Turbocharger Matching.pdf 5o1a17
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Engine Turbo/Super Charging
Super and Turbo-charging
Why super/ turbo-charging? • Fuel burned per cycle in an IC engine is air limited – (F/A)stoich = 1/14.6
Torq =
ηf m f QHV 2πnR
Power = Torq ⋅ 2πN m f = F η V ρa,0 VD A
( )
ηf,ηv– fuel conversion and volumetric efficiencies mf – fuel mass per cycle QHV– fuel heating value nR – 1 for 2-stroke, 2 for 4-stroke engine N – revolution per second VD – engine displacement ρa,0 – air density
Super/turbo-charging: increase air density
Super- and Turbo- Charging
Purpose: To increase the charge density • Supercharge: compressor powered by engine output – No turbo-lag – Does not impact exhaust treatment – Fuel consumption penalty
• Turbo-charge: compressor powered by exhaust turbine
– Uses ‘wasted’ exhaust energy – Turbo- lag problem – Affects exhaust treatment
• Intercooler – Increase charge density (hence output power) by cooling the charge
– Lowers NOx emissions
Charge-air pressure regulation with wastegate on exhaust gas end. 1.Engine, 2. Exhaust-gas turbochager, 3. Wastegate
Exhaust-gas turbocharger for trucks 1.Compressor housing, 2. Compressor impeller, 3. Turbine housing, 4. Rotor, 5. Bearing housing, 6. inflowing exhaust gas, 7. Out-flowing exhaust gas, 8. Atmospheric fresh air, 9. Pre-compressed fresh air, 10. Oil inlet, 11. Oil return
Images removed due to copyright restrictions. Please see illustrations of "Charge-air Pressure Regulation with Wastegate on Exhaust Gas End", and "Exhaust-gas Turbocharger for Trucks." In the Bosch Automotive Handbook. London, England: John Wiley & Sons, 2004.
From Bosch Automotive Handbook
Compressor: basic thermodynamics
Compressor efficiency ηc
2
W ideal ηc = W
W
m
actual
1
⎞ ⎛T ′ 2 ⎜ c p T1 Wideal = m − 1⎟ ⎟ ⎜ T1 ⎝ ⎠
T P2 Ideal process 2’
1
2
γ−1
T2′ ⎛ P2 ⎞ γ = ⎜⎜ ⎟⎟ T1 ⎝ P1 ⎠
P1
Actual process
s
γ−1 ⎛ ⎞ ⎜ ⎟ γ ⎛ ⎞ P 1 2 c p T1⎜ ⎜ ⎟ W m − 1⎟ actual = ⎜ ⎟ ηc ⎜⎜ ⎝ P1 ⎠ ⎟⎟ ⎝ ⎠ W T2 = T1 + actual m
Turbine: basic thermodynamics
Turbine efficiency ηt
4
W ηt = actual W
W 3
ideal
⎛ T ′⎞ ⎜1− 4 ⎟ W = m c T ideal p 3⎜ T3 ⎟ ⎝ ⎠
m
T P3
γ−1 ⎛ P4 ⎞ γ ⎜ ⎟
T4′ =⎜ ⎟ T3 ⎝ P3 ⎠
Ideal process P4
3
4 4’
Actual process
s
γ−1 ⎞ ⎛ ⎜ ⎛P ⎞ γ ⎟ c p T3 ⎜1− ⎜ 4 ⎟ ⎟ Wactual = ηt m ⎜ ⎟ ⎟⎟ ⎜⎜ ⎝ P3 ⎠ ⎝ ⎠ W T4 = T3 − actual m
Properties of Turbochargers
• Power transfer between fluid and shaft ∝ RPM3
– Typically operate at ~ 60K to 120K RPM
• RPM limited by centrifugal stress: usually tip velocity is approximately sonic • Flow devices, sensitive to boundary layer (BL) behavior – Compressor: BL under unfavorable gradient – Turbine: BL under favorable gradient
Typical super/turbo-charged engine parameters
• Peak compressor pressure ratio ≈ 3.5
• BMEP up to 22 bar • Limits: – compressor aerodynamics – cylinder peak pressure – NOx emissions
Compressor/Turbine Characteristics
• Delivered pressure P2 ,RT ,P ,N,D,μ, γ, geometric ratios) • P2 = f(m 1 1 • Dimensional analysis: – 7 dimensional variables → (7-3) = 4 dimensionless parameters (plus γ and geometric ratios)
⎛ P2 ⎞ N m ⎜⎜ ⎟⎟ = f( , ,Re, γ, geometric ratios) γRT1 / D ⎛ P1 ⎞ ⎝ P1 ⎠ ⎜⎜ ⎟⎟ RT1D 2 ⎝ RT1 ⎠
Velocity
Density
Velocity
High Re number flow →weak Re dependence For fixed geometry machinery and gas properties T1 ⎞
⎛ P2 ⎞ ⎛⎜ N m ⎟ ⎜⎜ ⎟⎟ = f , P1 ⎟ ⎝ P1 ⎠ ⎜
⎝ T1 ⎠
Compressor Map
3.4 7250
3.2
6960
3.0 72%
70%
2.6
2.0
65% 60%
it
2.4 2.2
6530
74%
Su rg el im
Pressure ratio
2.8
75%
6070
1.8
5550
1.6
4840
1.4 4025 N/ T1
1.2 1.0
2650
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
. "Corrected" Flow rate m T1/P1
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
T1= inlet temperature (K); P1= inlet pressure (bar); N = rev. per min.; m = mass flow rate (kg/s) (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Compressor stall and surge
• Stall – Happens when incident flow angle is too large (large Vθ/Vx) – Stall causes flow blockage
• Surge – Flow inertia/resistance, and compression system internal volume comprise a LRC resonance system – Oscillatory flow behave when flow blockage occurs because of compressor stall
¾ reverse flow and violent flow rate surges
Turbine Map 2.8 2.6
π tTS = .70
2.4
2.0 1.8 .65
Pressure ratio
2.2
1.6
.6 0
4000
3000 2500
.40
1.2 1.0
3500 5
N T03
1500 500
0.5
1.0
.5
1.4
.50
1.5
2.0
. Flow rate m T03/P03
2.5
3.0
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
T03=Turbine inlet temperature(K); P03 = Turbine inlet pressure(bar); P4= Turbine outlet pressure(bar); N = rev. per min.; m = mass flow rate (kg/s) (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Compressor Turbine Matching Exercise
• For simplicity, take away intercooler and wastegate • Given engine brake power ) and RPM, output (W E compressor map, turbine map, and engine map • Find operating point, i.e. air flow ( m a ), fuel flow rate ( m f ) turbo-shaft revolution per second (N), compressor and turbine pressure ratios (πc and πt) etc.
1
4 T
C
2
m f
3 Engine
W E
Q L
Compressor/ turbine/engine matching solution
Procedure : 1. Guess πc ; can get engine inlet conditions : γ−1 ⎡ ⎤ γ ⎥ T1 ⎢ P2 = πc P1 T2 = − 1⎥ ⎢(πc ) ηc ⎢ ⎥ ⎢⎣ ⎥⎦ 2. Then engine volumetric efficiency calibration a that can be ' swallowed' will give the air flow m
Compressor 3.4 7250
3.2
6960
3.0 72%
70%
2.6
6530
74%
2.4 it
65% 60%
rg e
lim
2.2 2.0
75%
Su
Pressure ratio
2.8
a and πc , the compressor speed N can be 3. From m
6070
1.8
5550
1.6
obtained from the compressor map f may be obtained from the 4. The fuel flow rate m
4840
1.4 4025 N/ T1
1.2 1.0
2650
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
. "Corrected" Flow rate m T1/P1
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
2.8 2.6 2.4
η tTS = .70
6. Guess π t , then get turbine speed Nt from turbine map 7. Determine turbine power from turbine efficiency on map ⎡ = η ⎢1− ⎛⎜ 1 W t t⎢ ⎜ ⎢ ⎝ πt ⎣
2.0
.65
1.8
4000 .60
1.0
3000 1500 500
0.5
1.0
⎟ ⎟ ⎠
⎥ ⎥ ⎦
=W and N = N 8.Iterate on the values of πc and π t until W t c t c
3500 2500
1.2
γ−1 ⎤ ⎞ γ ⎥
.5 5
N T03
.4 0
Pressure ratio
2.2
1.4
5. Engine exhaust temperature T3 may be obtained from energy balance (with known engine mech. eff. ηM ) a +m f )c p T3 = m a c p T2 + m f LHV − WE − Q (m L ηM
Turbine
1.6
engine map : =m ,A/F) f LHV ηf (RPM,W W E E
.50
1.5
2.0
. Flow rate m T03/P03
2.5
3.0
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
Compressor/ Engine/ Turbine Matching
0.65
3.0 0.67
2.5
0.60
0.70 0.55
ns
ad
Co nst
ant
Co
t lo tan
sp e ed
it el
im
2.0
Su rg
Pp/P1
0.72
• Mass flows through compressor, engine, turbine and wastegate have to be consistent • Turbine inlet temperature consistent with fuel flow and engine power output • Turbine supplies compressor work • Turbine and compressor at same speed
1.5
T
C 1.0
0
1
2
3
4
m T1 p1 Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
Compressor characteristics, with airflow requirements of a four-stroke truck engine superimposed. (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Inter-
Cooler
Wastegate
Engine
Advanced turbocharger development
Electric assisted
turbo-charging
• Concept
– Put motor/ generator on
turbo-charger
– reduce wastegate function
• Benefit
InterCooler
– increase air flow at low
engine speed – auxiliary electrical output
at part load
Motor/ Generator
T
C
Wastegate
Engine
Battery
Advanced turbocharger development
Electrical turbo-charger Battery
• Concept – turbine drives generator; compressor driven by motor
• Benefit – decoupling of turbine and compressor map, hence much more freedom in performance optimization – Auxiliary power output – do not need wastegate; no turbo-lag
C
Motor
InterCooler
Engine
T
Generator
Advanced turbocharger development
Challenges • Interaction of turbo-charging system with exhaust treatment and emissions – Especially severe in light-duty diesel market because of low exhaust temperature
• Cost
Super and Turbo-charging
Why super/ turbo-charging? • Fuel burned per cycle in an IC engine is air limited – (F/A)stoich = 1/14.6
Torq =
ηf m f QHV 2πnR
Power = Torq ⋅ 2πN m f = F η V ρa,0 VD A
( )
ηf,ηv– fuel conversion and volumetric efficiencies mf – fuel mass per cycle QHV– fuel heating value nR – 1 for 2-stroke, 2 for 4-stroke engine N – revolution per second VD – engine displacement ρa,0 – air density
Super/turbo-charging: increase air density
Super- and Turbo- Charging
Purpose: To increase the charge density • Supercharge: compressor powered by engine output – No turbo-lag – Does not impact exhaust treatment – Fuel consumption penalty
• Turbo-charge: compressor powered by exhaust turbine
– Uses ‘wasted’ exhaust energy – Turbo- lag problem – Affects exhaust treatment
• Intercooler – Increase charge density (hence output power) by cooling the charge
– Lowers NOx emissions
Charge-air pressure regulation with wastegate on exhaust gas end. 1.Engine, 2. Exhaust-gas turbochager, 3. Wastegate
Exhaust-gas turbocharger for trucks 1.Compressor housing, 2. Compressor impeller, 3. Turbine housing, 4. Rotor, 5. Bearing housing, 6. inflowing exhaust gas, 7. Out-flowing exhaust gas, 8. Atmospheric fresh air, 9. Pre-compressed fresh air, 10. Oil inlet, 11. Oil return
Images removed due to copyright restrictions. Please see illustrations of "Charge-air Pressure Regulation with Wastegate on Exhaust Gas End", and "Exhaust-gas Turbocharger for Trucks." In the Bosch Automotive Handbook. London, England: John Wiley & Sons, 2004.
From Bosch Automotive Handbook
Compressor: basic thermodynamics
Compressor efficiency ηc
2
W ideal ηc = W
W
m
actual
1
⎞ ⎛T ′ 2 ⎜ c p T1 Wideal = m − 1⎟ ⎟ ⎜ T1 ⎝ ⎠
T P2 Ideal process 2’
1
2
γ−1
T2′ ⎛ P2 ⎞ γ = ⎜⎜ ⎟⎟ T1 ⎝ P1 ⎠
P1
Actual process
s
γ−1 ⎛ ⎞ ⎜ ⎟ γ ⎛ ⎞ P 1 2 c p T1⎜ ⎜ ⎟ W m − 1⎟ actual = ⎜ ⎟ ηc ⎜⎜ ⎝ P1 ⎠ ⎟⎟ ⎝ ⎠ W T2 = T1 + actual m
Turbine: basic thermodynamics
Turbine efficiency ηt
4
W ηt = actual W
W 3
ideal
⎛ T ′⎞ ⎜1− 4 ⎟ W = m c T ideal p 3⎜ T3 ⎟ ⎝ ⎠
m
T P3
γ−1 ⎛ P4 ⎞ γ ⎜ ⎟
T4′ =⎜ ⎟ T3 ⎝ P3 ⎠
Ideal process P4
3
4 4’
Actual process
s
γ−1 ⎞ ⎛ ⎜ ⎛P ⎞ γ ⎟ c p T3 ⎜1− ⎜ 4 ⎟ ⎟ Wactual = ηt m ⎜ ⎟ ⎟⎟ ⎜⎜ ⎝ P3 ⎠ ⎝ ⎠ W T4 = T3 − actual m
Properties of Turbochargers
• Power transfer between fluid and shaft ∝ RPM3
– Typically operate at ~ 60K to 120K RPM
• RPM limited by centrifugal stress: usually tip velocity is approximately sonic • Flow devices, sensitive to boundary layer (BL) behavior – Compressor: BL under unfavorable gradient – Turbine: BL under favorable gradient
Typical super/turbo-charged engine parameters
• Peak compressor pressure ratio ≈ 3.5
• BMEP up to 22 bar • Limits: – compressor aerodynamics – cylinder peak pressure – NOx emissions
Compressor/Turbine Characteristics
• Delivered pressure P2 ,RT ,P ,N,D,μ, γ, geometric ratios) • P2 = f(m 1 1 • Dimensional analysis: – 7 dimensional variables → (7-3) = 4 dimensionless parameters (plus γ and geometric ratios)
⎛ P2 ⎞ N m ⎜⎜ ⎟⎟ = f( , ,Re, γ, geometric ratios) γRT1 / D ⎛ P1 ⎞ ⎝ P1 ⎠ ⎜⎜ ⎟⎟ RT1D 2 ⎝ RT1 ⎠
Velocity
Density
Velocity
High Re number flow →weak Re dependence For fixed geometry machinery and gas properties T1 ⎞
⎛ P2 ⎞ ⎛⎜ N m ⎟ ⎜⎜ ⎟⎟ = f , P1 ⎟ ⎝ P1 ⎠ ⎜
⎝ T1 ⎠
Compressor Map
3.4 7250
3.2
6960
3.0 72%
70%
2.6
2.0
65% 60%
it
2.4 2.2
6530
74%
Su rg el im
Pressure ratio
2.8
75%
6070
1.8
5550
1.6
4840
1.4 4025 N/ T1
1.2 1.0
2650
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
. "Corrected" Flow rate m T1/P1
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
T1= inlet temperature (K); P1= inlet pressure (bar); N = rev. per min.; m = mass flow rate (kg/s) (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Compressor stall and surge
• Stall – Happens when incident flow angle is too large (large Vθ/Vx) – Stall causes flow blockage
• Surge – Flow inertia/resistance, and compression system internal volume comprise a LRC resonance system – Oscillatory flow behave when flow blockage occurs because of compressor stall
¾ reverse flow and violent flow rate surges
Turbine Map 2.8 2.6
π tTS = .70
2.4
2.0 1.8 .65
Pressure ratio
2.2
1.6
.6 0
4000
3000 2500
.40
1.2 1.0
3500 5
N T03
1500 500
0.5
1.0
.5
1.4
.50
1.5
2.0
. Flow rate m T03/P03
2.5
3.0
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
T03=Turbine inlet temperature(K); P03 = Turbine inlet pressure(bar); P4= Turbine outlet pressure(bar); N = rev. per min.; m = mass flow rate (kg/s) (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Compressor Turbine Matching Exercise
• For simplicity, take away intercooler and wastegate • Given engine brake power ) and RPM, output (W E compressor map, turbine map, and engine map • Find operating point, i.e. air flow ( m a ), fuel flow rate ( m f ) turbo-shaft revolution per second (N), compressor and turbine pressure ratios (πc and πt) etc.
1
4 T
C
2
m f
3 Engine
W E
Q L
Compressor/ turbine/engine matching solution
Procedure : 1. Guess πc ; can get engine inlet conditions : γ−1 ⎡ ⎤ γ ⎥ T1 ⎢ P2 = πc P1 T2 = − 1⎥ ⎢(πc ) ηc ⎢ ⎥ ⎢⎣ ⎥⎦ 2. Then engine volumetric efficiency calibration a that can be ' swallowed' will give the air flow m
Compressor 3.4 7250
3.2
6960
3.0 72%
70%
2.6
6530
74%
2.4 it
65% 60%
rg e
lim
2.2 2.0
75%
Su
Pressure ratio
2.8
a and πc , the compressor speed N can be 3. From m
6070
1.8
5550
1.6
obtained from the compressor map f may be obtained from the 4. The fuel flow rate m
4840
1.4 4025 N/ T1
1.2 1.0
2650
0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
. "Corrected" Flow rate m T1/P1
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
2.8 2.6 2.4
η tTS = .70
6. Guess π t , then get turbine speed Nt from turbine map 7. Determine turbine power from turbine efficiency on map ⎡ = η ⎢1− ⎛⎜ 1 W t t⎢ ⎜ ⎢ ⎝ πt ⎣
2.0
.65
1.8
4000 .60
1.0
3000 1500 500
0.5
1.0
⎟ ⎟ ⎠
⎥ ⎥ ⎦
=W and N = N 8.Iterate on the values of πc and π t until W t c t c
3500 2500
1.2
γ−1 ⎤ ⎞ γ ⎥
.5 5
N T03
.4 0
Pressure ratio
2.2
1.4
5. Engine exhaust temperature T3 may be obtained from energy balance (with known engine mech. eff. ηM ) a +m f )c p T3 = m a c p T2 + m f LHV − WE − Q (m L ηM
Turbine
1.6
engine map : =m ,A/F) f LHV ηf (RPM,W W E E
.50
1.5
2.0
. Flow rate m T03/P03
2.5
3.0
Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
Compressor/ Engine/ Turbine Matching
0.65
3.0 0.67
2.5
0.60
0.70 0.55
ns
ad
Co nst
ant
Co
t lo tan
sp e ed
it el
im
2.0
Su rg
Pp/P1
0.72
• Mass flows through compressor, engine, turbine and wastegate have to be consistent • Turbine inlet temperature consistent with fuel flow and engine power output • Turbine supplies compressor work • Turbine and compressor at same speed
1.5
T
C 1.0
0
1
2
3
4
m T1 p1 Figure by MIT OpenCourseWare. Adapted from Haddad, Sam David, and Watson, N. Principles and Performance in Diesel Engineering. Chichester, England: Ellis Horwood, 1984.
Compressor characteristics, with airflow requirements of a four-stroke truck engine superimposed. (From “Principles and Performance in Diesel Engineering,” Ed. by Haddad and Watson)
Inter-
Cooler
Wastegate
Engine
Advanced turbocharger development
Electric assisted
turbo-charging
• Concept
– Put motor/ generator on
turbo-charger
– reduce wastegate function
• Benefit
InterCooler
– increase air flow at low
engine speed – auxiliary electrical output
at part load
Motor/ Generator
T
C
Wastegate
Engine
Battery
Advanced turbocharger development
Electrical turbo-charger Battery
• Concept – turbine drives generator; compressor driven by motor
• Benefit – decoupling of turbine and compressor map, hence much more freedom in performance optimization – Auxiliary power output – do not need wastegate; no turbo-lag
C
Motor
InterCooler
Engine
T
Generator
Advanced turbocharger development
Challenges • Interaction of turbo-charging system with exhaust treatment and emissions – Especially severe in light-duty diesel market because of low exhaust temperature
• Cost
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