Homework Ch5 Answers

Homework Ch5 Answers:

5.5 i. Max torque increases (CR up) ii. Max power decrease iii. Speed at Max power -= decrease (slow burn) iv. Economy – increase, efficiency up due to CR.

5.6 Derivation see p28.

Q  m f CV  (ma  m f )c p (T3  T2 ) ( 1) ( 1) T3  T2 ,T2  T1rv ;(T3  T2 )   1rv ( 1) m f CV  (ma  m f )c p  1rv

CV (ma  m f ) ( 1)   AFR  1 c p  1rv m f CV AFR  ( 1) 1 c p  1rv CV   ( 1)  1 (AFR  1)c p rv

Plug in the numbers and go from there….

BSFC = mf/W BMEP = W/Vs = N*m/m^3=Pa Mf=BSFC*BMEP*Vs Bsp = BMEP*vp/4 (only one power stroke per 4 cycles?)

m V   AFR= a  s v  v m f BSFC * BMEP *Vs BSFC * BMEP BSAC=BSFC*AFR p=ATM 105000Pa=N/m^2 R 287J/kgK=Nm/kgK T 290K AFR p=RT p/RT= 1.262 N/m^2 kg/m^3 Nm/kgK*K Speed m/s 4 6 8 10 12 BSFC kg/MJ 0.068 0.063 0.061 0.064 0.076 BMEP Bar 7.24 8 8.08 7.7 6.93 BMEP Pa=N/m^2 760,200 840,000 848,400 808,500 727,650 nV 89.80% 88.30% 82.10% 77.70% 76.20%

Power BSP MW/m^2 0.7602 1.2600 1.6968 2.0213 2.1830 BMEP*Vp/4

AFR 21.92 21.05 20.01 18.94 17.38

BSAC kg/MJ 1.49 1.33 1.22 1.21 1.32 BSFC*AFR 5.8 Max Power:

Omega Torque rad/sec Power (W) Power (HP) 1000 120 104.7197551 12,566 16.81 2000 140 209.4395102 29,322 39.23 3000 147 314.1592654 46,181 61.78 3250 146 340.3392041 49,690 66.47 3500 144 366.5191429 52,779 70.61 3750 142 392.6990817 55,763 74.60 4000 138 418.8790205 57,805 77.33

Max Power 57,805 W SFC 270 g/KWh Fuel Consumpotion 15607.35 g/hr 4.335375 g/second

Cycles per second (RPM/120) 33.3333333 Volume 2.5 litres Nv 82% Volume.Second 68.3333333 litres/second

p 105,000 v 0.068333333 litres

R 287 T 290

m=pv/RT 0.086206897 Chapter Seven 7.1 Vs 1.2 liters Speed 5500 RPM 91.66667 RPS Torque 108 N/m

Power 62.20353 kW kW=T*2*pi()*rps/1000 Work 678.584 J per cycle

BSFC 294 g/KWh CV 43 MJ/kg N* 1

Powerb = Vs*pb*N Nbrake

Calculate BMEP 565486.7 Work/Vs/N 5.654867 Bar

KWh=kW*3600 62.20353 Fuel Used/Hr 18287.84 Fuel Used Per 5.079955 g Second Fuel Used Per 0.055418 g Cycle CV/cycle 2382.961

W/CV 28.5%

T 27 W 169.646 J per cycle Power 15.55088 KW Power 1 hour 15.55088 KWh BSFC 325 g/KWh Fuel/Hour 5054.037 Fuel/Cycle 0.015315 g/cycle CV/Cycle 658.5564

W/CV 0.257603 7.3

a) Vc = 500/12 = 41.3 Vs eff = (1+cos60)*500/2 = 375 b) Rv = (375+41.6)/41.6 = 10 c) Using m = RT/PV and 416 cc m=484x10-6 kg.; ma = 333x10-6 kg a. Note: careful of notation here, this is actually mass of air not air plus fuel. d) Mi = 444x10-6 kg. a. N Deliv => Vs* p = 100,000*0.000500/(287*300) = 580.7 x 10-6 kg b. Nd = 444/580 = 76.5% e) See diagram f) Ma = 333 x 10-6 kg., mf = 22.7 x 10-6 kg. Mt = 356 x 10-6 kg. g) T1 = 300K (-1) a. T2 = T1 rv = 754 b. Qin = 22.7 x 10-6 kg x 44MJ/kg = 999 J i. DT = 999/(950*0.000356) = 2954 ii. DT = 44 x106/(950*AFR+1) = 2950K iii. T3 = 3704K (1-) c. T4=T3 rv = 3704 (0.398) = 1475K d. Qout = (T4-T1) *mt *Cv)=1175*950*356*10-6) = 397J h) Otto a. No = 1-(Qout/Qin) = 1-397/999 = 60.1% b. No = 1-rv(1-) = 60.1% c. No = 1-(T4-T1)/(T3-T2) = 1 -1175/2950 = 60.1% i) Ni = 30% SFC=1/(Cv*No) =1/(44MJ/kg * 30%) = 75.5g/MJ j) @6000 RPM = 100 cycles/sec = 22.7x10-4 kg/sec = 2.27g/sec. a. ma = 2.27*14.7 = 33.6g/sec k) Pb = mf/sfc = 0.00227/0.0775 MJ = 29.3KW = 39.4 HP