Question

1. (30P) A twenty-cylinder MTU 4 STROKE 20V 4000 M93 marine diesel engine is designed to develop 3490 kW brake power at 1800 rpm. The bore is 170mm and stroke length is 210 mm for each cylinder and bsfc is 205 g/kWh. The calorific value of the petrol is 42 800kJ/kg. The inlet air manifold conditions are 80 kPa and 313 K. and The engine has a volumetric efficiency of 0.82. Calculate: a) Displacement (litre) b) Bmep (kPa) c) Fuel consumption (kg/h) d) Thermal efficiency e) Average piston speed f) Air flow rate and fuel ratio. g) Thermal efficiency.

Answer 1

Solution Griven :-. no. of cylinder (K) . no. of power stroke (n) 20 N (for 4 stroke) (n) = 1800 300 pm Brave power (BP) 3490 Kla = 3480 x 103 W 0.170 m =210 mm e Bone (o) stroke length (1) 0.210m Brake specific full consumption (bufc) = 205 8/kWh Calorific value of the diesel (CV) 42800 kr/kg Inlet air condition pressure (PI 80 КРg temperature (T) = 313 K Erigine volumetric efficiency (Mv) = 0.82 * Ang (Q) :- Displacement Cva) ;- Vd = t/ud 2 l XK 17160.17) 2 x 0.21 x 20 Vd Vd = 0.0953 m Vd = 95.33 litre 5 Ans (6:Brake meomen effective pressure Comep): – GOX BP 60 x 3490 Bnep LANK 0,21% 7/4X(0.172X900X20 Bnep 2440.6 Kp9 = Ans (C):- Fuel consumption (W): - mis = (bsfc) x (BO) 205 (w) x ( 3490 kW) ins- 715450 g/h mf 715.45 to/h

And(d): - Brake thermal efficiency: (Math) :- BP nbeth mit X CV Nobth 3450 715 45x 1 x 42800 36005 Noth = 0.422 Mbth 42.2% → Ans (@:- Average piston speed (5.) :- Šp = 2LN 2 X 0.21X1800 Sp 756 m/min. And (f): Density of air: RT 0.287 X313 P 80 Pa = 1.1228875 My mo? volumetric efficiency (nel! - orv = sa Vaisp X N/ -> ma ma - 0.82 = (4.922887) C0.0953) x 1800/2 ma 78.97 Kg/min ma 4738.455 kg/h als flow rate ma Als flow rate & fuel ratio mf 4738.455 715.45 AFR = 6.623 1 (AFR) =

Thank you.