 2.47 + 0.634 3.104 m

(1)Hydraulic efficiency is given by

=0.798 = 79.8%.

(2)Discharge through turbine, Q = Area of flow x Velocity of flow

=30x3262=97.86

(3)Power developed by turbine

× Weight of water

x 1000 x 9.81 x 97.86

= 2378 kW

(4)Specific speed is given by 355.08 rpm

Problem.11. In a Francis turbine of very low specific speed, the velocity of flow from inlet to exit of the runner remains constant. If the turbine discharges radially, show that the degree of reaction p can be expressed as

where a and are the guide and runner vane angles respectively and the degree of reaction p is equal to the ratio of pressure drop to the hydraulic work done in the runner, assuming that the losses in the runner are negligible.

Solution. Applying Bernoulli’s equation between the inlet and exit of the runner and neglecting the potential difference, we get

(for radial discharge)

Where and are the pressure heads at the inlet aid the exit of the runner respectively.

Thus pressure head drop due to hydraulic work done in the runner is given by

Now

Or

Or … (1)

For radical discharge

Also

Or u =V [cos a-sin a cot ]

And

Thus, introducing these values in equation (i) above and simplifying it, we get.

Problem.12. A Francis turbine supplied through a 6 m diameter penstock has the following particulars.

Output of installation 63500 kW

Flow 117

Speed 150 r.p.m.

Hydraulic efficiency 92%

Mean diameter of turbine at entry 4 m

Mean blade height at entry 1 m

Entry diameter of draft tube 4.2 m

Velocity in tail race 2.4 m/s

The static pressure head in the penstock measured before entry to the runner is 57.4 m. The point of measurement is 3 m above the level of the tail race. The loss in the draft tube is equivalent to 30% of the velocity head at entry to it. The exit plane of the runner is 2 m above the tail race an the flow leaves the runner without swirl. Determine:

1. The overall efficiency,

2. The direction of flow relative to the runner at inlet,

3. The pressure head at entry to the draft tube.

Solution.

(a)The net head H for the turbine is given by equation

And 2.4 m/s.

Thus by substitution, we get

= 60.98 m

The overall efficiency is given by

= 0.907 or 90.7%

(b) Neglecting the vane thickness, the velocity of flow at inlet i given by equation

B=1m; and D=4m


= 9.31 m/s

31.42 m/s

Or 0.92

17.52 m/s

The direction of flow relative to the runner at inlet is given by

=0.6698

(c) The pressure head at entry to the draft tube is given by equation

2m =8.44m/s; =2.4m/s

And =1.09 m


+1.090—4.25m

Problem.13. A model of Francis turbine one-fifth of full size, develops 3 kW at 306 r.p.m. under a head of 1.77 m. Find the speed and power of full size turbine operating under a head of 5.7 m, if (a) the efficiency of the model and the full size turbine are same, (b) the efficiency of the model turbine is 76% and the scale effect is considered.

Solution: (a) For the same efficiency of the model and the prototype

Or

=109.8 r.p.m.

Further

Or

=433.43kW

(b) According to Moody’s equation,

Or

Or

=114.5 r.p.m.

We know that

Or

=491.09 kw

Problem.14. Show that in a turbine, with radial vanes at inlet and outlet, the hydraulic efficiency is given by:

Where is the guide blade angle. Assume the flow velocity to remain constant.

Solution. Neglecting losses with in the runner, the energy balance gives:

Head supplied = (work done or head utilized) + (kinetic head at exit)

For radial vanes at inlet and outlet

Hydraulic efficiency,

Problem.15. A Kaplan turbine develops 2250 kW under a net head of 5.5 m and with overall efficiency 87 percent. The draft tube has a diameter of 2.8 m at its inlet and has an efficiency of 78 percent. In order to avoid cavitation, the pressure head at entry to the draft tube must not drop more than 4.5 m below atmosphere. Calculate the maximum height at which the runner may be set above the tail race level.

Solution. Power available from the turbine shaft,

P = wQH x

2250 × = (9810 x Q x 55) x 0.87;

Q = 47.93

Now =7.79 m/s

Given: = 4.5m

=4.5-

Draft tube efficiency,

0.78 =

= 4.5 - x 0.78 = 2.087

Problem.16. An inward flow pressure turbine has runner vanes which are radial-at the inlet and inclined backward at 45° to the tangent at discharge. The guide vanes are inclined at 15° to tangent at inlet and velocity of water leaving the guides in 24 m/sec. Determine correct speed for runner and absolute velocity of water at point of discharge if diameter at entry is twice that at discharge and width at entry is 0.6 times that at discharge.

Solution.

Fig. Input and outlet velocity triangle

Given =15°

= 24 m/s

In outlet velocity triangle

= m/s Ans.

We know

x33.94

x 33.94 = 28.28 m/s

(For radial discharge)

= 105.94 m/s Ans.

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