126208289 Francis Turbine

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Transcript of 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.

• Examples

• Losses in Francis turbines

• NPSH

• Main dimensions

Francis turbines

X blade runnerTraditional runner

SVARTISEN

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

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

Outletdraft tube

Outletrunner

Inletrunner

Outletguide vane

Inletguide vane

Hydraulic efficiency

n

uuh Hg

ucuc

2211

3

23

31

21

1

3

23

31

21

1

z2c

hgz2c

hg

lossesz2c

hgz2c

hg

1

Losses in Francis Turbines

Draft tube

Output Energy

Head [m]

Hyd

raul

ic E

ffic

ienc

y [%

]

Losses in Francis TurbinesH

ydra

ulic

Eff

icie

ncy

[%]

Output Energy

Output [%]

Friction losses between runner and covers

Friction losses

Gap losses

Gap losses

Gap losses

Friction losses in the spiral casingand stay vanes

Guide vane losses

Friction losses

Runner losses

Draft tube losses

Relative path

Absolute path with the runner installed

Absolute path without the runner installed

Velocity triangles

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

NPSH

vasb hJg

cHhh

2

22

2

NPSH

s2b HhhNPSH

NB:HS has a negative value in this figure.

AsvabR NPSHHhhNPSH

NPSH required

g

ub

g

caNPSH m

R

22

22

22

25.0b2.015.0b05.0b

0.2a6.115.1a05.1a

PumpsTurbines

Main dimensions

D1

D2

• Dimensions of the outlet

• Speed

• Dimensions of the inlet

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

Diameter at the outlet

222m tanuc

D1

D2

22

4

mc

QD

22

2m2m

22

D

4Qcc

4

DQ

Connection between cm2and choose D2 :

Speed

Connection between n and choose u2 :

2

222 D

60un

60

nDu

Correction of the speed

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

Hz50fforz

3000n

p

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

Find D2 from:

2222 tan

44

u

Q

c

QD

m

m37.222tan40

5,714D2

Find speed from:

rpmD

un 322

37.2

604060

2

2

Correct the speed with synchronic speed:

rpm3339

3000n9zchoose

3.9n

3000z

Kp

p

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

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

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

Diameter at the inlet

D1

D2

n

60uD

2

D

60

2n

2

Du

11

111

Height of the inlet

22m11m AcAc

Continuity gives:

We choose: cm2 = 1,1 · cm1

4

D1.1DB

22

11

01 BB

Inlet angle

u1

w1c1

1

cu1

cm1

11

11tan

u

m

cu

c

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

Diameter at the inlet

smhguu 15,7554381,92728,0211

mn

uD 31,4

333

6015,756011

D1

Height of the inlet

B1

mD

DB 35,0

31,44

35.21,1

4

1,1 2

1

22

1

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

SVARTISEN

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

THANK YOU

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

ACCOMPLISH MINI PROJECT THIS