Analytical method

ANALYTICAL METHOD

Consider a reciprocating steam engine mechanism OCP as shown in Fig.11.1. Let crank OC rotates with angular velocity o rad/sec and the connecting rod PC makes angle & with the line of stroke PO. Let x be the displacement of piston from initial point P' to P, when the crank turns through an angle θ from I.D.C.

Let

r = Crank radius,

l = Length of the connecting rod,

θ = Angle made crank with I.D.C.,

ϕ = Inclination of connecting rod to the line of stroke PO, and

n = l/r = Ratio of length of connecting rod to the radius of crank, also known as obliquity ratio.

1. Velocity of the Piston (vP)

From the geometry of Fig.11.1, displacement of the piston is given by

By expanding the above expression by binomial theorem, we get

Substituting equation (iii) in equation (i), we get

Differentiating equation (11.4) with respect to θ, we get

Therefore, velocity of the piston or velocity of P with respect to O,

2. Acceleration of the Piston (aP)

We know that acceleration is the rate of change of velocity. So, acceleration of the piston P is given by

Differentiating equation (11.5) with respect to θ, we get

Note 

1. If the value of n is very large, then aP = ω2 r cos θ, as in case of BHM.

2. When θ = 0°, i.e., at IDC position,

3. When θ = .180°, i.e., at ODC position,

As the direction of motion is reversed at the outer dead centre (ODC) position, therefore, changing the sign of the above expression, we get

3. Angular Velocity of the Connecting Rod (ωPC)

From the geometry of the Fig.11.1, we find that

Differentiating both sides with respect to time t, we get

Since the angular velocity of the connecting rod PC is same as the angular velocity of point P with respect to C and is equal to d ϕ/ dt , therefore

4. Angular Acceleration of the Connecting Rod (αPC) 

We know that the angular acceleration of P with respect to C,

Differentiating equation (11.7) with respect to θ, we get

The negative sign indicates that the sense of angular acceleration of the rod is such that it tends to reduce the angle ϕ.

Note

1. Since sin2 θ is small as compared to n2, therefore it may be neglected. Thus equations (11.7) and (11.8) are reduced to

2. Also in equation (11.9), unity is small as compared to n2, hence the term unity can be neglected.

Example 11.1 

In a slider crank mechanism, the length of the crank and connecting rod are 100 mm and 400 mm respectively. The crank rotates uniformly at 600 rpm clockwise when the crank has turned through 45° from the inner dead centre. Find, by analytical method: 1. velocity and acceleration of the slider, and 2. angular velocity and angular acceleration of the connecting rod.

Given data: 

r = 100 mm = 0.1 m; l = 400 mm = 0.4 m; N = 600 rpm; θ = 45°

Solution:

ω = 2πN/60 = 2 π (600)/60 = 62.83 rad/s

Obliquity ratio, n = l/r = 0.4/0.1 = 4

1. Velocity and acceleration of the slider (vP and aP):

Velocity of the slider is given by

Acceleration of the slider is given by

2. Angular velocity and angular acceleration of the connecting rod (ωPC and αPC):

Angular velocity of the connecting rod is given by

Angular acceleration of the connecting rod is given by

Example 11.2 

A petrol engine has a stroke of 120 mm and connecting rod is 3 times the crank length. The crank rotates at 1500 rpm clockwise direction. Determine: 1. velocity and acceleration of the piston, and 2. angular velocity and angular acceleration of the connecting rod, when the piston has traveled one-fourth of its stroke from IDC. 

Given data: L = 120 mm = 0.12 m or r = 0.12/2 = 0.06 m; l = 3 r, or n = l/r = 3; N = 1500 rpm 

Solution:

ω = 2πN/60 Ν = 2 π (1500)/60 = 157.08 rad/s 

It is given that the piston has traveled 1/4th of its stroke from IDC. 

⸫ θ = 1/4 of stroke from IDC = 1/4 × 180° = 45°

1. Velocities and acceleration of the piston (vP and aP):

Velocity of the piston is given by

Acceleration of the piston is given by

2. Angular velocity and angular acceleration of the connecting rod (ωPC and αPC):

We know that angular velocity of the connecting rod,

Angular acceleration of the connecting rod is given by

Example 11.3 

In a reciprocating engine mechanism, the crank and the connecting rod are 300 mm and 1 m long respectively and the crank rotates at a constant speed of 200 rpm. Determine, analytically:

(i) the crank angle at which the maximum velocity occurs, and

(ii) the maximum velocity of the piston.

Given data: 

r = 300 mm = 0.3 m; l = 1 m; N = 200 rpm

Solution:

ω = 2πN/60 = 2 π (200)/60 = 20.94 rad/s

(i) Crank angle at which the maximum velocity occurs:

Let

θ = Crank angle from IDC at which the maximum velocity occurs 

We know that the velocity of the piston,

For maximum velocity of the piston,

(ii) Maximum velocity of the piston:

Substituting the value θ = 75° in the vP equation, we get

Example 11.4

The crank and connecting rod of a steam engine are 0.35 m and 1.55 m in length. The crank rotates at 180 rpm clockwise. Determine the velocity and acceleration of the piston, when the crank is at 40°from the inner dead centre position. Also determine the position of the crank for zero acceleration of the piston.

Given data: 

r = 0.35 m; l = 1.55 m; N = 180 rpm; θ = 40°

Solution:

ω = 2πN/60 = 2π (180)/60 = 18.85 rad/s

n = l/r = 1.55/0.35 = 4.428

1. Velocity and acceleration of the piston when θ = 40° (vP and aP);

Velocity of the piston is given by

Acceleration of the piston is given by

2. Position of the crank for zero acceleration of the piston:

Let θ1 = Position of the crank from IDC for zero acceleration of the piston.

We know that acceleration of the piston,

On solving this quadratic equation in cos θ1, we get