Turbocharge Design - Calculation Sample

Fluid properties - Assume constant properties: γ := 1.333

... Ratio of specific heat

Cp := 1147

J kg⋅ K

... Specific heat of air standard at constant pressure

kJ := 1000J kmol := 1000⋅ mol R := 287 ⋅

... Gas constant

J kg⋅ K

P1 := 1.6bar = 0.16⋅ MPa r := 18.5

ASSUMPTIONS rc := 2

... V3 per V2, cut off ratio

T1 := 300K = 26.85⋅ °C

... Temperature inlet of cylinder

η := 1 −

1 γ− 1

r

 rcγ − 1   = 0.569  γ⋅ ( rc − 1 )

⋅

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 υ1

m

kg

kg

= 1.858

3

m

STAGE 2:

υr2 :=

υr1 r

= 33.578

From table T-9

T2 := 900K +

h2 := 932.93⋅

P2 := P1⋅

υ2 :=

ρ2 :=

T2 T1

R ⋅ T2 P2 1 υ2

υr2 − 34.31 32.18 − 34.31

kJ kg

+

⋅ ( 920 − 900 ) ⋅ K = 906.87 K

υr2 − 34.31 32.18 − 34.31

⋅ r = 8.948 MPa 3

= 0.029

= 34.379

m

kg

kg 3

m

STAGE 3: T3 := rc⋅ T2 = 1813.739373 K

⋅ ( 955.38 − 932.93) ⋅

kJ kg

= 940.641

kJ kg

From table T-9 h3 := 2003.3⋅

kJ kg

T3 − 1800K

+

( 1850 − 1800) ⋅ K

T3 − 1800K

υr3 := 3.944 +

( 1850 − 1800) ⋅ K

⋅ ( 2065.3 − 2003.3) ⋅

kJ kg

= 2020.336823

⋅ ( 3.601 − 3.944) = 3.849748

P3 := P2 = 8.948 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 := 880K +

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 − 36.61 

kJ kg

+

υr4 − 36.61 34.31 − 36.61

= 0.474 MPa 3

= 0.538

= 1.858

m

kg

kg 3

m

Overall performance:

η_diesel := 1 −

η = 0.569

u4 − u1 h3 − h2

= 0.582

⋅ ( 674.58 − 657.95) ⋅

kJ kg

= 665.179

kJ kg

kJ kg

CENTRIFUGAL COMPRESSOR - TURBOCHARGER D := 92mm Stroke := 93.8mm N := 4

... Number of cylinders 2

V1_2 := π⋅

V2 :=

D

4

V1_2 r−1

⋅ Stroke⋅ N = 2494.183293 cm

= 142.525 cm

3

3

V1 := V2⋅ r = 2636.708052 cm V1 − V2 = 2494.183293 cm

3

3

2494cm = 2.494 L

3

RPM := 3000rpm

m_exhaust := ρ4⋅ ( 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

Svaneless := 5mm

... Pressure ratio of compresor

... Vaneless space (between impeller and diffuser)

rdt := 55mm ... The diffuser throat radius rdo := 65mm

... The diffuser outlet radius

Ndif := 15

... Number of channels of diffuser

Guess & suggestion data: rperR := 0.5

... (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 := 400K

... Suggested maximum Mach number at inlet impeller - W1 direction ... Maximum static temperature

1. CENTRIFUGAL COMPRESSOR calculaon using MATHCAD 1.1. 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. k := 1 − rperR

2

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

6

59

3

(

)

2

(

)

M1Wmax ⋅ sin β1 ° ⋅ cos β1 ° RHS :=

i

i

i

1

 ( γ − 1 ) ⋅ M1Wmax2⋅ cos β1 ° 2 ( i) 1 +  2  

( γ−1)

RHS =

+

3 2

6

59

7

60

8

61

9

65

10

70

RHSi 0.1

11

80

0.05

RHS = i

0.2

0.009764 0.15

0.03754 0.078411 0.123322 0.159069

0

0

20

40

60

80

0.168735 0.170895

β1i

0.170563 0.169864 0.163267 0.14624 0.08499

RHSmax := RHS

RHSmax = 0.170563

β1 := β1 °

β1 = 60⋅ °

7

7

Po1 := 1atm

... Ambient pressure

To1 := 15 °C

... Ambient temperature

ω :=

π⋅ k⋅ γ⋅ Po1⋅ γ⋅ R⋅ To1 mf

⋅ RHSmax

ω = 47510.874429⋅ rpm

Take centrifugal compressor speed become ... RPM := 47500rpm

RPM = 47500 ⋅ rpm

New ratio of hub and shroud at inlet become ... 2

RHSmax :=

RPM ⋅ mf π⋅ k⋅ γ⋅ Po1⋅ γ⋅ R ⋅ To1

RHSmax = 0.17

1.b. The eye tip diameter at inlet

ρo1 :=

Po1 R⋅ To1

ρo1 = 1.225

3

m

Choose the mach number at inlet, M1 := M1Wmax⋅ cos( β1)

kg

M1 = 0.4

To1

T1 :=

2

1+

T1 = 280.673 K

( γ − 1) ⋅ M1 2

C1 := M1⋅ γ⋅ R⋅ T1

C1 = 131.074

m s

Ca1 := C1 From the isentropic relationship at a point 1

ρ1 :=

  R⋅ To1  To1  ⋅ 

Po1

T1

γ− 1

ρ1 = 1.132

kg 3

m

γ

   To1 

P1 := Po1⋅ 

T1

γ−1

P1 = 91.203⋅ kPa

Therefore the eye tip diameter at inlet is:

R1t :=

mf π⋅ ρ1⋅ k⋅ Ca1

D1t := 2 ⋅ R1t

R1t = 45.630499 ⋅ mm

D1t = 91.261⋅ mm

And the hub diameter at inlet is: D1r := D1t ⋅ rperR

D1r = 45.63⋅ mm

Peripheral speed at the impeller eye tip (shroud) & β1t: U1t := π⋅ D1t ⋅ β1t := atan

RPM rev

Ca1 

  U1t 

m

U1t = 226.974699

s

β1t = 30.006⋅ °

Peripheral speed at the impeller eye root (hub) & β1r: U1r := π⋅ D1r⋅ β1r := atan

RPM rev

Ca1 

  U1r 

1.b. The impeller outlet diameter

U1r = 113.48735

m s

β1r = 49.113096 ⋅ °

From Eq (4-11) the stagnation temperature difference is

To1

To3 := To1 +

ηt

γ− 1     γ ⋅  PR13 − 1

To3 = 330.632 K

To3 − To1 = 42.482 K

From Eq (4-9) ( To3 − To1) ⋅ Cp

U2 :=

ψ⋅ σ

U2 = 236.877

m s

And Tip flow coefficient (ϕ2) at exit of impeller U2⋅ rev

D2 :=

D2 = 95.242⋅ mm

π⋅ RPM

The new value of To3 & ηt, U2 := D2⋅ π⋅

RPM rev

U2 = 236.877

m s

2

U2 ⋅ ψ⋅ σ

To3 :=

Cp

+ To1

γ− 1     γ To1⋅  PR13 − 1 ηt :=

To3 − To1

To3 = 330.632 K

ηt = 0.82

Therefore the tip flow coefficient is: D2

r2 :=

r2 = 47.621175 ⋅ mm

2 mf

ϕ2 :=

2

ϕ2 = 0.352109

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, 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 = 330.632 K

To2

T2 :=

T2 = 283.373 K

2

1+

M2 ⋅ γ⋅ R 2 ⋅ Cp γ

PR12 := 1 + ηc ⋅

( To2 − To1) 



To1

γ− 1

 

γ   γ− 1  T2   P2 := Po1⋅  ⋅ PR12   To2  

ρ2 :=

PR12 = 1.655

P2 = 90.456⋅ kPa

P2 ρ2 = 1.112

R⋅ T2

kg 3

m Po2 :=

P2 γ−1

 T2     To2 

Po2 = 94.009⋅ kPa

γ

And TOTAL enthalphy rise (Hrev) at exit of impeller To2s := ηc ⋅ ( To2 − To1) + To1

To2s = 326.808 K

H_adiabatic := Cp⋅ ( To2s − To1)

H_adiabatic = 44341.039968

2

m

2

s The head coefficient become ....

μ :=

H_adiabatic U2

μ = 0.79

2

The specific speed (ns) : ns := 1.773

ϕ2

ns = 1.255

3

μ

4

To find the flow velocity normal to the periphery of the impeller From the velocity triangles:

Cax2 := U2⋅ σ

Cax2 = 197.808

C2 := M2⋅ γ⋅ R⋅ T2

C2 = 329.257

Cr2 :=

2

( C2) − ( Cax2)

2

m s

m

Cr2 = 263.216

s m s

From the continuity equation of area: A2 :=

−3 2

mf

A2 = 2.487 × 10

ρ2⋅ Cr2

m

The depth of impeller at exit of impeller: b2 :=

A2

b2 = 8.311479 ⋅ mm

π⋅ D2

D_shroud := D1t

D_hub := D1r

The blade work input is: H_impeller := mf ⋅ ( U2⋅ Cax2)

H_impeller = 34.113851 ⋅ kW

1.d. Overall dimension of centrifugal compressor At inlet section (1): D_shroud = 91.261⋅ mm β1t = 30.00568⋅ ° D_hub = 45.63⋅ mm

β1r = 49.113096 ⋅ ° At outlet section (2): D2 = 95.242⋅ mm b2 = 8.311479 ⋅ mm β2 := atan

U2 − Cax2 



Cr2

Cr2

W2 :=

m s

m s

Cax2 = 197.808

Cr2 = 263.216 α2 := acos

β2 = 8.443⋅ °

W2 = 266.099

cos( β2)

C2 = 329.257

 

m s

m s

Cr2 

  C2 

α2 = 36.925⋅ °

Z = 12

RPM = 47500 ⋅ 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 requires that Cx.r = constant. 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 = 47.621⋅ mm r2v = 52.621175 ⋅ mm

Cx3i :=

C2⋅ r2

Cx3i = 297.972

r2v

m s

To find the radial velocity Cr at the diffuser vane entry start by assuming the value at the impeller exit, i.e. 263.216 m/s. Then Cr3i := Cr2 Cx3i = 297.972

C3i :=

Cr3i = 263.216

m

C3i = 397.58

m

s

m s

2

2

Cr3i + Cx3i

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 P0. Therefore ... PR12 = 1.655 Po3i := PR12⋅ Po1

T3i := To2 −

C3i

Po3i = 167.718 ⋅ kPa

2

T3i = 261.726 K

2 ⋅ Cp γ

P3i := Po1⋅ 

T3i 

  To2 

ρ3i :=

P3i R⋅ T3i

γ−1

⋅ PR12

P3i = 65.809⋅ kPa

ρ3i = 0.876

kg 3

m

With reference to Figure, the area of flow in the radial directioon at radius r2v = 52.621 mm (inlet diffuser) is

2

Ar3i := 2 ⋅ π⋅ r2v⋅ b2

Ar3i = 0.002748 m

mf

Cr3i :=

m

Cr3i = 302.407

ρ3i⋅ Ar3i

s

by ITERATION ... Cr3i := 302.407

m

... change this value for iteration

s

2

C3i :=

2

Cr3i + Cx3i C3i

T3i := To2 −

C3i = 424.543

m s

2

T3i = 252.063 K

2 ⋅ Cp γ

T3i 

P3i := Po1⋅ 

γ−1

⋅ PR12

  To2 

ρ3i :=

P3i

ρ3i = 0.783

R⋅ T3i

kg 3

m

2

Ar3i := 2 ⋅ π⋅ r2v⋅ b2 Cr3i :=

P3i = 56.608⋅ kPa

Ar3i = 0.002748 m

mf

Cr3i = 338.5755428

ρ3i⋅ Ar3i

m s

No further iterations are necessary. Thus at the inlet to the vanes Cr,3,i = 246.2013186 m/s. α3i := atan

Cx3i 

α3i = 41.350188 ⋅ °

  Cr3i 

Moving to the radius at the diffuser throat, at the throat radius, 343 mm Cx3t :=

C2⋅ r2

Cx3t = 285.084

rdt

m s

by ITERATION again to find propertes at diffuser throat section ... Start with assuming Cr,3 = Cr,3,i Cr3t := Cr3i

( Cr3t) = 338.576 m

Cr3t := 1308.0100793

C3t :=

2

m s

... change this value for iteration

s 2

Cx3t + Cr3t

3m

C3t = 1.339 × 10

s

C3t

T3t := To2 −

2

T3t = −450.608 K

2 ⋅ Cp γ γ−1

T3t 

P3t := Po1⋅ 

⋅ PR12

  To2 

ρ3t :=

P3t

ρ3t = ( −4.478 − 0.042i)

R⋅ T3t

kg 3

m 2

Ar3t := 2 ⋅ π⋅ rdt ⋅ b2 Cr3t :=

P3t = ( 579.136 + 5.464i) ⋅ kPa

Ar3t = 0.002872 m

mf

Cr3t = ( −56.5986266 + 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 α3t := atan

Cx3t 

α3t = ( −78.770868 − 0.103249i ) ⋅ °

  Cr3t 

A3t :=

Ar3t⋅ Cr3t

2

A3t = ( −0.000121 + 0.000001i ) m

C3t

As we have 15 diffuser vanes, the width of each throat is: Throat_width :=

A3t Ndif ⋅ b2

Throat_width = ( −0.974021 + 0.009189i ) ⋅ mm

2 ⋅ r2v = 105.242 ⋅ mm rdt ⋅ 2 = 110 ⋅ mm Svaneless = 5 ⋅ mm Moving to the radius at the diffuser outler, at the outlet radius, 550 mm Cx3 :=

C2⋅ r2

Cx3 = 241.225

rdo

m s

by ITERATION again to find propertes at diffuser outlet section ... Start with assuming Cr,3 = Cr,3,i Cr3o := Cr3t

Cr3o := 87.84762

Cr3o = ( −56.599 + 0.534i) m s

m s

... change this value for iteration

2

C3o :=

2

Cx3 + Cr3o

C3o

T3o := To2 −

C3o = 256.723

m s

2

T3o = 301.902 K

2 ⋅ Cp γ

P3o := Po1⋅ 

T3o 

γ− 1

  To2 

ρ3o :=

⋅ PR12

P3o

ρ3o = 1.345

R⋅ T3o

kg 3

m

2

Ar3o := 2 ⋅ π⋅ rdo⋅ b2

Ar3o = 0.003394 m

mf

Cr3o :=

P3o = 116.559 ⋅ kPa

Cr3o = 159.4398561

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 α3o := atan 

Cx3

 

α3o = 56.536926 ⋅ °

 Cr3o 

A3o :=

Ar3o⋅ Cr3o

2

A3o = 0.002108 m

C3o

As we have 15 diffuser vanes, the width of each throat is: Throat_width_outlet :=

A3o Ndif ⋅ b2

Overall dimension of DIFFUSER:

At inlet: r2v = 52.621⋅ mm α3i = 41.350188 ⋅ °

At throat: rdt = 55⋅ mm

Throat_width_outlet = 16.909629 ⋅ mm

α3t = ( −78.770868 − 0.103249i ) ⋅ ° Throat_width = ( −0.974021 + 0.009189i ) ⋅ mm

At outlet: rdo = 65⋅ mm α3o = 56.536926 ⋅ ° Throat_width_outlet = 16.909629 ⋅ mm

Ndif = 15

TURBINE - TURBOCHARGER P4 = 0.474 MPa T4 = 888.694 K From previous data - calculation: Po1 := P4 = 0.474 MPa

... Stagnation pressure at inlet to nozzles

To1 := T4 = 888.694 K

... Stagnation temperature at inlet to nozzles

m_flow := m_exhaust = 0.728

kg

... The mass flow of exhaust gas available to the turbine

s

Assumptions : P2 := .725⋅ Po1 = 0.344 MPa

... Static pressure at exit from nozzles

T2 := 0.9275⋅ To1 = 824.264 K

... Static temperature at exit from nozzles

P3 := 0.5⋅ Po1 = 0.237 MPa

... Static pressure at exit from rotor

T3 := 0.86325 ⋅ To1 = 767.165 K

... Static temperature at exit from rotor

To3 := 1.002⋅ T3 = 768.7 K

... Stagnation temperature at exit from rotor

r3av_r2 := 0.5

... Ratio r,ave / r1

RPM = 47500 rpm

... Rotational speed

Analysis: a). The total-to-static efficiency is given by

1− ηt_ts :=

 To3     To1  γ−1

1−

 P3     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 enthalphy loss coefficient for the nozzles and rotor rows Nozzle loss coefficient: γ− 1

   Po1 

T2s := To1⋅  ζN :=

P2

T2 − T2s To1 − T2

γ

= 820.093 K

= 0.064736

Rotor loss coefficient:

U3 := U2⋅ r3av_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)   γ 2   P3 m h3_h3s := Cp⋅ 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

β3av := acot

C3 

 = 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_ho3ss := Cp⋅ 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 = 59.328

r2 :=

D2 2

s

= 0.075 m

r3av := r3av_r2⋅

h :=

D2 2

Q3 2 ⋅ π⋅ r3av⋅ C3

= 0.037292 m

= 0.049 m

Hub diameter: r3h := r3av −

h

= 0.013 m

2

D3h := 2⋅ r3h = 0.026 m Tip diameter: r3t := r3av +

h 2

= 0.062 m

D3t := 2 ⋅ r3t = 0.123 m

h). The power developed by the turbine 2

Wturb := m_flow⋅ U2 = 100.205185⋅ kW

i). The rotor exit blade angles at the hub and tip At tip

U3t := U2⋅

r3t r2

β3t := atan

= 306.516

m s

U3t 

 = 79.046⋅ °  C3 

At mean - from previous calculation

β3av = 72.264⋅ °

At hub

r3h

U3h := U3⋅

r2

β3h := atan 

= 32.237

m s

U3h 

 = 28.519⋅ °  C3 

j). The nozzle exit angle

C2 :=

2 ⋅ Cp⋅ ( To1 − T2 ) = 384.452

α2 := asin

m s

U2 

 = 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 = 370.99

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 -

v = 52.621 mm (inlet