R006 Assignment Help

Unformatted text preview: Given N=1500 rpm → ω = 2*Pi*N/60 = 157.08 rad/s. Rod length l = 150 mm = 0.15 m; Crank r = 0.06 m; and Pin off-set d = 0. PISTON VELOCITY AND ACCELERATION:- From equations in slide#24, we have v = 0 for θ=0 o (intake TDC). Looks like some brackets are missing in the equation given there because it is dimensionally incorrect. Please correct the equation for acceleration as follows: a = 0.06*(157.08) 2 { 1 + [(0.06/0.15)* (cos 2- 0)/1] + [(0.06/0.15) 3 *(cos 2 0 * 0)/1] } = 2072.63 m/s 2 . RECIPROCATING FORCE:- Given that M REC = 3 kg and (e/l) ratio is 0.005, using equation in slide#25, we have F REC = 3 kg* 0.06*(157.08) 2 *[cos270 - (0.005*sin270) + (cos(2*270)*0.06/0.15)] = -1754 N or 1.75 kN magnitude force. *Notice that the θ value during compression BDC would be 180 o (for a value of 0 o at intake TDC) and hence θ = 270 o during the mid-way of compression. 5. Find the value of split pin angle for a V8 engine with a bank angle of 60 o . Also, perform an engine unbalance analysis for this engine by evaluating the first order and second order forces. Hint: Refer to slides 49-52 SPLIT PIN ANGLE:- Firing interval (FI) = 720/8 = 90 o . Bank angle (given) = 60 o . So, the split pin angle (SPA) = 90 - 60 = 30 o . ENGINE UNBALANCE ANALYSIS:- Crank Pin arrangement for this engine can be seen in slide#63 shown as below: As calculated before, SPA or δ = 30 o . Then the engine unbalance forces due to the reciprocating masses can be evaluated as follows (similar to that in slide#49).masses can be evaluated as follows (similar to that in slide#49)....
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