Hw1

ME781 Power tr ain D ynamic Homew or k #1

Ming Feng (Alex) Hsieh

Note: All the created Simulink blocks in this homework are in the attached file “HW1.rar”. The “ptsim3.mdl” and “Ptpara3.m” has been modified as “hw1.mdl” and “hw1m.m” according to the homework problems. The additional files “Pa.mdl” is the partial pressure for air block for Problem 1.5 and “EGR_Disturbance.mdl” is the block for generating step disturbance in EGR for Problem 1.5.

Problem 1.1

(1) According to the Dalton’s law and the ideal gas law, the mass flow rates of air and EGR can be individually derived from their pressures. By Delton’s law, the following mass flow diagram can be derived.

 ai m

Manifold (air) Pa, Vm, Ma

 ao m

dma ,m,air = m ai − m a 0 dt ma ,m,air = mass of air in the intake manifold According to the ideal gas law, PV M ma ,m,air = a m a RuT Combining two equations, dma ,m ,air V M P P d PV M = ( a m a ) = m a ( a − a2 T ) dt dt Ru T Ru T T PV M V M m a 0 = m ai + a m 2 a T − m a Pa Ru T Ru T (2) The same as the flow of air, the mass flow diagram of EGR is shown below

m egri dma ,m ,egr dt

Manifold (EGR) Pa, Vm, Megr

= m egri − m egr 0

ma ,m,egr = mass of EGR in the intake manifold According to the ideal gas law,

m egro

ma ,m ,egr =

( P − Pa )Vm M egr Ru T

Combining the two equations, dma ,m,egr d ( P − Pa )Vm M egr Vm M egr ( P − Pa ) ( P − Pa )  = ( )= ( − T) dt dt RuT Ru T T2 m egro = m egri +

( P − Pa )Vm M egr  Vm M egr T− ( P − Pa ) RuT 2 RuT

Problem 1.4

Combine (1) and (2) and (5) and apply the condition M egr = M a = M Combining (1) and (2): m ai − m ao + m egr i − m egro =

Vm M  PaVm M  Vm M   ( P − Pa )Vm M  Pa − T+ ( P − Pa ) − T 2 RuT RuT RuT RuT 2

Applying (5): m ai − m egro

V M PVm M  P + m egr i − m egro = m P − T P − Pa Ru T Ru T 2

 ( P − Pa )Vm M egr V M P  − m egr ( P − P )) = (m − Vm M ( P − PT ) + m (m egri + T a ai P − Pa Ru T Ru T T Ru T 2 Rearranging Pa , the equation of Pa is obtained. RT RT P Pa = Pa ( − u (m ai + m egr i )) + u m ai P Vm MP Vm M Converting to the crank angle domain, d d dθ d ω = =ω dt dθ dt dθ ; = angular velocity RT RT P ωPa = Pa ( ω − u (m ai + m egr i )) + u m ai P Vm MP Vm M P RuT RT Pa = Pa ( − (m ai + m egr i )) + u m ai P Vm MPω Vm Mω

Problem 1.5

The Pa block is shown below:

The inner connection of the Pa block is:

(1) Run the simulation with no EGR, the simulation results of Pm and Pa are shown below:

Combining the two pictures into one, we can see that the two pressures Pa and Pm are overlapped.

Subtract Pa from Pm , it is obvious that they are almost the same beside some computation error.

It can be concluded that since no EGR is applied, all gas in the manifold is air. By which the partial pressure of air will the same with the total pressure in the manifold. The simulation results verify the supposition. (2) Introducing 20% step disturbance in EGR. The 20% EGR disturbance is introduced during crank angle 1000~1500 as shown below.

The Pa and Pm plots are shown below:

Comparing the two pressures, as shown in the following two pictures, it is obviouse that the partial pressure of air Pa is smaller than the total manifold pressure Pm when the EGR is introduced during 1000~1500 because of the incoming recirculated gas. It is apparent that the difference pressure is the EGR pressure.

And the engine torque from the simulation is shown below.

Comparing with the no EGR condition, the following pictures show the differences.

It is obviously that under the same conditions, the same spark, throttle…etc., when the EGR is introduced, the manifold pressure and partial pressure are decrease. However, the engine torque is obviously increased (the two curves are overlapped besides crank angle from 1000 to 1500 where EGR is triggered, the 20% EGR disturbance cause the torque to increase from 75 to 3500). When looking at the “Combustion” block in the PTSIM, we can find that the increased torque is due the recirculated gas (input2 “EGR”).

Problem 1.6

The flowing diagram is the flowchart for idle speed control.

Loadings

Tload

ISC Valve

Air flow, m isc

Intake Manifold

Air flow, mcyl ,i

Combistion

Torque, T eng

Crankshaft

EGR, megr Temperature, T

Pressure, P

Idle Speed air mass flow rate control Yiscontrol (t) Spark Control

Engine Speed, ω

Controller

Crank Angle, θ

Fuel Injection Control EGR Control, Megr

Equations according to each block in the diagram will be introduced in the following. (1) ISC Valve The ISC valve equation can be expressed as: m isc = f1 (Yisctronol , T , P ) m isc = output of the ISC valve block, mass flow rate of air into the intake manifold T = intake air temperature, can be measure P = is the intake air pressure, can be measured Yisctronol = input of the ISC valve block, flow rate command from the controller.

The ISC valve receives the flow rate control command Yisctronol from the controller and than transform the command to the mechanical movement to control the air flow rate m isc into the manifold. And of course the final output mass m isc will be affected by different temperature and pressure according to the ideal gas law. The equation can be derived either from analytical method or experimental data. (2) Intake Manifold Under the assumption that the molecular weights of the intake air and EGR are the same, the intake manifold equation can be expressed as:

dPm η vVd N RT RT + Pm = (m isc + m egr ) m + m dt 2Vm Vm Tm Pm = output of the equation, pressure of the manifold

η v = engine breathing efficiency, can be obtained either from analytical method or experimental data Vd = displacement volume of the engine, 3.8L in this case N = engine speed (revolution per second), time related parameter, can be measured from

sensor Vm = intake manifold volume, depends on the car spec m isc = mass flow rate of air from ISC valve, input of the equation m egr = mass flow rate of EGR, input of the equation R = gas constant Tm = manifold temperature, can be measured or analytically derived according to the intake air temperature T and temperature of EGR, it is a variable in this case instead of constant because of the EGR Since the idle speed control is to control speed with small perturbation near the idle engine speed N 0 . The equation can also be linearized as:

dδPm RT RT = C1 N 0δPm + C1 P0δN + C2T0δu + (C2u0 + 0 − 20 )δTm dt To To C1 = − C2 =

η vVd 2Vm

R Vm

u = m isc + m egr

N 0 = engine idle speed (revolution per second) P0 = manifold pressure at idle speed T0 = manifold temperature at idle speed u0 =intake mass flow rate at idle speed According to the above equation the manifold pressure Pm can be obtained. By the ideal gas law, the mass flow rate of air into each cylinder m c yl ,i , the output the intake manifold block, can be obtained. η V NP m c yl ,i = v d m 2 RTm (3) Combustion The combustion module transform the intake air mc yl ,i , according to the spark control Vspark , fuel injection control minj , EGR control M egr , and engine speed ω , to the engine torque Teng . The equation can be expressed as: Teng = f (mc yl ,i , Vspark , minj , M egr , ω )

(4) Crankshaft The crankshaft transfers the engine torque Teng to the engine rotation θ , which can also be transferred to engine speed ω and engine acceleration α . dω Teng = Tload + J + Bω dt Tload = torque loading J = equivalent engine inertia

B = damping coefficient ω = engine speed= θ By Laplace transform, the engine rotation can be obtained:

1 (Teng − Tload ) Js + Bs ω = θ

θ=

2

(5) Controller The controller receives signals from sensors, ω , θ ,…etc., and than, according to the idle speed control algorithm, generates the ISC valve control signal Yisctronol to ISC valve and spark control Vspark , injection control minj , and EGR control M egr to other combustion components. The control functions can be expressed as: Yiscontrol = f 2 (ω , θ ) Vspark = f 3 (ω , θ ) minj = f 4 (ω , θ ) M egr = f 5 (ω , θ )