8/11/2019 solved problems and sheet.doc

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Q1//In an impulse stage the mean diameter of the blade ring is 800 mm and the speed of rotation is 3000

rpm. The steam issues from the nozzles with a velocity of 300 m/s and the nozzle angle is 20. The blades

are symmetric and the blade velocity coefficient !"# is 0.8$ . %hat is the power developed in the stage whenthe a&ial thrust on the blades is '(0 ).*

Solution:

smDN

b /$.'2+$0

3000,8.0,

$0===

o

ii

sma 20-/300 ==

From the velocity triangular:

$+$+.0$.'2+20cos300

20sin,300

cos

sintan =

=

=

o

o

ii

iii

ba

a

Since the blades are symmetric, then:o

oi 3.33==

Also from the velocity triangular:

-

smr

smr

r

ra

o

i

i

iiii

/'$''8,8$.0

/'8

3.33sin$'.'02

sinsin

==

=

=

=

The axial thrust is given by:

skgm

m

m

rrmF eeiiaxial

/(+$.

'(03$+(.'(,

#8$.0'!3.33sin'81

sinsin1

=

==

=

=

developedpowerthe

kW

rrmPowereeii

8(.3++

coscos1

=

=

Q//The nozzles of the impulse stage of a turbine receive steam at '+ bar and 300 and discharge it at

l0bar. The nozzle efficiency is + and the nozzle angle is 20. The blade speed is that re4uired forma&imum blade efficiency- and the inlet angle of the blades is that re4uired for entry of the steam without

shoc". The blade e&it angle is + less than the inlet angle. The blade friction factor is 0.. alculate for a

steam flow of '3+0 "g/h- !a# the a&ial thrust. !b# The diagram power- and !e# the diagram efficiency.

!roblems sheet '

b"ect: !o#er !lant Technology

a%ter: Steam Turbines

cturer:'r( S () (Faisal

ia

i

e

i

ir

e

erea

fe=r

esin

e

fi=r

isin

i

f=fi-f

e

i=r

icos

i

e=r

ecos

e

b

= i+

e

8/11/2019 solved problems and sheet.doc

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Solution:

3038.9= kgkJ/h'

The state *s* is in the su%erheated region

"5/"g2(3.'h-2+0T 2s's ==

smhhai snozz /$3.(2$#'.2(3.3038!,'000,+.0,2#!2 2' ===

For maximum efficiency:

(+.200

($.02

20cos

2

cos

=

===

b

a

b oi

i

oe

i

ii

iii

ba

a

3'+3$

3$

2.0(+.200

'$.'(+

cos

sintan

==

=

==

=

smari

iii /2+.2(8

sinsin ==

smkrr ie /(2.223, ==

smrr eeii /3+.32coscos1 =+=

smrrf eeii /8+.30sinsin1 ==

NfmFaxial +.''8+.30,3$00

'3+0===

NmFdriving '3.'(3+.32,3$00'3+0 ===

!roblems sheet 20

ia

i

e

i

ir

e

er

eafe=r

esin

e

fi=r

isin

i

f=f

i-fe

i=r

icos

i

e=r

ecos

e

b

= i+

e

8/11/2019 solved problems and sheet.doc

3/4

kW

mbdevelopedPower

(/2.2/

'0,(+.200,'3.'( 3

=

==

(.8$8$(.0

#$3.(2$!,3$00

'3+0,

2

'

2(2

2

===blade

++++++nsolved !roblems++++++Q1//the velocity of steam at inlet to simple impulse turbine is '000 m/s and the nozzle single is 20o. The

blade speed is (00 m/s and the blades are symmetrical. 6etermine the blade angle- tangential force-

diagrame power- a&ial thrust- and the diagram efficiency for frictionless blades.

If the relative velocity at e&it is reduced by friction to 80 of that at inlet- then recalculates the diagram

power- a&ial thrust- and the diagram efficiency. Ta"e mass flow rate ' "g/s .Ans#er .(o, 10(2 3,

4(2 56, 0 3, 7( 8,+ .7.(4. 56, 2(79 3, 2(28

Q// a single row impulse turbine receive 3 "g/s of steam with velocity (2+ m/s . The blade speed ratio is

0.( and the output power is ''.+8 "%. If the frictional losses in the moving blade amount to '(.2 "%-

determine the diagram efficiency and the blade velocity coefficient. The nozzles angle is '$o.

Ans#er: .(.8; 0(92

Q.//a simple impulse turbine !6e 7aval turbine# is supplied with steam at '+ bar- (00 osuperheat. The

steam e&pands in the nozzles- which have an efficiency of 0- to pressure of 'bar. 9ssuming the nozzle

angle is 20o- blade velocity coefficient is 0.8- and symmetrical blades: determine for ideally ma&imum blade

efficiency condition;

'.The blade speed. 2. The blade angle. 3.The power output for hrkgm /000= .

(.

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Q2//for 6e7aval turbine of symmetrical and frictionless blades degree -derive an e&pression for blade

efficiency showing the optimum operation of blade speed ratio !b/a i# and the corresponding ma&imum

efficiency power.

Ans#er: #!cos(i

i

i

ba

b

a

b= %here !

ia

b# is the blade speed ratio

by ma"ing 0#!

#!=

i

b

a

bd

d

we find2

cos i

optimumiab = : ma& ib

2cos= : power> 22 bm

The velocity diagram will be;

Q(7//in a single stage simple impulse turbine the steam flows ata rate of +"g/s. It has rotor of '.2m diameter

running at 3000 rpm. )ozzle angle is '8?- blade speed ratio is 0.( and velocity coefficient is 0.- outlet

angle of blade is 3? less than inlet angle. 6etermine blade angles and power developed.

Q(9//In a certain stage of an impulse turbine- the nozzle angle 20? with the plane of the wheel. The mean

diameter of the blade ring is 2.8 metres. Its develop ++"w at 2(00 rpm. =our nozzles- each of '0 mm

diameters e&pand steam isentropically from '+ bar and 2+0? to 0.+ bar. The a&ial thrust is 3.+ ).

alculate blade angle at entrance and e&it and power lost in blade friction.

!roblems sheet 22

b

ia

b

optimumia

b

ma&b

b

mri

re

ai

ae