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constant during the motion. The operating point has therefore moved up- ward from point a to b as shown in Fig. 3.5a. During the motion, dWe = J ei dt = r: i dA. = area abed dWr = area Obe - Oad dWm = dWe- dWr = area abed + area Oad - area Obe =area Oab (3.23) (3.24) (3.25) If the motion has occurred under constant-current conditions, the mechani- cal work done is represented by the shaded area (Fig. 3.5a), which, in fact, is the increase in the coenergy. dWm =dWi If fm is the mechanical force causing the differential displacement dx, fmdx = dWm = dWi fm = f!Wf(i,x)l ax i = constant X= Xz ~) ~) (3.26) X = Xz FIGURE 3.5 Locus of the operating point for motion in system of Fig. 3.2. (a) At constant current. (b) At constant flux linkage. Mechanical Force in the Electromagnetic System 103 Let us now consider that the movement has occurred very quickly. It may be assumed that during the motion the flux linkage has remained essentially constant, as shown in Fig. 3.5b. It can be shown that during the motion the mechanical work done is represented by the shaded area Oap, which, in fact, is the decrease in the field energy. Therefore, fm dx = dWm = -dWr t; = _ awr(A,x)l m ax A = constant (3.27) Note that for the rapid motion the electrical input is zero (i d,.\ = 0) because flux linkage has remained constant and the mechanical output energy has been supplied entirely by the field energy. In the limit when the differential displacement dx is small, the areas Oab in Fig. 3.5a and Oap in Fig. 3.5b will be essentially the same. Therefore the force computed from Eqs. 3.26 and 3.27 will be the same. EXAMPLE 3.2 The A-i relationship for an electromagnetic system is given by . _ ( Ag ) 2 l- 0.09 which is valid for the limits 0 < i < 4 A and 3 < g < 10 em. For current i = 3 A and air gap length g = 5 em, find the mechanical force on the moving part, using energy and coenergy of the field. Solution The A-i relationship is nonlinear. From the A-i relationship ,.\ = 0.09i 112 g The coenergy of the system is From Eq. 3.26 WI = ' A di = ' . z di f . f. o 09"1/2 0 0 g 0.09 2 .312 • I = ---z JOU es g 3 fm = aWi(i, g) I ag i ~ constant = -0.09 X ~i 312 ~~ 3 g i = consta:1t 104 chapter 3 Electromechanical Energy Conversion For g = 0.05 m and i = 3 A, fm = -0.09 X~ X 3312 X 0.~52 N · m = -124.7N·m The energy of the system is fA. fA ( Ag ) 2 Wr = 0 z dA = 0 0.09 dA From Eq. 3.27 r; = _ aWr(A, g) I m dg A ~ constant 3 X 0.092 For g = 0.05 m and i = 3 A, 0.09 X 3112 A= 0.05 = 3.12 Wb-turn and {; = _ 3.123 X 2 X 0.05 m 3 X 0.092 = -124.7N·m The forces calculated on the basis of energy and coenergy functions are the same, as they should be. The selection of the energy or coenergy function as the basis for calculation is a matter of convenience, depending on the variables given. The negative sign for the force indicates that the force acts in such a direction as to decrease the air gap length. • 3.3.1 LINEAR SYSTEM Consider the electromagnetic system of Fig. 3.2. If the reluctance of the magnetic core path is negligible compared to that of the air gap path, the A-i relation becomes linear. For this idealized system A= L(x)i (3.28) where L(x) is the inductance of the coil, whose value depends on the air gap length. The field energy is Wr = f idA (3.29) Mechanical Force in the Electromagnetic System 105 From Eqs. 3.28 and 3.29 From Eqs. 3.27 and 3.30 a ( A2 ) I t; = -- - m ax 2L(x) A~ constant For a linear system From Eqs. 3.26, and 3.33 A2 dL(x) ---- 2L2(x) dx _ 1 . 2 dL(x) -zl -- dx Wr = W[ = !L(x)i2 a I fm =- (!L(x)i2) ax i = constant _ 1 . 2 dL(x) -zl -- dx (3.30) (3.31) (3.32) (3.33) (3.34) Equations 3.32 and 3.34 show that the same expressions are obtained for force whether analysis is based on energy or coenergy functions. For the system in Fig. 3.2, if the reluctance of the magnetic core path is neglected, Bg Ni = H 2g = -2g g f.Lo From Eq. 3.16, the field energy is B2 Wr = - 2 g X volume of air gap f.Lo B2 = - 2 g XAg2g f.Lo where Ag is the cross-sectional area of the air gap. From Eqs. 3.27 and 3.36 a ( B 2 ) fm = - - 2 g X Ag X 2g ag f.Lo B2 = - 2 g (2Ag) f.Lo (3.35) (3.36) (3.37) ., r ..... 106 chapter 3 Electromechanical Energy Conversion The total cross-sectional area of the air gap is 2Ag. Hence, the force per unit area of air gap, called magnetic pressure Fm, is EXAMPLE 3.3 Bz F =-g N/m2 m 2~-to (3.38) The magnetic system shown in Fig. E3.3 has the following parameters: N = 500 i=2A Width of air gap= 2.0 em Depth of air gap = 2.0 em Length of air gap = 1 mm Neglect the reluctance of the core, the leakage flux, and the fringing flux. (a) Determine the force of attraction between both sides of the air gap. (b) Determine the energy stored in the air gap. Solution (a) (b) N ~-toNi Bg=-z- g = 47710-7(500 X 2)2 X 2.0 X 2.0 X 10-4 2 X 1 X 1 X 10-6 = 251.33 N FIGURE E3.3 p EXAMPLE 3.4 Mechanical Force in the Electromagnetic System 107 Bz = - 2 g XAg X lg JLo = 251.33 X 10-3 J = 0.25133 joules The lifting magnetic system shown in Fig. E3.4 has a square cross section 6 X 6 cm2• The coil has 300 turns and a resistance of 6 ohms. Neglect reluctance of the magnetic core and field fringing in the air gap. (a) The air gap is initially held at 5 mm and a de source of 120 V is connected to the coil. Determine (i) The stored field energy. (ii) The lifting force. (b) The air gap is again held at 5 mm and an ac source of 120 V (rms) at 60 Hz is connected to the coil. Determine the average value of the lift force. Solution (a) Current in the coil is FIGURE E3.4 i = 120 = 20A 6 108 chapter 3 Electromechanical Energy Conversion Because the reluctance of the magnetic core is neglected, field energy in the magnetic core is negligible. All field energy is in the air gaps. Field energy is B Ni=Hl =~lg g g f-1-o f-1-oNi Bg=lg 47710-7 X 300 X 20 2 X 5 X 10-3 = 0. 7 54 tesla Bz Wr = - 2 g X volume of air gap f-1-o 2 ~-~~~~- 7 X 2 X 6 X 6 X 5 X 10-7 J = 8.1434 J From Eq. 3.37 the lift force is Bz fm = - 2 g X air gap area f-1-o 2 ~-~~~~-7 X 2 X 6 X 6 X 10-4 N = 1628.7 N (b) For ac excitation the impedance of the coil is Z =R + jwL Inductance of the coil is L = N 2 = N 2J1-oAg Rg lg 3002 X 47710-7 X 6 X 6 X 10-4 2 X 5 X 10-3 = 40.7 X 10-3 H Z = 6 + j377 X 40.7 X 10-3 0, = 6 + j15.340 Current in the coil is I = 120 rms Y(62 + 15.342) = 7.29 A F Rotating Machines 109 The flux density is The flux density is proportional to the current and therefore changes sinusoidally with time as shown in Fig. E3.4. The rms value of the flux density is The lift force is }LoNlrms Brms = 2g = 47710-7 X 300 X 7.29 T 2 X 5 X 10-3 = 0.2748 T Bz fm = - 2 g X 2Ag }-to (3.38a) (3.38b) The force varies as the square of the flux density as shown in Fig. E3.4. Bzl fmlavg = -2 g X 2Ag }-to avg B~, . = - 2 X mr gap area J.to 0.27482 X 2 X 6 X 6 X 10-4 2 X 47710-7 = 216.3 N (3.38c) which is almost one-eighth of the lift force obtained with a de supply voltage. Lifting magnets are normally operated from de sources. • 3.4 ROTATING MACHINES The production of translational motion in an electromagnetic system has been discussed in previous sections. However, most of the energy converters, particularly the higher-power ones, produce rotational motion. The essential part of a rotating electromagnetic system is shown in Fig. 3.6. The fixed part of the magnetic system is called the stator, and the moving part is caUed the rotor. The latter is mounted on a shaft and is free to rotate between the 110 chapter 3 Electromechanical Energy Conversion Stator i, wm{~ ' I ' Rotor FIGURE 3.6 Basic configuration of a ro- tating electromagnetic system. poles