151-0567-00 Engine Systems (HS 2016) Exercise 6 Topic: Optional Exercises Raffi Hedinger (hraffael@ethz.ch), Norbert Zsiga (nzsiga@ethz.ch); November 28, 2016 Problem 1 (ECU Priority) Use the information on p. 152 of the book to assign appropriate clock rates to the task listed below. Differentiate between time and segmental controlled tasks. Ignition Injection Drive by wire throttle (electronic throttle control) EGR valve Waste gate Electronically actuated thermostat of the cooling system Adaption of the injection valve constant to aging (volume/ms opening time) Air mass flow sensor evaluation (summation) λ signal evaluation Knock detection Air conditioning Solution 1 The following abreviations are used below: S = segmental controlled; 0.1/10/100/1000 = time in [ms]. S Ignition S Injection S Drive by wire throttle (electronic throttle control) 10 EGR valve 10 Waste gate 100 Electronically actuated thermostat of the cooling system 1000 Adaption of the injection valve constant to aging (volume/ms opening time) S Air mass flow sensor evaluation (summation) S λ signal evaluation 0.1 Knock detection 1000 Air conditioning
Problem 2 (Ignition Angle Control) Mark the following points and regions on the curve in Fig. 1: The optimal ignition angle The region where knocking occurs The ignition angle when the engine is in idling mode T [Nm] ζ [ ] Before TDC TDC Figure 1: Qualitative torque production as a function of the ignition angle Solution 2 T [Nm] ζ opt ζidle ζ [ ] Before TDC TDC Figure 2: Qualitative torque production as a function of the ignition angle The gray area in Fig. 2 represents the region where knocking occurs. (It is only the general trend, it could also start slightly before or after the torque-optimal ignition angle.) To prevent
knock in critical operating regions, the spark timing is delayed from the position where it provides maximum brake torque. This reduces the peak pressure and temperature in the cylinder by moving the peak of the burn-rate trajectory beyond the top dead center which causes the combustion to burn into the expansion stroke leading to a sacrifice of efficiency! The ignition angle at which the torque production is the highest often coincides with the beginning of the knocking region or as some people put it: Slight knocking is the sound of fuel economy. In the idling mode, a part of this fuel economy is sacrificed for the sake of robustness: The ignition channel is very fast and permits an almost immediate reaction to load disturbances. Shifting the ignition angle slightly towards TDC allows for a reaction to more or less load torque, which is desirable. Problem 3 (EGR) a) EGR is usually not applied at full load, why? b) Why is an EGR cooler useful? c) Describe the influence of a higher EGR rate on the thermodynamic efficiency of both, SI and CI engines. Solution 3 a) The main reason is the resulting power losses resulting from EGR at full load in SI engines. (Since the restriction λ = 1 must hold to fulfill governmental legislations, introducing inert gases (EGR) leads to less fuel injected and to power losses eventually.) In CI engines the smoke-limit (λ < 1.3) is reached if EGR is used under full load conditions. This could be handled introducing a second turbocharger which would increase the available air and would therefore allow EGR to be used under full load conditions as well. b) The NO x emissions are sensitive to changes in the combustion starting temperature. Cooling the EGR (similar to the supercharged air) reduces the NO x emissions. c) Additional (external or internal) EGR reduces the formation of NO x, but negatively affects the thermodynamic efficiency of SI engines. In fact, increased EGR rates lead to slower burning speeds and, therefore, to reduced factors e egr (x egr ) (see the definition of the thermodynamic efficiency on p. 72 of the class textbook). In Diesel engines the EGR rate has - with the exception of very large ratios - almost no effect on the efficiency. The Diesel-specific diffusion-controlled burning mechanism and the always abundant oxygen render this type of combustion much more tolerant to high EGR, which is one of the reasons why EGR in modern Diesel engines is a standard feature. However, as stated in the introduction of the book, higher EGR rates lead to increased particulate matter formation, such that this trade-off is often an important limiting factor in Diesel engines. Problem 4 (Control-Oriented Models (COM)) Why are control-oriented models used to derive pollutant formation models? Solution 4 Pollutant formation in internal combustion engines is rather difficult to predict theoretically or by numerical simulation. Control-oriented models, therefore, often rely on the results of experiments which are summarized in appropriate maps. Further information can be found on p. 98 of the class textbook.
Problem 5 (Lambda) a) Explain how the λ value is measured in real cars. b) Derive a characteristic line (gradient) of a LSF-λ-sensor to determine the λ value with an accuracy of at least 0.1%. (Assign the appropriate x-values in Fig. 3.) Assume: 50 mv measurement inaccuracy Perfect control 800 mv (λ 1) and 200 mv (λ 1) Figure 3: Characteristic Line of an LSF. Solution 5 a) Usually, two lambda sensors are used in modern cars. One is placed between the engine and the catalytic converter (LSU type) and one after the catalytic converter (LSF type) as shown in Fig.??. This results in an accuracy in measuring the λ-value of 0.01%. LSU: This sensor-type has a linear output between a minimum and a maximum value that represent λ values of 0.65 and (pure air) respectively. LSF: This sensor-type has a non-linear output that switches between two thresholds that represent either lean or rich combustion. b) A change of 50 mv in the signal should not change the λ value more than 0.1%. This means that a difference of 600 mv must not result in a change of more than 1.2% of the λ value. Fig 4 shows a possible characteristic line. In real LSF-sensors, the gradient is even steeper to reach an accuracy of 0.01%.
Figure 4: Characteristic line of an LSF to reach a measurement accuracy of 0.1% with a measurement inaccuracy of 50 mv. Problem 6 (Feedforward Control) Feedforward control can be made as fast as technically possible, it will never render the system unstable (unless you add an unstable pole). Feedback control, however, is tricky because the feedback s zeros become new poles. Use Fig. 5 to derive the transfer function r y and analyze the poles of the a) feedforward loop b) feedback loop r + FF FB + P y Figure 5: Overview of a Feedforward and Feedback Control Loop Solution 6 Using Fig. 5, the system output y can be calculated as: y = P (F B (r y) + F F r) (1) The transfer function T (s) = y r finally results as: y r = P F B + P F F I + P F B (2)
a) There are only the poles of the open loop transfer function P F F since F B = 0. b) There are new poles I + P F B which can render the system unstable. Problem 7 (Minimum Idling-Mode Fuel Consumption) You are given the (idealized) efficiency map of a naturally aspirated 2.5-liter SI engine. You know that the efficiency map shown below was created using an affine relationship between the injected fuel mass per engine cycle and the resulting engine torque. Given the following data: the engine s idle speed is 700 [rpm]; the engine is a 5-cylinder engine; the bore (diameter) and the stroke of the cylinders are equal; calculate the minimum fuel consumption in [ l h] when the engine is idling. For the minimum fuel consumption, assume no auxiliary devices are in use which means that the engine does not produce any mean torque. Figure 6: Iso efficiency lines of a modeled naturally aspirated 2.5-liter SI engine.
Solution 7 1. Calculate the mean cylinder speed c m at the given idle speed ω e,idle = 700 [rpm]. c m = S ω e,idle. We need to determine the stroke S of the cylinder which is defined as: ( ) S2 2.5 4 S = 10 3 5 ( ) 4 0.5 10 3 (1/3) S = = 0.086 [m] Inserting into the definition of the mean cylinder speed, we get: c m = S ω e,idle 0.086 (700/30 ) = [ m ] = 2 s 2. Use the straight Willans lines in order to get the intersection with the p mϕ -axis: Read out of the map, at c m = 2 [m/s]: p me η e p mϕ 6 0.33 18 2 0.25 8 From which we can get the value of p mϕ when p me is zero: p mϕ,idle = 3 [bar]. 3. Calculate m ϕ,idle, the amount of fuel injected in one engine cycle: p mϕ,idle = H l m ϕ,idle V d m ϕ,idle = 3 105 2.5 10 3 42.5 10 6 = 1.76 10 5 [kg] 4. Calculate ṁ ϕ,idle, the fuel mass flow in [l/s]: ṁ ϕ,idle = m ϕ,idle ωe,idle 4 = 1.76 10 5 700/30 [ ] 4 kg = 1.03 10 4 s 5. Convert [kg/s] to [l/h]: Gasoline has a density of 0.75 [kg/l] and we finally get: ṁ ϕ,idle = 1.03 10 4 3600 [ ] 0.75 l = 0.49 h