A 50kw,250v ,dc shunt generator has a field circuit resistance of 60ω and an armature resistance of 0.02ω. Calculate i) load current ,field current and armature current. Ii) the generated armature voltage when delivering rated current at rated speed and voltage
Answers
Armature current I
a
= I
L
+ I
Sh
=
Output in watts
terminal voltage
=
500×10
3
400
=400A
In order that the current/path is not more than 200
A, a lap winding is to be used. The number
of parallel paths required is
A =
1250
200
= 6.25
Since the number of parallel paths cannot be a frac
tion as well an odd integer, A can be 6 or 8.
Let it be 6. (Note: A current/path of 200A need not
strictly be adhered)
Check: Frequency of the induced emf
f
=
PN
120
=
6
×
60
120
=30Hz and can be considered
as the frequency generally lies between 25 and 50 H
z
D
⡰
L =
kW
1.64 × 10
⡹⡲
B
⤑⤲
q N
=
500
1.64 × 10
⡹⡲
× 0.7 × 38400 × 600
≈ 0.189m
⡱
Note: Since the capacity is considerable, power dev
eloped in the armature can be taken as
power output for a preliminary design.
L = 1.1 pole arc
=1.1
ψ
τ
Since
ψ
lies between 0.6 and 0.7, let it be 0.7.
2
Since
τ
=
π
D
P
, L=1.1
×
0.7
×
πD
6
=0.4D
D
⡰
× 0.4D = 0.189
D
⡱
=
0.189
0.4
= 0.473m
⡱
D = 0.78m and L = 0.4 × 0.78 ≈ 0.31m
Check: Peripheral velocity v=
π
DN
60
=
π
×0.78×600
60
=24.2m/s
and is within the limits,i.e.30 m/s.
Answer:
(i) Load Current = 200 A, Field Current = 4.17 A, and Armature Current = 204.17 A
(ii) Generated Voltage emf = 254.08 V
Explanation:
Given:
P = 50kW
Vt = 250 V
Rsh = 60 Ohms
Ra = 0.02 Ohms
Solution:
(i)
Load Current:
Field Current:
Armature Current:
(ii)