Sub "ect: !o#er !lant Tec Technology hnology $ha%ter: Steam Turbines &ecturer:'r( S () (Faisal
Q1//In an impulse stage the mean diameter of the blade ring is 800 mm and the speed of rotation is 3000 rpm. The steam issues from the nozzles with a velocity of 300 m/s and the nozzle angle is 20. The blades are symmetric and the blade velocity coefficient !"# is 0.8$ . %hat is the power developed in the stage when the a&ial thrust on the blades is '(0 ).* Solution: e
β n i s e r = f e e f - i f = f
α i
β e
β i
i
b
α e ae
r i
r e
β n i s i r = f i
ai
ωi =r i cos β i
ωe=r e cos β e
ω= ωi + ωe π DN π , 0.8 , 3000 = = '2+.$ m / s $0 $0 ai 300m / s - α i 20o From the velocity triangular: o ai sin α i 300 , sin 20 tan β i = = ai cosα i − b 300 cos 20o − '2+.$ Since the blades are symmetric, then: β i = β o = 33.3o Also from the velocity triangular: ai sin α i = r i sin β i b=
=
=
=
0.$+$+
'02.$' = r i sin 33.3
-
⇒ r i = '8
m / s
∴ r o = 0.8$ ,'8 = '$'m / s
The axial thrust is given by: 1 r i sin α i − r e sin α e F axial = m 1'8 sin 33.3!' − 0.8$# = m ,'(.3$+( = '(0 = m = .(+$kg / s ⇒ m the power power developed
1r i cos β i Power = m =
−
r e cos β e
3++.8(kW
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 for ma&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
'
Solution:
α i
e
β n i s e r = f e
β e
β i
i
b
α e ae
e f - i f = f
r e
r i
β n i s i r = f i
ai
ωi =r i cos β i
ωe=r e cos β e
ω= ωi + ωe
h'
= 3038.9 kJ / kg
The state *s* is in the su%erheated region T's = 2+0 °- h 2s = 2(3.' "5/"g 2η nozz !h' − h2 s #
ai =
2 , 0.+ ,'000 , !3038. − 2(3 .'#
=
=
(2$.$3m / s
For maximum efficiency: cos α i b cos 20 o =
=
ai
=
2
2 ⇒
tan β i
r i
=
=
b
=
200 .(+
ai sin α i ai cos α i
ai sin α i sin β i
=
0.($
=
'(+.'$
−b
200.(+
=
0.2
⇒
β i
=
3$
⇒
β e
=
3$ − + = 3'o
2(8.2+m / s
r e = r i , k = 223.(2m / s ω = 1r i cos β i + r e cos β e = 32.3+m / s f = 1 r i sin β i
F axial
=
F driving
−
f = m
r e sin β e = 30.8+m / s
'3+0
, 30.8+ = ''.+ N 3$00 '3+0 ω = =m , 32.3+ = '(.'3 N 3$00
!roblems sheet
20
Power developed
=
ω bm
=
'(.'3 , 200.(+ , '0
=
η blade
=
2(2 ' '3+0 2 , , !(2$.$3# 2 3$00
3
−
2/.(/2kW
=
0.8$( = 8$.(
++++++nsolved !roblems++++++ Q1// the velocity of steam at inlet to simple impulse turbine is '000 m/s and the nozzle single is 20o. The blade speed is (00 m/s and the blades are symmetrical. 6etermine the blade angle- tangential forcediagrame power- a&ial thrust- and the diagram efficiency for frictionless blades. If the relative velocity at e&it is reduced by friction to 80 of that at inlet- then recalculates the diagram power- a&ial thrust- and the diagram efficiency. Ta"e mass flow rate ' "g/s .Ans#er .(o, 10(2 3, 4(2 56, 0 3, 7( 8,+ .7.(4. 56, 2(79 3, 2(28 Q// a single row impulse turbine receive 3 "g/s of steam with velocity (2+ m/s . The blade speed ratio is 0.( and the output power is ''.+8 "%. If the frictional losses in the moving blade amount to '(.2 "%determine the diagram efficiency and the blade velocity coefficient. The nozzles angle is '$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; = 000 kg / hr . '.The blade speed. 2. The blade angle. 3.The power output for m (.
(
Q// The nozzles of the impulse stage of turbine receive steam at '+ bar- 300 o and discharge it at '0 bar. The nozzle efficiency is + and the nozzle angle is 20o. The blade speed is that re4uired for ma&imum power and the inlet angle of the blades is that re4uired for entry of steam without shoc" .the blade e&it angle is +o less than the inlet angle. The blade velocity coefficient is 0.. alculate for steam flow rate '3+0 "g/hr; '. The diagram power. 2. The diagram efficiency. Ans#er .0(. 56 , 7(. 8 Q// the steam from the nozzle of single wheel impulse turbine discharges with velocity of $00 m/s at angle of 20o . The blade wheel rotates at 3000 rev/min and the mean blade radius is +0mm. the a&ial velocity of steam at e&it from the blade ! f e # is '$( m/s and the blades are symmetrical- calculate; '. The blades angles. 2. The diagram power. 3. The diagram efficiency . (. The blade velocity coefficient. Ans#er 7o ,1(. 56 , 208 ,0(299
!roblems sheet
2'
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/ai# and the corresponding ma&imum efficiency power. Ans#er:η b
=
(
b ai
d !η b # b by ma"ing d ! # ai
!cosα i
=
0
−
b ai
#
%here !
b
we find
ai
b ai
=
optimum
# is the blade speed ratio
cosα i 2
: ma&η b
=
2
cos α i : power> 2m b2
η b
The velocity diagram will be;
η b
ma&
b ae
mr i r e
b
ai
ai b ai
optimum
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