industrial electrical automation motor control technology

Sunday, February 26, 2012

Example of electrical computation formulas for electrical calculation involving the function and operation of DC motors

Example 1: The field winding of a shunt DC (Direct Current) motor has a resistance of 110 ohms, and the emf (electro-motive force) applied to it is 220 volts. What is the amount of power expended in the field excitation?

With reference to Ohms law, the formula to compute for the current through the field is:
If = Et ÷ Rf

Where:
If - field current
Et - EMF armature terminal
Rf - field resistance

If = 220 volts ÷ 110 ohms
If = 2 amperes

And the power expended = Et x If = 220 volts x 2 amperes = 440 watts.

Similar results can also be obtained directly by using the equation:
Pf = Et2 ÷ Rf = (220)2 ÷ 110 = 440 watts

Example 2: A shunt DC motor whose armature resistance is 0.2 ohm and whose terminal voltage is 220 volts requires an armature current of 50 amperes and runs at 1,500 RPM when the field is fully excited. If the strength of the field is decreased and the amount of armature current is increased, both by 50%, at what speed will the motor run?

The expression for the counter emf of the motor is:
Ea = Et - Ia x Ra
and;
Ea1 = 220 volts - (50 amperes x 0.2 ohm) = 210 volts
Similarly;
Ea2 = 220 volts - (75 amperes x 0.2 ohm) = 205 volts
Also;
Ea = N ɸ K
and;
Ea1 ÷ Ea2 = N1 ɸ1 K1 ÷ N2 ɸ2 K2
Since the field is decreased by 50%, then
ɸ1 = 1.5 ɸ2, and Z1 = Z2
It follows that;
210 volts ÷ 205 volts = 1,500 RPM x 1.5 ÷ N2
N2 = 1,500 x 205 x 1.5 ÷ 210 = 2,196 RPM

Example 3: A 7.5 horsepower 220 volts interpole DC motor has armature and shunt field resistances of 0.5 ohm and 200 ohms respectively. The current input at 1,800 RPM under no-load conditions is 3.5 amperes. What are the current and electromagnetic torque for a speed of 1,700 RPM?

Under no-load conditions at 1,800 RPM,
Ia = IL - If = 3.5 amperes - (220 volts ÷ 200 ohms) = 2.4 amperes
ɸ KNL = Et - (Ia Ra) ÷ N = 220 - (2.4 x 0.5) ÷ 1,800 = 0.1216
ɸ KNL = ɸ KFL
At 1,700 RPM,
Ia = Et - (N ɸ K) ÷ Ra = 220 - (1,700 x 0.1216) ÷ 0.5 = 26.6 amperes

IL = Ia + If = 26.6 + 1.1 = 27.7 amperes
Te (electromagnetic torque) = 7.05 K ɸ Ia = 7.05 x 0.1216 x 26.6 = 22.8 ft-lb.

Example 4: The mechanical efficiency of a shunt DC motor whose armature and field resistances are 0.055 and 32 ohms respectively, is to be tested by means of a rope brake. When running at 1,400 RPM, the longitudinal pull on the 6 inch diameter pulley is 57 lbs. Simultaneous readings on the line voltmeter and ammeter are 105 volts and 35 amperes respectively. Calculate (a) the counter emf developed; (b) the copper losses; (c) the efficiency.

Ia = IL - If = 35 - (105/32) = 31.7 amperes
(a) Ea = Et - (Ia x Ra) = 105 - (31.7 x 0.055) = 103.26 volts
(b) Pc = (If2 x Rf) + (Ia2 x Ra) = [(3.3)2 x (32)] + [(31.7)2 x (0.055)] = 404 watts
(c) Output = 2π x 1,400 x (3/12) x 57 ÷ 33,000 = 3.8 horsepower
Input = (105 x 35) ÷ 746 = 4.93 horsepower
Efficiency = 3.8 horsepower ÷ 4.93 horsepower = 0.771 or 77%


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