Home | | **Dynamics of Machines** | Forces on the Reciprocating Parts of an Engine, Neglecting the Weight of the Connecting Rod

The various forces acting on the reciprocating parts of a horizontal engine are shown in Fig. 15.8. The expressions for these forces, neglecting the weight of the connecting rod, may be derived as discussed below :

**FORCES** **ON** **THE** **RECIPROCATING** **PARTS** **OF** **AN** **ENGINE, NEGLECTING THE WEIGHT OF THE CONNECTING ROD**

The various forces acting on the reciprocating parts of a horizontal engine are shown in Fig. 15.8. The expressions for these forces, neglecting the weight of the connecting rod, may be derived as discussed below :

**1. **** Piston effort**. It is the net force acting on the piston or crosshead pin, along the line of

**Fig. 15.8. **Forces on the reciprocating parts of an engine.

*W*R* *= Weight of the reciprocating parts in newtons =* m*R.*g*

We know that acceleration of the reciprocating parts,

Accelerating force or inertia force of the reciprocating parts,

It may be noted that in a horizontal engine, the reciprocating parts are accelerated from rest, during the latter half of the stroke (*i.e.* when the piston moves from inner dead centre to outer dead centre). It is, then, retarded during the latter half of the stroke (*i.e.* when the piston moves from outer dead centre to inner dead centre). The inertia force due to the acceleration of the reciprocating parts, opposes the force on the piston due to the difference of pressures in the cylinder on the two sides of the piston. On the other hand, the inertia force due to retardation of the reciprocating parts, helps the force on the piston.

**2. Force acting along the connecting rod. **It is denoted by

**3. Thrust on the sides of the cylinder walls or normal reaction on the guide bars. **It is

**4. Crank-pin effort and thrust on crank shaft bearings. **The force acting on the

Resolving *F*Q perpendicular to the crank,

**5. Crank effort **or

Mathematically,

*Q. The crank-pin circle radius of a horizontal engine is 300 mm. The mass of the reciprocating parts is 250 kg. When the crank has travelled 60° from I.D.C., the difference between the driving and the back pressures is 0.35 N/mm**2**. The connecting rod length between centres is 1.2 m and the cylinder bore is 0.5 m. If the engine runs at 250 r.p.m. and if the effect of piston rod diameter is neglected, calculate : 1. pressure on slide bars, 2. thrust in the connecting rod, 3. tangential force on the crank-pin, and 4. turning moment on the crank shaft.*

**Solution. **Given:** ***r*** **= 300 mm = 0.3 m ;** ***m*R** **= 250 kg;** **θ** **= 60°;** ***p*1** **–** ***p*2** **= 0.35 N/mm2;** ***l*** **=

1.2 m ; *D* = 0.5 m = 500 mm ; *N* = 250 r.p.m. or ω = 2 π × 250/60 = 26.2 rad/s First of all, let us find out the piston effort (*F*P).

We know that net load on the piston,

**1. Pressure on slide bars**

Let φ = Angle of inclination of the connecting rod to the line of stroke.

We know that pressure on the slide bars,

FN = FP tan φ = 49.424 × tan 12.5° = 10.96 kN Ans.

**2. Thrust in the connecting rod**

We know that thrust in the connecting rod,

**3. ***Tangential force on the crank-pin*

We know that tangential force on the crank pin,

**4. Turning moment on the crank shaft**

We know that turning moment on the crank shaft,

**Q. ***The crank and connecting rod of a petrol engine, running at 1800 r.p.m.are 50 mm and*** ***200 mm respectively. The diameter of the piston is 80 mm and the mass of the reciprocating parts is 1 kg. At a point during the power stroke, the pressure on the piston is 0.7 N/mm*2*, when it has moved 10 mm from the inner dead centre. Determine : 1. Net load on the gudgeon pin, 2. Thrust in the connecting rod, 3. Reaction between the piston and cylinder, and 4. The engine speed at which the above values become zero.*

**Solution. **Given :** ***N*** **= 1800 r.p.m. or** **ω** **= 2π** **× 1800/60 = 188.52 rad/s ;** ***r*** **= 50 mm = 0.05** **m; *l* = 200 mm ; *D* = 80 mm ; *m*R = 1 kg ; *p* = 0.7 N/mm2 ; *x* = 10 mm

**1. Net load on the gudgeon pin**

We know that load on the piston,

When the piston has moved 10 mm from the inner dead centre, *i.e.* when *P**1**P* = 10 mm, the crank rotates from *OC**1* to *OC* through an angle θ as shown in Fig. 15.10.

By measurement, we find that *θ = 33°.

We know that ratio of lengths of connecting rod and crank,

*n *=* l*/*r *= 200 /50 = 4

and inertia force on the reciprocating parts,

*2.**Thrust in the connecting rod*

We know that thrust in the connecting rod,

**3. Reaction between the piston and cylinder**

We know that reaction between the piston and cylinder,

**4. Engine speed at which the above values will become zero**

A little consideration will show that the above values will become zero, if the inertia force on the reciprocating parts (*F**1*) is equal to the load on the piston (*F**L*). Let ω *1* be the speed in rad/s, at which *F**I* = *F**L* .

*Q. A vertical petrol engine 100 mm diameter and 120 mm stroke has a connecting rod 250 mm long. The mass of the piston is 1.1 kg. The speed is 2000 r.p.m. On the expansion stroke with a crank 20° from top dead centre, the gas pressure is 700 kN/m2. Determine: *

*1.Net force on the piston, 2. Resultant load on the gudgeon pin, 3.Thrust
on the cylinder walls, and 4. Speed above which, other things re-maining same,
the gudgeon pin load would be reversed in direction.*

Solution. Given: D = 100 mm = 0.1 m ; L = 120 mm = 0.12 m or r = L/2 =
0.06 m ; l = 250 mm = 0.25 m ; mR = 1.1 kg ; N = 2000 r.p.m. or

ω = 2 π × 2000/60 = 209.5 rad/s ; θ = 20°; p = 700 kN/m2

1. Net force on the piston

The configuration diagram of a vertical engine is shown in Fig. 15.11.

We know that force due to gas pressure,

We know that for a vertical engine, net force on the piston,

**2. ***Resultant load on the gudgeon pin *

Let φ = Angle
of inclination of the connecting rod to the line of stroke. We know that,

sin φ = sin θ / *n* = sin 20°/4.17 = 0.082 ∴ φ = 4.7°

We know
that resultant load on the gudgeon pin,

**3. Thrust
on the cylinder walls**

We know that thrust on the cylinder
walls,

FN = FN
tan φ = 2256.8 . tan 4.^{o} 7 = 185.5 N Ans.

4. Speed, above which, the gudgeon pin load
would be reversed in direction

Let N1 = Required speed, in r.p.m.

The
gudgeon pin load i.e. F _{Q} will be reversed in direction, if F_{Q}
becomes negative. This is only possible when F_{P} is negative.
Therefore, for F_{P} to be negative, F_{I} must be greater than
(F_{L} + W _{R}),

**Q. ***A horizontal steam engine running at 120 r.p.m.
has a bore of 250 mm and a stroke of*** ***400 mm. The connecting rod is 0.6 m and mass of
the reciprocating parts is 60 kg. When the crank has turned through an angle of
45° from the inner dead centre, the steam pressure on the cover **end side is 550 kN/m2 and that on the crank end side is 70 kN/m2. Considering the diameter of the piston rod equal to 50 mm, determine: **1. **turning moment on the crank shaft, 2. thrust on the bearings, and 3. acceleration of the flywheel, if the power of the engine is 20 kW, mass of *

**Solution. **Given :** ***N*** **= 120 r.p.m. or** **ω** **= 2π** **× 120/60 = 12.57 rad/s ;** ***D*** **= 250 mm = 0.25** **m ;

*L *= 400 mm = 0.4 m or* r *=*
L*/2 = 0.2 m ;* l *= 0.6 m ;* m*R*
*= 60 kg ;* *θ* *= 45° ;* d *= 50 mm = 0.05 m ;* p*1*
*= 550 kN/m^{2}* *= 550 × 10* *^{3}* *N/m^{2}* *;* p*2*
*= 70 kN/m^{2}* *= 70 × 10* *^{3}* *N/m^{2}

**1. Turning moment on the
crankshaft**

First of all, let us find the net load on the
piston (*F*P).

We know
that area of the piston on the cover end side,

**2. Thrust on the bearings**

We know
that thrust on the bearings,

**3. Acceleration of the flywheel**

Given: *P*
= 20 kW = 20 × 10 ^{3} W; *m* =
60 kg ; *k* = 0.6 m

Let α = Acceleration of the flywheel in rad/s^{2}.

We know
that mass moment of inertia of the flywheel,

*I *=* m.k*^{2}* *= 60 × (0.6)* *^{2}* *= 21.6 kg-m^{2}

∴ Accelerating torque, *T*A = *I*.α = 21.6 α N-m ...**( i)**

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