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TWO DIMENSIONAL CASCADEA Two Dimensional Cascade consists of a number of identical blades, equally spaced and parallel to one another.

Pitch wisedirection

Stream-wisedirection

α2

C2 Exit velocity profile

α2 Exit angle profile

Pitch-wise direction

C2

α2

Suction side Pressure side

One dimensional flow through cascade

Two dimensional flow through cascade

Boundary layer on blade surfacesC2

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Cascade consists of a number of identical blades, equally spaced and parallel to one another.

Blade shape:-

L.E.tmax

a

Chord length (l)

T.E.

x

0

t/2 Camber line

Thickness distribution of an aerofoil

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Cascade Nomenclature

γ

θ

s

Blade directionAir direction

α1’α1

i

α2

α2’δ

Camber line

α1, α2 = fluid flow angles α’1 α'2 = Blade angles γ = stagger angle θ = camber angle ί = incidence angle δ = Deviation angle s = pitch h = span

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Analysis of Cascade forces

y

x

C1 Cx1

Cy1

C2 Cx2

Cy2

p2 , p02

p1 , p01

Y

FR X

L

D

α1

αm

α2

Control surface

X,Y are forces on fluid

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For a unit depth of span and in-compressibility applying continuity equationis ρ A C = constant

Cx1 ( s.1 ) = Cx2 ( s.1 ) ---------------(1)hence

Cx1 = Cx2 = Cx = C1 cos α1 = C2 cos α2

For constant axial velocity, applying momentum equation in x- direction

X + p1. (s.1) - p2. (s.1) = m. (Cx2- Cx1)

X – (p2 – p1)s = 0 --------------(2)

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αm

90-αm

90-αm

αmX cosαm

Y sinαm

Y

XDL

X sin αm

Y cos αm

α2

αm

α1

Cm C2

C1

Cym

Cy1

Cy2

L

D

CC1

C2

αm

L

D

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Momentum equation in y- direction:-

Y = m {(- Cy2) – (- Cy1)} but we know that m = ρ A Cx hence Y = ρ (s.1). Cx (Cy1 – Cy2) -----------(3) or

Y = ρ s C²x ( tan α1 - tan α2 ) ----------(3a)

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Energy losses:-

Fluid experiences a total pressure loss ∆p0 due to skin friction and related effect. Thus (p01-p02)/ ρ = ∆p0/ ρ = { p1+(1/2) ρ C1²}/ ρ - {p2+(1/2) ρ C2²}/ ρ

= (p1 –p2)/ ρ + (1/2)(C1² - C2²)

∆p0/ ρ = (p1-p2)/ ρ +(1/2){(C²y1+C²x1)-(C²y2+C²x2)}

∆p0/ ρ = (p - p2)/ρ + (1/2)(Cy1+Cy2)(Cy1-Cy2)

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On substitution of above values, we get

∆p0/ρ = -X/ρs + ½(Cy1+Cy2)(Y/ρsCx) (Taking Cym = ½(Cy1+Cy2 ) = -X/ρs + Cym(Y/ρsCx) (Cym =Cx tan αm ) = (-X + Y tanαm) / ρs -------------(5)

where tan αm = (tan α1 + tan α2) / 2 -------------(6)

NON DIMENSIONAL PARAMETERS:

PRESUURE LOSS COEFFICIENT, ζ = ∆p0/(½ ρCx2) ----------(7)

PRESSURE RISE COEFFICIENT , Cp = (p2-p1)/(½ρCx2) = X/(½ ρsCx2) ---------- (8) TANGENTIAL FORCE COEFFICIENT, C f = Y/(½ ρs.1 Cx2) = 2(tan α1 - tan α2) -----(9)

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Relation between Cf , ζ , and Cp:-

Divide by (½C²x) to equation (5), we get ∆po/(½ρC²x) = -X/(½ ρsC²x) + Y tanαm/(½ ρsC²x)

ζ = - Cp + Cf tanαm

OR Cp = Cf tanαm - ζ -------------------(10)

LIFT AND DRAG FORCE:-

SINCE Cm = Cx/cosαm --------------------(11)

According to fig.

Lift force L = X sin αm + Y cos αm ------------(12)

Drag force D = Y sin αm - X cos αm -------------(13)

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From equation (5) D = cosαm(Y tan αm - X)

since ∆Po/ρ = (-X + Y tan αm)/ρs -------( From eq 5 )

D = s∆po cos αm ---------------------------(14) Substituting in eq. 12 ,the value of X from (5), we get L = (Y tanαm - ∆po.s) sinαm + Y cos αm (From Eq.5; X= Y tanαm - ∆po.s)

L = (Y/cosαm){sin²αm + cos²αm} - s∆po sin αm = (Y/cosαm) - s∆po sin αm

But Y = ρsC²x(tanα1-tanα2) ---(From 3a)

Hence L = ρsC²x(tanα1- tanα2)secαm - s ∆po sinαm -----------(15) But we know that, CL = L /(½ρAC²m) = L /[ ½ρ(l.1) C²m] {where,l=chord length}..(16a)

and CD = D /(½ρlC²m) (16b)

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Hence CD = s. ∆po. cos αm /½ρl C²m ---------------(from 14) CD = (s/l) ζ cos³αm -(From eq. 7 for ζ) ---------------(17)

On substitution of L from equation (15). we get CL = ρsC²x( tanα1 – tanα2) sec αm – s ∆po sinαm (1/2) ρ.l. C²m With Cm= Cx cos αm

CL = 2 (s/l) (tanα1- tanα2) cosαm – CD tanαm ------------------(18)

Within normal range of operation in cascade, the value of CD is much less than CL. i.e. CD<< CL and αm< 60º

Then CD tanαm can be neglected Therefore CL ~ 2 (s/l) (tanα1- tanα2) cosαm

Hence L / D = CL / CD = [2(s/l)cos αm (tanα1- tanα2)] /ζ(s/l)cos³αm

But (tanα1- tanα2) = Cf/2 from (9) Hence L/D= CL / CD = (Cf/ζ) sec²αm -------------------(19)

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EFFICIENCY OF A COMPRESSOR CASCADE:- Efficiency of compressor cascade can be defined in a same way as diffuser efficiency

ηD = ( p2 – p1) / (1/2)ρ(C²1 - C²2) Since ∆po = po1 – po2 = {p1+ (1/2)ρC1²} – {p2+ (1/2)ρC2²}

Hence, pressure loss

(p1 –p2) = {-∆p0 + (1/2)ρ(C1² - C2²)hence ηD = {-∆p0 + (1/2)ρ(C1² - C2²)}/(1/2)ρ(C1² - C2²)

ηD = 1- {∆p0} / {(1/2)ρ(C1² - C2²)}

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Since C 1² -C 2² = (C²x1 + C²y1) - (C²x2 + C²y2) = C²y1 - C²y2 =(Cy1 +Cy2 )(Cy1 -Cy2 ) But C ym = (Cy1 + Cy2) /2hence

C 1² -C 2² = 2 C ym .(Cy1 -Cy2 ) = 2C²x tanαm (tanα1-tanα2)so efficiency

ηD = 1 –[ ∆po /{ρC²x tanαm (tanα₁-tanα₂)}]

Substituting from (7a), the value of ∆po &Cf from(9),the value of tangential force coefficient ∆p0 = (1/2)ρC²x . ζ ----------------------(From 7)And (tanα1-tanα2) = Cf / 2 ----------------------(From 9)

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ηD = 1 –[ (1/2)ρCx².ζ / ρCx² tanαm.Cf/2]

ηD = 1- [ζ / Cf.tanαm] -------------------(20)

From equation (19) CL / CD = (Cf/ζ) sec²αm

ηD = 1 - [CD sec²αm/CL tanαm] CD

= 1 - CLcos2 αm . sin αm

cos αm

ηD = 1- [2CD/CL sin2αm] ---------(21)

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Maximum efficiency of a cascade:

Assuming CD/CL as constant and differentiating (21) with respect to αm to give optimum flow angle for maximum efficiency, we get

cos2αm = 0 cos2αm = cos90º 2αm = 90º αm = 45º So,

(ηD)max = 1- [ 2CD / CL]

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The Cascade Wind-tunnel

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Instrumented Blade

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Compressor blade cascade

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Probe Traversing

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Pressure and velocity distribution around a blade

Pressure distribution on blade pressure side

Pressure distribution on blade Suction side

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Secondary Losses

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Compressor cascade performance

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Losses and their estimation:-

Profile losses - Due to growth of boundary layer on blade profile.

Annulus loss – due to growth of boundary layer on end walls.Secondary losses –

Tip clearance losses -

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Secondary Losses

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Tip clearance losses

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Losses in compressor stage