Fuel and Air Flow in the Cylinder

Chapter 6 Fuel and Air Flow in the Cylinder 6.1) A four cylinder four stroke 3.0 L port-injected spark ignition engine is running at 00 rpm on a stoichiometric mix of octane and standard air at 100 kpa and 98 K. If the average octane flowrate is 3.0 g/s, a.) what is the mass of fuel entering each cylinder per cycle? b.) what is the volumetric efficiency, and c.) assuming complete combustion, what is the rate of heat release? a) The mass of fuel entering each cylinder per cycle for a four stroke engine is: ( )( ) ( )( ) 1 60 1 m f = ṁ f = (3.0) N 00 4 n c m f = 4.09 10 g b) The volumetric efficiency can be calculated knowing the stoichiometric air-fuel ratio A s = 15.03 (Table 3.5). Since the engine is port injected we obtain: e v = ṁa +ṁ f ρ i v d N We know that ρ i = P i RT I = Substituting back in = ṁf (AF s +1) ρ i v d N 100 (0.87)(98) = 1.17kg /m 3 (3)(15.03+1) e v = (1170)(3.0 10 3 ) ( ) 00 60 e v = 0.74 c) The lower heat of combustion q lhc = 44.43 MJ /kg (Table 4.1) Q = ṁ f q lhc = (3.0 10 3 )(44,430) Q = 133kW 1

CHAPTER 6. FUEL AND AIR FLOW IN THE CYLINDER 6.) A carburetor has a pressure drop of 0.05 bar and a fuel-air ratio FA = 0.06 at a demand D c = 0.4. a.) What is the fuel-air effective area ratio A f /A a? b.) If the demand changes to D c = 0.6, what is the change in the fuel-air ratio FA? The fuel-air ratio FA is given by Equation 6.5 FA = ṁf ṁ a = ( 1.73 D c )( )( ρf Af ρ A a )[ (P P ) (ρ f )(c ) a) Therefore, using reference values of Figure 6.3 ( ) A F FA = (D c ) A a 1.73 A F =.90 10 3 A a ( ρ ρ f ] 1/ )[ ρf (c ) ] 1/ ( ) ( )[ 0.06 1.17 (749)(346) = (0.4) (P P ) 1.73 749 (500) b) If the demand D c changes from 0.4 to 0.6, the pressure ratio P /P will change. The demand is: D c = 3.86 [ ( P P ) 1.43 ( ) ] 1.71 1/ P P The following Matlab root finding program is used to solve for P /P = 0.9038 function [ y ] = demand( pr ) % calculates roots of the demand equation (6.6) dc=0.6; y=dc-3.86*(pr.^1.43 - pr.^1.71)^(1/); end and in the command line z=fzero( demand, [.80.99]) z = 0.9038 The carburetor pressure drops so P P = P (1 P /P ) = (98,700)(1 0.9038) = 9495Pa Assuming AF /A a is constant, FA = ( 1.73 0.6 FA = 0.0778 This is an increase of about 30% )( ) [ ] 1/ 7.49 9495 (.90 10 3 ) 1.17 (749)(346) ] 1/

3 6.3) Carburetor venturis are sized assuming the maximum quasi-steady flow during the intake stroke is twice the average. Estimate the venturi throat diameter required for a four cylinder 5.0 L engine with a volumetric efficiency of 0.9, and maximum speed of 6000 rpm. State clearly the assumptions you need to make. Assume a demand D c = 1, so the maximum air flow is choked ṁ a,max = ṁ a,cv = (ρ )(A a )(c ) Rearranging one obtains A a = 1.73 = 0.579(ρ )(A a )(c ) (ṁa,max ρ c ) ( ) (γ+1) (γ 1) γ +1 Since the maximum intake flow is assumed to be twice the average ( ) ṁa,intake A a = 1.73 ρ c The airflow rate into the engine is found from the volumetric efficiency assuming direct injection into the cylinders, so ( ) N ṁ a,max = (e v )(ρ )(v d ) Assuming a four stroke cycle and a single carburetor with four cylinders there will be continuous intake flow through the carburetor serving the four cylinders so ṁ a,intake = ṁ a and ( ) N A a = (1.73)(e v )(v d ) c Since ( π (d A a = (C d ) 4) ) d = [ 4 π (1.73)(e ]1 v)(v d )(N) [4 = (C d )(c ) d = 6.18cm π (1.73)(0.9)(5 10 3 ) (0.75)(346) ( 6000 60 ) ]1 = 6.18 10 m Note that the diameter is proportional to the square root of the engine displacement and speed.

4 CHAPTER 6. FUEL AND AIR FLOW IN THE CYLINDER 6.4) What is the crank angle injection duration (deg) needed for a fuel injector in a single cylinder diesel engine operating at 1500 rpm with a bsfc of 0. kg/kwh so that it produces 50 kw? The injector effective area A f is 1.0 mm and pressure is 550 bar. The cylinder pressure is 50 bar. The injection duration (Equation 6.10) is given by: t = m f (ρ f P) 1 / A f and the crank angle function θ is θ = 360 N t The cycle average fuel consumption rate is ( ) m f = bsfc W 50 b = (0.) = 0.183 60 kg /min The mass of fuel injected per cycle is m f = m f ( N /) = 0.183 ( 1500 /) =.44 10 4 kg So the injection duration is t =.44 10 4 ( 840 500 10 5 ) 1 / (1 10 6 ) = 8.41 10 4 s The crank angle duration is θ = (360)(N)( t) = 360 θ = 7.5degrees ( ) 1500 (8.41e 4) 60

5 6.5) If the diesel injector in problem (6.4) has a nozzle diameter d n of 0.30 mm, what is the spray angle? Assume the cylinder temperature is 800 K. The spray angle is given by Equation 6.13 tan θ = 4π ( ) 1/ ( ) ρg 3 A 6 ρ l The cylinder gas density ρ g during injection is ρ g = P RT = 5000 (0.87)(800) = 1.8kg/m3 ( ) Ln A = 3.0+0.8 = 3.0+0.8 d n tan θ = 4π ( ) 1/ ( ) 1.8 3 = 0.149 3.93 840 6 θ = 16.9degrees ( 1.0 0.3 ) = 3.93 As the cylinder pressure and temperature increase during compression, the corresponding increase in gas viscosity will increase the spreading angle.

6 CHAPTER 6. FUEL AND AIR FLOW IN THE CYLINDER 6.6) For the engine fuel injector in problem (6.5), what is the spray tip penetration versus time? Determine the time and crank angle interval for the fuel to reach the cylinder wall. The engine bore is 150 mm. Assume a centrally located injector. The spray tip penetration is given by Equation 6.17 ( P S = 3.07 ρ g From the previous problem, T g = 800K P g = 5000kPa ) 1/4 (td n ) 1/ ( 94 T g ) 1/4 ρ g = P RT = 5000 (0.87)(800) = 1.8kg /m 3 ( ) 500 10 5 1/4 ( ) 1/4 94 S = 3.07 (0.3 10 3 ) 1/ (t) 1/ 1.8 800 Assuming a centrally located injector, then s = b ( ) ( ) b t = 1 150 10 3 = =.17 10 3 s 1.61 1.61 The crank angle interval is θ = 360 N T = 360 θ = 19.5degrees ( ) 1500 (.17 10 3 ) 60

7 6.7) A diesel fuel injector has a total orifice area of 0.15 mm with an average C d = 0.60. If the average pressure difference between the fuel injector and cylinder is 400 bar, a.) what is the amount of diesel fuel injected over a 8 ms period?. b.) If the engine has a 50% thermal efficiency and operates at 1500 rpm, estimate the power (kw) produced by the engine. a) The mass of fuel injected over a time t is m f = (ρ f P) 1 (C D )(A f )( t) = ( 840 400 10 5)1 (0.6) ( 0.15 10 6)( 8 10 3) m f = 1.87 10 4 kg = 0.187g b) The heating value is Q = (m f )(q lhv ) = ( 1.87 10 4) (4,940) Q = 8.03kJ c) The engine power is ( ) N Ẇ = W = η Q Ẇ = 50.kW ( ) N = (0.5)(8.03) ( ) 1500 60

8 CHAPTER 6. FUEL AND AIR FLOW IN THE CYLINDER 6.8) To illustrate the effect of combustion chamber geometry on swirl amplification consider an axisymmetric engine where at bottom center the velocity field of the air inside the cylinder is approximately v r = v z = 0 and v θ = V o (r/b). The cylinder has a bore, b, and the piston has a disk-shaped bowl of diameter, d, and depth, h. The motion is said to be solid body since the gas is swirling as though it were a solid. If at top dead center the motion is also solid body and angular momentum is conserved during compression, what is the ratio of the initial to final swirl speed, ω o /ω 1, as a function of the compression ratio and the cylinder geometry? The moment of inertia of solid body rotation of a disk of diameter, d, and depth, h, is I = πρhd 4 /3. At bdc the moment of inertia I 0 is I 0 = 1 ( π (b 8 (ρ 0) 4) s+d 4 h ) and at tdc I 1 is I 0 = 1 ( π (d 8 (ρ 1) 4) 4 h ) Since angular momentum is conserved I 1 w 1 = I 0 w 0 The swirl amplification is therefore w 1 w 0 = I 0 I 1 = ρ 0 ρ 1 [ b 4 s+d 4 ] h d 4 h = ρ 0 ρ 1 [ (b Since the mass of gas in the cylinder is constant So m = ρ 0 V 0 = ρ 1 V 1 ρ 0 ρ 1 = V 1 V 0 The compression ratio r is So r = V 0 = b s+d h V 1 d = h ( ) s d h = (r 1) b ( b d ) (s h d ) +1 ) ] 4 s h +1 and w 1 w 0 = 1 r [ (b d ) ] 4 s h +1 = 1 r [ (b ) (r 1)+1] d For r 1 w1 w 0 ( ) b d The swirl amplification is proportional to the square of the diameter ratio.

9 6.9) An engine has a mean piston speed Ūp of 10.0 m/s and a clearance volume height h of 5 mm. What is the characteristic length L, integral scale l, Taylor microscale λ, and Kolmogorov microscale η at the end of compression? Compare your calculations with Example 6., and explain the differences. Assume the fluid kinematic viscosity at the end of compression is 100 10 7 m /s and C η = C λ = 1, C l = 0.. At the end of the compression stroke, the characteristic length L = h, the clearance volume height. L = h = 5mm The integral scale l is l = (C 1 )(L) = (0.)(5) = 1mm The Taylor microscale λ is ( 15 λ = C λ )1/ (Re t ) 1 / (l) u t = u p = 10 = 5m /s so Re t = u t l µ = (5) ( ) 1 10 3 100 10 7 = 500 λ = ( )1/ 15 (500) 1 / (l) = 0.17mm 1 η = (C n ) 1 /4 (Re t ) 3/4 (l) = (1) 1 /4 (500) 3 /4 (1) = 0.0094mm = 9.4microns Compared with Example 6., the piston speed had doubled and the clearance height is halved. As a consequence, the integral scale is halved, the Taylor microscale and the Kolmogorov microscale are approximately halved, and the turbulent Reynolds number is unchanged.

10 CHAPTER 6. FUEL AND AIR FLOW IN THE CYLINDER 6.10) A single cylinder two stroke carbureted engine of 85 mm bore and 110 mm stroke is operating at 500 rpm. It has a compression ratio r = 8, is fueled with gasoline, and is running rich with an equivalence ratio φ = 1.. If its indicated power is 0 kw with inlet air temperature of 345 K, inlet pressure of 101 kpa and exhaust pressure of 105 kpa, compute its scavenging ratio S r. Use Figure 4.4 to estimate the indicated thermal efficiency η Otto of an equivalent fuel-air cycle, and assume η/η Otto = 0.80. The scavenging ratio S r is the ratio of the actual mass of charge to the ideal mass at T i and P e that would just fill the cylinder at bottom dead center. S r = ṁ i ρ s V c N From Example 3., the molecular mass M of a mixture of octane and air is 30.4, so the mixture density ρ s is ρ s = P em R u T i = (101)(30.4)/(8.314)(345)= 1.07 kg/m 3 The cylinder volume V c is V c = π r 4 b s r 1 = π 8 4 (0.085) (0.110) 8 1 = 7.13 10 4 m 3 Assuming and 4 stroke fuel-air cycles have approximately the same indicated efficiency, from Figure 4.4 or from the Ottofuel.m program, η fa is 0.33, so η i = (0.80)(0.334) = 0.7. The energy supplied by the gasoline is Q i = Ẇi/η i = 0/ 0.67 = 74.85 kw. The fuel flow rate is ṁ f = Q i /q c = 74.85/44,510= 1.68 10 3 kg/s. The air flow rate ṁ a is ṁ a = ṁ f /FA = ṁ f /(FA s φ) = (1.68 10 3 )/(0.0655)(1.) =.14 10 kg/s The scavenging ratio S r is thus S r = ṁa +ṁ f ρ s V c N = (.14 10 )+(1.68 10 3 ) (1.06)(7.13 10 4 )500/60 = 0.73