Download EE 328 _lect_04

Electrical Power and Machines
Lecture 4
<Dr Ahmed El-Shenawy>
<Dr Hadi El Helw>
DC Generators
Electric Equivalent Circuit of a DC Generator
Field Winding
Ra
ia
+
E
RL
Field winding is responsible of setting up a flux
Types of DC Generators
1. Permanent Magnet Generators
φ Constant Flux
Ra
+
E
Ia
Ia = IL
IL
RL
This type is characterized by being
smaller, lighter and more efficient
compared to wound generators
2. Electro-magnet Generators
a.
Separately Excited
Rf
If
+
-
Ia = IL
b.
VF
Ia
E
Ra
RL
Self Excited
IL
Rf
+
Ia
If
E
Ra
Ia = If + IL
RL
IL
Induced E.M.F Equation
e.m.f “EC” is generated in armature windings
EC = BLV
N
B: Flux density (T)
L
L: length of conductor (m)
r
V: velocity of conduction motion
V =Wr
B =φ / A = φ/ L Ap = φ/ [L(πD/P)]
V = Wr = (2πN/60)r
Then
(N: rev/min)

WD
 PW
 L

L ( D P )
2
2
Ec 
S
Machine e.m.f “Ea” = EC (total no. of conductors/no. of parallel paths)
Ea 
 PW
Z

 K a  pW
2
a
Where Ka = ZP/2 π a ……………….. Machine constant
And
Z = 2 C NC (C:no. of coils, NC : no. of turns)
Example
Calculate the voltage generated by a six pole dc machine if its windings are: 1)
lap wound 2) Wave wound, if the flux per pole is 0.05 wb. and the generator
speed is 120 rev/min, 200 armature coils with 15 turns each
Solution
Z = 2CNC = 2×200×15 = 6000 conductors
1)Lap wound
a= P=6
Ea = Ka φp w = [6000×6/2π ×6]×0.05×[2π×120/60] = 600 V
2)Wave wound
a=2
Ea = Ka φp w = [6000×6/2π×2]×0.05×[2π×120/60] = 1800 V
Torque Equation
P = Ea Ia = Ka φp w Ia = T × w
Then the developed torque is given by: Td = P/w = Ka φp Ia
Example
A 24 slot, 2 pole Dc machine has 18 turns per coil, the average flux density per
pole is 1T. The effective length is 20 cm and the armature radius is 10 cm. the
magnetic poles are designed to cover 80 % of the armature periphery. If the
armature velocity is 183.2 rad/sec and the armature current is 25 A, determine:
• the current in each conductor
• the developed torque
• the developed power
Solution
•As the machine is a 2 pole,
•then IC = Ia / 2 = 25/2 = 12.5 A
• Ap = 2πrL/p = 2×3.14×0.1×0.2/2 = 0.063 m2,
•as Ae = 0.8 Ap ,
•then Ae = 0.8×0.063 = 0.05 m2,
•then φp = B Ae = 1×0.05 = 0.05 wb
•and Ka = (Zp/2πa) = (2×24×18)×2/2П×2 = 137.51
•Td = Ka Ia φp = 137.51×25×0.05 = 171.89 N.m
• Pd = Td ×w = 171.89×183.2 = 31490 W
Magnetization characteristics of DC machine
As Ea α φp w, now if the armature circuit is left open, and armature is rotated at rated
speed, then Ea = K1 φp = Ka wm φp and since φp = Kf If (flux per pole depend on the
Induced E.M.F at no load
m.m.f provided by current If ) then Ea = K1 Kf If
Since Ea is an indirect measure of the
flux per pole and since If is a measure
of the applied m.m.f (Nf If ), then the
curve is similar to a B-H curve
Air gap line
Magnetic
circuit
Air gap
m.m.f
From previous operation of the machine Er
Field Current If
m.m.f required for magnetic material is almost negligible at small values of flux density
Separately Excited Dc Generator
VF
Ea = Vt + Ia Ra
IL = Ia
Rf
If
+
-
Vf = If Rf
Ia
E
Ra
RL
Ea = VN L
-
Vt
IL
+
Drop due to armature resistance and reaction
Vt
Speed
Shunt Generator
Rf
If = Vt /Rf
Ia = IL +If
Ia
Vt = Ea – IaRa =ILRL
If
+
E
Ra
Ia
IL
RL
-
Terminal Voltage
VNL
Er
Magnetization curve
el d
Fi
sis
re
ta
e
nc
lin
Vt
+
•Er induces an e.m.f in armature winding
(Er) and as the field is parallel then a small If
flows giving a m.m.f that sets up a flux
giving aid to the residual flux.
•An increase in the flux per pole will
increase the induced e.m.f thus increasing
If
•The shunt generator continues to build up
voltage until the point of intersection of the
field resistance line and the magnetization
saturation curve
e
If
Note :
A decrease in the field circuit resistance will cause the shunt generator to build up
faster to a higher voltage, the generator will not build up if the field resistance is
greater than or equal the critical resistance
RF2
RC
RF
RF1
Terminal Voltage
VNL
Er
RC >RF2 >RF >RF1
If
Series Generator
Rf
Vt = Ea – Ia Ra –Ia Rf
RL
+
TS
Ea
Vt
Ia = IL = If
IL
-
IL = Vt /RL
+
Ia
Ra
wm
•Series generators are used to supply a constant load current
•Rf should be small to reduce voltage drop (so it is made of thick wire)
Voltage Regulation
Voltage regulation is a measure of the terminal voltage drop at full load
V.R % = 100×(VNL – VFL )/VFL
VNL: NO load terminal voltage
VFL : Full load terminal voltage
Losses in DC Generators
Input
Mechanical Power
Rotational losses “Pr”
Mechanical Losses:
• Windage (drag on armature caused by air)
• Friction (brushes-commutator , bearing-shaft)
Magnetic Losses:
• Hysteresis losses
• Eddy losses
Developed Power “Pd” = EaIa
• Output Power “Po” =Vt IL
• Copper Losses “PCU” = I2R
r
P
(R
ot
at
io
na
l)
Power Flow Diagram
Pin = T SWm
Pd =TdWm= EaIa
P0 = Vt I L
P CU
=
I2 R
Efficiency “η”
η %= (P0 /Pin) ×100
Example:
The d.c Separately-excited generator shown in Fig produces an open circuit emf
E of 220 V. Calculate the Terminal voltage V when generator supplies a current
of 10 A. The armature resistance is 1 ohm
Vt=Ea – Ra Ia
Where Ea= 220 V
Vt=220 – (10)(1)
=210 V
Ia= 10 A
Ra= 1 ohm
Example:
A separately-excited d.c. generator produces an open-circuit voltage of 250 V
with field current of 1.5 A. If the field current is increased to 2 A, calculate
a) The new open-circuit emf.
b) The terminal voltage if the generator supplies a current of 5 A. Assume
armature resistance- 0.8 ohm and the speed remains constant.
Solution:
a) Since the speed is constant
b)
E1 I f 1
E2

If2
Where
E1=250 V, If1= 1.5 A and If2 = 2A
E2=250 (2/1.5)=333 V
b) Vt=Ea – Ra Ia
Where
Ea=333 V, Ia= 5 A and Ra = o.8 ohm
V= 333 – (5)(0.8)
= 333 – 4
= 329 V
Example
A separately excited DC generator has a field resistance of 50 ohm, armature
resistance of 0.125 ohm and brush drop of 2V. At no load the generated voltage
is 275 V, and the full load current is 95 A. the field excitation voltage is 120 V and
the friction windage and core losses are 1500 watt. Calculate :
• The rated terminal voltage and output power
• the efficiency at full load
Solution
VF
Vt = Ea –IaRa –VB = 275-95×0.125-2 = 261.13 V
P0 = Vt Ia = 26.13×95=24.81 Kw
Rf
If
= 24.81×103+1500+952×0.125+1202/50
+
-
Pin = P0 + Plosses
Ia
E
Ra
RL
= 27.913 Kw
η = (24.81/27.913)×100 = 88.88 %
-
Vt
IL
+