effect of partial balancing of locomotives

EFFECT OF PARTIAL BALANCING OF LOCOMOTIVES

• In Section 13.3, we have discussed that due to partial balance of reciprocating masses there is

(a) an unbalanced force along the line of stroke, and

(b) an unbalanced force perpendicular to the line of stroke.

• The effect of an unbalanced force along the line of stroke is to produce:

1. Variation in tractive force along the line of stroke, and

2. Swaying couple.

• The effect of an unbalanced force perpendicular to the line of stroke is to produce:

1. Variation of pressure on the rails, which is known as hammer blow.

Now we shall discuss these effects one by one.

1. Variation of Tractive Force

• Definition: The resultant unbalanced force due to the two cylinders along the line of stroke is known as tractive force.

• Derivation:

1. To find tractive force (FȚ):

Let the crank for the first cylinder be inclined at an angle ✪ with the line of stroke. The crank for the second cylinder will be inclined at an angle (θ +90°) with the line of stroke as the crank for the second cylinder is at right angle to the first crank.

Let

m = Mass of the reciprocating parts per cylinder, and

c = Fraction of the reciprocating parts to be balanced. 

⸫ Total unbalanced primary force or variation in tractive force is given by 

2. To find maximum and minimum tractive force:

The tractive force is maximum or minimum when (cos θ – sin θ ) is maximum or minimum. Therefore for maximum or minimum tractive force value,

Case (i): Minimum value of tractive force will be at θ = 135° and is given by

Case (ii): Maximum value of tractive force will be at θ = 315° and is given by

 (FT)max = (1 - c) m ω2r [cos 315° - sin 315°]

3. To find maximum variation in tractive force:

From equations (13.8) and (13.9),

2. Swaying Couple

• Definition: The two unbalanced forces acting at a distance between the line of stroke of two cylinders, constitute a couple in the horizontal direction. This couple is known as swaying couple.

• This swaying couple tends to sway the engine alternatively in clockwise and anticlockwise directions in vertical axis (and hence it is named so).

• Derivation: Refer Fig.13.5.

1. To find swaying couple (CS):

Let

a = Distance between the centre lines of the two cylinders, and 

CS = Swaying couple

Swaying couple CS = Moments of forces about the engine centre line.

2. To find a maximum and minimum swaying couple:

The swaying couple is maximum or minimum when (cos θ + sin θ) is maximum or minimum. Therefore, for maximum or minimum swaying couple value,

Case (i): Minimum swaying couple will be at θ = 225° and is given by

Case (ii): Maximum swaying couple will be at θ = 45° and is given by

3. Hammer Blow

• Definition: The unbalanced force along the perpendicular to the line of stroke produces variation in pressure on the rails which results in hammering action on the rails. The maximum magnitude of this vertical unbalanced force is known as hammer blow.

• At high speeds, the force of the hammer blow could exceed the static load on the wheels and the wheels can be lifted off the rail when the direction of the hammer blow will be vertically downwards.

• Derivation: We know that the unbalanced force along the perpendicular to the line of stroke due to balancing mass B at radius b is given by

• The effect of hammer blow is to cause the variation in pressure between the wheel and the rail. The variation is shown in Fig. 13.6 for one revolution of the wheel.

Let

P = Downward pressure on the rails due to static wheel load W. 

If (P – B ω2 b) is positive, then the wheel will not be lifted from the rails. If (P - B ω2 b) is negative, then the wheel will be lifted from the rails. So the limiting condition in order that the wheel does not lift from the rails is given by

P = B ω2 b

This is the expression for permissible value of angular speed of wheel to avoid lifting of wheel from the track.

PROBLEMS ON TWO CYLINDER UNCOUPLED LOCOMOTIVES

Example 13.2

A two-cylinder uncoupled locomotive has inside cylinders 0.6 m apart. The radius of each crank is 300 mm and are at right angles. The revolving mass per cylinder is 250 kg and the reciprocating mass per cylinder is 300 kg. The whole of the revolving and two-third of the reciprocating masses are to be balanced and the balanced masses are to be placed, in the planes of rotation of the driving wheels a radius of 0.8 m. The driving wheels are 2 m in diameter and 1.5 m apart.

If the speed of the engine is 80 km/hr, determine:

(a) Hammer blow, (b) Maximum variation in tractive force, and

(c) Maximum swaying couple.

Given data: 

Two-cylinder uncoupled locomotive, a = 0.6 m; r = rB = rC = 300 mm = 0.3 m; m1 = 250 kg; m = 300 kg; c = 2/3; b = 0.8 m; D = 2 m or R = 2/2 = 1 m; L = 1.5 m; v = 80 km/hr = 22.22 m/s

Solution: 

Let us assume the two crank planes are B and C, and the two driving wheel planes are A and D. We know that the equivalent mass of the rotating parts to be balanced per cylinder at the crank pin or total mass to be balanced per cylinder,

Magnitude and direction of balance masses:

(i) Let us draw the plane diagram first, which shows the position of planes as in Fig.13.7(a).

(ii) Assuming plane B be horizontal, draw OC perpendicular to OB to indicate the angular position of planes as shown in Fig.13.7(b).

(iii) Let us tabulate the various data as shown in Table 13.1, taking plane A as the reference plane.

(iv) From the data in column (6) Table 13.1, draw couple polygon as shown in Fig.13.7(c), to some suitable scale (say, 1 cm = 50 kg.m). We know that the closing side of the couple polygon i.e., vector c'o' represents the balancing couple and a proportional to 1.2 MD.

(v) To determine the angular position of plane D, draw OD parallel to vector c'o' in Fig.13.7(b). By measurement, we get

(vi) Now draw the force polygon from the data in column (4) of Table 13.1, as shown in Fig.13.7(d), to some suitable scale (say 1 cm = 50 kg.m). We know that the closing side of the force polygon i.e., vector do represents the balancing force and is proportional to 0.8 mA. By measuring the closing side, we have

(vii) To determine the angular position of plane A, draw OA parallel to vector od in Fig.13.7(b). By measurement, we get

(a) Hammer blow:

It may be noted that the calculated balancing mass accounts for both rotating and reciprocating masses. But the effect of hammer blow is only due to the unbalanced force due to reciprocating masses only. Therefore in order to find the hammer blow, first we have to find the portion of the total balancing mass that accounts for reciprocating masses only. 

⸫ Balancing mass for reciprocating masses only*

⸫ Hammer blow or fluctuation in rail pressure is given by 

(b) Maximum variation in tractive force:

(c) Maximum swaying couple:

Example 13.3 

The following data refer to a two cylinder locomotive:

Rotating mass per cylinder = 300 kg; Reciprocating mass per cylinder = 330 kg; Distance between wheels = 1500 mm; Distance between cylinder centres = 600 mm; Diameter of treads of driving wheels = 1800 mm; Crank radius = 325 mm; Radius of centre of balance mass = 650 mm; Locomotive speed = 60 km/hr; Angle between cylinder cranks = 90°; Dead load on each wheel = 40 kN.

• In the similar manner, the balancing mass for rotating masses only

Determine:

(i) The balancing mass required in the planes of driving wheels if whole of the revolving and two-third of the reciprocating mass are to be balanced,

(ii) The swaying couple,

(iii) The variation in the tractive force,

(iv) The maximum and minimum pressure on the rails, and

(v) The maximum speed of locomotive without lifting the wheels from the rails. 

Given data: 

Two-cylinder locomotive m1 = 300 kg; m = 330 kg; L = 1500 mm = 1.5 m; a = 600 mm; D = 1800 mm 1.8 m; rB = rC = 325 mm = 0.325 m; b = 650 mm = 0.65 m; v = 60 km/hr = 16.67 m/s; P = 40 kN; c = 2/3.

Solution:

(i) Balancing masses required:

Let us assume the two crank planes are B and C, and the two driving wheel planes are A and D. We know that the equivalent mass of the rotating parts to be balanced per cylinder at the crank pin or total mass to be balanced per cylinder.

The magnitude and direction of the balancing mass can be obtained in the same manner, using the same proçedure as discussed in the previous problem.

Fig.13.8(a) shows the position of planes and Fig.13.8(b) indicates the angular position of planes. Table 13.2 represents the various data.

By measuring the closing side of the couple polygon (Fig.13.8(c)), we get 

By measuring the angular position of plane D from Fig.13.8(b), we get 

By measuring the closing side of the force polygon (Fig. 13.8(d)), we get 

By measuring the angular position of plane A from Fig.13.8(b), we get

⸫ The balancing mass = mA = mD = 200 kg

(ii) Swaying couple:

(iii) Variation in tractive force:

(iv) Maximum and minimum pressure on the rails:

We know that balancing mass for reciprocating masses only,

(v) Maximum speed of locomotive without lifting the wheels from the rail:

We know that the permissible value of angular speed of locomotive to avoid lifting,

⸫ Maximum speed of locomotive, v = Rω

Example 13.4 

The following particulars relate to an outside cylinder of an uncoupled locomotive:

Revolving mass per cylinder = 300 kg; Reciprocating mass per cylinder = 450 kg; Length of each crank 350 mm; Distance between wheels = 1.6 m; Distance between cylinder centres = 1.9 m; Diameter of driving wheels = 2 m; Radius of balancing mass = 0.8 m; Angle between the cranks = 90°

If whole of the revolving and 2/3 of reciprocating masses are to be balanced in planes of driving wheels, determine:

(i) magnitude and angular positions of the balance masses,

(ii) speed at which the wheel will lift off the rails when the load on each driving wheel is 35 KN, and

(iii) swaying couple at speed arrived in (ii) above.

Given data: 

Outside cylinder uncoupled locomotive; m1 = 300 kg; m = 450 kg; r = rA = rD = 0.35 m; Distance between wheels 1.5 m; a = 1.9 m; D = 2 m; b = 0.8 m; Angle between cranks = 90°; c = 2/3; P = 35 kN.

Solution: 

Let us assume the two crank planes are A and D, and the two driving wheel planes are B and C. We know that equivalent mass of the rotating parts to be balanced per cylinder at the crank pin,

(i) Magnitude and angular position of the balance masses:

(i) Let us draw the space diagram first, which shows the position of planes as in Fig.13.9(a). It may be noted from Fig.13.9 that the given locomotive is outside cylinder type.

(ii) Assuming plane A be horizontal, draw OD perpendicular to OA to indicate the angular position of planes as shown in Fig.13.9(b).

(iii) Let us tabulate the data as shown in Table 13.3, taking plane B as the reference plane.

(iv) Using the data in column (6) of Table 13.3, draw couple polygon as shown in Fig.13.9(c), to some suitable scale (say, 1 cm = 50 kg.m2). We know that the closing side of the couple polygon i.e., vector d'o' represents the balancing couple and is proportional to 1.28 mC.

By measuring the closing side, we have

(v) To determine the angular position of plane C, draw OC parallel to vector d'o' in Fig.13.9(b). By measurement, we have

(vi) Now draw the force polygon from the data in column (4) of Table 13.3 as shown in Fig.13.9(d) to some suitable scale (say, 1 cm = 50 kg.m). We know that the closing side of the force polygon i.e., vector co represents the balancing force and is proportional to 0.8 mB. By measuring the closing side, we have

(vii) To determine the angular position of plane B, draw OB parallel to vector co in Fig.13.9(b). By measurement, we get

(ii) Speed at which the wheel will lift off the rails:

Given: P = 35 kN; D=2m or R=2/2 = 1 m

Let v and ω = Linear and angular speed at which the wheels lift off the rails respectively.

Balancing mass for reciprocating parts

(iii) Swaying couple at speed ω = 17.52 rad/s: