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ass1-1soln - John’s automobile has a three-liter SI V6...

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Unformatted text preview: John’s automobile has a three-liter SI V6 engine that operates on a four-stroke cycle at 3600 RPM The compression ratio 1s 9. 5, the length of connecting rods 1s 16. 6 cm, and the engine is square (B= S) At this speed, combustion ends at 20° aTDC. Calculate: » 1. cylinder bore and stroke length 2. average piston speed 4* t 3. clearance volume of one cylinder 4. piston speed at the end of combustion 5. distance the piston has traveled from TDC at the end of combustion 6. volume in the combustion chamber at the end of combustion a) ’1’)’ For one cylinder, using Eq. (2-8) with S: B: . . . Vd = Wow/6‘ -_3L/6“ — 0.5 L— -— 0.0005 m3 =(1'T/4)BZS = (7r/4)B3 B = 0.0860m = 8.600(1) 2 S 1,) 27 Using Eq. (2—2) to find average piston speed: Up = ZSN= (2 strokes/rev) (0 0860 m/stroke) (3600/60 rev/sec) = 10. 32 m/sec ' é) .3)’ Using Eq. (2-12) to find the clearance volume of one cylinder: rc = 9.5 = (V4 + Vc)/Vc = (0.0005 + Vc)/Vc Vc = 0.000059 m3 = 59 cm3 4) 4r Crank offset,a = 3/2 = 0.0430m = 4.30cm R = r/a = 16.6 cur/4.30m = 3.86 _ Using Eq. (2-5) to find instantaneous piston speed: UP/Up =‘(1r/2)sin8[1 + (case/m» -= (6/2) sin(20°){1 + [comm/Wu = 0.668 . ' U, '= 0.668 U = (0.668) (10.32 m/sec) = 6.89 m/sec e) ,5')‘ Using Eq. (2- -3) to find piston position: s=acos0 + Vr2 - azsinzo = (0.0430 In) cos (20°) + V‘(0.166 m)2 — (0.0430 m)2sin2 (20°) = 0.206 :11 Distance from TDC: x = r + a — s = (0.1661'n) + (0.043 m) - (0.206 m) = 0.003 m= 0.3 cm ,‘1 ,6 Using Eq (2-14) to find instantaneous volume: ' V/V.=1+1(r,—1){R+1—cose—\/m] = 1 + §(9.5 — 1)[3.86 + 1 - cos(20°)— W] = 1.32 I , . V = 1.32 V, = (1.32) (59 cm3) = 77.9 cm3 = 0.0000779 m3 This indicates that, during combustion, the volume in the combustion chamber has only increased by a very small amount and shows that combustion in an SI engine occurs at almost constant volume at TDC 'l‘ne engme in Example r'roolem 2-1 15 connected to a aynamometer wmcn gives a brake output torque reading of 205 N~m at 3600 RPM. At this speed air enters the cylinders at 85 kPa and 60°C, and the mechanical efficiency of the engine is 85%. Calculate: 1. brake power (H; 2 2. indicated power 3. brake mean effective pressure 4. indicated mean effective pressure 5. friction mean effective pressure 6. power lost to friction ‘7. brake work per unit mass of gas in the cylinder 8. brake specific power ' 9. brake output per displacement 10. engine specific volume 09 J) Using Eq. (2-43) to find brake power: Wb— — 27rNr — (271' radians/rev) (3600/60 rev/sec) (205 N~m) = 77,300N-m1’sec = 77 3 kW = 104 hp lb) 2) Using Eq. (2- 47) to find indicated power. W3: Wb/nm = (77. 3 kW)/(0. 85): 90.9 kW = 122hp C) 35’ Using Eq. (2-41) to find the brake mean effective pressure: bmep = 4777/14; = (471' radianslcycle) (205 N—m) / (0.003 m3/cycle) = 859,000 Ni‘m2 = 859 kPa = 125 psia d)“ Equation (2-37c) gives indicated mean effective pressure: ' imep = bmep/nm = (859 kPa)/(0.85) = 1010 kPa = 146 psia 5) Equation (2-37d) is used to calculate friction mean effective pressure: fmep =‘imep — bmepl'= 1010 -— 859 = 151 kPa = 22 psia C) 6) Equations (2-15) and (2-44) are used to find fiiction power lost: AP = (1r/4)Bz— ‘ (qr/ti) (-0. 086 m)2 = 0.0058] 1112 for one cylinder Wr='A(1/2n)(fmep) = (1/4) (151 kPa) (0.00581 mzlcyl) (10.32 misec) (6 cyl) = 13. 6 kW— - 18 hp 01‘, it can be obtained from Eq. (2- 49): ' vie: W.— W, = 90.9? 77.3 = 13.6 kW _ 7) First brake work 15 found for one cyiinder for one cycle using Eq. (2-29): = (bmep) Vd = (359 kPa) (0.0005 m3) = 0.43 1:] It can be assumed the gas entering the cylinders at BBC is air: ma =_ PVBDC /RT = P(Vd + Vc)/RT = (85 kPa) (0.0005 + 0.000059)m3 / (0.287 kakg-K) (333 K) ' = 0.00050 kg Brake specific Work per unit mass: = Wb/ma = (0.43 H) /_(0.00050 kg) = 860 kJ/kg = 370 B'I‘Uflbm ' 8) ‘ Equation (2-51) gives brake specific power: BSP = 77. 7A1, = (77. 3 kW) /[(7r/ 4)(0. 086 m)2(6 cylinders)] — -2220 kWiruz = 0.2220 lecmz— " 1 92 lip/in. 2 9) Equation (2-52) gives brake output per diSplacement: BOPD = Wb [VJ = (77.3 kW) / (3 L) = 25.8 kaL = 35 lip/L = 0.567 hplin.3 10) Equation (2—53) gives engine specific volume: st = V. Min, = llBOPD = 1725.8 The engine in Example Problem 2—2 is running with an air—fuel ratio AF = 15, a fuel heating value of 44;000 kJ/kg, and a combustion efficiency of 97%. Calculate: ' L rate of fuel flow into engine 2. brake thermal efficiency 7 3. indicated thermal efficiency 4. volumetric efficiency 5. brake specific fuel consumption 09 «13‘ From Example Problem 2-2, the mass of air in one cylinder for-one cycle is ma = . 0.00050 kg. Then: mf = ma / AF = 0.00050/ 15 = 0.000033 kg of fuel per cylinder per cycle Therefore, the rate of fuel flow into the engine is: I ”if = (0.000033 kg/cyl-cycle) (6 cyl) (3600/60 rev/sec) (1 cycle/2 rev) = 0.0060 kg/sec = 0.01321bm/sec ' . b) ,2 Using Eq. (2-64) to find brake thermal efficiencyi (1,0,, = Wb/meHv n; = (77.3 kW) / (0.0060 kg/sec)(44,000 kJ/kg) (0.97) = 0.302 =.30.2% ' Or, :using Eq. (2-68) for one cycle of one cylinder: (0)6 = Wb /meHV m. = (0.43 k1) /(0.000033 kg) (44,000 kJ/kg) (0.97) = 0.302 a) 55 Indicated thermal efficiency using Eq. (2-65): ' (71,).- = («nob/n... = (130210.85 :- 0355 = 35.5% cl) 4? Using Eq. ((2-69) with standard air density for volumetric efficiency: a... = m. /p., Vd = (0.00050 kg)/(1.181 kg/Im3)(0.0005 m3) = 0.847 = 84.7% e) 5? Using Eq. (2—59) for‘brake specific fuel consumption: 5 bsfc = ”if/WI, = (0.0060 kg/sec) / (77.3 W) = 7.76 X 10—5 kg/kW-sec : 279 gm/kW'hr = 0.459 lbmfhp—hr ...
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