Dr. Ir. Harinaldi, M.EngMechanical Engineering Department

Faculty of Engineering University of Indonesia

ENERGY TRANSFER BETWEEN FLUID AND ROTOR

Basic Laws and EquationsBasic Laws and EquationsContinuity Equation

222111 ACACm ρρ ==)/( skgrateflowmassm =

Basic Laws and EquationsBasic Laws and EquationsSteady Flow Energy Equation (First Law of Thermodynamics)

( ) ( ) ( )

−+−+−+

−=− 1212

21

22

1

1

2

22

1 uuZZgCCppmWQρρ

Basic Laws and EquationsBasic Laws and EquationsNewton’s Second Law of Motion - Linear

( )12 xxx CCmF −=∑

Basic Laws and EquationsBasic Laws and EquationsNewton’s Second Law of Motion - Angular

Torque

( ) ( ) ∑∑∫ =×=⋅× TdA contentcvcs

FrnCCr ρ

Basic Laws and EquationsBasic Laws and EquationsNewton’s Second Law of Motion - Angular

Power (Euler Equation for Turbomachinery)

W =

For Power Machine (Turbine)

W = < 0

For Working Machine (Pump)

W = > 0

Basic Laws and EquationsBasic Laws and EquationsEntropy

Turbine Pump

sTdq ∆=

In the absence of motion, gravity and any other effects pvhsT

vpusTdddddd

−=+=

Flow Idealization through a Flow Idealization through a TurbomachineryTurbomachinery a. Working Machinea. Working Machine

C = W + U

Flow Idealization through a Flow Idealization through a TurbomachineryTurbomachinery b. Power Machineb. Power Machine

C = W + U

• Energy is transferred from rotor to the fluid• The increase of tangential component of

absolute velocity in the same direction with rotation of rotor

a. Working Machine

b. Power Machine

Flow Idealization through a Flow Idealization through a TurbomachineryTurbomachinery

• Energy is transferred from fluid to the rotor• The increase of tangential component of

absolute velocity in the opposite direction to rotation of rotor

Relation of Absolute, Blade and Relative Velocity

Analysis of Velocity TriangleAnalysis of Velocity Triangle

Cx = tangential component of absolute velocity (whirl velocity)Cr = radial component of absolute velocity

Relation of Blade Angle and Relative Flow AngleAnalysis of Velocity TriangleAnalysis of Velocity Triangle

β = blade angleβ ’= relative flow

angle

Nomenclatures:

1 = inlet/entry2 = outlet/exit

Ideal Condition :- No schock at entry- No slip at exit

β1 = β1’β2 = β2’

Analysis of Velocity TriangleAnalysis of Velocity TriangleEuler Equation of Turbomachinery

( ) ( )111222 xx CUmCUmTW −=∑= ωPower

continuitymmm 21 →==

Specific Power ( ) ( )

( ) ( )1122

1122

xx

xx

CrCr

CUCUmWw

ωω −=

−==

ωω 2211 ; rUrU ==

Note:60

60

2 nDrUn πωπω ==→=ω (rad/s)n (rpm)

Analysis of Velocity TriangleAnalysis of Velocity Triangle222xr CCC −=

( ) 222 UCWC xr −−= 2

222 WUCUCx−+=

( ) ( ) ( )2

21

22

21

22

21

22 WWUUCCw −−−+−=

Another approachLarge triangle :

Small triangle :

Therefore :

Analysis of Velocity TriangleAnalysis of Velocity TriangleEnergy Head

( ) ( )

( ) ( ) ( )[ ]21

22

21

22

21

22

1122

21 WWUUCCg

gCUCUh xx

−−−+−=

−=∆

( ) ( ) hmgCUCUmW xx ∆=−= 1122

Relation between rotor power and fluid energy

Change of Fluid energy Head

1 2 3

Analysis of Velocity TriangleAnalysis of Velocity TriangleEnergy Head

1. Change of head caused by change of kinetic energy of the fluid

2. Change of head that develops across the impeller due to the centrifugal effect,

3. Chage of head caused by the diffusion of relative flow in the blade passages

An inward flow power turbomachinery, having an external diameter of 1.5 m and internal diameter of 0.5 m runs at 400 rpm. The radial component of absolute velocity of flow at inlet is 10 m/s. If the absolute flow angle is 15o find :(a) Absolute velocity of water(b) Tangential component of absolute velocity (whirl velocity) at inlet(c) Relative velocity at inlet(d) Relative flow angle at inlet

Analysis of Velocity Triangle in Turbomachinery

1.

Solution:Given D1 = 1.5 m ; n = 400 rpm; Cr1 = 10 m/s ; 1 = 15o

Velocity triangle:

Cr1 = 10 m/s

U1 = D1n/60 = 31.42 m/s

1 = 15o

C1 = ?

W1 = ?

Cx1 = ?

1’ = ?

Analysis of Velocity Triangle in Turbomachinery

Cr1 = 10 m/s

U1 = 31.42 m/s

1 = 15o

C1=

W1 =

1 =

Cx1=

C1 = Cr1 / sin 15o = 38.64 m/s (a)

Cx1 = C1 cos 15o = 37.32 m/s (b)

W1 = [Cr12 + (Cx1 – U1)2]1/2 = 11.61 m/s (c)

Sin 1 = Cr1 / W1 = 0.861 1 = 59.5o

38.64 m/s

37.32 m/s

11.61 m/s

59.5o

1’ = 180o – 59.5o = 120.5o (d)

1’ =120.5o

An outward flow working turbomachinery, having an external diameter of 0.6 m and internal diameter of 0.3 m. The water enters the impeller radially with absolute velocity at 2.5 m/s. The relative flow angle at inlet is 30o and at outlet is 45o. If the water leave the impeller with radial component of absolute velocity equal to absolute velocity at inlet, find :(a) Rotating speed of impeller(b) Specific work of the rotor shaft

Analysis of Velocity Triangle in Turbomachinery

2.

Solution:Given D1 = 0.3 m ; D2 = 0.6 m ; C1 = Cr2 = 2.5 m/s ; 1’ = 30o; 2’ = 45o

Velocity triangle at inlet:

C1 = 2.5 m/s

1’ = 30o

U1

W1

tan 1’ = V1 / U1 U1 = V1 / tan 30o = 4.33 m/s

= 4.33 m/s

U1 = D1n/60n = 60U1 / (D1) = 275.8 rpm (a)

Analysis of Velocity Triangle in Turbomachinery

Velocity triangle at outlet:

Cr2 = 2.5 m/s

2 = 45o

W2

U2 = D2n/60 = 8.66 m/s

C2

Cx2 =

Cx2 = U2 – (Cr2 / tan 45o) = 6.16 m/s

6.16 m/s

wshaft = U2Cx2 - U1Cx1

= (8.66)(6.16) – 0 = 53.35 kJ/kg (b)