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AXIAL-FLOW MACHINES further concepts Euler Turbine Equations Axial Flow Gas Turbines Axial Flow Compressors Losses / Zweifel Free-Vortex Flow Blade Stresses Re: Peng Ch. #8 & #7

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Axial Turbine Cascade Stator

Rotor

Stator

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TURBINE STAGE STATOR STAGE V1 W1 U ROTOR V2 W2 STATOR

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AXIAL TURBINE

W2 V1 V2 W1

100 percent reaction 0

U stator rotor

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TURBINE STAGE FLOW ANGLE TERMINOLOGY

ALPHA (α) ~ STATOR ANGLES BETA (β) ~ ROTOR ANGLES STATOR V1 W1 U ROTOR V2 W2 STATOR

Watch out for notation!

AXIAL DIRECTION

0o

Negative Positive Angle Angle

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Euler Turbine Equation(s) Torque

T = m ( VU1 r1 – VU2 r2) / gc Power W = m ( VU1U1 – VU2U2) / gc Energy per unit mass ΔE = W / m = ( VU1U1 – VU2U2 ) /gc

ΔE = Δh = change in specific enthalpy

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Right end: ‘free wheeling’ , runaway condition.

No torque on rotor….no power …no use. Left end: Lots of torque but rotor velocity goes to zero. Since power = torque x angular velocity = 0

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Parsons Turbines (RN = 50% )

each with φ = ½ V2 W2 V1 W1 U W2 V1 V2 W1 U

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MAXIMUM UTILIZATION Work per unit mass = W / m = U (VU1 - VU2 ) For maximum work per unit mass: VU2 = 0 i.e. V2 is in axial direction only

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MAXIMUM UTILIZATION

Work per unit mass = W / m = U (VU1 - VU2 ) W2 V1 W1 V2 = Vx U S R

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α V2 W2 V1 W1 U

MAXIMUM UTILIZATION AXIAL TURBINE (W/m) = 2U2 / gc V2 = Vx = Va = axial direction

cos α = 2U / V1

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EXAMPLE An axial-flow turbine has a mean diameter of 40 cm and a flow coefficient (φ = Va/U) of one-half. Flow enters the stator in the axial direction with a velocity of 50 m/s and exits the stator with a nozzle angle (α, measured between V1 and U) of 17 degrees. The fluid flows through the stage with constant axial velocity and produces an output torque of 1600 Nm.

a) Sketch the velocity vector diagram for this stage. b) Calculate the power output of the turbine stage.

c) What is the degree of reaction?

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V2 = 50 W2 V1 W1

α = 17o

U

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SOLVING FOR VELOCITY VECTORS: Φ= Va/U = ½ U = 100 m/s VU1 = 50 tan 73o = 164 m/s VU1 = 164 m/s WU1 = VU1 – U V1 = 171 m/s VU2 = 0 WU2 = U = 100 m/s W2 = 112 m/s W1 = 80.85 m/s EULER EQNS: Torque = m (VU1 R1 – VU2R2) m = 1600 / (164)(0.2) = 49 kg/s Power = m (VU1U1 – VU2U2) = (49)(164)(100) = 800 kW DEGREE OF REACTION: VU1 + VU2 R = 1 – = 1 – (163 + 0) / 2(100) = 0.28 = 28% 2U

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COMPRESSOR STAGE STATOR STAGE V1 W1 U ROTOR W2 V2 STATOR V1

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AXIAL COMPRESSOR

V1 V2 W1 W2

100 percent reaction 0

U stator rotor ~50% reaction, not a great design

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High Reaction Axial Compressor

V1 V2 W1 W2

100 percent reaction 0

U stator rotor

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Low reaction axial compressor

V2 W2

V1 V2 W1 W2

100 percent reaction 0

U stator rotor

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Turbine and Compressor Performance Unrecoverable Pressure Losses Allied Signal ASE120 Gas Turbine Engine Phoenix, AZ Designed for cogeneration, emergency power and mechanical drive. ~35% thermal efficiency ~10 MW m-dot~34kg/s Texhaust ~500oC Weight ~ 5,000 lb NOx ~ 10-25 ppm

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(NOT RELATED TO UNRECOVERABLE PRESSURE LOSSES LECTURE) ASE120 engine performance as a function of inlet air temperature Notice that maximum output power decreases about 50% when the air let increases from ~ -20C to +50C (~16,000hp to 8,000 hp)

Ou

tpu

t P

ow

er

(h

p)

-20 Inlet Air Temperature (oC) 50

Power

16,000

8,000

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Unrecoverable Pressure Losses

Profile drag (and fluid friction) across blades

Annular skin friction

Secondary drag / losses

Tip leakage

Seal leakage

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IDEAL TURBINE V2 V1 W2 W1 U S R

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∆CU Actual performance is always less than the ideal theoretical performance due to un-recoverable losses

IDEAL

ACTUAL

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Turbine Performance: Turbine Flow ~ Nozzle Flow Turbine performance follows the general behavior of ideal flow through a nozzle…Tot max delP ~ den x sq vel (Eqn W 2.12) Incompressible flow, conservation of energy-ss, no change in PE ∆P = Po – Pst = total – static = dynamic pressure Don’t forget gc when using USCS units. Notice the total maximum pressure drop ~ density x velocity2 Mass flow ~ flow area x rho x delP CD ~ discharge coefficient: correction coefficient for ‘real fluids’ with viscosity, friction losses. Usually empirically based, rooted in experiment.

∆P = ½ρC2 / gc

m = A 2ρ∆Pgc ½ CD

“discharge coefficient”

CD = ∑CD

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PROFILE DRAG ACROSS BLADES In addition to classic profile / skin friction drag….. VIBRATIONS: fluttering / vortex shedding

esp. problem w/ thin blades…fans, some compressors

2FD CD = ρV2A

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Flow Separation over Compressor Rotor Blades FLOW High Reaction Low Reaction (no separation) (separated flow)

Separation leads to COMPRESSOR LOSSES STALL …losses in efficiency. Separation leads to losses in pressure, enthalpy It is very complicated to accurately predict the conditions in which B.L separation occurs. Therefore, empirically-based ‘rules-of-thumb’ have been developed. As we’ll see, flow separation can be minimized with a ‘cascade’ of blades…..i.e. form flow passages

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ANNULAR SKIN FRICTION LOSS s = spacing Due to boundary layer formation within the annular flow passage. Empirical correlation.

h

CD’ ~ 0.02 (s/h)

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SECONDARY DRAG / LOSSES End wall boundary layers are convected inward along the suction side of the blades. Secondary currents are set up in a plane transverse to the flow, dissipating energy. Results in un-favorable flow redistribution, competing vortices, and loss in stagnation pressure. Can be a significant fraction of total P-loss. These currents also take place in the wake regions down stream from the trailing edges of the blades.

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Secondary Flow in Blade Passages

The constant 0.018 stems from a complex relationship between flow acceleration, aspect ratio (h/c), spacing & pitch. RE: Logan Ch 6 Because the trailing vortices are similar to wing vortices, it is expected that the corresponding drag is proportional to the LIFT COEFFICIENT CL.

CD” = 0.018CL2

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Wilson Text Effect of aspect ratio (h/c or h/b) on secondary losses.

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TIP LEAKAGE

TIP LEAKAGE

CD”’ = 0.29 (k/h)CL3/2

The difference in pressure on the two sides of the moving blades results in leakage around the tip. Shroud reduces leakage ~50% k=clearance gap = f(T, rpm, loading) h= blade length CL=lift coefficient

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LOSSES FROM SEALS Pressure changes across casing boundaries are another source of losses. Each place where the rotor penetrates a pressure boundary needs a seal. Typically LABYRINTH SEALS.

GENERATOR LPT1 LPT2 HPT

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LABYRINTH SEALS

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LABYRINTH SEALS

CD”” = f(∆P, h, ρ, D, ….)

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Clearances change with thermal expansion, radial stresses, axial stresses, fluid loadings, rotor flex, rotor unbalance, startup/shutdown transients,…. Rotor alignment is critical for rotor seals.

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UNRECOVERABLE PRESSURE LOSSES (PROFILE, SKIN FRICTION & SECONDARY)

COMPRESSORS, PUMPS & FANS

ΔP stage = ρ U2 φ ψ χ ηstage = ΔPactual / ΔPideal = φχ χ = (R - φδ) / (φ+ δR) + (1 – R - φδ) / (φ + δ(1 – R)) rotor stator φ = flow coefficient = Cx / U ψ = blade loading coefficient = ΔCu / U R =degree of reaction δ = drag to lift ratio U = rotor velocity ΔP = pressure gain

LOGAN P115

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δ = drag to lift ratio = ∑CD / CL = (CD + CD’ + CD” + CD”’ + CD””) / CL

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EXAMPLE: Axial compressor with Rn=50% and the velocity vectors shown. The lift and drag coefficients are the same for the rotor and stator and are 1.4 and 0.1, respectively. CD includes all losses. Calculate the stage pressure rise and required energy input if the entering air has a density of 1.2 kg/m3. Each square corresponds to 50 m/s x 50 m/s 50 m/s V1 V2 W1 W2 U stator rotor

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EXAMPLE (con’t)

ΔP stage = ρ U2 φ ψ χ = ΔPactual U = 300 m/s Vx = 150 m/s VU1 = 50 m/s VU2 = 250 m/s

ψ = (VU1 – VU2)/U = (50-250)/300 = -2/3 (neg=compressor)

φ = Vx/U = 150/300 = 1/2

δ = drag/lift = 0.1/1.4 = 0.071

χ = (R - φδ) / (φ+ δR) + (1 – R - φδ) / (φ + δ(1 – R))

χ = (0.5-0.5x0.071)/(0.5 + 0.071x0.5) + (1-0.5-0.5x0.071)/ (0.5+0.071(1-0.5))

χ = (0.46/0.54) + (0.46/0.54) = 1.7

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EXAMPLE (con’t)

ΔP stage = ρ U2 φ ψ χ= ΔPactual ΔPactual = (1.2)(300)2 (0.5) (0.66)(1.7) = 60.6 kN/m2

= 60.6 kPa

ηstage = ΔPactual / ΔPideal = φχ = (0.5)(1.7) = 0.85 ΔPactual / ΔPideal = 0.85 = 60.6 kPa/ΔPideal

ΔPideal = 71.3 kPa

Energy per unit mass (Euler Turbine Equation) [W /m]ideal = ( Vu1U1 - Vu2U2 ) / gc

=(300)m/s [50-250]m/s = -60 kJ/kg

[W /m]actual = [W /m]ideal / ηstage = (-60)/(0.85) = -70 kJ/kg

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FLOW SEPARATION

de Haller Number: dH

dH = Wexit / Win = W2 / W1

dH > 0.72 No Separation Swiss engineer Established criterion for flow separation in compressor cascades (for compressor cascades, RE: Wilson, Ch#4 p.182) But…also sometimes applied to turbine blades as well using V1 & V2

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De Haller: dH = Wexit / Win > 0.72 no separation dH = W2 / W1 COMPRESSOR V1 V2 W1 W2

U S R

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De Haller: dH = Wexit / Win > 0.72 no separation dH = W2 / W1 COMPRESSOR V1 V2 W1 W2

U S R Increase W2, Increase dH, decrease potential for separation?

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Pressure Rise Coefficient (compressor cascade)

Cpr = (Pst,ex – Pst,in)/(Po,in – Pst,in) = rise in static pressure / inlet dynamic pressure

For an ideal compressor Cpr = 1 – (Wex / Win)2

Cpr < 0.5 Consider each stage of a compressor as a compressor The pressure rise at each stage can only increase so much, Limits on pressure rise Cpr ~< 0.5

VELOCITY VECTORS SUMMARY

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Axial Flow Compressor and Turbine

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DEGREE OF REACTION

Vu2 Vu1 Rn = ∆hrotor / ∆hstage = (h2 – h3) / (h1 – h3) = ∆Protor / ∆Pstage V2 V1 W2 W1 = (P2 – P3) / (P1 – P3) = 1 – (Vu1 + Vu2) / 2U

(Text eqn 5.8)

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Zweifel’s Correlation OPTIMUM SOLIDITY

Axial Turbine Cascade Stator

Rotor

Stator

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Lift and Drag Forces

FL = CL (ρAV2) / 2gc

FD = CD (ρAV2) / 2gc Comments on LIFT and DRAG forces: Lift and drag forces are difficult to model exactly due to complex flow phenomena including: viscous effects, boundary layer separation, etc. Hence, ‘empirically determined correction coefficient’ is typically used. CL and CD are lift and drag coefficients, usually determined empirically. CD typically accounts for both pressure and skin friction drag. ‘A’ is some characteristic area. ‘V’ is some representative velocity: free stream, average or mean velocity, etc.

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Isolated Flat Plate

Cascade of Blades

S C K=cascade coefficient = f(beta, s/c) S= pitch or spacing C = chord

β CL = 2πsinβ

CLcas=K(CL)

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BLADE SPACING / SOLIDITY (σ = c / s) Is there an option spacing?

Low solidity High solidity Losses by separation ~No separation Losses by skin friction s = pitch or spacing c = chord

s

c

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β1

β2 NOTE ERROR IN PENG’S TEXT: PAGE 227

CL = 2(s/b)cos2β2(tan β1 - tan β2) σ = c/s = solidity Wilson 7.4

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Zweifel (Peng Text, Ch 8 p 227) (Brown Boveri, Swiss, 1945)

CL,op ~ 0.8

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α = - 30o

β = + 60o

EXAMPLE The spacing (pitch) between turbine or compressor blades influences the machine’s performance.

Shown below is a cascade of blades for an axial compressor. Flow in U c = 12 cm b = 10 cm Flow out

(a) Determine the optimum blade spacing, s = ?

Optimum Spacing

Zweifel Correlation

(b/s) opt = |(2/CL,opt) cos2αex [tanαin – tan αex]| = |(2/0.8) cos 2 (60o) [tan (-30o) – tan (60o)]| = 1.44 thus, s = b/1.44 = 6.94 cm

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AXIAL TURBINES NUMBER OF BLADES Wind: 1-24 Water: 3-30 Gas / Steam: 11-110 Rotors usually have an even number of blades (for balance) Stators usually have an odd number of blades (for vibration)

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VORTEX FLOW Radial Equilibrium A tornado is nature’s efficient vortex flow. Characteristic of air / gas at higher velocities (200+ mph ~ 300 ft/s ~ 100 m/s) It pumps air from the higher pressure ground to overhead cloud. We can capture that flow characteristic in the design of axial compressor (and turbine) blade design that encourage free vortex flow.

UF6 Axial Compressor

Tornado

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Velocity Distribution in a Tornado

R = “eye” of tornado

R

Forced Vortex (rotational) r < R Vu = ωr

Vurn = constant

Free Vortex (irrotational) r > R Vu = ωR2/r

Vur = constant

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Pressure Distribution in a tornado Patm

R -P Minimum pressure is in center of eye. The entire field pressure is sub-atmospheric pressure (AKA Under-pressure)

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Uniform Circular Motion r

Fc v θ

F = ma

Fc = mac = mv2/r Fc = centripetal force (in +) ac = centripetal acceleration v = tangential speed = vu

θ = angular position (rad) ω = angular velocity (rad/s) ac = centripetal acceleration (m/s2) 2∏ radians ω = dθ/dt =

time for one revolution 2∏ =

2∏r/v

= v/r

ac = d2θ/dt2 = v2/r

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Radial Pressure Forces on Fluid Element (F=PA) P + dP P + ½ dP P + ½dP P dr r dθ Fc For Radial Equilibrium: The radial components of the pressure forces (F=PA) must be balanced by the centrifugal force on the fluid mass.

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For Radial Equilibrium

∑Fradial = Fc + ∑FP =0

Free-Vortex Flow: Vur = constant

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TIP

HUB

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TIP Vu1 HUB

Constant axial velocity Axial V2 = maximum utilization Cur = const r∆Vu = const

Don’t want hub to fall below 0% rxn

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TIP (RXN) HUB (IMP)

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Station A is entry to nozzle / stator Station B is entry to blade / rotor Station C is exit from blade Nozzle blade is twisted to give overall free vortex flow. The Blade is straight. Nozzle sets up flow in radial decreasing velocity profile….upon passage over uniform rotor blade, re-establishes uniform vel profile. Illustrates other losses: Wall friction losses (skin friction) and tip-clearance leakage (PsB > PsC)

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Euler Turbine Equation(s) Torque

T = m ( Vu1 r1 - Vu2 r2) / gc Power W = m ( Vu1U1 - Vu2U2) / gc Energy per unit mass W / m = ( Vu1U1 - Vu2U2 ) / gc

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TURBINE V2 V1 W2 W1 U S R W ~ Vu1U1 - Vu2U2 V2 V1 W2 W1

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FORCES & STRESS ON BLADES

“frozen” isochromatic fringes on three-dimensional photoelastic model of gas turbine blade hub.

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Forces acting on turbine & compressor blades

Fluid forces: Lift & Drag

Centrifugal forces Thermal forces / stresses Fluid forces: Lift & Drag Of special interest are the tangential components of the lift and drag forces

since these forces directly affect the power and efficiency. Produce bending forces / stresses due to changes in fluid pressure and momentum. Total Drag force = pressure drag (form drag, shape drag) + Viscous Drag (viscous drag) Due to complex flows, losses occur. Excitation and vibrations may result from fluctuating pressure gradients. Centrifugal forces Produces radial & bending stresses

(bending when centroids of all cross sections do not lie along a radial line) Limits the design length and rpm Easiest force to model Thermal forces / stresses When its not at a uniform temperature, thermal stresses arise.

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Lift and Drag Forces RADIAL LIFT AXIAL DRAG V1 Vm V2

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Lift and Drag Forces FL = CL (ρAV2) / 2gc FD = CD (ρAV2) / 2gc Comments on LIFT and DRAG forces: Lift and drag forces are difficult to model exactly due to complex flow phenomena including: viscous effects, boundary layer separation, etc. Hence, ‘empirically determined correction coefficient’ is typically used. CL and CD are lift and drag coefficients, usually determined empirically. CD typically accounts for both pressure and skin friction drag. ‘A’ is some characteristic area. ‘V’ is some representative velocity: free stream, average or mean velocity, etc.

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Isolated Flat Plate

Cascade of Blades

S C K=cascade coefficient = f(beta, s/c) S= pitch or spacing C = chord

β CL = 2πsinβ

CLcas=K(CL)

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80

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Fluid Forces: Vibration & Excitation Upstream variations in the flow can lead to blade vibrations. Can lead to ‘high frequency’, ‘high cycle’ fatigue.

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Vibration & Excitation Campbell diagram illustrating example of natural frequencies of a rotor blade. Centrifugal forces on the blade causes the blade to stiffen, ‘centrifugal stiffening’ and increase the natural vibration frequency with rotor rpm.

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CENTRIFUGAL STRESS ON STEAM TURBINE BLADE R r

CASING

HUB

BLADES

HUB

h

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Centrifugal Force on Uniform Blade

F = ma Fc = m(ω2r) = ρAh(2πN/60)2[(R + r)/2] Centrifugal Stress (σ = F/A) on Uniform Blade

σmax = ρh (2πN/60)2 (r + R) / 2gc Maximum Stress at Hub of Blade

σmax, hub = ½ ρ (Utip)2 [1-(Rhub/Rtip)2]

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EXAMPLE: Centrifugal Stress Given: ρ ~ 8,000 kg/m3 ~ 650 lbm / ft3 Uniform density and cross-section h= length of blade = 20 inches = 1.67 ft N= rate of rotation = 1800 rpm r= hub radius = 1.5 ft

σmax = ρh (2πN/60)2 (r + R) / 2gc

σmax = 19,400 psi

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CENTRIFUGAL FORCE: “Liberated Blade” If a blade ‘broke loose - liberated’, how high could it go? N = 1800 rpm K.E P.E. ½ m V2 = mgH V = centroid velocity

=2π(r+h/2)N/60 = 440 ft/sec = 645 mph H = ½ V2 /g = 3,006 ft

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Thermal Stresses

σ=Eε ∆L/L= α∆T

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BLADE COATINGS

“thermal & corrosion protection”

Materials:

• NiCrAlY alloy ( nickel, chromium, aluminum, yttium)

• Zirconia (ceramic, ZrO2)

• Alumina (Al2O3)

• Silicon carbide

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HOW COATINGS ARE APPLIED Electroplating Plasma spray Vapor-phase deposition Electron beam depositio

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TE

MP

ER

ATU

RE

BLADE COATINGS FOR THERMAL AND CORRESION CONTROL SUBSTRATE BOND COAT TOP COAT HOT COMBUSTION GASES OPERATING TEMPERATURE SUBSTRATE MELTING TEMPERATURE

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HIGH TEMPERATURE MATERIALS MATERIAL ~MELTING TEMPERATURE

oF oC

Nickel 2650 1450 Nickel-Chromium Alloy 2100 – 2300 1150 – 1260 Cobalt – Chromium Alloy 2400 – 2550 1315 – 1400 Titanium 3035 1668 Titanium-6Al-4V Alloy 3000 1650 For most metals used in blades, creep becomes significant at about one-half the melting point.

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