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ICE II Exercise WS16/17 Exercise 10 – Engine Management EXERCISE 10 – ENGINE MANAGEMENT Task 1 a) Searched: Required pressure and stationary mass flow Assumption: volumetric efficiency referring to intake manifold conditions 𝜆a = 1 ⇒ no scavenging charging efficiency 𝜆l = 1 The indicated power for t...

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ICE II Exercise WS16/17

Exercise 10 – Engine Management

1

EXERCISE 10 – ENGINE MANAGEMENT

Task 1 a) Searched: Required pressure and stationary mass flow

Assumption: volumetric efficiency referring to intake manifold conditions 𝜆a = 1 ⇒ no scavenging charging efficiency 𝜆l = 1 𝐻u ⋅ 𝜌G 1 + 𝜆z ⋅ 𝐿St

The indicated power for the high pressure cycle: 𝑝mi,HD = 𝜂i,HD ⋅ 𝜆l ⋅ 𝐻G,z = 𝜂i,HD ⋅ 𝜆⏟l ⋅

Multipoint injection ⇒ 𝜆z = 𝜆

𝑝mi,HD = 𝑝me + 𝑝mr − 𝑝m,LW

=1

𝑝mi,HD = 𝑝me + 𝑝mr − (𝑝E − p𝐴 ) = 𝜂i,HD ⋅ Here: 𝑝E = 𝑝SR and 𝑇E = 𝑇SR:

𝑝SR ⋅ 𝐻u 𝑅G ⋅ 𝑇E ⋅ (1 + 𝜆 ⋅ 𝐿St )

𝑝me + 𝑝mr + 𝑝A 𝜂i,HD ⋅ 𝐻u +1 𝑅G ⋅ 𝑇E ⋅ (1 + 𝜆 ⋅ 𝐿St )

Transposing to intake manifold pressure pSR leads to:

𝑝SR =

Gas constant of mixture 𝑅G = 𝜉B ⋅ 𝑅B + 𝜉L ⋅ 𝑅L =

𝑚B 1 𝑚L 𝑅m 𝑅m ⋅𝑅 = + + ⋅ ⋅ 𝑚L + 𝑚B 𝑀B 𝑚L + 𝑚B L 𝜆 ⋅ 𝐿St + 1 𝑀B

= 273.984

J kg K

With 𝑝SR it is possible to calculate the air mass flow:

𝑚󰇗B ⋅ 𝜂i,HD ⋅ 𝐻u = 𝑖 ⋅ 𝑛 ⋅ 𝑉H ⋅ 𝑝mi,HD 𝑚󰇗L = 𝑚󰇗B ⋅ 𝜆 ⋅ 𝐿St 𝑚󰇗L =

𝑝mi,HD ⋅ 𝑖 ⋅ 𝑛 ⋅ 𝑉H ⋅ 𝜆 ⋅ 𝐿St (𝑝me + 𝑝mr − (𝑝SR − 𝑝A )) ⋅ 𝑖 ⋅ 𝑛 ⋅ 𝑉H ⋅ 𝜆 ⋅ 𝐿St = 𝜂i,HD ⋅ 𝐻u 𝜂i,HD ⋅ 𝐻u

1 ⋅ 𝑅L 1 1+𝜆⋅𝐿 St

ICE II Exercise WS16/17

Exercise 10 – Engine Management

For operating point 1: 𝑝SR,1 =

2 bar + 0.9 bar + 1.1 bar kJ 0.35 ⋅ 42000 kg J 273.984

= 317 mbar

𝑝mi,HD = 2 + 0.9 − (0.317 − 1.1) = 3.683 bar+ 1 (1 ) kg K ⋅ 298 K ⋅ + 1 ⋅ 14.5 2000 3.683 bar ⋅ i ⋅ 60 ⋅ 0.002 m3 ⋅ 1 ⋅ 14.5 kg 𝑚󰇗L = = 0.0121 kJ s 0.35 ⋅ 42000 kg

For operating point 2: 𝑝SR,2 = 905 mbar

𝑝mi,HD = 10 + 1.01 − (0.905 − 1.3) = 11.405 bar 𝑚󰇗L = 0.034

kg s

b) Throttle behavior: Searched: desired steady state value of the isentropic throttle diameter

For operating point 1: Flow characteristic of throttle flap can be described with isentropic flow model: 𝑚󰇗DK = 𝑚󰇗L = 𝐴s,red ⋅ 𝑐s ⋅ 𝜌s

Index “1“: “before throttling device”  p1 and T 1

Index “2”: “after throttling device”  pSR and T E 𝜅 𝑝2 𝑐s = √2 ⋅ ⋅ 𝑅 ⋅ 𝑇1 ⋅ [1 − ( ) 𝑝1 𝜅−1 1

𝑝2 𝜅 𝜌s = 𝜌1 ⋅ ( ) 𝑝1

⇒ 𝑚󰇗L = 𝐴s,red ⋅ √

⏟

2

𝜅−1 𝜅

]

𝜅 𝑝2 𝜅 𝑝2 ⋅ [( ) − ( ) 𝑝1 𝑝1 𝜅−1

𝜅+1 𝜅

𝑝 flow coefficient 𝜓=𝑓( 2 ) 𝑝1

] ⋅ √2 ⋅ 𝑝1 ⋅ 𝜌1

2

ICE II Exercise WS16/17

Exercise 10 – Engine Management

3

Maximum value of flow rate function at critical pressure ratio:

d𝜓 𝑝 d( 2 = 0 𝑝1 ) 𝜅 𝜅−1 𝑝2 2 ⇒( ) =( = 0.528 → Chapter 10.1 in ICE2 lecture notes ) 𝑝1 krit 𝜅+1

D

air: κ =1.40

0.484

pSR / pU 𝜓max = 0.484 with κ = 1.4

0.528

1.0

With a pressure ratio smaller than 0.528 the value of the flow rate function remains constant at 0.484. Reason: The speed of sound is reached in smallest cross-section and a further increase of mass flow through increasing pressure is impossible.

𝑝2 0.317 = = 0.317 → below critical pressure ratio 𝑝1 1 Operating point 1: ⇒ 𝜓 = 0.484 𝜌1 =

𝑝1 1 ⋅ 105 kg kg = ⋅ 3 = 1.189 3 m 𝑅L ⋅ 𝑇1 287 ⋅ 293 m

⇒ 𝐴s,red =

𝑚󰇗L

𝜓 ⋅ √2 ⋅ 𝑝1 ⋅ 𝜌1

=

0.012

0.484 ⋅ √2 ⋅ 1.189 ⋅ 1 ⋅ 105

𝑝2 0.905 = = 0.905 → 𝑎𝑏𝑜𝑣𝑒 𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑟𝑎𝑡𝑖𝑜 𝑝1 1 Operating point 2:

⇒ 𝜓 = 0.2921

⇒ 𝐴s,red = 2.423 ⋅ 10−4 m2

m2 = 5.128 ⋅ 10−5 m2

ICE II Exercise WS16/17

Exercise 10 – Engine Management

4

c) Load step Searched: time until 95 % of the steady state intake pressure is reached

pSR pU,TU

TSR pA

.

mDK

.

mZyl

VSR

Under transient conditions the air mass flow into the cylinder differs from the flow passing the throttle. Mass flow balance for intake plenum: Simplification: 𝑅𝐺 = 𝑅𝐿

d𝑚SR = 𝑚󰇗L,DK − 𝑚󰇗L,Mot d𝑡

𝑚SR =

𝑝SR ⋅ 𝑉E 𝑅𝐿 ⋅ 𝑇E

d𝑝SR 𝑉SR d𝑚SR ⋅ = 𝑅𝐿 ⋅ 𝑇E d𝑡 d𝑡 Air mass flow through throttle flap: 𝑚󰇗L,DK = 𝐴s,red ⋅ 𝜓 ⋅ √2 ⋅ 𝑝1 ⋅ 𝜌1

Air mass flow into the cylinder:

𝑚󰇗G = 𝑚󰇗B + 𝑚󰇗L,Mot = 𝑚󰇗L,Mot ⋅ (1 +

𝑚󰇗G = 𝜆⏟l ⋅ =1

𝑖 ⋅ 𝑛 ⋅ 𝑉H ⋅𝑝 𝑅L ⋅ 𝑇SR SR

𝑚󰇗L,Mot = 𝜆⏟l ⋅ =1

𝑖 ⋅ 𝑛 ⋅ 𝑉H

1 𝑅L ⋅ 𝑇SR ⋅ (1 + 𝜆 ⋅ 𝐿 ) St

1 ) 𝜆 ⋅ 𝐿St ⋅ 𝑝SR

Differential equation for the pressure in the intake plenum:

ICE II Exercise WS16/17

⇒

𝑉E

𝑅L ⋅ 𝑇E

⋅

Exercise 10 – Engine Management

5

𝑖 ⋅ 𝑛 ⋅ 𝑉H 1 ⋅ 𝑝SR 𝑅L ⋅ 𝑇SR ⋅ (1 + ) 𝜆 ⋅ 𝐿St

d𝑝SR = 𝐴s,red ⋅ 𝜓 ⋅ √2 ⋅ 𝑝1 ⋅ 𝜌1 − d𝑡

Problem: flow rate function is nonlinear

Solution: Piecewise linearization of the flow rate function and piecewise solution of the differential equation Start: Operating point 1 with 𝑝SR = 0.317 bar 𝑝2

Target: 95 % of steady state pressure of OP2: 0.95∙0.905bar = 0.860 bar

𝜓i = 𝐷i + 𝑚i ⋅ Interval i

𝑝1 1. (start)

2. (end)

1. (start)

2. (end)

gradient mi

Di

pSR/p1

pSR/p1

ψ

ψ

1

0.317

0.528

0.484

0.484

0.0

0.484

2

0.528

0.6

0.484

0.479

-0.077

0.525

3

0.6

0.7

0.479

0.451

-0.273

0.642

4

0.7

0.8

0.451

0.396

-0.549

0.836

5

0.8

0.86

0.396

0.345

-0.857

1.082

0.5

Durchflussfunktion psi

0.4

0.3

0.2

0.1

0.0 0.0

0.1

0.2

0.3

0.4

0.5 p2/p1

0.6

0.7

0.8

0.9

1.0

ICE II Exercise WS16/17

Exercise 10 – Engine Management

Introduction of intermediate variables which are constant for all intervals: 𝑉E = 𝐴 ⏟ ⋅ √2 ⋅ 𝑝1 ⋅ 𝜌1 ⇒ 𝑅 ⏟L ⋅ 𝑇E d𝑝 d𝑡SR s,red ⋅ 𝐶3

𝐶1 = 3.817 ⋅ 10−7

kg s Pa

𝐶3 = 4.677 ⋅ 10−8

kg Pa

𝐶2 = 0.118 ⇒ 𝐶3 ⋅

kg s

𝐶2

𝑖 ⋅ 𝑛 ⋅ 𝑉H 1 𝑅L ⋅ 𝑇SR ⋅ (1 + ⋅𝜓−⏟ 𝜆 ⋅ 𝐿St ) ⋅ 𝑝SR 𝐶1

d𝑝SR 𝑝SR ) − 𝐶1 ⋅ 𝑝SR = 𝐶2 ⋅ (𝐷i + 𝑚i ⋅ 𝑝1 d𝑡 ⏟ 𝜓

Linear differential equation for interval „i“ 1 ⋅ d𝑝SR 𝑝SR 𝐶1 𝐶2 𝐶2 ⋅ 𝑚 ⋅ ⋅ 𝑝 ⋅ 𝐷 + − i SR i 𝐶3 𝐶3 𝑝1 𝐶3

Solution by separation of variables: ⇒ d𝑡 =

1

Further simplification for every interval: ⇒ d𝑡 =

𝐶 𝑚i 𝐶1 𝐶2 +( 2 ⋅𝑝 𝐶⏟3 ⋅ 𝐷i ⏟𝐶3 ⋅ 𝑝1 − 𝐶3 ) SR 𝐴

𝐵

⋅ d𝑝SR

d𝑥 1 1 1 𝐴 + 𝐵 ⋅ 𝑥2 ) = ⋅ ln(𝐴 + 𝐵 ⋅ 𝑥2 ) − ⋅ ln(𝐴 + 𝐵 ⋅ 𝑥1 ) = ⋅ ln ( 𝐴 + 𝐵 ⋅ 𝑥1 𝐴+𝐵⋅𝑥 𝐵 𝐵 𝐵

Solution of integral: ∫

Calculate integral for interval „i“: Δ𝑡i = 𝑡2,i − 𝑡1,i =

1 𝐴i + 𝐵i ⋅ 𝑝SR,2,i ) ⋅ ln ( 𝐵i 𝐴i + 𝐵i ⋅ 𝑝SR,1,i

That means:

𝑝SR1,i : Intake manifold pressure at start (left end) of interval

𝑝SR2,i : Intake manifold pressure at end (right end) of interval

6

ICE II Exercise WS16/17

Exercise 10 – Engine Management

7

Interval i

Ai

Bi

Δti

1

1223411

-8.161

0.0241

2

1326280

-10.108

0.0095

3

1623225

-15.058

0.0156

4

2111723

-22.036

0.0222

5

2734037

-29.815

0.0239

Total time: ∑ Δ𝑡i = 0.0953 s...