DC Machine Example Problems

8/10/2019 DC Machine Example Problems

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Andy Knight (/~aknigh/index.html) - Electrical Machines/~aknigh/electrical_machines/machines_main.html)

DC Machine Exampleseparately Excited

separately excited DC motor is rotated at 1000rpm, The variation of armature terminal

oltage as a function of field current is measured under no-load conditions and tabulated

elow:

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0 30 60 85 102 115 124 130 134

he field winding supply and the field resistance is adjustable. The armature

inding resistance and the armature terminal voltage .a. Calculate the field current if the motor is operated with no-load at 1000 rpm

b. The motor drives a load at 1200 rpm. Calculate the armature voltage at 1200 rpm if the

field resistance

c. Calculate the torque for the above condition

d. The motor supplies a mechanical load of 4000W at 1450rpm. The mechanical rotational

losses are 160W, calculate the efficiency

omments

his question is similar to many DC machine questions and falls into two parts. At the start of

he question, you are given a reasonable amount of data about a specific operating condition.

ou need to then take useful information from this operating condition and apply it to theew operating conditions specified in the rest of the question.

olution

his is a separately excited motor problem:

here are two important pieces of information in the beginning of the question:The information is being given for "no-load" conditions. No-load in machines means

no useable power flow out of the machine. If there is no power flow in a dc machine,

there is no armature current, and therefore the terminal voltage equals the

armature voltage:

You are being given data on the induced armature voltage at a given speed

The data for armature voltage at a given speed allows you to find the nonlinear

relationship between flux and field current, which is independent of speed.

nowing how the data in the question is useful is a significant part of the solution process for

C machine questions.

I

F

V

T

= 2 4 VV

F

= 0 . 2 R

A

= 1 3 0 VV

T

= 6 0 R

F

= 0

I

A

=

E

A

V

T
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a. This question requires you to read data from the table provided. Under no-load

conditions, occurs when

b. Armature voltage is given by

i.e. voltage is a function of and speed. Flux is a function of field current, and since

field voltage and resistance are specified in the question, a first step is to find field

current:

gives

From the table in the question, when: and then

. We need to find for the case when and

There are two possible approaches:

i. In both cases, the field current is constant, therefore flux will be constant.

Armature voltage will be proportional to speed:

Giving

ii. Using the armature voltage equation, find :

Now, at 1200 rpm armature voltage can be found directly from the armature

voltage equation:

Givingc. To find the torque, there are three possible approaches. Two of the approaches require

the calculation of if it has not already been done, two of the approaches require the

calcualtion of the armature current. require the armature current, which can be found

from the armature loop equation

I

i. Using power equations:

= 1 3 0 VV

T

= 0 . 7 AI

F

= E

A

= V

F

I

F

R

F

= 0 . 4 A

I

F

= 1 0 0 0 r p m

= 0 . 4 AI

F

= 1 0 2 VE

A

E

A

= 1 2 0 0 r p m

= 0 . 4 AI

F

= E

1 2 0 0

E

1 0 0 0

1 2 0 0

1 0 0 0

= 1 2 2 . 4 VE

A

= 0 . 4

I

F

= 0 . 4

I

F

=

E

A

= =

2

6 0

1 0 0

3

= 0 . 9 7 4

E

A

1 2 0 0

=

= 1 2 0 0

2

6 0

= 1 2 2 . 4 VE

A

= + V

T

E

A

I

A

R

A

= 3 8 . 0 A

A

P

c

= = E

A

I

A


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ii. Using the torque equation directly: . This approach requires the

calculation of , if not already done in the armature voltage calculation.

iii. Rearranging the torque-speed equation:

d. The final part of the question includes mechanical losses. In this case

No information is available on the flux in the machine, so armature voltage or current

cannot be directly determined from the armature voltage equation or torque equation.

The preferred approach is to consider the power in the armature circuit:

Solving the quadratic for armature current results in two values, or

. The correct answer will be the one that results in the lowest power

loss, . The armature voltage can be found by solving the armature

circuit equation

At this point, the armature losses can be found, but this is a separately excited machineand it is important to remember to account for power flow in the field circuit. To find

the field current that gives at 1450 rpm, it is necessary to calculate

armature voltage that would be induced at the same field current with a rotational

speed of 1000 rpm.

gives which, from the table, corresponds to a field current

Finally, efficiency can be found from

giving

eries Motor Exam le

P

c

= = E

A

I

A

=

1 2 2 . 4 3 0 I

A

1 2 0 0

= 3 7 N

= I

A

=

V

T

R

A

(

)

2

= P

c

=

+ P + a a P

4 0 0 0 + 1 6 0 = 4 1 6 0 W

= + V

T

I

A

E

A

I

A

I

2

A

R

A

+ = 0 I

2

A

R

A

V

T

I

A

P

c

= 6 1 6 AI

A

I A = 3 3 . 7 5 A

= 3 3 . 7 5 AI

A

= +

V

T

E

A

I

A

R

A

= 1 2 3 . 2 5 AE

A

= 1 2 3 . 2 5 A

E

A

= E

A

1 0 0 0

E

A

1 4 5 0

1 0 0 0

1 4 5 0

= 8 5 . 0 VE

A

1 0 0 0

= 0 . 3 AI

F

P

=

P

+

P

P

= + + = 3 9 5 WI

2

A

R

A

I

2

F

R

F

P

a a

=

4 0 0 0

4 3 9 5

= 9 1 . 0 %


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Andy Knight (mailto:[email protected])

series DC motor has combined armature and field resistance of . When

onnected to a supply of at standstill, the motor develops a torque of 1500Nm.

a. Calculate the armature current at standstill and combined motor constant

b. Calculate the torque when the speed is 500 rpm

c. Calculate the output power and efficiency when operating at 500 rpm (neglect

mechanical losses)

olution

his is a series motor problem:

a. At standstill, the angular velocity is zero, therefore the armature voltage is zero

. The full terminal voltage is dropped across the winding

resistances and gives

Using the torque equation for a series excited DC machine, , the combined

motor constant can be found:b. To find the torque it is possible to either substitute directly into the series motor torque-

speed equation, or to first find the armature currents:

i.

Re-arranging the torque speed equation and submitting for kc and speed gives

ii. Alternately, substitute into in the armature loop equation to

obtain and solve for armature current.

and use the torque equation .

c. To find efficiency neglecting rotational losses:

+ = 1 . 2 R

A

R

S

= 4 8 VV

T

c

( = c = \ 0 ) E

A

I

A

= + ( + ) V

T

I

A

R

A

R

S

= 4 0 AI

A

= c

I

2

A

c= 1 5 0 0 / 1 6 0 0 = 0 . 9 3 7 5

N

A

2

=

V

T

c

1

+ R

A

R

S

c

= =

2

6 0

5 0

3

= 0 . 8 5 4 N

= c E

A

I

A

= ( c + + ) V

T

I

A

R

A

R

S

= 0 . 9 5 5 AI

A

= c I

2

A

= = =

P

c

P

E

A

I

A

V

T

I

A

E

A

V

T

a = 9 7 . 0 %
mailto:[email protected]

DC Machine Example Problems

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