Republic of Yemen
Sana'a University
Factually of Engineering
Mechatronics Department
A report submitted in periodic experiments of the requirements for Electrical Machine laboratory by
Osama Hassan M. Al-Mohaia 2017-200
Group II
Supervised by
Eng. Osama As'ham
February 2019

To connect and run the machine as a self excited generator with the required and suitable instrument
To demonstrate and plot the external and internal load characteristics of a
DC series generator .
2x shunt wound machine LEYBOLD-DIDACTIC GMBH.
1x Tacho generator (1V./1000rpm).
Variable resistance as virtual load (0-1000 ohm).
4x Analog AVO-meter.
1x DC power supplies range of (0-250V).
Required Number of connection wires.
Power:0.3 KW
Armuture rated Voltage: 220 V
Armuture rated Current: 0.63 A
Rated Speed : 2000 rpm (min -1 )
Field Voltage: 220 V
Field Current: 0.08 A
Figure1 load charactristic of Dc shunt generator

The general procedure is detailed below:
1.
The motor generator set is arranged and connected to the dc supply as shown in the figure .
2.
The generator is run up to its normal speed by means of the motor .
3.
The speed is kept constant for all reading by varying the motor supply voltage.
4.
To obtain the external load characteristic R
L
is connected and a series of reading between zero load current and 100% load current is taken.
5.
Plot the relation between VL and IL.
6.
Plot the variation of power input ,power output and efficiency with the load current.
At speed of 2000rpm n(rpm) Vin (V)
2000
2000
2000
2000
2000
2000
2000
195
208
223
235
246
250
250
Iin(A)
0.18
0.31
0.44
0.53
0.6
0.72
0.84
Ifg(A)
0.084091
0.077273
0.072727
0.068182
0.061818
0.047727
0
IL(A)
0
0.1
0.2
0.3
0.4
0.5
0.63
Ia(A)
0.084091
0.177273
0.272727
0.368182
0.461818
0.547727
0.63
Vt(V)
185
170
160
150
136
105
0
Table 1 Results of generator voltage at speed 2000 rpm

load charactristic of D.C shunt generator 1 Figure

According to the result data and graph it conclude that:

The generator e.m.f becomes constant when the field currents is saturated.
-

The generator builds up the voltage only when the speed is more than critical speed and field circuit resistance is less than the critical field resistance.
There are residual value of the generator which make the relationship between generator voltage and field current diffrenet in increasing about decreasing.this residual value due to the flux stored in the armature .This agree with the theoretical concept which call hysteresis curve.
1.
Comment on the shapes of curves you have obtained.
It is notic that from the concluded curve , As the load on the generator is increased, I
L increases and so l a = IF + I
L
, also increases. An increase in l a increases the armature resistance voltage drop
Ia*Ra causing Vt= Ea – (Ia*Ra )to decrease. This is precisely the same behavior observed in a separately excited generat.or. However, when Vr decreases, the field current in the machine decreases with it. This causes the flux in the machine to decrease, decreasing
Eg- Decreasing Eg causes a further decrease in the terminal voltage Vt= Ea – (Ia*Ra ).
Notice that the voltage drop-off is steeper than just the lARA drop in a separately excited generator. In other words, the voltage regulation of this generator is worse than the voltage regulation of the same piece of equipment connected separately excited.
2.
Show how you can obtain the internal characteristic of the generator if the armature resistance is
50ῼ and the field resistance is 2.2 Kῼ.
To obtain the answer apply the equation which illustrate the load internal characteristic of shunt DC generator which is relationship between the generated voltage E g
and the load current I
L
.
E g
=V t
+I a
*(R a
) where I a
=I
L
+I
F
I
F
=V t
/R
F
. then we obtain I a
=I
L
+I
F
.
Ia(A)
0.084091
0.177273
0.272727
0.368182
0.461818
0.547727
0.63
Eg(V)
189.2045
178.8636
173.6364
168.4091
159.0909
132.3864
31.5
Table 1 the internal characteristic data of shunt generator

Figure 2 the internal characteristic of shunt dc generator
3.
Given that the value of load resistance R
L to be 1Kῼ.What is the Load voltage and Load current at the terminal of the generator?
First obtain the value of resistance load at any point on the curve or the table
R
L
=V t
/I
L
so, R
L
=160/0.2 =800ohm
From the load characteristic we obtain the value at point (160,0.2) and make linear line from the center it will be equal to(165,0.17)
So the current =0.17A and voltage 165V.
4.
Calculate the value of percentage regulation at I=0.5A.
The voltage regulation formula is V%=((V
NL
-V
F
)/V
F
)*100%
V%=((185-105)/105)*100=76.19%
5.
What is the value of no load current and no load power of the motorgenerator set?
At no load the current will be I=0 A
The power equal to 195*0.18=35.1W
6.
Calculate approximately the mechanical plus iron losses of the set if the shunt field resistance of the motor is 2.4Kῼ and that of the generator is 2.2Kῼ.
From the equtions of power flow diagram (p iron losses
+P mechanical losses
) =P in
- P conv since
P in
=V t
*I
L, and P conv
=E a
*I a
.
SO for the genertor at field resistance =2.2Kohm
P=189.02*0.0843=15.89W