Prime Movers and Fluid Machines 

BITS PilaniK K Birla Goa Campus

Dr. Varinder SinghDepartment of Mech Enggp p gg

Lecture outline• Francis Turbine construction• Application of Euler’s equation to Francis turbinepp q• Efficiency expression and condition for its maximum value• Design procedure

N i l l• Numerical examples

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FRANCIS TURBINE

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INTRODUCTIONNamed after the engineer Sir J.B FrancisReaction turbine- available head is supplied as pressure headReaction turbine available head is supplied as pressure head

to the runnerMixed flow turbine- as water enters radially and exits axiallySuitable for medium heads of about 30-400m

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SCHEMATIC DIAGRAMSCHEMATIC DIAGRAM

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Spiral Casing– Cross-sectional area gradually decreases around the circumference of

wheel – water enters the runner at constant velocity at all points.

Guide vanes– Aerofoil structure– Behave like nozzle– Form a part of governing mechanismForm a part of governing mechanism– Also used for starting and stopping the turbine

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Guide vanes controlling water inputp

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Construction

Runner– Water enters radially and leaves axiallyy y– Classified as slow, medium, fast– static pressure at the runner outlet is much less, sometimes less

than atmospheric pressurethan atmospheric pressure– Absolute velocity of water leaving the runner will be high– In order to reduce leakage due to the clearance between the

rotating runner and fixed guide blades forged steel ring seals arerotating runner and fixed guide blades, forged steel ring seals are provided.

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Slow runner

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Medium runner

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Fast runner

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Construction

Draft tube– Kinetic energy of water after passing through runner is converted to gy p g g

potential energy– Water velocity at the exit of draft tube is about 1m/s – free end of the draft tube is submerged to 0 8m below tail race levelfree end of the draft tube is submerged to 0.8m below tail race level– Important constituent of a reaction turbine

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Draft tube functions– Makes possible the installation of the turbine above the tail race

level without loss of head– Velocity of water discharged from the runner outlet is reduced– Prevents splashing of water coming out of the runner and guides

the water to the tail race

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POWER AND EFFICIENCYPower developed is given by

1 2( )w wP Q V V uρ= ±

Since the discharge is axial,                 hence2 0wV =

1wP QV uρ=

And input power isQgHρ

Where H is the applied head on the runner

Hydraulic efficiency,

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y y,1 1w

HV ugH

η =

POWER AND EFFICIENCYIf the losses in the runner are neglected, then the energy utilized in the

runner can also be expressed as (When no draft tube installed):2

2( )2output

VP Q gHρ= −

where,        is the velocity of water at runner outlet2V

Thus hydraulic efficiency,2 2

1 1 2 22 12

wH

V u gH V VH H H

η −= = = −

2gH gH gH

When draft tube with 100 percent efficiency is installed,2 2

1 1 4 42V u gH V V−

Overall efficiency,

η = Shaft output

1 1 4 42 12

wH

V u gH V VgH gH gH

η = = = −

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0η =QgHρ

Q1. A Francis turbine develops 15MW under a head of 200m. The specific speed is 100, overall efficiency is 0.85 and the velocity of flow is constant p , y yand is equal to 10m/s. The hydraulic efficiency is 0.88, the ratio of width to diameter of wheel at the inlet is 0.1 and the area occupied by the thickness of the blades is 5% the area of the water way. Estimate the area, guide blade angle, peripheral velocity and velocity of whirl at the i l t A i l di hinlet. Assume axial discharge.

35/4 ; 15 10s

N PN P kWH

= = ×Specific speed,

H( )

( )

5/4

1/23

100 200614

15 10N rpm

N

×= =

×

Overall efficiency.50 60f 600

60 5syn

syn

pNas N ×

= ⇒ = =6

0 3

15 100.8510 9 81 200Q

η ×= =

× × ×10 9.81 200Q× × ×38.9 /Q m s=

Also Q = area of flow X velocity of flowAlso, Q = area of flow X velocity of flow

28.9 0.8910f

Q mV

= =Net area of flow

10fV

1 1 fQ D bVπ=Again, discharge,

1 1.73 600 54 3 /D Nπ π × ×

1 18.9 0.95 0.1 10D Dπ= × × ×

1 1.73D m=

h l d l 11

s

54.3 /60 6054.3 / 2 9.81 200 0.87(inappropriate for N 100)

u m s

ϕ

= = =

⇒ = × × = =

Peripheral speed at inlet,

Recommended value of 0.65ϕ =

1

11

2 40.7 /60 1.296

u gH m suD mN

ϕ

ϕ

π

⇒ = =

= =

21 1

2

8.9Recommended value of 17.7 /0.95 (0.1 ) 0.095 (1.296)f

NQV m s

D D

V

π

π π= = =

1 1, wH

V uAlsogH

η =

2

2 0.92f

H

VgH

gHη

−= =

Velocity of whirl at inlet,

Guide blade angle α is given by

1

0.92 9.81 200 44.3 /40.7wV m s× ×

= =

11 1

17.7tan 21.7844.3

fVV

α α= = ⇒ =

Guide blade angle, α is given by

144.3wV

1

1

1 11

17.7tan 77.2644.7 40.7

f o

w

Vu V

β β= = ⇒ =− −

2

24 0.82

0.95 f

QD mVπ

= =

17 7V

122 25.6

60D Nu msπ −= =

22 2

2

17.7tan 34.6625.6

f oVu

β β= = ⇒ =

Operating characteristics

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Deriaz turbine

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Example 7.9p

A low head turbine works under a head of 5.5 m and produces 2.4MW, when the discharge through the turbine is 63.5 m3/s. The draft tube has an inlet diameter of 3.5 m and an outlet diameter of 4 m The draft tube efficiency is 80%m and an outlet diameter of 4 m. The draft tube efficiency is 80%.

(i) Find the increment in hydraulic efficiency and power output that would result if the original draft tube had an efficiency of 100% instead of 80%.

(ii) If the draft tube had been cylindrical with a constant diameter of 3.5m, the head, speed and discharge being the same as above, find the hydraulic efficiency and the new power as a percentage of the previous onenew power as a percentage of the previous one.

THANK YOUTHANK YOU

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