CEBU INSTITUTE OF TECHNOLOGY
UNIVERSITY
Name:______Bastian Louis S. Misa______ Date Performed:_____11/20/21________
Program & Year: ______BSME-3________ Date Due:___________________
Group:_____________
EE385
DC and AC Machineries, Lab
Laboratory Experiment No. 4
DC SHUNT GENERATOR CHARACTERISTIC
OBJECTIVE
1. Record the characteristics of the dc shunt generator
2. Calculate the efficiency of a dc shunt generator
SIMULATION SOFTWARE
NI Multisim version 14.1
FIGURE
PROCEDURE
1. Construct the circuit as shown above.
2. Set the machine parameters using the values given below.
RESULTS, DATA, CALCULATIONS
Terminal
Load
Voltage
Load
Power
(W)
(V)
Current(mA)
Output(W) Speed(rad/s)
-6
0
100
100.009 x10
0.01
1400
-3
25
99.967
62.48 x10
6.2459
1400
-3
50
99.942
124.928 x10
12.4856
1400
-3
75
99.917
187.345 x10
18.179
1400
100
99.892
249.731 x10-3
24.9461
1400
-3
125
99.867
312.085 x10
31.167
1400
-3
150
99.842
374.409 x10
37.3817
1400
200
99.793
498.963 x10-3
49.7930
1400
Power
Input(W)
1.96
8.204
14.42
20.72
26.88
33.18
39.34
51.8
Torque(Nm)
1.4x10-3
5.86 x10-3
10.5 x10-3
14.8 x10-3
19.2 x10-3
23.7 x10-3
28.1 x10-3
37 x10-3
GRAPH
100.05
100
99.95
99.9
99.85
99.8
99.75
0
1
2
3
4
5
6
7
IL
TERMINAL VOLTAGE Vs. LOAD CURRNET
8
9
Efficiency
(%)
0.
76.13
56.59
87.74
92.81
93.93
95.02
96.13
POUT
EFFICIENCY Vs POWER OUTPUT
COMPUTATIONS
Power Output
P0 = 99.992 ∗ 100.009 ∗ 10−6 = 0.01 𝑊
P1 = 99.967 ∗ 62.48 ∗ 10−3 = 6.2459 𝑊
P2= 99.942 ∗ 124.928 ∗ 10−3 = 12.4856 𝑊
P3 = 99.917 ∗ 187.345 ∗ 10−3 = 18.179 𝑊
P4 = 99.892 ∗ 249.731 ∗ 10−3 = 24.9461 𝑊 (125) = 99.867 ∗
312.085 ∗ 10−3 = 31.167 𝑊
P5= 99.842 ∗ 374.409 ∗ 10−3 = 37.3817 𝑊
P6 = 99.793 ∗ 498.963 ∗ 10−3 = 49.7930 𝑊
Power Input
POWER OUTPUT = (2π(1400*(30/π))(N)/60)
P0= (2π(1400*(30/π))( 1.4x10-3 )/60) = 1.96 W
P1= (2π(1400*(30/π))( 5.86 x10-3 )/60) = 8.20 W
P2=(2π(1400*(30/π))( 10.3 x10-3 )/60) = 14.42 W
P3=(2π(1400*(30/π))( 14.8x10-3 )/60) = 20.72 W
P4=(2π(1400*(30/π))( 19.2x10-3 )/60) = 26.88 W
P5=(2π(1400*(30/π))( 23.7 x10-3 )/60) = 33.18 W
P6=(2π(1400*(30/π))( 28.1 x10-3 )/60) = 39.34 W
P7=(2π(1400*(30/π))( 32.6x10-3 )/60) = 45.46 W
P8=(2π(1400*(30/π))( 37.0x10-3 )/60) = 51.8 W
Efficiency
η0 = 0.01/1.9 x 100 = 0.5%
η1= 6.25/8.20 x 100 = 76.13%
η2= 12.48/14.42 x 100 = 86.54%
η3= 18.17/20.72 x 100 = 87.69%
η4= 24.94/26.88 x 100 = 92.81%
η5= 31.167/33.18 x 100 = 93.93%
η6= 37.38/39.34 x 100 = 95.02%
η7= 49.79/51.8 x 100 = 96.13%
OBSERVATIONS AND INTERPRETATION OF DATA
In this experiment it can be observe that as the terminal voltage decreases, the load also increases
this signify a relationship between the two. The speed and torque utilized in the experiment is the same
with experiment #3. In relation to the pervious experiment, the power that is being inputted to DC
generator is the power that is produced by the DC motor. This shows the efficiency both of the motor
and the generator working together.
CONCLUSION
In conclusion base from the data and results from the previous experiment, we are able to understand
the characteristics of a Dc shunt generator through graphs. The terminal voltage is less than the generated Emf
Eg due to the voltage drop in the armature circuit. Whenever the load resistance is reduced the load current
increases. However as we go on reducing the load resistance, the terminal voltage also falls. This show an
inversely proportional relationship between the two. Just like the previous experiment, this characteristics are
important in determining the efficiency of generator. In terms of efficiency, we are able to graph the efficiency
of the generator which shows a good result. Upon understanding the characteristics of a DC generator we are
able to understand its strength and limitations.