No. 8 problem   A shunt generator has the following open-circuit characteristic at 800 r.p.m. Field amperes : 0.5    1.0    1.5    2.0    2.5    3.0    3.5 E.M.F. Volt : 54    107    152    185    210    230    245 Armature and shunt field resistances are respectively 0.1 Ω and 80 Ω. The terminal p.d. falls to 175 V when the armature current is 100 A. Find the O.C. volts and the volts lost due to (i) reduction in the field current (ii) armature resistance  (iii) armature reaction.

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter12: Power System Controls
Section: Chapter Questions
Problem 12.4P
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No. 8 problem

 

A shunt generator has the following open-circuit characteristic at 800 r.p.m.

Field amperes : 0.5    1.0    1.5    2.0    2.5    3.0    3.5
E.M.F. Volt : 54    107    152    185    210    230    245

Armature and shunt field resistances are respectively 0.1 Ω and 80 Ω. The terminal p.d. falls to 175 V when the armature current is 100 A. Find the O.C. volts and the volts lost due to
(i) reduction in the field current
(ii) armature resistance 
(iii) armature reaction.

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P 28.1.
Characteristics of
8. A shunt generator has the following open-circuit characteristic at 800 r.p.m.
D.C. Generators
Field amperes
E.M.F. Volt
0.5
1.0
1.5
2.0
2.5
3.0
3.5
ET 28.2.
54
107
152
185
210
230
245
Armature and shunt field resistances are respectively 0.1 2 and 80 2. The terminal p.d. falls to 175 V
when the armature current is 100 A. Find the O.C. volts and the volts lost due to (i) reduction in the field
current (ii) armature resistance (iii) armature reaction.
Separately-excited
Generator
[220 V (1) 27 V (ii) 120 V (iii) 8 V]
W Fig. 28.1
I Fig. 28.2
T Fig. 28.3
E P 28.3. No-load
The open-circuit characteristic of a shunt generator when driven at normal speed is as follows :
Field current
9.
0.5
1.0
1.5
2.0
2.5
3.0
3.5 A
O.C. volts
The resistance of armature circuit is 0.1 2. Due to armature reaction the effective field current is given
by the relation I (eff.) = I-0.003 I, Find the shunt field circuit resistance that will give a terminal voltage
of 220 V with normal speed (a) on open circuit (b) at a load current of 100 A. Also find (c) number of series
turns for level compounding at 220
1200 and (d) No. of series tuns for over-compounding giving a terminal voltage of 220 V at no-load and
230 V with 100 A armature current.
54
107
152
185
210
230
240 V
Curve for
Self-excited
with 100 A armature current ; take number of shunt turns per pole as
Generator
f Fig. 28.4
W Fig. 28.5
W Fig. 28.6
[(a) 80 2 (b) 66 Q (c) 6.8 Q (d) 9.1 turns]
10. Find how many series turns per pole are needed on a 500-kW compound generator required to give
450 V on no-load and 500 V on full-load, the requisite number of ampere-turns per pole being 9,000 and
6,500 respectively. The shunt winding is designed to give 450 V at no-load when its temperature is 20°C. The
final temperature is 60°C. Take a,= 1/234.5 per °C.
W 28.4 How to Find
Critical Resistance
[ 2.76] (Electrical Technology, Allahabad Univ. 1977)
Rc ?
OBJECTIVE TESTS – 28
EP 28.5 How to Draw
O.C. at Different v
1. The external characteristic of a shunt generator
(a) shunt
(6) series
10:41 PM
P Type here to search
A 4) G
24
2/22/2021
出
Transcribed Image Text:2 28. GENERATOR CHARACTERISTICS.pdf - Adobe Acrobat Pro File Edit View Window Help Оpen PA Create Customize 991 (25 of 28) 111% Tools Fill & Sign Comment Bookmarks Generator Characteristics 991 P 28.1. Characteristics of 8. A shunt generator has the following open-circuit characteristic at 800 r.p.m. D.C. Generators Field amperes E.M.F. Volt 0.5 1.0 1.5 2.0 2.5 3.0 3.5 ET 28.2. 54 107 152 185 210 230 245 Armature and shunt field resistances are respectively 0.1 2 and 80 2. The terminal p.d. falls to 175 V when the armature current is 100 A. Find the O.C. volts and the volts lost due to (i) reduction in the field current (ii) armature resistance (iii) armature reaction. Separately-excited Generator [220 V (1) 27 V (ii) 120 V (iii) 8 V] W Fig. 28.1 I Fig. 28.2 T Fig. 28.3 E P 28.3. No-load The open-circuit characteristic of a shunt generator when driven at normal speed is as follows : Field current 9. 0.5 1.0 1.5 2.0 2.5 3.0 3.5 A O.C. volts The resistance of armature circuit is 0.1 2. Due to armature reaction the effective field current is given by the relation I (eff.) = I-0.003 I, Find the shunt field circuit resistance that will give a terminal voltage of 220 V with normal speed (a) on open circuit (b) at a load current of 100 A. Also find (c) number of series turns for level compounding at 220 1200 and (d) No. of series tuns for over-compounding giving a terminal voltage of 220 V at no-load and 230 V with 100 A armature current. 54 107 152 185 210 230 240 V Curve for Self-excited with 100 A armature current ; take number of shunt turns per pole as Generator f Fig. 28.4 W Fig. 28.5 W Fig. 28.6 [(a) 80 2 (b) 66 Q (c) 6.8 Q (d) 9.1 turns] 10. Find how many series turns per pole are needed on a 500-kW compound generator required to give 450 V on no-load and 500 V on full-load, the requisite number of ampere-turns per pole being 9,000 and 6,500 respectively. The shunt winding is designed to give 450 V at no-load when its temperature is 20°C. The final temperature is 60°C. Take a,= 1/234.5 per °C. W 28.4 How to Find Critical Resistance [ 2.76] (Electrical Technology, Allahabad Univ. 1977) Rc ? OBJECTIVE TESTS – 28 EP 28.5 How to Draw O.C. at Different v 1. The external characteristic of a shunt generator (a) shunt (6) series 10:41 PM P Type here to search A 4) G 24 2/22/2021 出
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