Fluid properties - Assume constant properties:
γ := 1.333
... Ratio of specific heat
Cp := 1147
J
... Specific heat of air standard at constant pressure
kg⋅ K
kJ := 1000J kmol kmol := 1000 1000⋅ mol R := 287 ⋅
... Gas c onstant onstant
J kg⋅ K
P1 := 1.6bar = 0.16⋅ MPa r := 18.5
ASSUMPTIONS rc := 2
... V3 per V2, cut off ratio
T1 := 300 K = 26.85⋅ °C
... Temperature Temperature inlet of c ylinder ylinder
η := 1 −
1 γ− 1 r
⋅
γ rc − 1
γ⋅ ( rc − 1)
= 0.569
STAGE 1: P1 = 0.16 MPa T1 = 300 K kJ
u1 := 214.07
kg
υr1 := 621.2
υ1 :=
ρ1 :=
3
R⋅ T1
= 0.538
P1 1
kg
kg
= 1.858
υ1
m
3
m
STAGE 2:
υr2 :=
υr1 r
= 33.578
From table T-9
T2 := 900 K +
h2 := 932.93⋅
P2 := P1⋅
υ2 :=
ρ2 :=
T2 T1
R⋅ T2 P2 1
υ2
υr2 − 34.31 32.1 32.18 8 − 34.3 34.31 1
kJ
+
kg
906.87K ⋅ ( 920 − 900) ⋅ K = 906.87K
υr2 − 34.31 32.1 32.18 8 − 34.3 34.31 1
⋅ r = 8.948MPa 3
= 0.029
= 34.379
m
kg
kg 3
m
STAGE 3: T3 := rc⋅ T2 = 1813.739373 K
955.38 38 − 932. 932.93 93) ⋅ ⋅ ( 955.
kJ kg
= 940.641
kJ kg
From table T-9 h3 := 2003.3⋅
kJ
T3 − 1800K
+
( 1850 1850 − 1800 1800) ⋅ K
kg
T3 − 1800K
υr3 := 3.944 +
2065.3 .3 − 2003 2003.3 .3) ⋅ ⋅ ( 2065
kJ kg
= 2020.336823
⋅ ( 3.60 3.601 1 − 3.94 3.944 4) = 3.849748
( 1850 1850 − 1800 1800) ⋅ K
P3 := P2 = 8.948 MP MPa
υ3 :=
ρ3 :=
R⋅ T3 P3 1
3
= 0.058
= 17.189
υ3
m
kg
kg 3
m
STAGE 4:
υr4 :=
r rc
⋅ υr3 = 35.61
From table T-9
T4 := 880 K +
u4 := 657.95⋅
P4 := P1⋅
υ4 :=
ρ4 :=
T4 T1
R⋅ T4 P4 1
υ4
υr4 − 36.61 ⋅ ( 900 − 880) ⋅ K = 888.694 K 34.31 1 − 36.6 36.61 1 34.3
kJ
+
kg
υr4 − 36.61 34.3 34.31 1 − 36.6 36.61 1
= 0.474MPa 3
= 0.538
= 1.858
m
kg
kg 3
m
Overall performance:
l 1− η _diese:=
η = 0.569
u4 − u1 h3 − h2
= 0.582
674.58 58 − 657. 657.95 95) ⋅ ⋅ ( 674.
kJ kg
= 665.179
kJ kg
kJ kg
CENTRIFUGAL COMPRESSOR - TURBOCHARGER D := 92mm Stro Stroke ke := 93.8 93.8mm mm N := 4
... Number Number of c ylinders ylinders 2
V1_2 := π⋅
V2 :=
D
4
V1_2 r−1
3
⋅ Stroke⋅ N = 2494.183293 cm
3
= 142.525 cm
3
3
V1 := V2⋅ r = 2636.708052 cm
2494cm = 2.494L
3
V1 − V2 = 2494.183293 cm
RPM := 3000 3000rrpm
m_exhaust := ρ4 ⋅ ( V1 V1 − V2) ⋅
RPM 2
= 0.728
kg s
Given (request) data: mf := m_exhaust = 0.728 PR13 :=
P1 1atm
kg
... The mass flow rate
s
= 1.579
Svan Svanel eles esss := 5mm
... Pressure ratio of compresor
... Vaneless space (between impeller and diffuser)
rdt := 55mm
... The diffuser throat radius rdo := 65mm Ndif:=
15
... The diffuser outlet radius ... Number of channels of diffuser
Guess & suggestion data: rperR := 0.5
... (r/R) (r/R) The ratio of hub to shroud diameter at the eye
ηt := 0.82
... Overall efficiency
μ := 0.525
... Head coefficient, from Aungier Fig. 1-9
ψ := 1.04
... Power input factor
σ := 0.835
... The slip factor
ηm := 0.95
... Mechanical efficiency of compressor
limitedness: W3max := 90
m
αa3max := 11° Umax := 460
... Maximum outlet velocities
s
m
... The maximum included angle of the vaned diffuser passage ... Maksimum tip speed
s
M1Wmax := 0.8 Tmax := 400 K
... Suggested maximum Mach number at inlet impeller - W1 direction ... Maximum static temperature
1. CENTRIFUGAL COMPRESSOR calculaon using MATHCAD 1.1.
IMPELLER DESIGN
1.a. The optimum speed of rotation for the maximum mass flow rate condition if the mass flow is 0.728 kg/s and Using Eq (4-22) the appropriate known data are subtituted, nothing that all conditions apply at the eye tip or shroud. 2
k := 1 − rperR
k = 0.75
i := 0 .. 11
... index
β1 :=
... Blade angle with respect to tangent 0 0
10
1
20
2
30
3
40
4
50
5
55
3
(
)
2
(
)
M1Wmax ⋅ sin β1 ° ⋅ cos β1 ° RHS :=
i
i
i
1
1+
2
(
)
( γ − 1 ) ⋅ M1Wmax ⋅ cos β1 ° 2
i
2
( γ−1)
+
3 2
RHS = i
0.2
7
60
8
61
9
65
10
70
RHSi 0.1
11
80
0.05
0.009764
0.15
0.03754 0.078411 0.123322 0.159069
0
0
20
40
60
80
0.168735 0.170895
β1 i
0.170563 0.169864 0.163267 0.14624 0.08499
RHSmax Smax := RHS
RHSma RHSmax x = 0.1705 0.170563 63
β1 := β1 °
β1 = 60⋅ °
7
7
Po1 := 1 at atm
... Ambient pressure
To1 := 15 °C °C
... Ambient temperature
ω :=
π⋅ k ⋅ γ⋅ Po1⋅ γ⋅ R ⋅ To1 mf
⋅ RHSmax
ω = 47510.874429⋅ rpm
Take centrifugal compressor speed become ... RPM := 4750 47500r 0rpm pm
RPM RPM = 4750 47500 0 ⋅ rpm rpm
New ratio of hub and shroud at inlet become ... 2
RHSmax :=
RPM ⋅ mf
π⋅ k ⋅ γ⋅ Po1⋅ γ⋅ R ⋅ To1
RHSm RHSmax ax = 0.17 0.17
1.b. The eye tip diameter at inlet
ρo1 :=
Po1 R⋅To1
ρo1 = 1.225
3
m
Choose Choo se t he mach number at inlet, M1 := M1W M1Wmax max⋅ cos cos( β1)
kg
M1 = 0.4
To1
T1 :=
2
1+
T1 = 280. 280.67 673 3 K
( γ − 1) ⋅ M1 2
C1 := M1⋅ γ⋅ R ⋅ T1
C1 = 131. 131.07 074 4
m s
Ca1 := C1
From the isentropic relationship at a point 1
T1 ρ1 := ⋅ R⋅To1 To1 Po1
γ− 1
ρ1 = 1.132
kg 3
m
γ
T1 P1 := Po1⋅ To1
γ− 1
P1 = 91.2 91.203 03⋅ kPa kPa
Therefore There fore the eye tip diameter at inlet is:
R1t :=
mf
π⋅ ρ1⋅ k ⋅ Ca1
D1t := 2 ⋅ R1t
R1t R1t = 45.6 45.630 3049 499 9 ⋅ mm
D1t D1t = 91.2 91.261 61⋅ mm
And the hub diameter at inlet is: D1r := D1t⋅ rper rperR R
D1r D1r = 45.6 45.63 3⋅ mm
Peripheral speed at the impeller eye tip (shroud) & β1t: U1t := π⋅ D1t ⋅
RPM rev
Ca1 β1t := atan U1t
U1t U1t = 226. 226.97 9746 4699 99
m s
30.006 006⋅ ° β1t = 30.
Peripheral speed at the impeller eye root (hub) & β1r: U1r := π⋅ D1r ⋅
RPM rev
Ca1 β1r := atan U1r
1.b. The impeller outlet diameter
U1r U1r = 113. 113.48 4873 735 5
m s
49.113 1309 096 6⋅ ° β1r = 49.1
From Eq (4-11) the stagnation temperature difference is
To1
To3 := To1 +
ηt
γ− 1
⋅ PR13
γ
− 1
To3 To3 = 330. 330.63 632 2 K
To3 − To1 = 42.482 42.482 K
From Eq (4-9) ( To3 − To1) ⋅ Cp
U2 :=
U2 = 236. 236.87 877 7
ψ⋅ σ
m s
And Tip flow coeff coefficient icient (ϕ2) at exit of impeller U2⋅rev
D2 :=
D2 = 95.2 95.242 42⋅ mm
π⋅ RPM
The new value of To3 & ηt, U2 := D2⋅ π⋅
RPM
U2 = 236. 236.87 877 7
rev
m s
2
U2 ⋅ ψ⋅ σ
To3 :=
ηt :=
Cp
+ To1
γ− 1
To1 To1⋅ PR13 PR13
γ
To3 To3 = 330. 330.63 632 2 K
− 1
To3 − To1
ηt = 0.82
Therefore the tip flow coefficient is: D2
r2 :=
r2 = 47.6 47.621 2117 175 5 ⋅ mm
2 mf
ϕ2 :=
2
0.35210 52109 9 ϕ2 = 0.3
ρo1⋅ π⋅ r2 ⋅ U2 The Number of Blades: Z :=
0.63⋅ π 1 − .83
Z = 11.642
Z := 12
σ := 1 −
0.63⋅ π Z
σ = 0.835
... Will be used re-calculate
1.c. Velocity at exit and the the losses in the impeller and diffuser are the same, same, The axial depth of the impeller is: First guess M2 = 0.8 and iteration. The overall loss is proportional to (1 - ηc) = (1 - 0.82). Half of the overall loss is therefore 0.5(1 - 0.82) = 0.09 and therefore the effective efficiency of the impeller in compressing from Po1 to Po2 is (1 - 0.09) ... M2 := 1
ηc := 1 − 0.5⋅ ( 1 − ηt )
ηc = 0.91
From Eq (4-11) afer rearranging the subscripts, and To3 = To2:
To2 := To3
To2 To2 = 330. 330.63 632 2 K
To2
T2 :=
T2 = 283. 283.37 373 3 K
2
1+
M2 ⋅ γ⋅ R 2 ⋅ Cp γ
( To2 − To1) PR12 := 1 + ηc ⋅ To1
γ− 1
γ T2 γ−1 ⋅ PR12 P2 := Po1⋅ To2
ρ2 :=
PR12 PR12 = 1.65 1.655 5
P2 = 90.4 90.456 56⋅ kPa kPa
P2
ρ2 = 1.112
R⋅ T2
kg 3
m Po2 :=
P2 γ−1
T2 To2
Po2 Po2 = 94.0 94.009 09⋅ kPa kPa
γ
And TOTAL enthalphy rise (Hrev) at exit of impeller To2s := ηc ⋅ (To T o2 − To1) + To1
To2s To2s = 326. 326.80 808 8 K 2
H_adi _adiab abat atic ic := Cp⋅ ( To2 To2ss − To1)
H_adiabati H_adiabaticc = 44341.0399 44341.039968 68
m
2
s
The head coefficient become ....
μ :=
H_adiabatic
μ = 0.79
2
U2
The specific speed (ns) : ns := 1.773
ϕ2
ns = 1.255
3
μ
4
To find the t he flow velocity velocit y normal normal to the periphery periphery of the impeller From the velocity triangles:
Cax2 := U2⋅ σ
Cax2 Cax2 = 197. 197.80 808 8
C2 := M2⋅ γ⋅ R ⋅ T2
C2 = 329. 329.25 257 7
Cr2 :=
2
( C2) − ( Cax2)
2
m s
m
Cr2 Cr2 = 263. 263.21 216 6
s m s
From the c ontinuity equation of area: A2 :=
−3 2 A2 = 2.487 × 10 m
mf
ρ2⋅Cr2
The depth of impeller at exit of impeller: b2 :=
A2
b2 = 8.31 8.3114 1479 79⋅ mm
π⋅ D2
D_shr _shrou oud d := D1t
D_hub := D1r
The blade work input is: H_im H_impe pell ller er := mf ⋅ ( U2⋅ Cax2 Cax2)
H_impe H_impelle llerr = 34.113 34.113851 851⋅ kW
1.d. Overall dimension of centrifugal compressor At inlet section (1): D_sh D_shro roud ud = 91.2 91.261 61⋅ mm 30.005 0568 68⋅ ° β1t = 30.0 D_hu D_hub b = 45.6 45.63 3⋅ mm
49.113 1309 096 6⋅ ° β1r = 49.1
At outlet section (2): D2 = 95.2 95.242 42⋅ mm b2 b2 = 8.31 8.3114 1479 79⋅ mm U2 − Cax2 β2 := atan Cr2 W2 :=
Cr2
W2 = 266. 266.09 099 9
cos( β2)
C2 = 329. 329.25 257 7
β2 = 8.443⋅ °
m s
m s
Cax2 Cax2 = 197. 197.80 808 8
Cr2 Cr2 = 263. 263.21 216 6
m s
m s
Cr2 α2 := acos C2
α2 = 36.925⋅ °
Z = 12
RPM RPM = 4750 47500 0 ⋅ rpm rpm
ϕ2 = 0.352 σ = 0.835
2. DIFFUSER In the vaneless space between the impeller outlet and diffuser vanes the flow is that of a free vortex which at any radius radius requires requires that Cx.r = c onstan onstant. t. At the diffuser vane leading edge the radius is (r2 + 10)mm = (47.62 + 10) mm = 57.62 mm
r2 :=
D2 2
r2v := r2 + Svaneless
r2 = 47.6 47.621 21⋅ mm r2v r2v = 52.6 52.621 2117 175 5⋅ mm
Cx3i :=
C2⋅ r2
Cx3i Cx3i = 297. 297.97 972 2
r2v
m s
To find the radial velocity Cr at the diffuser vane entry start by assuming the value at the impelle exit, i.e. 263.216 m/s. m/s. Then Cr3i := Cr2 Cx3i Cx3i = 297. 297.97 972 2
C3i :=
m
Cr3i Cr3i = 263. 263.21 216 6
s
m s
2
2
Cr3i + Cx3i
C3i C3i = 397. 397.58 58
m s
If we assume that no losses across he vaneless space, the other half of the total losses takes place in the diffuser itself. The P0,2 at the impeller tip equals the stagnation pressure at the diffuser vane inlet P 0. Therefore ... PR12 PR12 = 1.65 1.655 5 Po3i o3i := PR12⋅ Po1
T3i := To2 −
C3i
Po3i Po3i = 167. 167.71 718 8 ⋅ kPa kPa
2
T3i T3i = 261. 261.72 726 6 K
2 ⋅ Cp γ
T3i P3i := Po1⋅ To2 ρ3i :=
P3i R⋅T3i
γ−1
⋅ PR12
P3i P3i = 65.8 65.809 09⋅ kPa kPa
ρ3i = 0.876
kg 3
m
Wit h reference With reference to Figure, the area of flow in t he radial radial directioon at radius r2v = 52.621 mm (inle diffuser) is
Ar3i := 2 ⋅ π⋅ r2v⋅ b2
Ar3i Ar3i = 0.00 0.0027 2748m 48m
mf
Cr3i :=
m
Cr3i Cr3i = 302. 302.40 407 7
ρ3i⋅Ar3i
2
s
by ITERATION ... Cr3i Cr3i := 302. 302.40 407 7
m
... change this value for iteration
s
2
C3i :=
2
Cr3i + Cx3i
T3i := To2 −
C3i
C3i C3i = 424. 424.54 543 3
m s
2
T3i T3i = 252. 252.06 063 3 K
2 ⋅ Cp γ
T3i P3i := Po1⋅ To2 ρ3i :=
γ−1
⋅ PR12
P3i
ρ3i = 0.783
R⋅T3i
kg 3
m
Ar3i := 2 ⋅ π⋅ r2v⋅ b2 Cr3i :=
P3i P3i = 56.6 56.608 08⋅ kPa kPa
Ar3i Ar3i = 0.00 0.0027 2748m 48m
mf
2
Cr3i Cr3i = 338.57 338.57554 55428 28
ρ3i⋅Ar3i
m s
No further iterations are necessary. Thus at the inlet to the vanes Cr,3,i = 246.2013186 m/s m/s.. Cx3i α3i := atan Cr3i
41.350 5018 188 8⋅ ° α3i = 41.3
Moving to the radius at the diffuser throat, at the throat radius, 343 mm Cx3t :=
C2⋅ r2
Cx3t Cx3t = 285. 285.08 084 4
rdt
m s
by ITERATION again to find propertes at diffuser diffuser throat section section ... .. . Start with assuming Cr,3 = Cr,3,i Cr3t := Cr3i
( Cr3t) = 338.576
Cr3t Cr3t := 1308 1308.0 .010 1007 0793 93
C3t :=
2
m
2
Cx3t + Cr3t
m s
... change this value for iteration
s 3m
C3t = 1.339 × 10
s
T3t := To2 −
C3t
2
T3t = −450.608 K
2 ⋅ Cp γ γ−1
T3t P3t := Po1⋅ To2
ρ3t :=
⋅ PR12
P3t
ρ3t = ( −4.478 − 0.042i)
R⋅T3t
kg 3
m
Ar3t := 2 ⋅ π⋅ rdt ⋅ b2 Cr3t :=
P3t = ( 579.136 + 5.464i) ⋅ kPa
Ar3t Ar3t = 0.00 0.0028 2872m 72m
mf
2
Cr3t r3t = ( −56.5 56.598 9862 6266 66 + 0.5339793i)
ρ3t⋅Ar3t
m s
It may be seen that there is no change in the new values so the radial velocity at the diffuser throat = 184.0268017 m/s Cx3t α3t := atan Cr3t A3t :=
α3t = ( −78.770868 − 0.103249i) ⋅ °
Ar3t⋅ Cr3t
A3t = ( −0.000121 + 0.000001i) m
C3t
2
As we have 15 diffuser vanes, the width of each throat is: Throat_width :=
A3t Ndif Ndif ⋅ b2
Thro Throat at_w _wid idth th = ( −0.97 0.9740 4021 21 + 0.009189i) ⋅ mm
2 ⋅r2v = 105.242⋅ mm rdt⋅ 2 = 110⋅ mm Svan Svanel eles esss = 5 ⋅ mm
Moving to the radius at the diffuser outler, at the outlet radius, 550 mm Cx3 :=
C2⋅r2
Cx3 Cx3 = 241. 241.22 225 5
rdo
m s
by ITERATION again to find propertes at diffuser diffuser outlet section section . .. Start with assuming Cr,3 = Cr,3,i Cr3o := Cr3t
Cr3o Cr3o := 87.8 87.847 4762 62
Cr3o = ( −56.599 + 0.534i) m s
m s
... change this value for iteration
2
C3o :=
2
Cx3 + Cr3o
C3o C3o = 256. 256.72 723 3
m s
2
T3o := To2 −
C3o
T3o T3o = 301. 301.90 902 2 K
2 ⋅ Cp γ
T3o P3o := Po1⋅ To2 ρ3o :=
γ− 1
⋅ PR12
P3o
ρ3o = 1.345
R⋅T3o
kg 3
m
Ar3o := 2 ⋅ π⋅ rdo⋅ b2
Ar3o Ar3o = 0.00 0.0033 3394m 94m
mf
Cr3o :=
P3o P3o = 116. 116.55 559 9 ⋅ kPa kPa
2
Cr3o Cr3o = 159.43 159.43985 98561 61
ρ3o⋅Ar3o
m s
It may be seen that there is no change in the new values so the radial velocity at the diffuser throat = 87.84762 m/s Cx3 α3o := atan Cr3o A3o :=
56.536 3692 926 6⋅ ° α3o = 56.5
Ar3o⋅ Cr3o
A3o A3o = 0.00 0.0021 2108m 08m
C3o
2
As we have 15 diffuser vanes, the width of each throat is: Throat_width_outlet :=
A3o Ndif Ndif ⋅ b2
Overall dimension of DIFFUSER:
At inlet: r2v r2v = 52.6 52.621 21⋅ mm 41.350 5018 188 8⋅ ° α3i = 41.3
At throat: rdt = 55⋅ mm
Throat_wi Throat_width_ou dth_outlet tlet = 16.909629 16.909629⋅ mm
α3t = ( −78.770868 − 0.103249i) ⋅ ° Thro Throat at_w _wid idth th = ( −0.97 0.9740 4021 21 + 0.009189i) ⋅ mm
At outlet: rdo = 65⋅ mm 56.536 3692 926 6⋅ ° α3o = 56.5 Throat_wi Throat_width_o dth_outlet utlet = 16.909629 16.909629⋅ mm
Ndif= 15
TURBINE - TURBOCHARGER P4 = 0.47 0.474MP 4MPaa T4 = 888. 888.69 694 4 K
From previous data - calculation: Po1 := P4 = 0.474 MP MPa
... Stagnation pressure at inlet to nozzles
To1 := T4 = 888.694 K
... Stagnation temperature temperature at inlet to nozzles
m_fl m_flow ow := m_ex m_exha haus ustt = 0.72 0.728 8
kg
... The mass flow of exhaust gas available to the turbine
s
Assumptions : P2 := .725⋅ Po1 = 0.344MPa
... Static pressure at exit from nozzles
T2 := 0.92 0.9275 75⋅ To1 = 824.264 K
... Static temperature at exit from nozzles
P3 := 0.5⋅ Po1 = 0.237MPa
... Static pressure at exit from rotor
T3 := 0.86 0.8632 325 5 ⋅ To1 = 767.165 K
... Static temperature at exit from rotor
To3 := 1.00 1.002 2⋅ T3 = 768.7K
... Stagnation temperature at exit from rotor
r3av_ 3av_r2 r2 := 0.5
... Ratio r,ave / r1
RPM = 47500rp 7500rpm m
... Rotational speed
Analysis: a). The total-to-static efficiency is given by
1−
ηt_ts :=
To3 To1 γ−1
P3 1 − Po1
= 0.849
γ
b). The outer diameter of rotor, inlet diameter
Cx3 := 0
U2 :=
D2 :=
Cx2 = U2
Cp⋅ ( To1 − To3) = 370.99
U2⋅rev RPM⋅ π
m s
= 149.166⋅ mm
c). The T he enthalphy enthalphy loss coeffic ient for the nozzles and rotor rows rows Nozzle loss coefficient: γ− 1
P2 T2s := To1⋅ Po1
ζ N :=
T2 − T2s To1 − T2
γ
= 820.093 K
= 0.064736
Rotor loss coefficient:
U3 := U2⋅ r3a r3av_r2 v_r2 = 185.495
C3 :=
W3 :=
m s
2 ⋅ Cp⋅ ( To3 − T3 ) = 59.328
2
2
C3 + U3 = 194.752
m s
m s
2
2
C3 = 3519.754213
m
2
s 2
W3 = 37928.205778
( γ− 1) 2 γ P3 m h3_h3s := Cp⋅ T3 T3 − ⋅ T2 = 18314.248869 2 P2 s
2
m
2
s
h3_h3s
ζR :=
r3av_r2⋅ W3
= 0.966
2
d). The blade outlet angle at the mean diameter β 3,av, and
C3 β3av := acot = 72.264⋅ ° U3
e). The total-to-total efficiency of the turbine 1
ηt_t :=
1
−
ηt_ts
1 2
= 0.858558
⋅ ( r3av_r2⋅ cot( β3av) )
2
f). The volume flow rate at rotor exit P3
Po3 :=
= 238.888⋅ kPa
γ
T3 To3
γ−1
γ−1 Po3 γ kJ ho1_ ho1_ho ho3s 3sss := Cp⋅ To1 To1⋅ 1 − = 160.353 ⋅ kg Po1
ρ3 :=
Q3 :=
P3 R⋅ T3
= 1.076
kg 3
m
m_flow
ρ3
3
= 0.676
m s
g). The hub and tip diameters of the rotor at exit
(
2
)
2
Q3 = π⋅ r3t − r3h ⋅ C3 Q3 = 2 ⋅ π⋅ r3av⋅ h ⋅ C3
h = r3t − r3h r3av =
r3t + r3h 2 m
C3 = 59.3 59.32 28
r2 :=
D2 2
s
= 0.075m
r3av r3av := r3av_ 3av_rr2⋅
h :=
D2 2
Q3 2 ⋅ π⋅ r3av⋅ C3
0.037292 m = 0.037292
= 0.049m
Hub diameter: r3h := r3av −
h 2
= 0.013m
D3h := 2⋅ r3h = 0.026m
Tip diameter: r3t := r3av +
h 2
= 0.062m
D3t := 2 ⋅ r3t = 0.123m
h). The power developed by the turbine 2
Wtur Wturb b := m_fl m_flow ow⋅ U2 = 100.205185⋅ kW
i). The rotor exit blade angles at the hub and tip At tip
U3t := U2⋅
r3t r2
= 306.516
m s
U3t β3t := atan = 79.046⋅ ° C3
At mean - from previous calculation
3av = 72.2 72.264 64⋅ ° β3av
At hub
U3h := U3⋅
r3h r2
= 32.237
m s
U3h = 28.519⋅ ° β3h := atan C3
j). The nozzle exit angle
C2 :=
2 ⋅ Cp⋅ ( To1 − T2 ) = 384.452
m s
U2 α2 := asin = 74.793176⋅ ° C2
k). The ratio of rotor width at inlet to its inlet tip diameter m_flow = ρ2⋅ A2⋅ Cr2 = ρ2⋅ π⋅ D2⋅ W2⋅ b2 b2 D2
m_flow
=
2
π⋅ ρ2⋅ D2 ⋅ W2
U2 = 370. 370.99 99
m s
α2 = 74.793⋅ ° W2 := U2⋅ cot( α2) = 100.843
ρ2 :=
P2 R⋅ T2
b2_D2 :=
= 1.453
m s
kg 3
m m_flow 2
= 0.071103
π⋅ ρ2⋅ D2 ⋅ W2 b2:= b2_D2⋅ D2 = 10.606131⋅ mm
- END of CALCULATION -
