solved problems and sheet.doc

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  = 32.3+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 , 32.3+ = '(.'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(92 Q.//a simple impulse turbine !6e 7aval turbine# is supplied with steam at '+ bar- (00 osuperheat. 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 (.