A maior rede de estudos do Brasil

Grátis
634 pág.

## Pré-visualização | Página 33 de 50

```whereas larger machines have a smaller percentage of losses.
The efficiency of the machine is
EXAMPLE 4.6
Eff = Poutput
Pin put
The de machine (12 kW, 100 V, 1000 rpm) of Example 4.2 is connected to
a 100 V de supply and is operated as a de shunt motor. At no-load condition,
the motor runs at 1000 rpm and the armature takes 6 amperes.
(a) Find the value of the resistance of the shupt field control rheostat (Rrc).
(b) Find the rotational losses at 1000 rpm.
(c) Find the speed, electromagnetic torque, and efficiency of the motor
when rated current flows in the armature.
(i) Consider that the air gap flux remains the same as that at no load.
(ii) Consider that the air gap flux is reduced by 5% when rated current
flows in the armature because of armature reaction.
(d) Find the starting torque if the starting armature current is limited to
150% of its rated value.
(i) Neglect armature reaction.
(ii) Consider armature reaction, Ir(AR) = 0.16 A.
Solution
(a) No load, Ia = 6 A.
= 100- 6 X 0.1
= 99.4 v
From the magnetization curve (Fig. 4.24), to generate E. = 99.4 V
at 1000 rpm requires Ir = 0.99 A.
vt 100 Rr = Rr + Rfw = - = -- = 101 a
c Ir 0.99
F'
T
Rrc = 101 - Rrw
= 101- 80
= 21 n
DC Motors 171
(b) At no load the electromagnetic power developed is lost as rotational
power.
Protational = Eala = 99.4 X 6 = 596.4 W
(c) The motor is loaded and I. = Ialrated = 120 A (Example 4.2).
(i) No armature reaction, that is, <I>NL = <l>fL.
EaiNL = 99.4 V
EaiFL = vt- I.R. = 100- 120 X 0.1 = 88 v
EaiFL Ka<f>FLWFL UlfL
EaiNL Ka<f>NLWNL WNL
EaiFL 88
WFL = EaiNL WNL = 99.4 X 1000 = 885.31 rpm
885.31
wm = 60 X 27T = 92.7
Pout= Eafa- Protational
= 10,560- 596.4
= 9963.6W
= 100(120 + 0.99)
= 12,099W
Eff = ;:t = ~~~~g~ X 100% = 82.35%
(ii) With armature reaction, <I>FL = 0.95<1>NL·
EaiFL _ Ka<f>FLWFL
EaiNL - Ka<f>NLWNL
~=0.95WFL
99.4 WNL
88 1
WFL = 99.4 X 0 _95 X 1000 = 931
.91 rpm
,.
I
,,
'" ',
172 chapter 4 DC Machines
Note that the speed increases if flux decreases because of arma-
ture reaction.
931.91 Wm =~X 21T = 97.59 rad/sec
88 X 120 T = 97 _59 = 108.21 N · m
Eff =~~~;~:X 100% = 82.35%, assuming rotational losses do
not change with speed.
(d) T = Ka<I>Ia.
(i) If armature reaction is neglected, the flux condition under load
can be obtained from the no-load condition.
1000 EaiNL = 99.4 V = Ka<I>Wm = Ka<I> ~X 27T
Ia = 1.5 X 120 = 180A
Tstart = 0.949 X 180 = 170.82 N · m
(ii) / 1 = 0.99 A. When Ia = 180 A
lr(eff) = lr- lr(ARl = 0.99- 0.16 = 0.83 A
From the magnetization curve (Fig. 4.24) the corresponding gen-
erated voltage is
EXAMPLE 4.7
Ea = 93.5 V ( = Ka<I>wm) at 1000 rpm
Ka<I> =
9~~5 = 10009~·~1T160 = 0.893 V/rad/sec
Tstart = 0.893 X 180 = 160.71 N · m •
The de machine of Example 4.2 runs at 1000 rpm at no load Ua = 6 A,
Example 4.6) and at 932 rpm at full load Ua = 120 A, Example 4.6) when
operated as a shunt motor.
(a) Determine the armature reaction effect at full load in ampere-turns of
the shunt field winding.
(b) How many series field turns per pole should be added to make this
machine into a cumulatively compound motor (short-shunt) whose
speed will be 800 rpm at full load? Neglect the resistance of the series
field winding.
DC Motors 173
(c) If the series field winding is connected for differential compounding,
determine the speed of the motor at full load.
Solution
(a) From Example 4.6,
It=0.99A
E.= 100-120X0.1 =88Vat932rpm
The effective field current Ut(ern) at full load can be obtained from the
magnetization curve (Fig. 4.24) of the machine, if we first find E. at
1000 rpm.
1000
E.lwoo = 932 X 88 = 94.42 V

From the magnetization curve, for E. = 94.42 V at 1000 rpm,
It(etf) = 0.86 A= It- It(AR)
It(AR) =It- 0.86 = 0.99- 0.86
= 0.13 A
The corresponding ampere-turns = Nrft(AR)
= 1200 X 0.13
(b) E. = 88 Vat 800 rpm.
= 156 At/pole
1000
E.lwoo = 800 X 88 = 110
V
From the magnetization curve for E. = 110 V at 1000 rpm,
N,r
1.32 = 0.99 + 1200 (120 + 0.99)- 0.13
N,v = 4.56 turns/pole
(c) For differential compounding,
4.56 X 120.99 It(eff) = 0.99- 1200
- 0.13
= 0.99-0.46-0.13
= 0.4A
1~,
"' j···
c
, •.
I
~il!<l
,,.
,,
r·
...
·,,
17 4 chapter 4 DC Machines
From the magnetization curve, at 1000 rpm and Ir = 0.4 A, E. = 65 V.
But E. = 88 V at full load (parts a and b). If the operating speed is n
rpm,
or
65 = K<l>1200
88 = K<l>n
88
n = 65 X 1000 = 1343.9 rpm •
Torque-Speed Characteristics
In many applications de motors are used to drive mechanical loads. Some
applications require that the speed remain constant as the mechanical load
applied to the motor changes. On the other hand, some applications require
that the speed be controlled over a wide range. An engineer who wishes to
use a de motor for a particular application must therefore know the relation
between torque and speed of the machine. In this section the torque-speed
characteristics of the various de motors are discussed.
Consider the separately excited de motor shown in Fig. 4.50. The voltage,
current, speed, and torque are related as follows:
E. = K.<Pwm = V1 - I.R.
T = K.<l>I.
From Eqs. 4.39, the speed is
From Eqs. 4.40 and 4.41
(4.39)
(4.40)
(4.41)
(4.42)
fJ + v, - FIGURE 4.50 Separately excited de motor.
p
E
3
.,;
~
(/)
DC Motors 17 5
-\ Ra
Slope <K.~l 2
FIGURE 4.51 Torque-speed charac-
teristics of a separately excited de
'-------------• T. Ia motor.
If the terminal voltage V1 and machine flux <P are kept constant, the torque-
speed characteristic is as shown in Fig. 4.51. The drop in speed as the applied
torque increases is small, providing a good speed regulation. In an actual
machine, the flux <P will decrease because of armature reaction as Tor I.
increases, and as a result the speed drop will be less than that shown in Fig.
4.51. The armature reaction therefore improves the speed regulation in a
de motor.
Equation 4.42 suggests that speed control in a de machine can be achieved
by the following methods:
1. Armature voltage control (V1).
2. Field control (<P).
3. Armature resistance control (R.).
In fact, speed in a de machine increases as V1 increases and decreases as <P
or R. increases. The characteristic features of these different methods of
speed control of a de machine will be discussed further.
Armature Voltage Control
In this method of speed control the armature circuit resistance (R.) remains
unchanged, the field current / 1 is kept constant (normally at its rated value),
and the armature terminal voltage (V1) is varied to change the speed. If
armature reaction is neglected, from Eq. 4.42,
(4.43)
where K 1 = 1/K.<l>
K2 = R.I(K.<l>)2
For a constant load torque, such as applied by an elevator or hoist crane
load, the speed will change linearly with V1 as shown in Fig. 4.52a. If the
terminal voltage is kept constant and the load torque is varied, the speed
can be adjusted by V1 as shown in Fig. 4.52b.
In an actual application, when speed is changed by changing the terminal
voltage, the armature current is kept constant (needs a closed-loop opera-
c:.
, ...
I
~I·" !
r ..
176 chapter 4 DC Machines
wm wm
~~
-
I ~I
-
I -~2
I -~3
I -~4
~ T
{a) {b)
{c) {d)
FIGURE 4.52 Armature voltage control of a de motor. (a) Variable speed. (b) Ad-
justable speed. (c) Operation under constant torque. (d) Operation with R. = 0.
tion). From Eq. 4.39, if I. is constant,
Therefore, as V1 increases, the speed increases linearly (Fig. 4.52c). From
Eq. 4.40, if I. remains constant, so does the torque (Fig. 4.52c). The input
power from the source (P = V1Ia) also changes linearly with speed (Fig.
4.52c). If R. is neglected, the values of VuE., and Pare zero at zero speed
and change linearly with speed (Fig. 4.52d).
The armature voltage control scheme provides a smooth variation```