Pumps Turbines

Vapor-cycle Power Plants

Ideal Regenerative Rankine Cycle with Open Feedwater Heater

Open Systems - Control Volume

Axial-flow Hydraulic Turbine

Compressors, pumps, and fans

Compressor

Side view of pump

End view of pump

Inflow and Outflow Velocity Triangles

Radial Flow Pump (Fan) U=ω r W

V

Relative Velocity – tangent to blade

Radial Flow Turbine Absolute velocity – tangent to stator blade

V

W

U=ω r

Relative velocity – tangent to rotor blade

Axial Flow Compressor (Pump) Relative Velocity – tangent to rotor blade U=ω r

Axial Flow Turbine

Relative Velocity – tangent to rotor blade Absolute velocity – tangent to stator blade

U=ω r

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 eˆθ   nozzle veloecity : U = ω r2 eˆθ     relative velocity : W = −W 2 eˆθ = V − U = −(V 2 + ω r2 ) eˆθ

Continuity equation

∫

CS

 ρW ⋅ nˆdA = ρ (V1 eˆ z ) ⋅ ( − A1 eˆ z ) + 2 ρ ( − W 2 eˆθ ) ⋅ ( − A2 eˆθ ) = 0

∴ m = ρQ = ρV1 A1 = 2 ρW 2 A2

Relative velocity W2

 slugs  gal  min  ft 3  slugs    =  1.94 16 = 0 . 0692     3   7.48 gal  min 60 s s ft        

Q 16 ( gal / min )  min  ft 3  144 in 2  W2 = =   2 2 2 A2 2 0.04 in  60 s  7.48 gal  ft

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  = 64.17 ft / s 

Lawn Sprinkler Moment-of-momentum equation  Tshaft = ( − Tshaft ) eˆ z = ∫

CS

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   r × V ρW ⋅ nˆdA   dm

= ( r1 eˆ r × V1 eˆ z ) ρ (V1 eˆ z ) ⋅ ( − A1 eˆ z ) + [ r2 eˆ r × ( − V 2 eˆθ ) ] 2 ρ ( − W 2 eˆθ ) ⋅ ( − A2 eˆθ )        m

m

= 0 − m 2 r2V 2 eˆ z = − m r2V 2 eˆ z

(a) U2 = 0 , V2 = W2

Tshaft

slugs  8 ft  ft   = m r2V 2 =  0.0692   64.17  s  12  s  slugs − ft 2 = 2.96 = 2.96 lb - ft 2 s

Lawn Sprinkler (b) ω = 500 rpm

rad  8  ft  U 2 = ω r = 500  2π ft = 34 . 91   s 12 s    ft ft 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

(c) Tshaft = 0

ft s W 64.17 ft / s  s  1 rev  rev ∴ω = 2 = 60 = 920    ( 8 / 12 ) ft  min  2π rad  r2 min W 2 = U 2 = ωr = 64.17

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 eˆ r + Vθ eˆθ   nozzle velocity : U = ω r2 eˆθ   relative velocity : W = (W 2 cos β ) eˆ r − (W 2 sin β ) eˆθ

Continuity equation

∫

CS

 ρW ⋅ nˆdA = − ρQ + 2 ρ (W 2 nˆ ) ⋅ ( A2 nˆ ) = 0

m = ρQ = 2 ρW 2 A2  ft 3   2 1 min  Q Q 2Q  ∴W 2 = = = = 2 2 2 2 A2 2 πD /4 πD  1  π ft   48 

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ft  1 min  = 24 . 45   60 s s  

Lawn Sprinkler Moment-of-Momentum equation

   V 2 = U 2 + W 2 = (W 2 cos β ) eˆ r − (W 2 sin β − ω r2 ) eˆθ     Tshaft = ( − Tshaft ) eˆ z = ∫ r × V ρW ⋅ nˆdA     CS

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dm

  m  = r2 eˆ r × [ (W 2 cos β ) eˆ r − (W 2 sin β − ω r2 ) eˆθ ] 2 ρ (W 2 eˆθ ) ⋅ ( A2 eˆθ ) = − m r2 (W 2 sin β − ω r2 ) ( eˆ r × eˆθ ) = − m r2 (W 2 sin β − ω r2 ) eˆ z

Tshaft = m r2 (W 2 sin β − ω r2 ) = ρQr2 (W 2 sin β − ω r2 )

 slugs  1 ft 3  12   ft  rad  12        =  1.94 ft 24 . 45 sin 25 − 2 π ft         3   s s  12   ft  60 s  12     slugs ⋅ ft 2 = 0.131 = 0.131 lb - ft 2 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 Rotating speed

 rev  1 min   U = ω r = 130    2π 1  1 min  60 s     U = ω r =  130 rev  1 min  2π 2  2 min  60 s  

(

rad  m ( 1.5 m ) = 20.4 s  s rad  ( 0.85 m ) = 11.6 m s  s

)

  m = ρV ⋅ ( Anˆ ) = ρ V ⋅ eˆ r A = ρVR A kg  m 3  kg      m1 = m 2 = ρQ =  1000 3  30 = 30 ,000  s  s m  

Continuity equation

 Q Q 30 m 3 / s m V = = = = 7 . 07  R, 1 A 2πr1 h1 2π ( 1.5 m )( 0.45 m ) s  1  30 m 3 / s m V = Q = Q = = 12 . 5  R , 2 A2 2πr2 h2 2π ( 0.85 m )( 0.45 m ) s

Velocity Triangles U1 = ω r1 U2 = ω r2 U2 < U1

Relative velocity – tangent to turbine blade

Velocity Triangle Inflow

100o

V1

10o

W1

Outflow

50o

VR,1

U1 = ω r1 Vθ ,1 Vθ ,1 = V1 cos β 1 = U 1 + W 1 sin 10 o  V R ,1 = V1 sin β 1 = W 1 cos 10 o

V2

40o

VR,2

W2

Vθ ,2 U2 = ω r2 Vθ ,2 = V 2 cos β 2 = U 2 − W 2 sin 40 o  o V R ,1 = V 2 sin β 2 = W 2 cos 40

Turbine blade root Turbine blade tip Relative velocity – tangent to turbine blade

Shaft Power Inflow

100o

V1

10o

W1

Outflow

50o

V2

VR,1

U1 = ω r1 Vθ ,1

40o

VR,2

W2

Vθ ,2 U2 = ω r2

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) ms = 21.6 ms

 o o o V = U + W sin 10 = ω r + V tan 10 = 20 . 4 + 7 . 07 tan 10 θ , 1 1 1 1 R , 1  ∴ V = U − W sin 40 o = ω r + V tan 40 o = 11.6 − 12.5 tan 40 o 2 2 2 R ,2  θ ,2

(

) ms = 1.11 ms

kg   m  m  m  m   W shaft = m (U 2Vθ ,2 − U 1Vθ ,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 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?

V1=

Velocity Triangles U1 = U2 W2

W1

U1=ω rm 45 θ

U2=ω rm

o

60o

V1

V x ,1 = V1 = W 1 cos 60 o  Vθ ,1 = 0

2

V2

V x ,2 = V 2 cos θ 2 = W 2 cos 45 o  Vθ ,2 = V 2 sinθ 2

 Continuity V ⋅ n ˆ = V1 cos θ 1 = V 2 cos θ 2 ⇔ V x ,1 = V x ,2 = 3m/s equation

Velocity Triangle s

V1 3m / s m  W = = = 6  1 cos 60 o cos 60 o s   U 1 = W 1 sin 60 o = V1 tan 60 o =  3 m  tan 60 o = 5.20 m  s  s V x ,2  V1 3m / s m W = = = = 4 . 24  2 o o o s cos 45 cos 45 cos 45  m  U = U = 5 . 20 (same arithmetic mean radius)  2 1 s   m  m m o o V = U − W sin 45 = 5 . 20 − 4 . 24 sin 45 = 2 . 20   2 2  θ ,2 s  s s 

V1

 Vθ ,2   = tan − 1  2.20 m / s  = 36.25 o (outflow direction) θ 2 = tan    3m / s   V x ,2  V 3m / s m V 2 = x ,2 = = 3 . 72 cos θ 2 cos 36.25 o s W shaft  m  m  N N ⋅m   w shaft = = U 2Vθ ,2 =  5.20  2.20  1 = 11 . 44 m s  s  kg ⋅ m / s 2  kg  −1

Shaft Power

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 ρ = 1000 kg/m3 at 10oC.

Continuity equation

m 1 = m 2 = m = ρQ

Energy equation

∫

CS

 eρV ⋅ nˆdA = Q net + W net in

in

 p V2  No loss ⇒ m out  + + gz  2 ρ  out

     loss = uout − uin − q net = 0  in    p V2  − m in  + + gz  = W shaft 2 net in ρ  in

Turbine Power Output  p1 = p 2 = 0 (Head water and tail water) ⇒  V1 = V 2 = 0 W net = m ( gz 2 − gz 1 ) = ρQg ( z 2 − z 1 ) in

kg  =  1000 3 m 

 10 ,000 m  60 s 

m2 = −1.962 × 10 kg ⋅ 2 s 8

3

 m  9.81 2 ( 0 m − 120 m ) s  

  N 8   = − 1 . 962 × 10 N ⋅ m = −196.2 Mw 2   kg ⋅ m / s 

Turbine output power

W t = ηW 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)

Energy equation

  p 2 − p1 V 22 − V12  m  u2 − u1 + + + g ( z 2 − z 1 )  = Q net + W shaft ρ 2 in net in   p 2 − p1 V 22 − V12 + + g ( z 2 − z 1 ) = w shaft − 1 loss 2 ρ 2 net in  p1 = p 2 = patm ⇒  V1 = 0 , V 2 = 2 m / s w shaft = − w shaft net out

Shaft power

net in

V 22 = g( z1 − z 2 ) − − 1 loss 2 2

( loss = ghL )

2   V 2  ∴ W shaft = m w shaft = ρQ  g ( z 1 − z 2 − hL ) −  2 net out net out   2 ( kg  m 3   m 2m / s )     9.81 2 ( 100 m − 20 m ) − =  999 3  30  s   2 m  s    = 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 abovementioned and additional conditions indicated in the figure, determine the amount of shaft power involved.

Solution: Continuity Equation ∫

CS

 ρV ⋅ nˆdA = − ρV1 A1 + ρV 2 A2 + ρV 3 A3 = − m 1 + m 2 + m 3 = 0

  slugs   15 m 1 = ρV1 A1 =  2 3    ft   m = ρV A =  2 slugs  45 3 3  ft 3   3   

ft  30 2  slugs ft  = 6.25  s  144 s  ft  5 slugs 2 ft  = 3.125  s  144 s 

⇒ m 2 = m 1 − m 3 = ( 6.25 − 3.125 )

slugs slugs = 3.125 s s

 2 ≠ m 3 Note: in general m

Solution: Energy Equation  ∫CS eρV ⋅ nˆdA = − ρV1 A1 e 1 + ρV2 A2 e 2 + ρV3 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

 p 3 V 32   p1 V12   p 2 V 22     + m 2   + m 3  W shaft = − m 1  + + +  2  2  2  net in  ρ  ρ  ρ 2 2 2     slugs 80 lb / in in 1 ft   lb     12  +  15   1 = − 6.25  3  2  s   2 slugs / ft  ft  2 s   slugs ⋅ ft / s      2 2  slugs   50 lb / in 2  in  1 ft   lb    12  +  35   1 +  3.125  3  2  s ft 2 s    2 slugs / ft     slugs ⋅ ft / s    2 2  slugs   14.7 lb / in 2  in  1 ft   lb    12  +  45   1 +  3.125  3  2  s   2 slugs / ft  ft  2 s   slugs ⋅ ft / s       Net shaft power lb − ft  1 hp   = −31.03hp =  − 17067.5  is out of the CV s  550 lb − ft / s  

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