Shri Ramdeobaba College of Engineering & Management, Nagpur NUMERICALS ON DC MOTOR UNDER THE GUIDANCE OF PROF. SURESH BALPANDE PRESENTED BY : PAYAL NIMJE(Roll No.97)(VIII-B)

Calculating mechanical power requirements in DC motor In DC motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to frictional losses, there are power losses in Joules/sec .(Iron losses in coreless DC motors are negligible). Pel = Pmech + Pj loss

Prot = M x ω ωrad = ωrpm x (2∏)/60 Physically, power is defined as the rate of doing work. For linear motion, power is the product of force, multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. Prot = M x ω Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. ωrad = ωrpm x (2∏)/60

It is important to consider the units involved when making the power calculation. A reference that provides conversion tables is very helpful for this purpose. Example: Determine the power required to drive a torque load of 3 oz-in (ounce-inch)at a speed of 500 RPM. The product of the torque, speed, and the appropriate conversion factor from the table is: 3oz-in x 500rpm x 0.00074 = 1.11 Watts

Thermal calculation Power loss: I^2 *R For example: If the total motor current was 0.203 A and the armature resistance 14.5 Ohms the power lost as heat in the windings is: power loss = (0.203)^2 x 14.5 = 0.59 Watts The heat resulting from I^2R losses in the coil is dissipated by conduction through motor components and airflow in the air gap. Motor manufacturers typically provide an indication of the motor’s ability to dissipate heat by providing thermal resistance values.  Thermal resistance is a measure of the resistance to the passage of heat through a given thermal path.

The total I^2R losses in the coil (the heat source) are multiplied by thermal resistances to determine the steady state armature temperature. The steady state temperature increase of the motor (T) is given by: Tinc = I2R x (Rth1 + Rth2) Where: Tinc = temperature increase I = current through motor windings R = resistance of motor windings Rh1 = thermal resistance from windings to case Rh2 = thermal resistance case to ambient

For example: A motor running with a current of 0 For example: A motor running with a current of 0.203 Amps in the motor windings, with an armature resistance of 14.5 Ohms, a winding-to-case thermal resistance of 8 °C/Watt, and a case-to-ambient thermal resistance of 39 °C/Watt. Solution: The temperature increase of the windings is given by: T = 0.203^2 x 14.5 x (8 + 39) = 28°C

PROBLEMS ON DC SERIES MOTOR

STEPS TO SOLVE SERIES MOTOR PROBLEMS Field coil being in series with armature, in general field current If and armature current Ia are same. The back emf can be calculated as Eb=V-Ia(Rse+Ra) Eb is also given as 4. Torque developed by motor is given by 5. One should be careful for situations when field current and armature current may not be same. One such situation occurs when diverter resistance is connected across the field coil controlling speed.

Question : A 220 V d.c series motor has armature and field resistances of 0.15 Ω and 0.10 Ω respectively. It takes a current of 30 A from the supply while running at 1000 rpm. If an external resistance of 1 Ω is inserted in series with the motor, calculate the new steady state armature current and the speed. Assume the load torque remains constant. Solution:

Since the load torque remains constant in both the cases, we have: Now equations involving back emfs:

In the second case: Thus taking the ratio of Eb2 and Eb1 we get:

Problem on DC shunt motor

Question: A 220 V shunt motor has armature and field resistance of 0 Question: A 220 V shunt motor has armature and field resistance of 0.2 Ω and 220 Ω respectively. The motor is driving a constant load torque and running at 1000 rpm drawing 10 A current from the supply. Calculate the new speed and armature current if an external armature resistance of value 5 Ω is inserted in the armature circuit. Neglect armature reaction and saturation. Solution:

For initial operating point: IL1= 10 A, ra = 0 For initial operating point: IL1= 10 A, ra = 0.2 Ω ,rf=220 ohm and supply voltage V = 220 V. Now we write down the expressions for the torque and back emf.

Since field resistance remains unchanged If2=If1=1 A Since field resistance remains unchanged If2=If1=1 A. Let the new steady armature current be Ia2 and the new speed be n2. In this new condition the torque and back emf equations are Taking the ratios of Te2 and Te1 we get,

Now taking the ratio emfs It may be noted that, for constant load torque the steady state armature current does not change with change in the value of the armature resistance.

STEPPER MOTOR NUMERICALS Steps per Revolution Typical stepper motors are 1.8 degrees per step, which is 200 steps per revolution. Example: A stepper motor moves 0.9 degree per step . So the steps per revolution will be

2. Steps vs frequency Example: If we want our motors to travel at a maximum rate of 200 inches per minute, and our step per inch rate is 8000 . Our Steps per second can be found as 26.7 KHz.

Where θs is the stator angle θr is the rotor angle ns is the number of stator poles nr is the number of rotor poles n is the number of steps per revolution Δθ is the stepping angle 3. Stepping angle : The number of steps per revolution is related to the stepping angle as follows:

4. rpm calculation: Example: A stepper motor that turns at 5000 PPS and has a step angle of 1.8º.What would be these pps into rpm? Solution: With a step angle of 1.8 degrees and 360 degrees in a single revolution, no. of steps per revolution = 200 steps per revolution (i.e. 360 / 1.8). No. of revolution per second= no. of pulses per second/steps per revolution 5000 / 200= 25 revolutions per second 25 * 60= 1500 rpm

Torque calculation: Calculating load torque is determine the force in the system and the logical distance between the motor shaft and the where the force is acting. T=Fr Example: The drum radius is 0.05m , 0.009m for the cable, the cable hanging off it has 12Kg mass worst case, and g is approx. 9.8. So the torque you need is 0.059 * 9.8 * 12 = 6.94Nm

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