Upload
others
View
31
Download
4
Embed Size (px)
Citation preview
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)