Question

Help with "B" please

Question 2 (a) A six-cylinder four-stroke internal combustion engine, running at 2,200 rpm, outputs 120 kW of brake power. The stroke is equal to the bore. The brake mean effective pressure is 5.5 MPa. Calculate the torque, total displacement volume, stroke and bore. [6] (b) In a car, the engine described in (a) above drives the wheels through the multispeed gearbox shown in Figure Q2. When the engine runs at 2,200 rpm, and the gears are selected as shown in Figure Q2, the car's wheels should rotate at approximately 900 rpm. Gears 1 and 3 have 40 and 32 teeth, respectively. All gears should have the same circular pitch. How many teeth should be on gears 2 and 4? What is the exact speed and torque of the car's wheels? [8] 2 driving shaft (from engine) driven shaft (to wheels) 3 selector lever Figure Q2 Simplified schematic diagram of gearbox (not to scale).

Answer 1

An: (2) No. of cylinder (K) = 6 Rpm (N) = 2200rpm Power (60) = 120 kW stroke = length; (L=0) Girent) word 2 P=TXW = (T) (2*N) (60) (120) = (T) (2** 2200) (60) T= 520.87 Nom also; P= (benep) (Vg nept) Vercept (120). (KK) (bnep (5.5 x103) (Kpa) tewept = 0.02 18 m 68 2 0.02.18 18 = dºx 2205 х 60x2 d=1= 63.208 mm t (6) circular pitch a Td (Where di diameter of gear t: teeth. also; + 24 = dgth, de + da de + d₂ atz & ta tytty = 40 +32= 72

ttt4=72 til also; N₂ N4 도 ta ( ^₂ = 2200 rpm goorom ta XA 22 tz 11 23 te 72 ; tz (9122-) = 72 z +22tz t₂ = 729 ~ 20.90=/21 teeth; at 31 51 teeth P = TXW (120) = (T) (2** N4) ; 21 2200 -X Nq = 2 x N₂ اگ ty 905.88rpm IN 1.2.65KW = Torque TE 120x60 2ax 905.88