EE 332 Lab 3
Students’ Name: Raya Zhu Jie Ren (Section H7) Xiaoang Ge ID:
Section: Due Date:
1229969 1212947
H8 March, 15. 2012
Tables Initial Tests Ra(W )
Rsf (W )
Rf (W )
t err (Nm)
8.2
2.3
286
0
Separately Excited No Load: 400mA Nominal Field Current
No Load: 80V Nominal Terminal Voltage
Vt
nm
w
kf
(mA)
(V)
(rpm)
(rad/s)
400 400 400 400
20.38 39.78 60.40 80.50
218 439 673 896
22.83 45.99 70.48 93.83
400
101.60
1138
119.17
0.853
Nominal
If
V t (V) 20 40 60 80 100
Nominal
Vt
If
nm
w
kf
(Vs/rad)
If (mA)
(V)
(mA)
rpm
(rad/s)
(Vs/rad)
0.893 0.865 0.857 0.858
400 350 300 250
80 80 79 80
400 350 300 250
898 925 962 1065
94.04 96.87 100.74 111.53
0.852 0.822 0.780 0.717
Separately Excit ed Load Test: Nominally 100V terminal voltage, 400mA field c urrent Nominal
Measurements
Calculated
Ia
If
Vt
Ia
Pt
tout
nm
E
Pout
Pfield
PRa
Pin
PF&W
tdev
h
kf
(A) 0.0 0.5 1.0 1.5 2.0 2.5 3.0
(mA) 400 400 400 400 400 400 400
(V) 100.2 100.5 103.3 101.2 100.0 100.3 100.6
(A) 0.363 0.530 1.000 1.519 2.050 2.496 3.054
(W) 36.2 52.9 103.3 152.9 204.4 250.2 309
(Nm) 0.00 0.14 0.50 0.86 1.26 1.57 1.96
(rpm) 1131 1116 1095 1035 977 951 906
(V) 97.22 96.15 95.10 88.74 83.19 79.83 75.56
(W) 0.0 16.4 57.3 93.2 128.9 156.4 186.0
(W) 45.8 45.8 45.8 45.8 45.8 45.8 45.8
(W) 1.08 2.30 8.20 18.92 34.46 51.09 76.48
(W) 82.0 98.7 149.1 198.7 250.2 296.0 354.8
(W) 35.1 34.2 37.8 40.8 41.0 42.8 46.6
(Nm) 0.30 0.44 0.83 1.24 1.67 2.00 2.43
(%) 0.00 16.58 38.46 46.92 51.53 52.83 52.42
(Vs/rad) 0.821 0.823 0.829 0.819 0.813 0.802 0.796
Series Excited Nominal
Measurements
Calculated
Ia
Vt
Ia
Pt
tout
nm
E
Pout
PRa+Rsf
PF&W
tdev
h
kc
(A) 3.0 2.5 2.0 1.5
(V) 81.1 80.8 80.4 80.2
(A) 2.960 2.576 2.035 1.490
(W) 240.4 210.3 163.9 120
(Nm) 1.42 1.12 0.74 0.39
(rpm) 808 916 1113 1432
(V) 50.02 53.75 59.03 64.56
(W) 120.32 107.43 86.25 57.88
(W) 92.00 69.68 43.48 23.31
(W) 28.08 33.19 34.17 38.80
(Nm) 1.75 1.44 1.03 0.64
(%) 50.05 51.09 52.62 48.24
(Nm/A ) 0.200 0.218 0.249 0.289
Graphs
2
Sample calculation Separately Excited DC Motor: 1. Using the no- load test data, calculate the value of kφ when V t=100V and If = 400mA (for no load test) π π ω
V2 t = E2 × 1138 = 60 n = 60 = 119.17rad/s 6 k = Vt = 101. 119.17 = 0.853Vs/rad E =PvVt −=IERI == 100. 6 − 3. 0 54 ×8. 2 = 75. 5 6V 75. 5 6× 3. 0 54 = 230. 7 5W 2 2 × 906 = n = 60 60 = 94.876rad/s PoutP=filout= If R=f =1.90.64×94. 8 76 = 185. 9 57W × 286 = 45.76W Pi = PtPout+ Pfi l1 85.= 30995 + 45.76 = 354.76W = Pi = 354.76 × 100% = 52.42% φ
ω 2. Using the separately excited load data, calculate Ea, Pdev, Pout, and ηfor the case where Ia=3.0A.
π
ω
τ
π
ω
η
3. A speed of 1400rpm is required to be obtained in the separately excited DC moor using field-weakening (reducing If ), while the armature voltage and current are held constant. Determine the τdev available at the speed of 1400rpm for the case of I a=3.0A.
E = Vt −2IR = 100.2 ×6 1400 − 3.054 ×8.2 = 75.56V = 60 n = 60 = 146.61rad/s 56 ×3.61 0 = 1.55Nm v = EI = 75.146. π
ω
τ
π
ω
Series DC Motor: 4. Using the series experiment data, calculate Ea, Pout, Pdev, τdev, η and the constant kc for the case where Ia=1.5A. Ω
R = R + R = 8. 2 + 2. 3 = 10. 5 f E = Vt − I2Rn = 80.2 2 ×−1432 1.490 × 10.5 = 64.555V = 60 = 60 = 149.96rad/s PPoutv== EoutI = =64.0.535586×1.×149.49096==96.57.187W 88W P 9 6. 1 87 v = = v 149.96 = 0.64Nm π
ω
τ
τ
π
ω
ω
= PoutPt = 5120.7.880 ×100% = 48.24% kc = Iv = 0.1.46904 = 0.289Nm/A
η
τ
5. Assuming the test series DC motor is mechanically coupled to a fan load for the cas e of Ia=3.0A, the speed is required to reduce to 600rpm by inserting an exte rnal resistance R ex in series with R s. Vt is constant while changing R sf . The torque required by a fan is proportional to the square of the speed. Neglect armature reaction and rotational losses, i.e. you can assume c is a constant in φ =cIa and τdev equals load torque. Determine the values of R ex and Ia required for the new speed. π π ω Set a Constant A,
= 260n = 2 ×60600 = 62.83rad/s v = lo = A A = lo = ( 2 1.×6048082 ) = 1.983e − 4 lo = A = 1.983e − 4 × 62.83 = 0.783Nm I =√ lkco = 00..728300 = 1.979A E = kcI Vt −=E0. 200 81.×11.−97924.×8662.83 = 24.86V Rll = I = 1.979 = 28.418 Rx = Rll − R = 28.418 − 10.5 = 17.918 τ
For the 3.0A case,
τ
ω
τ
π
ω
For the 600rpm case, τ
ω
τ
ω
Ω
Ω
Questions
1. Vt is a function of the speed in rpm, and Vt has a range of 20V to 100V. When current It is constant the rpm is increasing with Vt. The Vt vs rpm is a line with positive slope. 2. Yes, this line is linear. The line form can be indicated from the graph of Separately Excited Motor No load. When I f is constant, the line is a linear line. V t = Ea in the no
ᵠ
load test and E a k fw m . Therefore, Vt = k ὠm and, k and
ᵠ are all constant. Vt
is
ᵠ
proportional to ὠm . The slope is k . 3. The relationship of the speed and the field current is nonlinear. When If varies, through f f ( I f ) Note that
1 f ( I f )
E k fw m and a , we can get the relationship of Ea and If which is nonlinear. Vt=Ea, in no load test, the equation can be transform into this equation:
k w m V t
.
If Vt is a constant, the rpm will be inverse proportional to If .
4. The calculation value of torque is 0.66Nm and the measure value is 0.69. These two values are closed with error 4.5%. These values are closed matched. 5. The shape of torque-speed curve will be affected by an armature reaction. In a motor, when the load increases the flux-weakening effects reduce its flux. From the equation
w m
V t K
Ra
( K ) 2
t ind , the effect of a reduction in flux will increase the motor speed at
any given load over the speed it would run at without armature reaction. 6. The separately excited motor is not effected by armature voltage and current. In the application of a conveyor belt drive, this will keep a constant speed and it also easy to control. 7.
The flux in the machine is controlled by the field current. In series motor, due to non-saturating field happens at the most time, then it forms a linear relation , where c is a constant. As a result, flux and current are proportional.
8. There are two major components of losses in the system, the first one is the copper 2 losses and the second one is the rotational loss. From the equation I R, the losses can be found in the current and the resistance which are the field resistance and the armature resistance. The rotational losses are from core and mechanical loss. The core losses are the eddy current and hysteresis losses, and the mechanical loss is from
friction in the system. 9. From the speed and torque diagram of series excited motor, the speed and torque on the diagram have a relation of We can conclude that the series excited motor can provide a large amount of torque when the speed is low. It is useful when it is applied for the railway traction drive, because it can provide a faster acceleration at the start-up. Thus the series motor is suitable for a railway traction drive. 10. The direction of rotation can be changed by changing the direction of the current. The purpose can be also be obtain by changing the polarity of the armature winding or field winding.
