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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
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