Engineering Mechanics

(^466) „„„„„ A Textbook of Engineering Mechanics
22.8.VELOCITY DIAGRAM METHOD FOR THE VELOCITY OF PISTON OF A
RECIPROCATING PUMP
Fig. 22.13. Velocity diagram method.
Consider the mechanism of a reciprocating pump, in which AB be the crank, BC the connecting
rod and C the piston as shown in Fig. 22.13 (a). Now let us draw the velocity triangle as shown in Fig.
22.13 (b) and as discussed below :

1. First of all, take some suitable point a and draw a horizontal line ac representing the
   direction of motion of the piston. (i.e. vC).;


2. Through a draw another line ab representing the direction of motion of B (i.e. vB). It is at
   right angles to the crank AB.


3. Now cut off ab equal to the velocity vB to some suitable scale (such that vB = ω.r) where ω
   is the angular velocity of the crank and r is the radius of the crank.


4. Through b draw a line bc perpendicular to the connecting rod BC of the space diagram.


5. Now ac of the velocity diagram gives the velocity of piston vC to the scale.
   Note. The length bc to the scale gives the velocity of the connecting rod. The angular velocity
   of the connecting rod may now be found out from the relation bc/BC, where BC is the length of the
   connecting rod.
   Example 22.5. In a reciprocating pump, the lengths of connecting rod and crank is 1125
   mm and 250 mm respectively. The crank is rotating at 420 r.p.m. Find the velocity with which the
   piston will move, when the crank has turned through an angle of 40° from the inner dead centre.
   *Solution. Length of connecting rod (l) = 1125 mm = 1.125 m ; Length of crank (r) = 250
   mm = 0.25 m ; Angular rotation of crank (N) = 420 r.p.m and angle traversed by the crank (θ ) = 40°
   We know that angular velocity of crank,
   22420
   44 rad/s
   60 60

ππ×N

ω= = =

and velocity of B (vB)=ωr = 44 × 0.25 = 11 m/s

Fig. 22.14.

* We have already solved this example analytically as 22.3.