## ABSTRACT

Equation (10.29) is a second-order linear differential equation with constant coefficients, which can be solved by use of the differential operator D. Equation (10.29) can be rewritten as

and the solution is

(10.30)

(10.31)

The coefficient 'Y of the variable y in the two exponents may be written as

where

B = a{cosh[(1 + j)oy] + sinh[(1 + j) by]} + b{cosh[(1 + j)oy] - sinh[1 + j)oy]}

B = M cosh[(1 + j)oy] + N sinh[(1 + j)oy]

Chapter 10

(10.34)

(10.35)

(10.36)

Equation (10.26) can be solved in an analogous manner, to give

and eq. (10.36) then becomes

41T~ sinh[ (1 + j)oy] 2 B = -- webers/meter

107w sinh[(1 + j)bdb]

(10.38)

(10.39)

(10.40)

(10.41)

and

10 pw

wk sinh[ (1 + j) 0 db]

Substituting eq. (10.47) into eq. (10.42)

(1 + j)O\ cosh[(l + j)oy] 2 j = amperes/meter b wk sinh[(l + j)O~]

Chapter 10

(10.47)

(10.48)

(10.49)

r wk sinh[(l + j)o db]

The corresponding reactive component of the impedance drop is

(10.50)

(10.51)

db

t/> = f B dy webers y

(10.52)

Substituting eq. (10.39) into eq. (10.52) and performing the indicated integration gives us

= 47T\ cosh[(1 + j)o\] - cosh[(1 + j)oy]

v = V + V = V + j27Tfc/J volts r x r

and substituting eqs. (10.50) and (10.53) into eq. (10.54) yields

wk sinh[(l + j)OdbJ

wk sinh[(l + j)odbJ

. 4ikf \P cosh[(1 + j)O~J - cosh[(l + j)oy J + 2J -- -- volts

107 P wk (1 + j)o sinh[(l + j)odbJ (10.56)

Remembering that v'2j = 1 + j and remembering the definition of 0 in eq. (10.33) gives us

Ibp cosh[(1 + j)oy J V = (1 + j)o -

2 2 \P cosh[ (1 + j)o\J - cosh[ (1 + j)oy J + (1 + j) 0 - volts

\P ~ cosh[(1 + j),sy J V = (1 + j)1l -- -=--------wk~ sinh[(1 + j),s~J

+ (1+j)2,s2 \P db{cosh[(1+j),s~J-cosh[(1+j),sy]) wk~ (1 + j),s sinh[ (1 + j),sdbJ volts

(10.58)

ohms (10.59)

Substituting eq. (10.59) into eq. (10.58) and then simplifying the expression yields

dc sinh[(1 + j),sdbJ (10.60)

V = \Rdc,s~ sinh 2,s~ + sin 2c5~ cosh 2Mb - cos 21l~

sinh 2,s~ - sin 2,s~ + jI R .sd volts

b dc b cosh 2,s~ - cos 2,s~ (10.62)

(10.63)

sinh 2bdb + sin 2b~ R = Rd bdb h 'd 2'ci ac c cos 2u b - cos U -b

ohms

Rac sinh 2b~ + sin 2bdb Rdc = b~ cosh 2bdb - cos 2b~ numeric

(10.64)

(10.65)

As stated above, the product bdb is dimensionless, provided that consistent units are used for the rotor-bar resistivity p, which is one of the

0.2 2.0 6d = d j41f2 kf

(10.66)

The value of eq. (10.66) is not changed if the numerator and denominator are multiplied by 8db. Then for 8db greater than 1. 5,

volts (10.67)

volts (10.68)

Multiplying the numerator and denominator of eq. (10.68) by 3 and by 2 yields

471" ~ 3 V r = j27Tf\ 107 3w 2o~ volts (10.69)

V r = j271"~ (3.19) ~ 20~ volts (10.70) It will be shown in Chapter 13 that the permeance for that portion of the rotor slot which is occupied by the bar in Fig. 10.8 is equal to db/3w. Then eqs. (10.69) and (10.70) may be written as

292 Chapter 10

volts (10.72)

v = jl X volts r -0 ac

(10.73)

Equation (10.74) may be written as

for Ii ~ > 1. 5 numeric

(10.74)

(10.75)

Case 1:

Case 2:

Aluminum 0.300 in. wide, 2.00 in. deep 0.315 in.