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Motion of Piston and Crank of a Reciprocating Engine

Fig. 4.4 (a) shows different parts of a reciprocating engine. Point A executes linear motion along line AC and point B moves in a circle with ro as angular velocity.

Fig. 4.4 (b) shows the position of instantaneous centre 0.

v A= w0 AO

 

where ro0 is the angular velocity of the connecting rod AB.

 

Also                 V B = C0 . BO

or                     wr =w0  BO. w0 wr / BO

v A= w0 AO = (wr ) AO /BO

 

Thus the velocity of A (VA) can be measured from (4.6) by knowing the ratio AO/ BO, w and r .

Analytical Method to Find VA [Fig. 4.4 (c)]

Method 1.       Draw CD     AC

BE AC

∆AOB and ∆BCD

AOB = 90° – θ= BCD

  OAB = 90°- ϕ = BDC

And                    OBA = LCB (opposite angle)

 

AO / BO = CD /BC = CD /r

tan  ϕ = CD / AC

CD = AC tan ϕ = (AE + EC ) tan ϕ

= (L  cos ϕ + r cos θ) tan ϕ

= L cos ϕ x sin ϕ /cos ϕ + r cos θ tan ϕ

=L sin ϕ + r cos θ tan ϕ

We have                      VA = (wr) AO /BO = (wr) CD / r

= (wr) L sin ϕ + r cos θ tan ϕ / r

VA = w [ L sin ϕ   r cos θ tan ϕ]

 

Method 2. When the crank is at inner dead centre, the position of B is at B1 and A is at A1. Thus, A1B1 = AB = L and B1C = r.

AO / BO = CD /BC = CD /r

tan  ϕ = CD / AC

CD = AC tan ϕ = (AE + EC ) tan ϕ

= (L  cos ϕ + r cos θ) tan ϕ

= L cos ϕ x sin ϕ /cos ϕ + r cos θ tan ϕ

=L sin ϕ + r cos θ tan ϕ

We have                      VA = (wr) AO /BO = (wr) CD / r

= (wr) L sin ϕ + r cos θ tan ϕ / r

VA = w [ L sin ϕ   r cos θ tan ϕ]

 

Method 2. When the crank is at inner dead centre, the position of B is at B1 and A is at A1. Thus, A1B1 = AB = L and B1C = r.

Let                               A1 A=x

Al C-AC = X

( Al Bl + Bl C)- ( AE + EC) = X

(L + r)- (L cos <1> + r cos e)

dx  / dt = d /dt {(L +r ) – (L cos ϕ + r cos θ)}

= 0 – { – L sin ϕ dϕ /dt –r sin θ dθ /dt}

VA = L sin ϕ d ϕ /d r sin θ dθ /dt

VA = L sin ϕ d ϕ /dt +wr sin θ

Again              BE = L sin ϕ = r sin θ

Sin ϕ = r /L sin θ

cos ϕ d ϕ /dt = r /L . cos θ. dθ/dt = wr cos θ / L

d ϕ / dt = wr cos θ/ L cos ϕ

 

Substituting the value of dϕ /dt from ( 4.9) in ( 4.8) we get

 

VA = L sin ϕ w cos θ / L cos ϕ wr  sin θ

= wr cos θ tan ϕ+ wr sin e

= (J) [r cos θtan ϕ+ r sine]

= ro [r cos θ tan ϕ + L sin ϕ]

V A = (J) [L sin ϕ+ r cos e tan ϕ]