The firing order is a sequence of giving spark/firing to the cylinders of multi-cylinder inline engines. Firing order is provided in multi-cylinder inline engines in order to avoid vibrations (If firing order is not provided in multi-cylinder inline engines, then all the pistons will move in upward direction at the same time and same way they move in the downward direction at the same time. This creates a lot of vibration.) For avoiding these vibrations, firing order is provided in multi-cylinder inline engines.
Numerical
A four stroke five cylinder inline engine has a firing order of 1-4-5-3-2-1. The centre lines of cylinders are spaced at an equal interval of 15 cm, the reciprocating parts per cylinder have a mass of 1.5 kg, the piston stroke is 10 cm and the connecting rods are 17.5 cm long. The engine rotates at 600 rpm. Determine the value of maximum primary and secondary unbalanced forces and couples about the central plane.
Solution:
Given data:
mass of reciprocating parts mStroke SRadius of crank rLength of connecting rod LObliquity ratio nSpeed of engine Nω=1.5kg=10cm=0.10m=2S=20.10=0.05m=17.5cm=0.175m=rL=0.050.175=3.5=600rpm=602πN=602π×600=62.85srad
The force and couple data is given in table:
Plane
Mass (m), kg
Radius (r), m
Centrifugal Force (m x r), kg.m
Distance From R.P.(l), m
Couple (m x r x l), kg.m2
Primary Crank Position 'θ'
Secondary Crank Position '2θ'
1
1.5
0.05
0.075
-0.3
-0.022
0°
0°
2
1.5
0.05
0.075
-0.15
-0.011
576° i.e. 216°
1152° i.e. 72°
3 (R.P.)
1.5
0.05
0.075
0
0
432° i.e. 72°
864° i.e. 144°
4
1.5
0.05
0.075
0.15
0.011
144°
288°
5
1.5
0.05
0.075
0.3
0.022
288°
576° i.e. 216°
fig.2 (a) Position of planes
fig.2 (b) Primary crank positions
Assuming the engine to be vertical, the position of cylinders are shown in fig.2(a).
The angles are measured from cylinder 1.
Cylinder 3 is considered as a reference plane (R.P.)
The engine is working on a four stroke cycle, the angle between two crank =5720=144°
For drawing a primary force polygon, take a suitable scale of 1 cm = 0.075 kg.m. From fig.2(c) it is seen that, primary force polygon is closed and hence there is no unbalanced primary force.
fig.2 (c) Primary force polygon
2. Primary couple polygon
fig.2 (d) Primary couple polygon
Draw the primary couple polygon by taking a suitable scale of 1cm=0.011kgm2. The vector od′ represents the unbalanced primary couple Cp.
Now from fig.2 (d), measure od′;
od′=1.5cm
Hence, the unbalanced primary couple is,
Cp=od′×scale×ω2=1.5×0.011×(62.85)2=65.18N⋅m
3. Secondary force polygon
fig.2 (e) Secondary crank positions
fig.2 (f) Secondary force polygon
Draw the secondary force polygon by taking a suitable scale of 1cm=0.075kg.m.
From fig.2 (f), it is seen that the secondary force polygon is closed, hence there is no unbalanced secondary force.
4. Secondary couple polygon
fig.2 (g) Secondary couple polygon
Draw the secondary couple polygon by taking suitable scale of 1cm=0.011kgm2. The vector od′ represents the unbalanced secondary couple Cs
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