Balancing of multi-cylinder in-line engines

BALANCING OF MULTI-CYLINDER IN-LINE ENGINES

• If an engine consists of two or more cylinders, then it is called as multi-cylinder engine. 

• If the cylinder centre lines of the multi-cylinder engines are in the same plane and also in the same side of the centre line of the crank shaft, then the multi-cylinder engines are known as in-line engines.

• As discussed already, it is impossible to completely balance the single-cylinder reciprocating engine. This drawback has been overcome by using multi-cylinder engines. 

• Fig.13.11 illustrates a typical four-cylinder in-line engine having two inner cranks and two outer cranks. In this engine, the inner cranks 2 and 3 are 180° from the outer cranks 1 and 4.

1. Balancing of Primary Forces of Multi-cylinder In-line Engines

For the complete primary balance of the reciprocating parts of a multi-cylinder engine, the following two conditions must be satisfied:

1. The algebraic sum of the primary forces must be equal to zero i.e., primary force polygon must close.

2. The algebraic sum of the primary couple must be equal to zero i.e., primary couple polygon must close.

As discussed already in Section 13.3, the primary unbalanced force due to a reciprocating mass is equal to the component parallel to the line of stroke of the centrifugal force produced by an equal mass attached to, and revolving with, the crank pin.

In order to give the primary balance of the reciprocating parts of a multi-cylinder engine, it is convenient to imagine the reciprocating masses to be transferred to their respective crankpins and to treat the problem as one of revolving masses.

Note 

1. The closing side of the primary force polygon gives the maximum unbalanced primary force and the closing side of the primary couple polygon gives the maximum unbalanced primary couple.

2. If the primary force polygon is a closed figure, then there is no primary unbalanced force. If the primary couple polygon is a closed figure, then there is no primary unbalanced couple.

2. Balancing of Secondary Forces of Multi-cylinder In-line Engines 

We know that the secondary force due to the reciprocating mass is given by

The equation (i) can also be written as

Now consider two cranks of an engine. One is actual crank and the other one is imaginary with the following specifications.

Thus when the actual crank has turned through an angle θ, the imaginary crank would have turned an angle of 2θ.

From the equations (ii) and (iii), it is clear that the effect of the secondary force is equivalent to an imaginary crank of length (r/4n) rotating at twice the speed of the actual crank (i.e., 2ω), as shown in Fig.13.12.

For the complete secondary balance. of the reciprocating parts of a multi- cylinder engine, the following two conditions must be satisfied:

1. The algebraic sum of the secondary forces must be equal to zero i.e., secondary force polygon must close.

2. The algebraic sum of the secondary couples must be equal to zero i.e., secondary couple polygon must close.

Note

1.The closing side of the secondary force polygon gives the maximum unbalanced secondary force and the closing side of the secondary couple polygon gives the maximum unbalanced secondary couple.

2. If the secondary force polygon is a closed figure, then there is no unbalanced secondary force. If the secondary couple polygon is a closed figure, then there is no unbalanced secondary couple.

3. Conditions for Complete Balancing of Multi-cylinder Engines

For complete balancing of multi-cylinder engines, the following conditions must be satisfied:

1. Primary force must balance, i.e., primary force polygon must close. 

2. Primary couple must balance i.e., primary couple polygon must close. 

3. Secondary force must balance i.e., secondary force polygon must close.

4. Secondary couple must balance i.e., secondary couple polygon must close.

4. Firing Order

• Firing order is nothing but the sequence in which charge is ignited inside the engine cylinder.

• In multi-cylinder Engines, there are several possibilities of the order in which firing takes place. To overcome the problems of vibration, fuel distribution, exhaust distribution, etc., the designers are selecting different firing orders.

Example 13.5 

The crank and connecting rods of a four-cylinder in-line engine running at 1800 rpm are 60 mm and 240 mm each respectively and the cylinders are spaced 150 mm apart. If the cylinders are numbered 1 to 4 in sequence from one end, the cranks appear at intervals of 90° in an end view in the order 1-4-2-3. The reciprocating mass corresponding to each cylinder is 1.5 kg.

Determine: (i) unbalanced primary and secondary forces, if any and (ii) unbalanced primary and secondary couples with reference to central plane of the engine.

[A.U., Nov/Dec 2010] 

Given data: 

4-cylinder in-line engine; N = 1800 rpm; r = 60 mm = 0.06 m; l = 240 mm = 0.24 m; Distance between cylinders 150 mm = 0.15 m; m = 1.5 kg.

Solution: 

ω = 2лN/60 = 2π (1800)/60 = 188.5 rad/s 

The position of the cylinder planes is shown in Fig.13.13(a). Taking central plane of engine as reference, the various data may be tabulated as shown in Table 13.4.

(i) Unbalanced primary forces and primary couples

(a) Primary force polygon: For the given firing order, i.e., 1–4–2–3, the primary crank position is shown in Fig.13.13(b). Draw primary force polygon, using the data in column (4) of Table 13.4 and considering primary crank positions, as shown in Fig.13.13(c), to some suitable scale (say, 1 cm = 0.03 kg.m).

Since the obtained primary force polygon is closed, therefore there is no unbalanced primary force. Ans.

(b) Primary couple polygon: Draw primary couple polygon, using the data in column (6) of Table 13.4 and considering primary crank positions, as shown in Fig.13.13(d), to some suitable scale (say, 1 cm = 0.005 kg.m2). We know that the closing side of the primary couple polygon (shown by dotted line) represents the magnitude of unbalanced primary couple i.e., CUP. By measurement, we get

(ii) Unbalanced secondary forces and secondary couples

(a) Secondary force polygon: First draw the secondary crank positions, using the data in column (8) of Table 13.4, as shown in Fig.13.13(e). Now draw secondary force polygon, using the data in column (4) of Table 13.4 and considering secondary crank positions, as shown in Fig.13.13(f), to some suitable scale (say, 1 cm = 0.03 kg.m).

Since the obtained secondary force polygon is closed, there is no unbalanced secondary force. Ans.

(b) Secondary couple polygon: Draw secondary couple polygon, using the data in column (6) of Table 13.4 and considering secondary crank positions, as shown in Fig.13.13(g), to some suitable scale (say, 1 cm = 0.005 kg.m2). We know that the closing side of the secondary couple polygon (shown by dotted line) represents the magnitude of unbalanced secondary couple CUS. By measurement, we get

Example 13.6 

An air compressor has four vertical cylinders 1, 2, 3 and 4 in line and the driving cranks at 90° intervals reach their upper most positions in this order. The cranks are of 150 mm radius, the connecting rods 500 mm long and the cylinder centre line 400 mm apart. The mass of the reciprocating parts for each cylinder is 22.5 kg and the speed of rotation is 400 rpm. Show that there are no out-of-balance primary or secondary forces and determine the corresponding couples, indicating the positions of No.1 crank for maximum values. The central plane of the machine may be taken as reference plane. 

Given data: 

r = 150 mm = 0.15 m; l = 50 mm = 0.5 m; Distance between cylinder = 0.4 m; m = 22.5 kg; N = 400 rpm.

Solutions 

ω = 2πN/60 = 2π (400)/60 = 41.88 rad/s

The position of cylinder planes is shown in Fig.13.14(a). Taking central plane of engine as reference, the various data may be tabulated as shown in Table 13.5. Since the firing order is not given, let us assume it as 1-2-3-4.

1. Unbalanced primary forces: Draw primary force polygon, using the data in column (4) of Table 13.5 and considering primary crank positions, as shown in Fig.13.14(c), to some suitable scale (say, 1 cm = 1 kg.m).

Since the obtained primary force polygon is closed, therefore there is no unbalanced primary force. Ans.

2. Unbalanced primary couples: Draw primary couple polygon, using the data in column (6) of Table 13.5, and considering primary crank positions, as shown in Fig.13.14(d), to some suitable scale (say, 1 cm = 1 kg.m2). We know that the closing side of the primary couple polygon (shown by dotted line) represents the magnitude of unbalanced primary couple i.e., CUP. By measurement, we get

This couple makes 45° and (180° + 45°) with direction of crank 1. Ans.

3. Unbalanced secondary forces: First draw the secondary crank positions, using the data in column (8) of Table 13.5, as shown in Fig.13.14(e). Now draw secondary force polygon, using the data in column (4) of Table 13.5 and considering secondary crank positions, as shown in Fig.13.14(f), to some suitable scale (say, 1 cm = 1 kg.m).

Since the obtained secondary force polygon is closed, therefore there is no unbalanced secondary force. Ans.

4. Unbalanced secondary couples: Draw secondary couple polygon, using the data in column (6) of Table 13.5 and considering secondary crank positions, as shown in Fig.13.14(g). We know that the closing side of the secondary couple polygon (shown by dotted line) represents the magnitude of unbalanced secondary couple i.e., CUS.

⸫ Unbalanced secondary couple, CUS = 2.025 + 2.025 - 0.675 - 0.675

This couple makes an angle 90° with crank 1 at all its positions. Thus it can occur at 0°, 90°, 180o, and 270° with crank 1. Ans.

Example 13.7

A four-cylinder in-line marine oil engine has cranks at angular displacement of 90°. The outer cranks are 3 m apart and inner cranks are 1.2 m apart. The inner cranks are placed symmetrically between the outer cranks. The length of each crank is 450 mm. If the engine runs at 90 rpm and the mass of reciprocating parts for each cylinder is 900 kg, find the firing order of the cylinders for the best primary balancing force of reciprocating masses. Determine the maximum unbalanced primary couple for the best arrangement.

Given data: 

r = 450 mm = 0.45 m; N = 90 rpm; m = 900 kg

Solution: 

ω = 2лN/60 = 2 π (90)/60 = 9.42 rad/s

Given that the cranks are placed at an angular displacement of 90° to each other, so the primary force will be balanced irrespective of the order of firing.

The position of cranks when the resultant couple is least, is the best arrangement for firing. Taking plane 1 as reference plane, the various data may be tabulated as shown in Table 13.6.

1. First disposition of cranks (firing order 1—2—3—4):

Let us assume that the cranks are initially put in order 1-2-3-4. The corresponding primary crank position is shown in Fig.13.15(b).

Draw the primary couple polygon; using the data in column (6) of Table 13.6, as shown in Fig.13.15(c), to some suitable scale (say, 1 cm = 400 kg.m2). We know that the closing side of the polygon represents the unbalanced primary couple, i.e., CUP1. By measuring the closing side, we get

Unbalanced primary couple, CUP = 1200 kg.m2 = 1200 ω2 N.m 

= 1200 (9.42)2 = 106.48 kN.m

2. Second disposition of cranks (firing order 1-4-2-3):

Let 1-4-2-3 be the order of firing. The corresponding primary crank positions is shown in Fig. 13.15(d). Draw the primary couple polygon, using the data in column (6) of Table 13.6, as shown in Fig.13.15(e), to some suitable scale (say, 1 cm 400 kg.m2). We know that the closing side of the polygon represents the unbalanced primary couple, i.e., CUP2. By measuring the closing side, we get

Unbalanced primary couple, CUP2 = 534 kg.m2 = 534 × ω2 N.m

= 534 (9.42)2 = 47.38 kN.m

3. Third disposition of cranks (firing order 1–2–4–3):

Let 1-2-4-3 be the order of firing. The corresponding primary crank positions is shown in Fig. 13.15(f). Draw the primary couple polygon, using the data in column (6) of Table 13.6, as shown in Fig.13.15(g), to some suitable scale (say, 1 cm = 400 kg.m2). We know that the closing side of the polygon represents the unbalanced primary couple i.e., CUP3. By measuring the closing side, we get

Unbalance primary couple, CUP3 = 1310 kg.m2 = 1310 × ω2 N.m 

= 1310 (9.42)2 = 116.24 kN.m

Any other possible crank arrangement other than above three positions will yield the same arrangements.

Now the best arrangement among above for firing order is 1–4–2–3, which gives the least primary unbalance couple of 47.38 kN.m Ans.

Example 13.8 

The firing order of a six-cylinder, vertical, four-stroke, in-line engine is 1-4-2-6-3-5. The piston stroke is 80 mm and length of each connecting rod is 180 mm. The pitch distances between the cylinder centre lines are 80 mm, 80 mm, 120 mm, 80 mm and 80 mm respectively. The reciprocating mass per cylinder is 1.2 kg and the engine speed

• Taking crank 1 along the horizontal axis (i.e., 0°) always, the other three cranks can be arranged in the following six ways:

(i) 1-2-3-4 (ii) 1-2-4-3 (iii) 1-3-2-4 (iv) 1-3-4-2 (v) 1–4–2–3 (vi) 1-4-3-2

But the following pairs of firing orders would produce the same amount of primary couple.

⸫ 1–2–3–4 = 1-4-3-2; 1-2-4-3 = 1-3-4-2; 1-3-2-4 = 1-4-2-3

Therefore, in real sense, there are only three different firing orders as follows:

(1) 1-2-3-4; (2) 1-4-2-3; and (3) 1-4-2-3.

is 2400 rpm. Determine the out-of-balance primary and secondary forces and couples on the engine taking a plane mid-way between the cylinders 3 and 4 as the reference plane. 

Given data: 

L = 80 mm or r = 40 mm 0.04 m; l = 180 mm = 0.18 m; m = 1.2 kg; N = 2400 rpm.

Solution: 

ω = 2πN/60 = 2 π (2400)/60 = 251.33 rad/s

The position of the cylinder planes and crank planes are shown in 13.16(a) and (b) respectively. Taking a plane mid-way between the cylinders 3 and 4 as the reference, the various data may be tabulated as in Table 13.7.

1. Primary force polygon: Draw the primary force polygon, using the data in column (4) of Table 13.7 and considering primary crank positions, as shown in Fig.13.16(c), to some suitable scale (say, 1 cm = 0.048 kg.m).

Since the obtained primary force polygon is closed, therefore there is no unbalanced primary force. Ans.

2. Primary couple polygon: Draw the primary couple polygon, using the data in column (6) of Table 13.7 and considering primary crank positions, as shown in Fig.13.16(d), to some suitable scale (say, 1 cm = 0.005 kg.m2).

Since the obtained primary couple polygon is closed, therefore there is no unbalanced primary couple. Ans.

3. Secondary force polygon: First draw secondary crank positions using the data in column (8) of Table 13.7, as shown in Fig.13.16(e). Now draw the secondary force polygon, using the data in column (4) of Table 13.7 and considering the secondary crank positions, as shown in Fig.13.16(f), to some suitable scale (say, 1 cm = 0.048 kg.m).

Since the obtained secondary force polygon is closed, therefore there is no unbalanced secondary force. Ans.

4. Secondary couple polygon: Draw secondary couple polygon, using the data in column (6) of Table 13.7 and considering the secondary crank positions, as shown in Fig.13.16(g), to some suitable scale (say, 1 cm = 0.005 kg.m2).

Since the obtained secondary couple polygon is closed, therefore there is no unbalanced couple polygon. Ans.