126208289 Francis Turbine

41
MINI PROJECT An Academic Project submitted by JAGANNATH.C I.Mech.E., A.M.I.Mech.E (I). CHARTERED ENGINEER (INDIA) MEM.NO ME/KH/272870IME THE INSTITUTION OF MECHANICAL ENGINEERS (INDIA) CHENNAI.

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Transcript of 126208289 Francis Turbine

Page 1: 126208289 Francis Turbine

MINI PROJECT

An Academic Project submitted by

JAGANNATH.C I.Mech.E., A.M.I.Mech.E (I).

CHARTERED ENGINEER (INDIA)

MEM.NO ME/KH/272870IME

THE INSTITUTION OF MECHANICAL ENGINEERS (INDIA)

CHENNAI.

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• Examples

• Losses in Francis turbines

• NPSH

• Main dimensions

Francis turbines

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X blade runnerTraditional runner

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SVARTISEN

P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm

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P = 169 MWH = 72 mQ = 265 m3/sD0 = 6,68 mD1e = 5,71mD1i = 2,35 mB0 = 1,4 mn = 112,5 rpm

La Grande, Canada

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Outletdraft tube

Outletrunner

Inletrunner

Outletguide vane

Inletguide vane

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Hydraulic efficiency

n

uuh Hg

ucuc

2211

3

23

31

21

1

3

23

31

21

1

z2c

hgz2c

hg

lossesz2c

hgz2c

hg

1

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Losses in Francis Turbines

Draft tube

Output Energy

Head [m]

Hyd

raul

ic E

ffic

ienc

y [%

]

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Losses in Francis TurbinesH

ydra

ulic

Eff

icie

ncy

[%]

Output Energy

Output [%]

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Friction losses between runner and covers

Friction losses

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Gap losses

Gap losses

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Gap losses

Friction losses in the spiral casingand stay vanes

Guide vane losses

Friction losses

Runner losses

Draft tube losses

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Relative path

Absolute path with the runner installed

Absolute path without the runner installed

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Velocity triangles

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Net Positive Suction Head, NPSH

1

g

cz

g

chhz

g

ch b

222

23

3

23

32

22

2'22 hhh b

23

23

3 2zz

g

chHs

g2

czHhz

g2

ch

23

2sb2

22

2

J

JHhg2

ch sb

22

2

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NPSH

vasb hJg

cHhh

2

22

2

NPSH

s2b HhhNPSH

NB:HS has a negative value in this figure.

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AsvabR NPSHHhhNPSH

NPSH required

g

ub

g

caNPSH m

R

22

22

22

25.0b2.015.0b05.0b

0.2a6.115.1a05.1a

PumpsTurbines

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Main dimensions

D1

D2

• Dimensions of the outlet

• Speed

• Dimensions of the inlet

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Dimensions of the outlet

g

ubua

g

ub

g

caNPSH m

R

2

tan

22

22

222

22

22

We assume cu2= 0 and choose 2 and u2 from NPSHR:

13o < 2 < 22o (Lowest value for highest head)

35 < u2 < 43 m/s (Highest value for highest head)

1,05 < a < 1,150,05 < b < 0,15

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Diameter at the outlet

222m tanuc

D1

D2

22

4

mc

QD

22

2m2m

22

D

4Qcc

4

DQ

Connection between cm2and choose D2 :

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Speed

Connection between n and choose u2 :

2

222 D

60un

60

nDu

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Correction of the speed

The speed of the generator is given from the number of poles and the net frequency

Hz50fforz

3000n

p

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ExampleGiven data:

Flow rate Q = 71.5 m3/sHead H = 543 m

We choose:a = 1,10b = 0,102 = 22o

u2 = 40 m/s

m8,22

g2

401,022tan401,1NPSH

22

R

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Find D2 from:

2222 tan

44

u

Q

c

QD

m

m37.222tan40

5,714D2

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Find speed from:

rpmD

un 322

37.2

604060

2

2

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Correct the speed with synchronic speed:

rpm3339

3000n9zchoose

3.9n

3000z

Kp

p

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We keep the velocity triangle at the outlet:

u2

w2c2

c2K

u2K

2

mn

DnD

DnDn

Dn

DQ

Dn

DQ

u

c

u

c

KK

KK

KK

K

K

mKm

35.2333

322373.2

60

4

60

4

tan

33

32

2

32

32

2

22

2

22

22

22

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Dimensions of the inlet

2u21u12u21u1

h cucu2Hg

cucu

At best efficiency point, cu2= 0

1u11u1

h cu2Hg

cu96,0

11

h1u u2

96,0

u2c

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We choose: 0,7 < u1 < 0,75

Diameter at the inlet

D1

D2

n

60uD

2

D

60

2n

2

Du

11

111

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Height of the inlet

22m11m AcAc

Continuity gives:

We choose: cm2 = 1,1 · cm1

4

D1.1DB

22

11

01 BB

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Inlet angle

u1

w1c1

1

cu1

cm1

11

11tan

u

m

cu

c

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Example continuesGiven data:

Flow rate Q = 71.5 m3/sHead H = 543 m

We choose:h = 0,96

u1 = 0,728cu2 = 0

66,0728,02

96.0

22

1111

uccu huuh

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Diameter at the inlet

smhguu 15,7554381,92728,0211

mn

uD 31,4

333

6015,756011

D1

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Height of the inlet

B1

mD

DB 35,0

31,44

35.21,1

4

1,1 2

1

22

1

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Inlet angle

u1

w1c1

1

cu1

cm1

sms

mm

m

ucc 9,13

1,1

9,4022sin

1,1

sin

1,1222

1

135,054381,92

9,13

hg2

cc 1m

1m

9.62659,0728,0

135,0arctanarctan

11

11

u

m

cu

c

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SVARTISEN

P = 350 MWH = 543 mQ* = 71,5 m3/SD0 = 4,86 mD1 = 4,31mD2 = 2,35 mB0 = 0,28 mn = 333 rpm

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THANK YOU

I AM THANKFULL TO MY H.O.D AND FRIENDS WHOSE TIMELY ASSISTANCE MADE ME TO

ACCOMPLISH MINI PROJECT THIS