Vapor-cycle

Power Plants
Ideal Regenerative Rankine Cycle
with Open Feedwater Heater
Open Systems - Control Volume
Axial-flow
Hydraulic
Turbine
Compressors, pumps, and fans

Side view End view

Compressor
of pump of pump
Inflow and Outflow Velocity Triangles
 Radial Flow Pump (Fan)
U = r
W
V

Relative Velocity – tangent to blade

 Radial Flow Turbine
Absolute velocity – tangent to stator blade
V W

U = r

Relative velocity – tangent to rotor blade

Axial Flow Compressor (Pump)

Relative Velocity – tangent to rotor blade

U = r
 Axial Flow Turbine

Relative
Velocity –
tangent to
rotor blade
U = r
Absolute
velocity –
tangent to
stator blade
 Lawn Sprinkler
Water enters a rotating lawn sprinkler
through its base at the steady rate of 16
gal/min as shown in the figure. The exit
cross-sectional area of each of the two
nozzles is 0.04 in.2, and the flow leaving
each nozzle is tangential. The radius
from the axis of rotation to the centerline
of each nozzle is 8 in. (a) Determine the
resisting torque required to hold the
sprinkler head stationary. (b) Determine
the resisting torque associated with the
sprinkler rotating with a constant speed
of 500 rpm. (c) Determine the angular
velocity of the sprinkler if no resisting
torque is applied.
 Lawn Sprinkler

absolute velocity : V  V 2 ê
 
nozzle veloecity : U   r2 ê
   
relative velocity : W  W 2 ê  V  U  V 2   r2  ê

Continuity equation

 W  n̂dA   V1 ê z     A1 ê z   2    W 2 ê     A2 ê   0
CS

 m  Q  V1 A1  2 W 2 A2 Relative velocity W2

 slugs  gal  min  ft 3  slugs
  1.94 
3 
 16     
 7.48 gal   0 . 0692
 ft   min   60 s    s
Q 16  gal / min   min  ft 3  144 in 2 
W2        64.17 ft / s
2 A2 2 0.04 in 
2

 60 s  7.48 gal  ft
2

 Lawn Sprinkler
Moment-of-momentum equation

Tshaft    Tshaft  ê z    
   
r  V W  n̂dA
CS    
dm

  r1 ê r  V1 ê z   V1 ê z     A1 ê z    r2 ê r    V 2 ê   2    W 2 ê     A2 ê 

               
m m

 0  m 2 r2V 2 ê z   m r2V 2 ê z

 slugs  8 ft  ft 
Tshaft  m r2V 2   0.0692   64.17 
(a) U2 = 0 ,  s  12  s
slugs  ft 2
V2 = W2  2.96 2
 2.96 lb - ft
s
 Lawn Sprinkler
 rad  8  ft
U 2   r  500  2  ft   34.91
 s  12  s
ft ft
(b)  = 500 rpm V 2  W 2  U 2   64.17  34.91  29.26
s s
 slugs  8 ft  ft 
Tshaft  m r2V 2   0.0692   29.26   1.35 lb - ft
 s  12  s

Tshaft  0  V 2  0
ft
(c) Tshaft = 0 W 2  U 2  r  64.17
s
W 64.17 ft / s  s  1 rev  rev
  2   60    920
r2  8 / 12  ft  min  2 rad  min
 Lawn Sprinkler
The two-nozzle law sprinkler discharge water at a rate of 1 ft3/min
and rotates at 1 rev/s. The length from pivot to nozzle exit is 12
in., and the nozzles produce jets that are ¼ in. in diameter.
Determine the resisting torque due to friction in the bearing.

W = Vrel
 Lawn Sprinkler

absolute velocity : V  Vr ê r  V ê
 
nozzle velocity : U   r2 ê
 
relative velocity : W  W 2 cos   ê r  W 2 sin   ê
Continuity equation

 W  n̂dA   Q  2  W 2 n̂    A2 n̂   0
CS

m  Q  2 W 2 A2
 ft 3 
2 1 
Q Q 2Q  min   1 min  ft
W 2        24. 45
2 A2 2 πD /4 D
2
2

 1 
 ft 
2
 60 s  s
 48 
 Lawn Sprinkler
Moment-of-Momentum equation
  
V 2  U 2  W 2  W 2 cos   ê r  W 2 sin    r2  ê
 
Tshaft    Tshaft  ê z   r  V W  n̂dA
   
CS    
dm

    m   
 r2 ê r   W 2 cos   ê r  W 2 sin    r2  ê  2  W 2 ê    A2 ê 
  m r2 W 2 sin    r2   ê r  ê    m r2 W 2 sin    r2  ê z
Tshaft  m r2 W 2 sin    r2   Qr2 W 2 sin    r2 
 slugs  1 ft 3  12   ft   rad  12  
  1.94 3 
  
  ft   24.45  sin 25 
  2  ft  
 ft  60 s  12   s  s  12  
slugs  ft 2
 0.131 2
 0.131 lb - ft
s
 Hydraulic Turbine
A simplified sketch of a hydraulic
turbine runner is shown in the
figure. Relative to the rotating
runner, water enters at section (1)
(cylindrical cross section area A1
at r1 = 1.5m) at an angle of 100o
and leaves at an angle of 50o from
the tangential direction. The blade
height at sections (1) and (2) is
0.45 m and the volume flow rate
through the turbine is 30 m3/s.
The runner speed is 130 rpm in
the direction shown. Determine
the shaft power developed.
 Hydraulic Turbine
  rev  1 min  rad  m
U
 1   r   130   2  1.5 m   20.4
Rotating 

1
 min  60 s  s  s
speed U   r   130 rev  1 min  2 rad 
 0.85 m   11 .6 m
 2 2
 min  60 s  s  s

 
 
m  V   Anˆ    V  eˆ r A  VR A
 kg  m 3  kg
m1  m 2  Q   1000 3  30
   
  30 ,000
 m  s  s
 Q Q 30 m 3 / s m
Continuity V
 R, 1 A    7 . 07
 2r1 h1 2  1.5 m  0.45 m  s
equation 
1

V  Q  Q  30 m 3 / s m
 12 .5
 R , 2 A2 2r2 h2 2  0.85 m  0.45 m  s
 Velocity Triangles
U1 = r1
U2 = r2
U2 < U1

Relative velocity – tangent to turbine blade

 Velocity Triangle
Inflow 100o 50o Outflow
10o 40o
V1 V2 W2
W1 VR,1 VR,2

U1 = r1 V,2
V,1 U2 = r2

V ,1  V1 cos  1  U 1  W 1 sin 10 o V ,2  V 2 cos  2  U 2  W 2 sin 40 o

 
V R ,1  V1 sin  1  W 1 cos 10 V R ,1  V 2 sin  2  W 2 cos 40
o o

Turbine blade tip Turbine blade root

Relative velocity – tangent to turbine blade
 Shaft Power
Inflow 100o 50o Outflow
10o 40o
V1 V2 W2
W1 VR,1 VR,2

U1 = r1 V,2
V,1 U2 = r2

 V , 1  U 1  W 1 sin 10 o
  r1  V R , 1 tan 10 o

 20. 4  7 .07 tan 10 o
 ms  21.6 ms

2 2 2 R ,2 
V  U  W sin 40 o   r  V tan 40 o  11 .6  12.5 tan 40 o
  ,2
 ms  1.11 ms
 kg   m  m  m  m 
W shaft  m U 2V ,2  U 1V ,1    30 ,000   11 .6  1.11    20.4  21.6  
 s   s  s  s  s 
 1.283  10 7 W  12.83 MW
 Prob. 5.75: Axial Flow Pump
An axial flow gasoline pump consists
of a rotating row of blades (rotor)
followed downstream by a stationary
row of blades (stator). The gasoline
enters the rotor axially (without any
angular momentum) with an absolute
velocity of 3 m/s. The rotor blade inlet
and exit angles are 60o and 45o from
the axial direction. .The pump annulus
passage cross-sectional area is constant V1=
Consider the flow as being tangent to
the blades involved. Sketch velocity
triangles for flow just upstream and
downstream of the rotor. How much
energy is added to each kilogram of
gasoline?
 Velocity Triangles

U1 = U2
W2
W1
U1=rm U2=rm
45o

2
60o V2
V1

V x ,1  V1  W 1 cos 60 o V x ,2  V 2 cos  2  W 2 cos 45 o

 
V ,1  0 V ,2  V 2 sin  2
 
Continuity V  n̂  V cos   V cos   V  V  3 m/s
1 1 2 2 x ,1 x ,2
equation
 V1 3m / s m
W 
 1 cos 60 o cos 60 o   6
 s

U  W sin 60 o  V tan 60 o   3 m  tan 60 o  5.20 m
 1 1 1
 s s
Velocity
Triangle  V x ,2 V1 3m / s m
W
 2  o
 o
 o
 4 . 24
s  cos 45 cos 45 cos 45 s
 m
U
 2  U 1  5 .20
s
(same arithmetic mean radius)

 m  m m
V
  ,2  U 2  W 2 sin 45 o
 5 . 20   4 .24  sin 45 o
 2 . 20
 s  s s
V1
 V ,2 
 2  tan 
1
  tan  1  2.20 m / s   36.25 o (outflow direction)

 V x ,2   3m / s 
V 3m / s m
V 2  x ,2   3 . 72
cos  2 cos 36.25 o s
Shaft W shaft  m  m  N  N m
w shaft   U 2V ,2   5.20  2.20  1   11 . 44
Power m  s  s  kg  m / s 2  kg
 Turbine Power Output
The hydraulic turbine has an efficiency of 90 percent. The 10oC
water flow rate is 10,000 m3/min. Determine the turbine output
power in watts for frictionless pipe flow.
(1)

Control
Volume
Analysis
(2)
 Turbine Power Output
Assume steady state and constant water density  = 1000 kg/m3 at 10oC.

Continuity equation m 1  m 2  m  Q

Energy equation
    

CS
eV  n̂dA  Q net  W net
in in
 loss  uout  uin  q net  0 
 in 
 p V2   p V2 
No loss  m out    gz   m in    gz   W shaft
 2  out  2  in net in
 Turbine Power Output
 p1  p 2  0
  (Head water and tail water)
V1  V 2  0
W net  m  gz 2  gz 1   Qg  z 2  z 1 
in

 10 ,000 m
 3
 kg m
  1000 3   9.81 2  0 m  120m 
 m  60 s s 
m2  N 
 1.962  10 kg  2 
8

2 
  1 .962  10 8
N  m  196.2 Mw
s  kg  m / s 

Turbine output power

W t  W net  0.9 196.2 Mw   176.6 Mw

in
 Head Loss
Problem 5.20R: A hydroelectric power plant operates under the
condition illustrated in the figure. The head loss associated with
flow from the water level upstream of the dam, section (1), to the
turbine discharge at atmospheric pressure, section (2), is 20 m.
How much power is transferred from the water to the turbine blade?
(1)

(2)
  p 2  p1 V 22  V12 
Energy  g  z 2  z 1    Q net  W shaft

m  u 2  u1  
equation   2  in net in

p 2  p1 V 22  V12
  g  z 2  z 1   w shaft  1 loss 2
 2 net in

 p1  p 2  patm
 
V1  0 , V 2  2 m / s
V 22
w shaft   w shaft  g z 1  z 2    1 loss 2  loss  ghL 
net out net in 2
 V 2

Shaft  W shaft  m w shaft  Q  g  z 1  z 2  hL  
 2

 2 
power net out net out

 kg 
  999 3  30
 m 3
   m 
  9.81 2  100 m  20 m  
 2 m / s  2


 m  s   s  2 
 23.5  10 6 N  m / s  23.5 Mw
Head loss hL = 20 m, reduces the available elevation
head from 100 m to 80 m.
 Energy Equation
Problem 5.120: A liquid enters a fluid machine at section (1) and
leaves at sections (2) and (3) as shown in the figure. The density
of the fluid is constant at 2 slugs/ft3. All of the flow occurs in a
horizontal plane and is frictionless and adiabatic. For the above-
mentioned and additional conditions indicated in the figure,
determine the amount of shaft power involved.
 Solution: Continuity Equation

CS
V  n̂dA   V1 A1  V 2 A2  V 3 A3   m 1  m 2  m 3  0
  slugs  ft  30 2  slugs
m 1  V1 A1   2 3 
 15  ft   6.25
  ft  s  144  s

m  V A   2 slugs  45 ft  5 2
ft   3.125
slugs
 3 3 3  ft 3  
s  144 s
   
slugs slugs
 m 2  m 1  m 3   6.25  3.125   3.125
s s

 2  m 3
Note: in general m
Solution: Energy Equation

CS eV  n̂dA   V1 A1 e 1  V2 A2 e 2  V3 A3 e 3  m 1 e 1  m 2 e 2  m 3 e 3  0
  
Q net  0 & loss  m 1 u1  m 2 u2  m 3 u3  0 (adabatic & frictionless)
in

 p1 V12   p 2 V 22   p 3 V 32 
W shaft   m 1 
    m 2     m 3   
    2 
net in   2    2   

  6.25
slugs 

  80 lb / in 
2
 in 
2
1
 12    15   1
2
ft   lb 

3  2 
 s   2 slugs / ft  ft  2 s   slugs  ft / s  
 

  3.125 
 
slugs   50 lb / in 2  in 
2
1
 12    35   1
2
ft   lb 

3  2 
 s   2 slugs / ft  ft  2  s   slugs  ft / s  


  3.125 
 
slugs   14.7 lb / in 2  in 
2
1
 12    45   1
ft  
2
lb 

3  2 
 s   2 slugs / ft  ft  2 s   slugs  ft / s  
 
 lb  ft  1 hp  Net shaft power
   17067.5    31.03hp
 s  550 lb  ft / s  is out of the CV