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Chapter 7 Balancing of Reciprocating Machines We have already discussed how to obtain inertia forces of linkages (the four-bar and reciprocating engine mechanisms). Such inertia forces cannot be avoided and we have to consider these forces in designing of linkages. (These inertia forces increase stresses in members and loads in bearing). Inertia force αω 2 , where ω is the angular velocity of links. For linkages with very high speed, inertia forces (or loading) results in vibrations, noise emission and sometimes machine failure due to fatigue. Inertia force of a reciprocating mass: y A 2 ω l 3 r φ O2 B x x r+l Figure 1 A reciprocating mechanism The displacement of the piston is given as 1/ 2 sin 2 θ x = r cosθ + l 1 − 2 n with n = l/r (1) The velocity of the piston can be obtained as s in 2 θ 1 x = − ω r s in θ + 0 .5 2 n 1 − n − 2 s in 2 θ (2) The acceleration of the piston can be obtained as ( n 2 − 1)cos 2θ + cos 4 θ x = −ω 2 r cosθ + 1.5 n3 1 − n −2 sin 2 θ In equation (1) the term 1 − (sin 2 θ ) / n 2 (3) 0.5 is expanded bi-nominally to give a harmonic analysis. i.e x, x and x can be represented as the cosine and sine terms of harmonics. The frequencies representing integral multiples of the fundamental frequency are called harmonics, with the fundamental frequency being the first harmonic. For example: ω0 2ω fundamental harmonic nω … 0 0 n th harmonic second harmonic A series of harmonic functions are known as the Fourier series. A harmonic function of any arbitrary frequency ω is periodic i.e. it repeats itself at equal intervals of time T = 2π / ω , where T is the period of excitation. If a function is periodic, it is not necessary that it will be harmonic.Let x = A cos ωt is a harmonic function with period 2π / ω or frequency ω . Any arbitrary periodic functions can be represented by a convergent series of harmonic functions whose frequencies are integral multiples of certain fundamental frequency ω 0 (= 2π / T ) . Fourier Series: f (t ) = 12 a0 + ∑ r =1 (ar cos rωt + br sin rω0t ) n (4) with T /2 ar = 2 f (t ) cos rω0 t dt T −T∫/ 2 ω 0 = 2π / T ; T /2 and br = 2 f (t )sin rω0 t T −T∫/ 2 r = 1, 2,3, ⋅⋅⋅ where T is the period of function f(t). Shaking force: Bearing loads due to inertia forces of rotating/reciprocating masses of mechanism causes shaking of the mechanism due to its dynamic nature i.e. it changes with time. This bearing loads due to inertia forces of mechanism are called shaking forces. Shaking force can be minimised by balancing the inertia forces in opposition to each other such that little or no force is transmitted to machine supports. The degree to which this shaking force is undesirable depends upon the frequency, ω , of shaking force and the natural frequency, ωnf , of flexible members (such as shafts, bearings, supports, etc.) through which the force is transmitted. For ω = ωnf we have undesirable resonance condition. Primary and secondary forces can be replaced by rotating masses as: 1. A mass mb at radius r at A and rotating with crank with ω the component in horizontal direction will be same as by reciprocating mass mb of piston i.e. primary force. 122 2. A, mass mb at radius ( r / 4n) and rotating at twice the crank angular velocity ω . The component in the horizontal direction will be same as that due to piston mb i.e. secondary force. The acceleration of piston can be obtained up to second harmonics as: A p = −ω 2 r (cos θ + where, cos 2θ ) n (5) n=l/r primary force F32 cosφ =mbrω 2cosθ A φ F32 F32Sin φ θ θ mbω2cosφ mb O2 mbω2 bsin φ mbω2 b Figure 2 A crack with a counter balance mass ( F32 sin φ ) r cos θ + (mb ω 2 b sin φ )b cos θ + T12 = 0 with T12 = rF32 sin φ cos θ − mb ω 2 b 2 sin φ cos θ For a single cylinder engine the inertia force due to the reciprocating mass, m B = m3 B + m4 , will be cos 2θ F = mBω 2 r cos θ + n (6) where m3 B is one of the equivalent mass of connecting rod at piston and m 4 is the mass of piston. For F → + ve the inertia force will be directed away from main bearing and for F → −ve the inertia 123 force towards the main bearing. Bearing load due to this inertia force will be in the same direction as the inertia force. In equation (6), various terms can be defined as follows: First term: mb rω 2 cos θ (7) is called primary force. It reaches its maximum value of mb rω 2 twice per revolution, when θ = 0 and π . Second term: mBω 2 r cos 2θ n (8) is called secondary force. It reaches its maximum value of mBω 2 r / n four times per revolution of crank when θ = 0, π / 2, π and 3π / 2 . (i) Harmonic motion (Fundamental) is given as x = A cos ω t , where A is the amplitude and ω is the x = − ω 2 A cos ω t , where T = 2π / ω is the period and frequency so that x = −ω A sin ω t and θ =ωt . The primary inertia force Fi p = + mω 2 A cos ω t (ii) Second harmonic motion is given as x = A cos 2ω t , so that x = −2ω A sin 2ω t and x = −4ω 2 A cos 2ω t . The secondary inertia force Fi s = 4mω 2 A cos 2ω t = (4m)ω 2 A cos 2ω t Both the primary and secondary inertia forces represent a force, which is constant in direction but varying in magnitude, and hence cannot be completely balanced by a rotating mass (or which inertia force is varying in the direction but is constant in the magnitude). Assumption: (i) The replacement of the connecting rod by two masses at its ends maintains complete dynamic equivalence (ii) The crank is so designed that the combined centre of mass of the crank and the apportioned connecting rod mass, m3 , at the crank pin lies on the axis of rotation, and (iii) the engine speed is constant. For multi-cylinder engines, we shall apply these assumptions to each constituent cylinder. 124 Bearing load: In reciprocating engine the bearing load due to the piston inertia can be obtained with help of following free body diagram, which is self-explanatory. From O2 AB q = x sin φ T = F34 sin φ x = F34 . q = F12 . q = T21 T = T21 Figure 3 (a) A reciprocating engine Figure 3 (c) Free body diagram of the piston Figure 3(b) Free body diagram of the connecting rod Figure 3 (d) Free body diagram of the crank Figure 3 (f) Equivalent system of Figure (e) Figure 3 (e) Free body diagram of bed 125 The full effect on the engine frame of the inertia of the reciprocating mass is equivalent to the force, mω 2 r (cos θ + n −1 cos 2θ ) , along the line of the stroke at O. Figure 4 (a) (top) Bearing load due to rotating mass (bottom) reaction force on the engine frame Figure 4 (b) (top) Centrifugal forces on disc (bottom) Free body diagram of disc From Figure 4, it can be seen that the frame will experience forces: (i) horizontal direction mω 2 e cos θ and (ii) vertical direction mω 2 e sin θ . The primary inertia force can be balanced by a counter mass mb at radius b such that mb b = mB r . Figure 5 A reciprocating machine with a counter balance mass The component of inertia force mb along the line of stroke is mb bω 2 cos θ . But a force mb bω 2 sin θ is then introduced in a direction ⊥ (perpendicular) to line of stroke. In most practical applications, the reciprocating mass is partially balanced by a rotating mass mb , reducing the inertia force in the line of stroke and ⊥ direction, such that cmB r = mbb 126 where c is the fraction of reciprocating mass to be balanced and mB is the mass of the slider. The unbalanced primary force along the line of action is given by = (mB r cosθω 2 − mbbω 2 cosθ ) or mb .rω 2 cosθ − cmB rω 2 cosθ or (1 − c)mB rω 2 cos θ (9) The unbalance force in the vertical direction mbbω 2 sin θ = cmB rω 2 sin θ (10) Resultant unbalance force at any instant = mB rω 2 (1 − c) 2 cos 2 θ + c 2 sin 2 θ (11) If the revolving and reciprocating (primary force) masses to be balanced simultaneously then, mbb = mAr + cmB r = (mA+ cmB )r (12) where mA is the mass of rotating components. The direction of resultant force at any instant is given as tan φ = cmB rω 2θ c tan θ = 2 (1 − c)mB rω cosθ 1 − c (b) (a) (13) (c) Equivalent of Figure (c) Figure 6 Frame reaction in the horizontal direction Resultant R will be rotating in opposite direction to crank (i.e. cw). For c = 12 ; R is constant and rotating with ω in cw direction. So primary force can be balanced as shown in Figure 7. 127 Figure 7 Conditions of the balance in a multi cylinder engine: c1 c2 2 2 3 m2 m3 m4 m1 1 3 m3 m2 m1 4 c3 stroke c4 m4 Figure 8 Multi-cylinder in-line engine In Figure 8 θ1 , θ 2 and θ 3 are θ 4 are angular positions of the multi cylinder engine (Four engines in Figure 8) and φ 2 , φ3 and φ 4 are the phase difference of cranks 2, 3, 4 with respect to 1. The line of stroke of each cylinder is assumed to in the horizontal plane through the crankshaft axis. The inertia force of reciprocating masses will be situated in same horizontal plane. For each cylinder: r is the crank radius, l is the connecting rod length, m is the total reciprocating mass (the piston and the connecting rod) and φi is the angle between cranks, which depends upon the firing order. Subscript i represent the cylinder number (i = 1, 2, 3, 4). Equilibrium of Primary forces: Conditions for the complete equilibrium of primary forces are: 1. The algebraic sum of primary forces is zero and 2. The algebraic sum of their moments with respect to any of the transverse plane ( ⊥ to crankshaft) is equal to zero. 128 The resultant primary shaking force will be F p = m1r1ω 2 cos θ1 + m2 r2ω 2 cos θ 2 + m3 r3ω 2 cos θ 3 + ⋅ ⋅ ⋅ ⋅ + mn rnω 2 cos θ n n = ∑ mi ri ω 2 cos θ i (14) i =1 with θ i = θ1 + φi i = 1,2,3 ⋅ ⋅ ⋅ n (15) where θ1 is the variable and φi are constants and φ1 = 0 . Substituting (14) into (15), we get F p = ∑ mi ri ω 2 cos(θ1 + φi ) or or (16) n n i =1 i =1 = ∑ mi ri ω 2 cos θ1 . cos φi − ∑ mi ri ω 2 sin θ1 sin φi n n i =1 i =1 ω 2 cos θ1 ( ∑ mi ri cos φi ) − ω 2 sin θ1 ( ∑ mi ri sin φi ) The primary force, F p , will be maximum, when δF p δθ1 (17) = 0 i.e. ∑ mr sin φ ∑ mr cos φ θ1 = tan −1 − (18) For F p to be zero for all values of θ1 , then from eqn. (17), we get n ∑ mr cos φ = 0 i =1 (19) n mr sin φ = 0 ∑ i =1 (20) and Similarly, for moments, we can write n ∑ mra cos φ = 0 i =1 (21) n mra sin φ = 0 ∑ i =1 (22) and where a is the distance from the center line (line of stroke) of any cylinder to an arbitrarily selected reference plane ( ⊥ to the crank shaft axis). Thus, for complete balancing of primary forces and moments, eqns. (19) to (22) should be satisfied. 129 Equilibrium of Secondary Forces: Conditions for complete equilibrium of the secondary forces are: 1. The algebraic sum of the secondary forces is zero and 2. The algebraic sum of their moments with respect of any transverse plane is also zero. Thus the resultant of secondary forces: n r Fs = ∑ mi riω 2 cos 2θ i i i −1 with 1 ri = ni li (23) n r or Fs = ∑ mi riω 2 cos 2 (θ1 + φi ) l i i −1 or mr 2 mr 2 Fs = ω 2 cos 2θ1 ∑ cos 2φ − ω 2 sin 2θ1 ∑ sin 2φ l l (i) Fs is maximum, when (24) ∂Fs = 0 which gives ∂θ1 mr 2 sin 2φ ∑ 1 l θ1 = tan −1 − 2 mr 2 cos 2φ ∑ l (25) (ii) For Fs = 0 i.e. criteria for the balancing of the secondary force is (from eqn. 24) ∑ mr 2 cos 2φ = 0 l (26) ∑ mr 2 sin 2φ = 0 l (27) and (iii) For balancing of moments, we have: ∑ mr 2 .a sin 2φ = 0 l (28) and ∑ mr 2 .a cos 2φ = 0 l (29) 130 Thus for complete secondary balancing eqns. (26)-(29) has to be satisfied. Fundamental eqns. (19) to (22) and eqns. (26) to (29) express completely the primary and secondary balances for reciprocating engine with n-cylinders arranged in a line on one side of the crank shaft with the cylinder centre lies all in the plane containing the crankshaft axis. Firing order affects the balance of multi cylinder engine: For example Two-stroke engine: One cycle of operation in one revolution of crankshaft. Interval between the crank= 2π / n . Four- stroke engine: One cycle of operation in two revolutions of crankshaft. Interval between the crank= 4π / n , where n is numbers of cylinder in the multi-cylinder engine. (A) Crank arrangement for two-stroke engine: (1) The crank arrangement can be so chosen to make the cylinder firing not in a sequential order but distribute this on either side of the crankshaft with respect to its middle plane. (2) Once crank arrangement is fixed there may be several choice on firing order itself (see specially four- stroke engine). Cranks are rotating in the cw direction. Firing will start say from engine 1. Next cylinder, which is π / 2 in phase in ccw, will be ready for firing (because of completion of compression stroke). Next firing can be done in engine 4 then in 2 and finally in 3. Hence, the firing order will be: 1-4-2-3. Another version can be 1-3-2-4. For given crank arrangement (see in Figure 9 for 4, 5 and 6 cylinders) for two- stroke engine only unique firing order is possible i.e. for 4 cylinder: 1-4-2-3, 5 cylinder: 1-52-4-3 and 6 cylinder: 1-6-2-5-3-4. 1 1 2 3 4 ° 4 90 3 2 Firing order 1-4-2-3 (a) Four cylinder engine (interval= 131 2π π = ) 4 2 5 2 1 1 ° 3 4 5 3 72 2 4 Firing order 1-5-2-4-3 (b) Five cylinder engine: interval= 1 1 6 2 3 2π = 72o 5 4 5 6 4 60° 3 2 5 Firing order 1-6-2-5-3-4 (c) Six cylinder engine: interval= 2π = 60o 6 Figure 9 Two-stroke engine (B) Four stroke engine: For a given crank arrangement there may be several firing order possible. Refer to Figure 10. For 4- cylinder the firing order could be: 1-4-2-3 or 1-3-2-4; for 5-cylinder the firing order could be: 1-5-3-4-2 (only one) and for 6- cylinder the firing order could be: 1-6-2-5-3-4, 1-6-4-5-3-4 or 1-3-4-5-6-2. 1 2 4 3 1 2 3 4 Firing order 1-4-2-3 (a) Four cylinder engine: internal = 4π =π 4 132 4 144° 1 1 2 3 4 5 5 3 4 2 3 1-0 3 5-144 2 2-288 Firing order 1-5-2-4-3 (b) Five cylinder: interval = 1 1 5 2 3 4-432 ≡ 72° 4π = 144 5 4 3-576 ≡ 216° 5 6 4 6 3 2 Firing order 1-6-2-5-3-4 (c) Six cylinder: interval = 4π = 120 6 Figure 10 Four-stroke engine Balancing of Radial Engines (Direct and reverse crank method) mr 2 Direct crank A A mr 2 B O A Reverse crank Fig. 1(a) O2 Figure 11(b) Primary direct and reverse crank Figure 11(a) Slider-crack mechanism Reverse crank OA′ is an image of direct crank OA. Direct crank and reverse crank rotates in opposite direction. Let mr is the reciprocating mass at B, then primary force will be mr ω 2 r cos θ . This is 133 equivalent to the component of the centrifugal force produced by mass mr at A. Let total reciprocating mass m r is divided into two parts i.e. m r / 2 for each in the direct crank and in the reverse crank as shown in Fig. 11(a). The centrifugal force acting on primary direct and reverse crank is mr 2 m ω r . Total component in the direction of line of stroke = 2 r ω 2 r cos θ = mrω 2 r cosθ . The 2 2 component in vertical direction cancel each other i.e. mr 2 ω r sin θ . 2 2 r The secondary force is mr ( 2ω ) cos 2θ , where mr is replaced by the mass of mr / 2 each 4n 4r placed at c and c ′ as shown in Figure 11(c) with oc = oc′ = = crank radius of secondary crank. n Secondary crank oc rotates at 2ω rad/sec . Secondary crank oc ′ rotates at −2ω rad/sec . c direct crank 2 o 2 c reverse crank Figure 11(b) Secondary direct & reverse crank Three cylinders radial engine: Line of stroke is 120º. All three lines of stroke are in one plane. Shaking forces from reciprocating parts of each cylinder will be acting along the line of strokes in a single plane. Let m r be the mass of reciprocating part of each cylinder. Cylinder no. Direction of rotation Crank angle Direction of rotation Crank angle 1 ccw 0º cw 0º 2 ccw 240º cw 120º 3 ccw 120º cw 240º Primary unbalance: mass = 3mr 3m ; crank radius = r; speed= ω cw; Fp = r ω 2 r . It can be balanced 2 2 by a balancing mass mb at radius b i.e. mb b = 3mr and placed at opposite to the crank. 2 134 mr 1 Line of stroke connecting rod for cylinder 1 120 120 120° connecting rod for cylinder2 mr connecting rod for cylinder 3 line of stroke 2 3 m r Figure 12(a) Three cylinders radial engine 1,2,3 mr 2 mr 3× 2 1 mr 2 2 3 mr 2 Direct primary cranks Reverse primary cranks Figure 12(b) 1 mr m 3× r 2 2 mr 2 3 2 mr 2 2 Direct Secodary cranks (Balanced by itself) 1,2,3 2 Reverse secondary cranks Figure 12(c) Secondary unbalance:mass = 3m r r l , radius = , n= 4n r 2 and speed 2 r 3m Fs = r ( 2ω ) ; by gear train we have to obtain secondary balancing. 4n 2 135 = +2ω ccw . Hence, Tutorial Problems: (1) A single cylinder horizontal oil engine has a crank 190.5 mm long and a connecting rod 838.2 mm long. The revolving parts are equivalent to 489.3 N at crank radius and the piston and gudgeon pin weigh 400.32 N. The connecting rod weighs 511.52 N and its center of gravity is 266.7 mm from the crank pin centre. Revolving balance weights are introduced at a radius of 215.9 mm on extensions of the crank webs in order to balance all the revolving parts and one-half of the reciprocating parts. Find the magnitude of the total balance weight and neglecting the obliquity of the connecting rod. ( i.e. l / r 〉〉 0 ) , the nature and magnitude of the residual unbalanced force on the engine. RPM=300 cw. (Answer: 987.456 N, 5390.976 N and revolving at 300 rpm ccw). (2) A four cylinder Marin oil engine has the cranks arranged at angular interval of 90º. The inner cranks are 1.22 m apart and are placed symmetrically between the out cranks, which are 3.05 m apart. Each crank is 457.2 mm long, the engine runs at 90 rpm; and the weight of the reciprocating parts for each cylinder is 8006.4 N. In which order should the cranks be arranged for the best balance of the reciprocating masses, and what will then be the magnitude of the unbalanced primary couple? (Answer: 1,4,2,3;42.82 kN-m) (3) In a three-cylinder radial engine all three connecting rods act on a single crank. The cylinder center lines are set at 120º , the weight of the reciprocating parts per line is 22.24 N, the crank length is 76.2 mm, the connecting rod length is 279.4 mm and the rpm are 1800 cw. Determine with regard to the inertia of the reciprocating parts (a) the balance weight to be fitted at 101.6 mm radius to give primary balance (b) the nature and magnitude of the secondary unbalanced force (c) whether the fourth and sixth order forces are balanced or unbalanced. (Answer: (a) 25 N (b) Constant magnitude 2508.67 N, 3600 rpm ccw (c) 4 th harmonic is unbalanced, 6 th harmonic is balanced). Primary Balance of Multi-cylinder In-line Engines: The conditions in order to give primary balance of the reciprocating masses are: 2 ∑ mrec .ω r cosθ = 0 (i.e. Algebraic sum of primary forces shall be zero) and 2 ∑ mrec. .ω ra cosθ = 0 (i.e. Algebraic sum of the (primary force) couple about any point in the plane of force shall also be zero). where a is the distance of the plane of rotation of the crank from a parallel reference plane. Above two equations should satisfy for the angular position of the crankshaft relative to the dead center (reference line). The primary force due to the reciprocating mass is identical with the component parallel to the line of stroke of the centrifugal force produced by an equal mass attached to and 136 revolving with the crank pin. Let the reciprocating masses are attached to crank pin at A, B, C and D are revolving with crank as shown in Figure 13(a). The force polygon (centrifugal forces) may be drawn. (i.e. oa, ab, bc and cd). Primary forces of the individual cylinder are equal to the component of oa, ab, bc and cd along the line of stroke (say PQ). Now ef, fg ,gh and he are primary forces and since algebraic sum of these are equal to zero (for this configuration), the engine is balanced for this configuration. But suppose cranks turns clockwise through an angle γ , then the effect will be same as crank shaft is fixed and line of stroke is rotating ccw by γ angle. Let us say SP is now new line of stroke. Now kl , lm, mn and nk1 are primary forces and these algebraic sum is not zero, but it is equal to kk1 (primary force is not balanced). The primary force will be only balanced when k1 coincides with k or point d with o in force polygon. i.e. the centrifugal force polygon is a closed one. Q b a A S D O f h c B C o l mk to line of stroke e d γ k1 n p Force polygon of the centrifugal forces produced when revolving masses respectively equal to reciprocating masses at crank pin Four cylinder engine OA,OB,OC,OD are four cranks Figure 13(b) Force polygon Figure 13(a) Four cylinder engine In similar way the primary couple can only be balanced if the couple polygon for the corresponding centrifugal forces is closed. Hence, if a system of reciprocating masses is to be primary balance, the system of revolving masses, which is obtained by substituting an equal revolving mass at the crank pin for each reciprocating mass, must be balanced. The problem of the primary balance of reciprocating masses may therefore be solved by using the method for revolving masses. Secondary Balance of Multi-cylinder In-line Engine: The conditions for the complete secondary balance of an engine are ∑ m ( 2ω ) rec 2 r cos 2θ = 0 4n and 137 ∑ m ( 2ω ) rec 2 r a cos 2θ = 0 4n for all angular positions of the crank shaft relative to the line of stroke. As for the primary balance, these conditions can only be satisfied if the force and couple polygons are closed for the corresponding system of revolving masses. If to each imaginary secondary crank (angular velocity= 2ω and crank length ( r / 4n ) ) a revolving mass is attached, equal to the corresponding reciprocating mass, then the system thus obtained must be completely balanced. Example 1: A shaft carries four masses in parallel planes A, B, C and D in this order, along it. The masses at B and C are 18 kg and 12.5 kg, respectively, and each has an eccentricity of 6 cm. The masses at A and D have an eccentricity of 8 cm. The angle between the masses at B and C is 1000, and that between the masses at B and A is 1900 (both angles measured in the same direction). The axial distance between the planes A and B is 10 cm. And between B and C is 20 cm. If the shaft is in complete dynamic balance, determine (i) the masses at and D (ii) the distance between the planes C and D, and (iii) the angular position of the mass at D. Solution: B (Reference) 1000 ld D 10cm A 20cm B 0 lCD C 190 φ0 D A Figure 14 Multi-cylinder engine Given data are: mB = 18 kg Aim is to obtain: mA = ? rB = 6 cm; mD = ?; lCD = ? mC = 12.5 kg rC = 6 cm; rA = rD = 8 cm θ D = ? (with respect to B) Condition of dynamics balancing are ∑ mr cos φ = 0; ∑ mr sin φ = 0; ∑ mrl cos φ = 0; ∑ mrl sin φ = 0 Taking reference as the crank B for the phase measurement (i.e. φB = 0) and for the moment the cylinder A as reference and are as follows 138 mB rB lB + mC rC lC cos φC + md rd cos φd = 0; mC rC lC sin φC + md rd ld sin φd = 0 m A rA cos φ A + mB rB + mC rC cos φC + md rd cos nφd = 0; mA rA sin φ A + mC rC sin φC + md rd sin φd = 0 Hence, 1080 − 390.7 + 8md ld cos φd = 0; 2215.82 + 8md ld sin φd = 0 − 7.88mA + 108 − 13 + 8md cos φd = 0; or − 1.39m A + 73.86 + 8md sin φd = 0 8md ld cos φd = −689.3 (1) 8md ld sin φd = −2215.82 (2) − 7.88mA + 8md cos φd = −95 (3) − 1.39 mA + 8md sin φd = −73.86 (4) Equations (1) and (2) will give tanφd = 3.2146 ∴ φd = 252.72 so in third quadrant. So that md ld = 90 (5) Equations (3) and (4) gives −7.88mA − 2.3763md = −95; ∴ m A = 9.67 kg; − 1.39mA − 7.639md = −73.86 md = 7.91 kg; ld = 36.66 cm; lcd = 6.66 cm Using Tabular method calculation are much easier. Take moments about plane A and plane D with phase reference as crank B. Choosing plane D as first reference as advantage that mD and φD will not appear in two moment equation and mA and lCD can be calculated from this two equation directly. So choice of reference plane should be such that maximum number of unknown can be eliminated (at least two). Second reference plane can be chosen as A to eliminate mA from equation. Example 2: By direct and reverse cranks method, consider the W-engine in which there are three rows of cylinders. An engine of this type has four cylinders in each row and the crankshaft is of the normal “flat” type with one connecting rod from each row coupled directly to each crankpin. The middle row of cylinders is vertical and the other two rows are inclined at 600 to the vertical. The weight of the reciprocating parts is 26.69 N per cylinder; the cranks are 7.6 cm long. The connecting 139 rods 27.9 cm long and the rpm 2000 cw. Find the maximum and minimum values of the secondary disturbing force on the engine (Figure 15). Figure-15 W-engine Solution: wrec = 26.69 N; r = 7.6 cm; n = l = 3.67; mrec = 2.72 kg; l = 27.9 cm; ω = 2000 rpm cω = 209.43 rad/sec r 2 1 3 600 600 ω Figure-16 W-engine Secondary cracks 2 (m/2) (m/2) 2 600 600 (a) Direct (b) reverse 2ω 2ω (m/2) (m/2) Resultant Unbalance m m + 2 × 0. 5 = m 2 2 m r r m + ( 2ω) 2 = ω 2 2 4 4n 2 2 r mω m(2ω) 2 = n 4n m/2 2ω 2ω 140 Total 4 rows: 4 mω2 r = 9882.2 N n Total 4 rows: 2 mω2 r = 4941.1 N n (i). At present position Resultant is minimum: 2ω 2ω m r + (2ω)2 ; 2 4n 0 0 0 ↑ θ2 = 0, θ2 = 90 ↓; ↑ θ2 = 180 , θ2 = 270 ↓ Minimum unbalance = 4941.1 N ↑= (ii). For θ2 = 450 (or θ1 = 600 + 450 or θ3 = 3000 + 450 ) the resultant is maximum. 2ω m m/2 2ω 3m r (2ω)2 = 14823.3 N 2 4n ↑ θ4 = 450 , θ4 = 1350 ↓; ↑ θ4 = 2250 , θ4 = 3150 ↓ maximum unbalance = Exercise Problems: 1. A four-cylinder marine oil engine has the cranks arranged at angular intervals of 900. The inner crank are 1.2 m apart and are placed symmetrically between the outer cranks, which are 3.0m apart. Each crank is 45.7 cm long, the engine runs at 90 rpm, and the weight of the reciprocating parts for each cylinder is 8006 N. In which order should the cranks be arranged for the best balance of the reciprocating masses, and what will then be the magnitude of the unbalanced primary couple? 2. The reciprocating masses for three cylinders of a four-crank engine weight 3047, 5078 and 8125 kgs and the centerlines of the se cylinders are 3.66 m, 2.59 m. and 1.07 m. respectively from that of the fourth cylinder. Find the fourth reciprocating mass and the angles between the cranks so that they may be mutually balanced for primary forces and couples. If the cranks are each 0.61 m long, the connecting distributing rods 2.75 m long and the rpm 60, find the maximum value of the secondary distributing force and crank positions at which it occurs. ------------------------*********************-------------------------- 141