We have studied in earlier video that balancing is done on a locomotive. But because of balancing there are 3 effects that occurs on the locomotive, which are:
Variation of tractive force
Swaying couple
Hammer blow
In this notes we will study the effect of variation of tractive force and also derive the equation to calculate the tractive force.
The first thing to remember here is that the Locomotive engines are reciprocating engines and we know that we cannot completely balance a reciprocating engine, we partially balance it. And partially balancing generates horizontal force and vertical force.
In the video of partial balancing of reciprocating of engines ( Video Part-1 & Video Part-2 ) we have studied that the unbalanced force acting in horizontal and vertical direction is given by:
FHFV=(1−C)⋅mrω2⋅cosθ−−−(1)=C⋅mrω2⋅sinθ
Now, let us imagine the horizontal force FH in three dimensions by looking at this diagram
Unbalanced horizontal force acting on cylinders of locomotive
In the above diagram, let's represent the horizontal unbalanced force acting on cylinder 1 as FHU1 and similarly the horizontal unbalanced force acting on cylinder 2 as FHU2.
From this, it becomes easier to understand what is traction force. It is the resultant of the two horizontal unbalanced force. Hence we get the following equation:
FT=FHU1+FHU2−−−(2)
Traction force definition
The resultant unbalanced horizontal force due to the two cylinder along the line of stroke of cylinder is called traction force.
Remember:
Try to write the definition by explaining the things that we did in the calculation. This way it becomes easier for you to understand it.
Now for cylinder 1, and using equation (1) from above, we can say that:
FHU1=(1−C)⋅mrω2⋅cosθ−−−(3)
Similarly for cylinder 2:
FHU2=(1−C)⋅mrω2⋅cosθ−−−(4)
As we have studied in the last video, if cylinder 1 angle is at θ then the cylinder 2 angle will be at θ+90°. Therefore from equation (4) we can say that:
FHU2=(1−C)⋅mrω2⋅cos(θ+90°)−−−(5)
From equation (2)
TFwhere,=FHU1+FHU2=(1−C)⋅mrω2⋅cosθ+(1−C)⋅mrω2⋅cos(θ+90°)=(1−C)⋅mrω2⋅[cosθ+cos(θ+90°)]=(1−C)⋅mrω2[cosθ−sinθ]−−−(6)m=mass of the reciprocating partr=radius of the crankω=angular velocity of the crankC=portion of the reciprocating part which is to be balanced
Now we need to find the maximum and minimum of this tractive force. To do this we need to differentiate equation (6) with respect to the variables. Here our variable is θ. So we need to differentiate FT with respect to θ.
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