5-0567-00 Engine Systems (HS 205) Exercise 3 Topic: Lectures 3+4 Raffi Hedinger (hraffael@ethz.ch), Norbert Zsiga (nzsiga@ethz.ch); October 9, 205 Problem (Willans Approximation) A useful simplification of the engine s torque and efficiency characteristics can often be made according to a so called Willans Approximation. a) What does the break mean effective pressure p me represent? Derive it s mathematical expression (Eq. (2.94) of the book). b) What does the fuel mean effective pressure p mϕ represent? Derive it s mathematical expression (Eq. (2.95) of the book). c) What is the advantage of using the normalized expressions p me and p mϕ when describing engine properties? d) You are given the Willans Line Parameters (p me0 and e) for a defined (gasoline) engine speed, and a certain load torque T e (in Nm) that acts on the crankshaft. You further have knowledge of the engine s displacement volume. How can you determine the corresponding engine efficiency graphically? e) Using the same Willans Line, how can you determine the idling consumption (in liter/s)? Solution a) The break mean effective pressure p me represents a normalized torque provided by the engine. It is the pressure that has to act on the piston during one full expansion stroke to produce the same amount of work as the real engine does in two engine revolutions (assuming a four-stroke engine). The mathematical expression of p me is derived as follows: p me dv = expansion stroke engine cycle p me = T e 4π T e dϕ p me = const., T e = const. b) The fuel mean effective pressure p mϕ represents a normalized amount of fuel burnt during one engine cycle. It is the same amount of energy as the chemical energy stemming from the fuel. Assuming an engine with an efficiency of (00%), p me and p mϕ would take on the same value. However, since in reality this is never the case, the value of p me is substantially lower and the overall engine efficiency η e can be expressed as the ratio pme p mϕ. The mathematical expression is derived similar to the one of p me : p mϕ dv = ṁ ϕ H l dϕ p mϕ = const. expansion stroke engine cycle p mϕ = m ϕ H l, where ṁ ϕ is the fuel mass flow and m ϕ the total fuel mass injected in one engine cycle.
c) Using the normalized expressions p me and p mϕ allows for a direct comparison of engines with different displacement volumes. Working with either Diesel or gasoline engines, the values of e and p me0 are comparable. d) Given T e and, p me,given = Te 4π results. The value of p mϕ,graph can be determined graphically. The engine efficiency finally results as p me,given p mϕ,graph. p me p me,given 2 0 p mϕ,idle 3 e p mϕ,graph. p mϕ p me0 Affine Willans Model of the Engine Efficiency. e) Extracting the value of p mϕ,idle (where the Willans line crosses the p mϕ -axis), the fuel consumption in liter/s results as p mϕ,idle, where ρ gasoline is the density of gasoline H l ρ gasoline ωe,idle 4π ( 0.75 kg ) and H dm 3 l is the lower heating value of gasoline ( 42.5 MJ kg, remember these values for the exam! ). Problem 2 (Maximum Value the Break Mean Effective Pressure p me ) a) Approximate the maximum value of p me of a naturally aspirated SI engine. b) Why is it not that easy to determine the maximum p me of a turbocharged engine? Hint: Which parameters in the equations you used in subproblem (a) need to be changed? Solution 2 a) The maximum value of the break mean effective pressure p me of a naturally aspirated SI engine can be approximated using the relation η e = pme p mϕ and inserting reasonable values
for the parmeters: p me,max = η e,max p mϕ,max = η e,max Hl m ϕ = η e,max Hl mair λ σ 0 = η e,max Hl Vd ρ air λ σ 0 4.25e7.2 0.36 4.7 = 2.5 [bar] As shown in the calculation above, the maximum p me value is independent of the engine s displacement volume. This is the key point why the torque and fuel mass flow are normalized: Engines of different sizes are made comparable! b) To determine the maximum p me of a turbocharged engine, the same approach as in subproblem (a) can be used. However, it is not possible to accurately determine the density of the air in the intake manifold ρ a since it depends on the maximum boost ratio Π e,max (Assuming a general upper value of 2.5 would be reasonable.). Problem 3 (Maximum Torque of a Naturally Aspirated Engine) In most modern, naturally aspirated engines the maximum torque is closely proportional to the displacement volume. a) Validate the statement above by explaining how an engine works. b) Validate the statement above mathematically, assuming an engine can be modelled using a Willans Approach. Solution 3 a) The amount of air in the cylinder of a naturally aspirated engine is determined by the pumping properties of the engine itself. These naturally aspirated engines (both CI and SI) are limited in their power by their volumetric efficiency. Consequently, in most modern engines the maximum torque is closely proportional to the swept volume. b) Using a Willans Approximation, this can be visualized as follows: p me,max = e p mϕ,max p me0 T max 4π = e Hl m ϕ,max p me0 T max 4π = e Hl Vd ρ air p me0 λ σ ( 0 e Hl ρ air p me0 λ σ 0 T max = 4π ), only for λ
According to the equation above, the maximum torque produced by an engine is proportional to its displacement volume. (The influence of the displacement volume on the engine losses p me0 is neglected in the equations above which is a good approximation when p mϕ = p mϕ,max.) Problem 4 (Engine Effiency Map) a) Use a Willans Approximation to draw a map of the the iso-efficiency lines of a naturally aspirated 4-cylinder, 2-liter SI engine. Make the following simplifications: The temperature of the engine corresponds to the reference temperature ϑ. Since the engine is naturally aspirated the maximum boost ratio Π e,max in Eq. (2.2) is equal to. The engine always runs at λ =. The ignition angle corresponds to the optimal (maximum break torque) ignition angle ζ 0. EGR is not considered. The volumetric efficiency λ l is equal to. The engine s bore and its stroke have the same lengths. Use the maximum p me curve of Fig. 2.27 of the book. Make appropriate assumptions for all parameters that are not mentioned in this list. When modelling p me0f according to Eq. (2.2), set the value of the speed dependent parameter k 3 to 2 of the one suggested in the book. Eq. (2.2) is a good average model for a wide range of engines, however, in this specific case, a smaller value fits better. b) Compare your result to Fig. 2.27 on p. 66 of the book. Point out the most obvious difference between your map and Fig. 2.27. Can you find a reasonable explanation? Hint: Which one of the simplifications above does hold over the whole speed and torque range in common engines? (Of course there is not only one that does not, but you will get a decent solution adjusting only one of them!) It is helpful to draw the Willans Lines of the map in Fig. 2.27 for specific piston speeds (e.g. c m = {6, 0, 4}) and compare them to the ones of your model! The following Matlab commands might be useful: meshgrid, contour, polyval, interp Solution 4 a) The map in Fig. 2 was created using Eqs. (2.97), (2.00), (2.03), (2.07), (2.09) and (2.2), assuming a thermodynamic efficiency e(ω e ) according to Fig. 2.32. Eq. (2.00) was solved for p mϕ = f(ω e, p me ), which was then used to plot the engine efficiency η e = pme p mϕ.
Figure : Iso efficiency lines of a modelled naturally aspirated 2-liter SI engine. b) The main difference when comparing Fig. 2 to Fig. 2.27 of the book is that Fig. 2 has its maximum efficiencies always at the maximum p me boundary for any given piston speed c m. This makes perfect sense for a model with straight Willans Lines: According to Fig. 2.29 of the book, the relative influence of the engine losses p me0 drops with increasing p mϕ and thus the efficiency must increase as well. In real engines, however, the efficiency is not at the maximum torque line but slightly below it. The reason for this is the use of fuel enrichment at high torques. Fuel enrichment, i.e., running the engine at air/fuel ratios lower than stoichiometry introduces additional mass into the cylinder, increasing the total heat capacity and therefore lowering the overall temperature level. In addition, the evaporation withdraws heat (the evaporation enthalpy) from the cylinder charge. Fuel enrichment is used to increase the full-load power and cool the engine at high loads. Cooling the engine at low speeds helps prevent knocking (unwanted self-ignition of the injected gazoline due to high temperatures - this is more likely to happen at low speeds because there is more time available). Furthermore, in old engines, cooling was necessary at high speeds because the allowed temperature levels of old TWCs were a restricting factor. New TWCs can work at much higher temperatures and therefore fuel enrichment at high speeds is not mandatory anymore. Fig.?? compares the Willans Lines of both the model and the real engine map. It reveals a very good match between model and reality (which is also achieved by changing the factor k 3 in Eq. (2.2)) for low to medium load torques and clarifies the section above: The Willans Lines of the real engine decrease at high loads with respect to the model s Willans Lines. With knowledge of the amount of fuel enrichment as a function of p the engine speed and the ratio me p me,max, this behavior could be modeled using an approach according to Fig. 2.33 of the book.
Figure 2: Iso efficiency lines of a modeled naturally aspirated 2-liter SI engine. Problem 5 (Engine Efficiency) Why is the overall efficiency of SI engines in part load conditions substantially lower than the one of CI engines? Solution 5 The torque generated by an SI engine is air-controlled. The throttle is actuated such that the pressure in the inlet manifold results in the desired amount of air induced into the combustion chamber. In contrast, the torque generation of a CI engine is fuel-controlled. The air massflow into the cylinder is specific to the engine speed whereas the amount of injected fuel is defined by the driver via the gas pedal. Therefore the pumping losses p me0g as defined in equation (2.0) on page 69 of the book do not occur in that case. The relative influence of the pumping losses on the engine efficiency is higher in part load conditions and therefore SI engines perform inferior. Problem 6 (Thermodynamic Efficiency) Explain why the thermodynamic efficiencies of both SI and CI engines (under nominal conditions) drop at high as well as at low engine speeds. Solution 6 At very low speeds, the relatively large heat losses through the wall reduce engine efficiency, while at very high speeds, the combustion times become unfavorably large compared to the available interval in the expansion stroke.
Problem 7 (Moore-Greitzer Surge Model) The Moore-Greitzer model is used to examine the non-linear phenomena of surge in a compressor. Based on first principles, the model approximates the extended compressor s characteristic that defines the pressure-ratio to mass-flow relation beyond the surge limit. a) Draw the pressure-ratio to mass-flow characteristics of a centrifugal compressor with a constant speed. Draw the same characteristics of a throttle in the same figure. The throttle should be drawn for two different configurations. In the first case the throttle area is supposed to be large enough to maintain a stable operation, whereas in the second case the throttle area is too small and leads to an unstable equilibrium of the system. b) Explain qualitatively why the equilibrium points are stable/unstable. c) Analyze the stability of these two equilibria determining the eigenvalues of the linearized system. Assume that the throttle is operated in a non-choked condition and use the following parameters: Π c,min :.5 Π c, : 0.5 µ c0 : 0.2 kg/s τ c : 0.00 s τ m : 0.00 s ϑ : 400 K R : 287 J/kg.K p in : 0 5 Pa A alpha : 0.002 m2 c d : The Moore-Greitzer model consists of the following differential equations: d dt µ c(t) = [ Π c ( µ c (t)) Π c (t)] τ c () d dt Π c(t) = [ µ c (t) µ t (Π c (t), A α (t))] τ m (2) with the approximation of the pressure ratio across the compressor and the throttle mass-flow Π c ( µ c ) = Π c,min + Π c, [ 2 + 3 2 ( µ c µ c,0 2 ) 2( µ c µ c,0 2 )3 ] (3) µ t (Π c (t), A α (t)) = { cd A α (t) c d A α (t) p in R ϑin 2 if Π c < 2 p in R ϑin 2 Π c [ Π c ] if Π c 2
Solution 7 Π c Π t Throttle 2 Throttle Equilibria 0 µ t µ c Figure 3: Extended compressor map shown with two throttle with different throttle areas. a) Figure 3 shows the compressor characteristic for a fixed speed and additionally the characteristics of two independent throttles. The steady state operating points are defined by the intersection of compressor and throttle characteristics, hence by the points where the pressure rise across the compressor equals the pressure loss across the throttle. To avoid any confusion, please consider Figure 2.4 in the book that displays the three main components of the model (compressor, receiver and throttle). Note that in this simplified Moore-Greitzer model the turbine (of the turbocharger) is approximated by a throttle. Therefore the term throttle corresponds in this context not to the throttle positioned upstream of the intake manifold, but to the turbine. b) Case : Compressor is running at a stable operating point (first equilibrium). Suppose that due to disturbances the throttle mass-flow µ t is increased. The higher outgoing throttle mass-flow leads to a decreased compressor pressure-ratio Π c of the plenum. Since the compressor characteristic has a negative slope around the first equilibrium, the decreased pressure-ratio goes along with an increased compressor mass-flow µ c. A bigger compressor mass-flow on the other hand brings the compressor pressure-ratio Π c back to a higher value. Therefore we can conclude qualitatively that equilibrium is stable since the compressor returns to operate at this point after small disturbances (this is referred to as a negative feedback). Case 2: Compressor is running at an unstable operating point (second equilibrium). Again suppose that due to disturbances the throttle mass-flow µ t is increased. The higher outgoing throttle mass-flow leads to a decreased compressor pressureratio Π c. The positive slope of the compressor characteristic at this operating point causes the compressor mass-flow µ c to decrease as well. The lower compressor mass-flow yields again lower pressure-ratio. In the region of equilibrium 2, small disturbances lead to an unstable behavior of the pressure-ratio and mass-flow, i.e. they decrease or increase without returning to initial condition (this is a positive feedback).
c) The equilibria by definition represent steady states and can therefore be found with Π c = Π c ( µ c) µ c = µ t (Π c). Since the two equations are coupled, there is no easy way to find the steady state points analytically. See the provided MATLAB code to get an idea of how this can be solved by an iterative method as fminsearch. Starting from a general nonlinear system ẋ = f(x), the linearized system will have the form ẏ = Ay, where A = f x x=x is the Jacobian of the system at the equilibrium x. Setting x = µ c and x 2 = Π c and using the RHS of the two model equations () and (2) as f and f 2, respectively, we obtain A = [ f f x x 2 f 2 f 2 x x 2 ] x=x = [ τ c Π c µ c τ c τ m τ m µt Π c ] µc= µ c, Π c=π c, where Π c µ c = 3 ( 2 Πc, 6 Πc, µ c ) 2 µ c,0 µ c,0 µ c,0 2 ( µ t p in Π c = c d A α 2 R θ 2 Π ) ( 2 c Π 2 c Π c 2 + 4 ) Π c 3. The eigenvalues of A can be found using Matlab. computed as For the first equilibrium they are λ =.46e4 λ 2 = 0.05e4. Since both eigenvalues are negative we can conclude that the steady-state operation of the compressor with a throttle area of 20e 4m 2 leads to a stable behavior. This result agrees with the fact, that the critical throttle area that denotes the transition from a stable compressor operating point to a compressor state where an attracting limit-cycle occurs (i.e. surge), was found to be equal to 9.58e 4m 2.