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MIT OpenCourseWare http://ocw.mit.edu Solutions Manual for Electromechanical Dynamics For any use or distribution of this solutions manual, please cite as follows: Woodson, Herbert H., James R. Melcher, and Markus Zahn. Solutions Manual for Electromechanical Dynamics. vol. 3. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). License: Creative Commons Attribution-NonCommercial-Share Alike For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ELECTRO:ECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS PROBLE4 13.1 In static equilibrium, we have = 0 -Vp - pgi, Since p = pRT, (a) may be rewritten as RT Solving, - + pg = 0 dx we obtain p = p -L x RT 1 a e PROBLEM 13.2 Since the pressure is a constant, Eq. (13.2.25) reduces to Ov dv = dz JB (a) y where we use the coordinate system defined in Fig. 13P.4. Now, from Eq. (13.2.21) we obtain Jy = (Ey + vB) (b) If the loading factor K, defined by Eq. (13.2.32) is constant, then Thus, J y Then KvB = (c) avB(1-K) = pv dv dv or + E SP dv jdz avB (1-K) _- = (d) 2 (e) B GB2(1-K) _ = - o(I-K) Ai A(z) (f) From conservation of mass, Eq. (13.2.24), we have PiviAi = pA(z)v (g) Thus PiviAi v 2 dv -a(l-K)Bi Ai dz = (h) Integrating, we obtain - a(l-K)Bi 9n v = z + C (i) -a v= whereZ d vie d Oi vi Pi Vi a(l-K)B v = v. at z = 0. 1 (j) and we evaluate the arbitrary constant b y realizing that ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.3 Part a We assume T, Bo, w, a, cp and c Since the electrodes are short- E = 0, and so circuited, J are constant. = y vB (a) o, We use the coordinate system defined in Fig. 13P.4. Applying conservation of energy, Eq. (13.2.26), we have v2) = 0, where we have set h = constant. v Thus, v is a constant, v = vi. Conservation of momentum, Eq. (13.2.25), implies 2 - ViBo io dz Thus, p = (b) (c) - viBz + p (d) The mechanical equation of state, Eq. (13.1.10) then implies v.B 2 z +i V B 2 z RT RT - = RT i (e) From conservation of mass, we then obtain (- Piviwdi = + vi wd(z) (f) Thus d(z) = pld ( (g) viBoz Part b Then v.B p(z) = Pi 2 z 1RT (h) RT i PROBLEM 13.4 Note: There are errors in Eqs. (13.2.16) and (13.2.31). 1 d(M 2 ) S dx They should read: {(Y-1)(1+y1M2 )E 3 + y[2 +(y-1)M 2 ]v 1B 2 } J3 -M r (13.2.16) and i2 M2 2 ) d(M d ) dx (1M) ) (1- [2 + S ( [(y-I)(I1+yM2)E + 3 '2 + 2 } 3 (y-1)M v1 2] P 1 2 ypv (y-1)M2 ]dA A dx (13.2.31) Part a We assume that 0, y , Bo, K and.M are constant along the channel. the corrected form of Eq. (13.2.31), we must have Then, from ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.4 (continued) 2 [(y-l) (l+M2)(-K) + y(2+(y-1)M )] 1- 0 = -K) [2 +(y-1)M ] (a) Now, using the relations and v2 M2 yRT p pRT v M2 Yp pv we write (b) B 2 0C(l-K)M 2 Thus, we.obtain [(y-)(l+yM2 )(-K) + y(2+ (y-1)M2 )] 2 + (y - 1)M 2 1 dA A2 dz pVA (c) From conservation of mass, pvA = (d) piviAi A Using (d), we integrate (c) and solve for Ai to obtain A(z) Ai 1 1 - ýz (e) where )]B2M2 (1-K) 2 yM12 )(-K) + y(2 +(y-l)M [(y-1)(1+ Sivi[ 2 + (y-l)M2 ] We now substitute into Eq. (13.2.27) to obtain vB 2 (1-K)o 1 1 dv S (-M) -v dz o [(y-l)(-K) + y] (-M2) Yp 1 dA A dz (f) A dz Thus may be rewritten as d = v dz (-M 2 ) [(y-l)(-K) + y] iiAi (1-M2) p viAi A (g) Ai Solving, we obtain £n v = or v(z) - 82 (1 - = Vi where 8 z) + en vi 8 z)2/ (h) 1 (i) 1 i 82 = n(l - (-i 2 ) 2 (1-K)M 2 [(y-l)(-K) +y]Bo i vi 01 Now the temperature is related through Eq. (13.2.12), as ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.4 (continued) M2 yRT = Thus v2 (j) 2 T(z)i (k) From (d), we have P(z) Pi vi Ai v A = ( Thus, from Eq. (13.1.10) T Vi Ai p(z) = Pi v A Ti (m) Since the voltage across the electrodes is constant, w( or w(z) Thus, - Kv(z)B KviBoi Kv(z)B Kv(z)B0 = w(z) wi (n) Vi wi v(s) (o) vi v(z) Then d(z) A(z) di i Ai (q) () w(z) Part b We now assume that a, y, Bo, K and v are constant along the channel. Then, from Eq. (13.2.27) we have 0 = 1-) Cl-M 2 ) But, from Eq. -- C(y-l) (-K) + y]viB2 1l i0 1 (l-K)v. B2 z = ov. i 3z (s) 2 B 1 0 Pi Substituting the results A(z) 1- Pi (l-K) = 1 A A dz (13.2.25) we know that Pi where (-K) yp _ i A(z) of b), (-Ka) pi into and solvin for we obtain (u) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.4 (continued) and so, from Eq. (13.1.10) = T(z) T. p(z) Pi Pi p(z) (v) As in (p) w(z) wi vi = v(z) = 1 (w) Thus d(z) di A(z) A = (x) Part c We wish to find the length Z such that C T(i) + ]2 1 [v C p z C T(o) + .9 = (y) [v(o)] For the constant M generator of part (a), we obtain from (i) and (k) 2 1 C C [ Ti p () ]2 1 Ti C p T(z) i 1 2 Vi 1 221 -28 1 /Bl 9 +1 2 (Z 2i Reducing, we obtain - ) (1 - 26 /8 2 .9 (aa) Substituting the given numerical values, we have ,1 = We then solve 2/8 and .396 (aa) for 2 1 = - 7.3 x 10 -2 Z, to obtain % 1.3 meters k For the constant v generator of part (b), we obtain from (s), CT i pI + ) C v i 2 (1 - B ) (u) and (v) i + S= 2 T p 2 (t), (bb) .9 i or C T. C (1- 8 4/ ) 3 1 2 v2 1 2 C T + L v 1 2 p i = .9 Substituting the given numerical values, we have 83 (cc) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.4 (continued) = B and .45 = B /B .857 Solving for Z, we obtain Z 1.3 meters. PROBLEM 13.5 We are given the following relations: E(z) wi di B. E. w(z) d(z) I /2 (z) Ai' B(z) 0, y, and K are constant. and that v, Part a From Eq. (13.2.33), J = (1-K)ovB (a) For constant velocity, conservation of momentum yields dp dz = (1-K)ovB 2 - (b) Conservation of energy yields pvC pdz = - K(1-K)G(yB) 2 (c) Using the equation of state, = p pRT (d) we obtain or T Thus, (-K) (1-K)ovB 2 + dz C p 2 vB (1-K) = T (l-K) -K OvBB R p dT dz T dP -- + T dz + dz (1-K)OvB 2 R (f) (g) + (g) B.i2 (A.) 1 1 A(z) 2 and .A. = and (e) (e) + R Also 2 p(z)A(z) 11 d = T d- Therefore and pc dT d = p dz ovB'fdi(1-K)(-2 + --- ) P(!) Bp -K(1-K)av (i) Pi ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.5 (continued) and so K(1-K)OvB2 dT i dz () pic p Therefore T 2 OvB i - K(1-K) z + T PiCp = Let (k) 2 Let -K(1-K)avBi = Pi Cp (i) Then T = az C- Ti + 1) (m) 1 + ovB (1-K)( i S K cp i R dz (n) pi (az + Ti) We let S vB(-K)( K 1 Cp R P. a Pi e KR Integrating (n), we then obtain in p = 8 in(az + Ti) + constant or p= =z i - + 1 ) 8 (o) Therefore A A(z) = () Ti Part b From (m), T() at T(a) + T i .8 Ti Ti or -l- . -. 2 Ti Now K(1-K)viBi E Ti Pic Ti But R Ti c T pi 2 = (1) Pi = y pi(1 i y = 2.5 x 10 6 ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.5 (Continued) Thus Thus - .5(.5)50(700)16 Ti -2 8.0 x 10-2 - .7(2.5 X 10 6) Solving for R, we obtain •2x 102 z= 1.25 meters = 8 Part c pi p= ( (Xz 1) •.+ Numerically 8= - 1 KR 1 1 (l1)K 6. % Thus p(z) .7(1 - = .08z) 6 Then it follows: 7 Ti(l - = T(z) Pi(1 - pRT = p(z) = 7 = 5 x 10S(l - .08z) .08z) .08z) From the given information, we cannot solve for Ti, only for v Pi RT. = I Pi Now = 2 -i s 7 x 10 yM vi yRT(z) v2 yp(z) .5 1 - .08z Part d The total electric power drawn from this generator is VI = p = - E Thus pe -E(z)w(z)J(z)£d(z) E(z)(1-K)GvB(z)£d(z)w(z) = - Eiwi(1-K)ovB diz = -KvB i = K(vB )2 w diO (-K) = .5(700)216(.5)50(.5)1.25 = 61.3 x 106 watts = 61.3 megawatts 86 ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS T (z) T (z) Ti ST (1 - .08z) 1.25 p(z) p(z) = p(z) 7ý(1 - .08z) 1.25 p(z) = 5 x 10 5 (1 - 1.25 m 6 (z) M2 (z) = 1.25 (1- .5 .08z) .08z) 7 ELECTROILECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.6 Part a We are given that 4 Vo E--i x3 x X Lj/3 and 4 e Eo VO 9 L'3 x The force equation in the steady state is dv ix pv m xx d• = Since pe/p pe E = q/m = constant, we can write Vd O 2 v2A dx2Vx Solving for v /3 X!3 V4 = L3 m) we obtain x Part b The total force per unit volume acting on the accelerator system is F = peE Thus, the total force which the fixed support must exert is f = total - FdVi x EV 2 -16 E0 0 =-f L8 27 -3 x Adx Adx i x 0 f total 8 = 8 9 oo 2 Ai x PROBLEM 13.7 Part a We refer to the analysis performed in section 13.2.3a. The equation of motion for the velocity is, Eq. (13.2.76), 2 v 2 a 2(a) a ax 2 •-22V 3-ý = The boundary conditions are v(-L) = V v(O) 0 = cos Wt We write the solution in the form (a) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.7 (continued) Re[A ej(wt-kx )+ B ej(wt+kx l v(x t) = ) ] (b) a where a Using the boundary condition at x 1 = 0, we can alternately write the solution as Re[A sin kx e j v = t I Applying the other boundary condition at x = - L, we finally obtain V v(xt) = - sin kL sin kx cos wt. (d) 1 sin kL 1 The perturbation pressure is related to the velocity throughEq. (13.2.74) Pov'3 at -x, a(e)' Solving, we obtain 0Vo oL s sin kx sin wt = sinkL 1 x (f) 1 p V 0 k k 0 sin kL cos kx 1 sin wt or p f) - = (g) where po 0 is the equilibrium density, related to the speed of sound a, through Eq. (13.2.83). Thus, the total pressure is p Vw + k sin kL cos kx P= + p and the perturbation pressure at x, = p'(-L, t) = sin at (h) - L is 00Voa sin k cos kL sin wt sin kL (i) Part b There will be resonances in the pressure if sin kL = 0 n = 1,2,3.... n7 kL = or (j) (k) Thus w (a) a L PROBLEM 13.8 Part a We carry through an analysis similar to that performed in section 13.2.3b. We assume that E = iE 2 (xl,t) J= i2J2 (x, t) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.8 B = i [P 00 H + 3 i H' (x 1 ,t)] 03 Conservation of momentum yields Dv D 1 = Dtax .x +J2 o(Ho + H3) 20 1 (a) 0 Conservation of energy gives us D p Dt 2 1 (u + v ) = 1 2 ax pv 1 (b) + J E 22 We use Ampere's and Faraday's laws to obtain aH' = - - JJ [email protected] ax and (c) 2 P oaH DE a2 x (d) at 3x, while Ohm's law yields J Since = 2 -+ ([E - v B 2 13 i (e) oo E = vB 2 (f) 1 3 We linearize, as in Eq. (13.2.91), so E 2 ' v p H 100 O Substituting into Faraday's law av, alH'3 Linearization of the conservation of mass yields at = o )v (h) Thus, from (g) ,H' 3 ooy_ p o at at () Then Ho Po p' H'3 Linearizing Eq. (13.2.71), we obtain = p Dt o p D' Dt (k) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.8 (continued) Defining the acoustic speed as \ 0o/ where p is the equilibrium pressure, (oH 2 = P0o p 2 we have p' = as22 (p) Linearization of convervation of momentum (a) yields av p o at 1 = - 9H' H o(m) x3 o oHo T 3xl (m 1 or, from (j) and (Z), Iv, ao 0 a2 (n) -s Po at= Differentiating (n) with respect to time, and using conservation of mass (h), we finally obtain °H S2v 2V= 2v ( a2 (0 ) Defining 0 1 2 a2 a2 p1 H +-0 s we have (p) p o a2 v 3 v I 1 2(V) a2V Part b We assume solutions of the form ( Re [A ej (Ct-kx )+ A e j wt+ V = 1 where k = 1 (r) a V(-L,t) = V = - L is cos wt = s V Re e jwt (s) s = 0 M dViot) = p dt 'A' + poH H oo 1=0 From (h), )] The boundary condition at x and at x kx 2 (j) and 1-g__ = a2 s 1 A (t) 1=0 (t), P v at (o u) ax ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.8 (continued) H' a Ho Po Thus dv,(0,t) dt M 2 aoH =A From (u), we solve for p'j + 2 az-T- I p' =A to obtain: x =0 Substituting into (s) and (t), we have a 2/Poak Mjw(A + A) 1 2 = A(- ~)• \ ) (A 1 / a'\W A2) Ale+jk£+ A2e-jki Solving for A A 1 1 and A , we obtain 2 (Mjw + Aap o)V s = 2(-Mw sin kZ + Aap cos k£) (Aap - Mjw)V s 2(-Mw sin kZ + Aa %cos kt) A2 Thus, the velocity of the piston is v (O,t) = [A Re j wt + A ]e S1 v (0,t) 2 Aap V w s + Aaos Wt -Mw Elsin kk +Aap cos kZCOWt = 1 (aa) PROBLEM 13.9 Part a The differential equation for the velocity as derived in problem 13.8 is 32v 1 Ftwhere a 2 2 a - = a 2 + s a a with = S a2v 1o aX 1 2 (a) 00 Po 2 (YPo2 where p -•oP Po = 0 Part b We assume a solution of the form p 1 p H 2 2 2 ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.9 (continued) where k = [De J(t-kx1)] Re V(x ,t) = 1 a We do not consider the negatively traveling wave, as we want to use this system as a delay line without distortion. V(-L,t) = and at x The boundary condition at x = - L is Re V ejWt s = 0 is dV BV (O,t)+ p'(O,t)A - = dt ;, (b) A joHoH' From problem 13.8, (h), (j) and (9) H' av pa s p ', - =t at o ax and 1 - H = a o S 0 Thus, (b) becomes ta 2 t -BDej + (-) where =0 p'A S PoD(- jk) pI =w x0O j (c) t 2 (d) se w (d) Thus, for no reflections 2 Ap0a2 - B + (a) a 0 a (e) S or B = Aapo (f) PROBLEM 13.10 The equilibrium boundary conditions are T[- (L 1 (L T[- + L + A),t] 2 + A),t]As = = T 0 - PoAc Boundary conditions for incremental motions are T[- (L 2) - T[- (L 1) 1 + L 2 + A),t] = + A),t]A v (-Ll ,t) = 3) and 4 ) v (O,t) = - p(-L Ve[-(L 0 T s (t) d ,t)Ac = M + A),t] - v (-L ,t) since the mass is rigid since the wall at x=0 is fixed. PROBLEM 13.11 Part a We can immediately write down the equation for perturbation velocity, using equations (13.2.76) and (13.2.77) and the results of chapters 6 and 10. 93 ELECTROMECHANTCS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.11 (continued) We replace 3/at by 3/at + v-V (-+ Letting V t )v' o = x to obtain a 2 sax 2V = Re V j(t-kx) v' we have (W- kV S)2 = a2k 2 as Solving for w, we obtain = W k(V ± as ) Part b Solving for k, we have k = V±a V+a o s For Vo > as, both waves propagate in the positive x- direction. PROBLEM 13.12 Part a We assume that E = i J = iz z z i y B= E (x,t) z J (x,t) z o1[H + H'(x,t)] y We also assume that all quantities can be written in the form of Eq. (13.2.91) . Vx p' - Jz Ho (conservation of momentum (a) linearized) The relevant electromagnetic equations are aH' and 3E 3x (b) J z ax Yx = aH' 0o---- z at (c) and the constitutive law is Jz = z (E + VxpoH ) z xoo We recognize that Eqs. (13.2.94), (13.2.96) and (13.2.97) are still valid, so 1 I3' P at ax (d) eav (e) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM' 13.12 (continued) and a 2 P' p' (f) S Part b We assume all perturbation quantities are of the form v Using (b), x = ^ Re[v e j(wt-kx) ] (a) may be rewritten as (g) + jkp + p0HojkH pojwv = and (c) may now be written as (h) jojwH - jkE = Then, from (b) and (d) o(E - jkH = + V oHo (i) ) Solving (g) and (h) for H in terms of v, we have vOVoH H = (j) - jk+po 9 From (e) and (f), we solve for p in terms of v to be p= (k) kP asv Substituting (j) and (k) back into (g), we find 2 jk(po Ho)2 v L2 - Lk 0) 00jk+ 'w Thus, the dispersion relation is k j(MoHo)20 k, (02 - k2a2) -_ (+ 2 = (m) 0 + j11w)po We see that in the limit as a + m, this dispersion relation reduces to the lossless dispersion relation _-2 k2(a2 + = (n) 0 Part c If a is very small, we can approximate (m) as W2- k2a2 _ j( w s fo for which we can rewrite (o) as 2 O kr )H = (0o) ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.12 (continued) kl s k2 - O 2 2 ( = 2 , we obtain Solv ing for k 2 0p ()02_ j 2 0 k2= 2 22a OHo Po 2 ) w° 0 02 2) O+ o a s 2 2a s a s Since a is very small, we expand the radical in (q) to obtain 2 [12 O A oo _ 2 a . O2] S L 2 Thus, our approximate solutions for k EW are (IIoHo)2 k2 a ( o0 0 O s H ) a (11 2 The wavenumbers for the first pair of waves are approximately: while for the second pair, we obtain + o 0 k~+~lo~op The wavenumbers from (u) represent a forward and backward traveling wave, both with amplitudes exponentially decreasing. Such waves are called 'diffusion waves'. The wavenumbers from (v) represent pure propagating waves in the forward and reverse directions. ELECTROMECHANICS OF COMPRESSIBLE, INVISCID FLUIDS PROBLEM 13.12 (continued) Part d If 0 is very large, then (m) reduces to H2 W2 - k 2 a 2 - o 0 ; k p0 a2 = a 2 Ow s H2 O + (w) Po This can be put in the form a 2 a where H 2 k0 f(w,k), =- As (x) f(wk) k2= (Y) 2 pOwa a becomes very large, the second term in (x) becomes negligible, and so k 2 - 2 However, it is this second'term which represents the damping in space; that is, k + ,a f(w,k) j2a (z) Thus, the approximate decay rate, ki, is ki ki 2a w 2 or i H2 k o f(w,k)a 2p a0 a 2poaw• k H2 w a 2pa 0 (aa)