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

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Improving design (f_dc.html) Separate& Shunt Field (f_dc_sep.html)

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Voltage and Torque (f_dc_volts.html) Series Field (f_dc_series.html)

DC Machines (f_dc.html)

Commutation (f_dc_comm.html) Power (f_dc_power.html)

AC Machines (f_ac.html) Real Machines (f_dc_real.html) Examples (f_dc_examples.html)

DC Machine Examples Separately Excited

A separately excited DC motor is rotated at 1000rpm, The variation of armature terminal voltage as a function of field current is measured under no-load conditions and tabulated below: IF VT

0.0 0

0.1 30

0.2 60

0.3 85

0.4 102

0.5 115

0.6 124

0.7 130

0.8 134

The field winding supply V F = 24V and the field resistance is adjustable. The armature winding resistance R A = 0.2Ω and the armature terminal voltage V T = 130V . 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 R F = 60Ω 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 Comments

This question is similar to many DC machine questions and falls into two parts. At the start of the question, you are given a reasonable amount of data about a specific operating condition. You need to then take useful information from this operating condition and apply it to the new operating conditions specified in the rest of the question. Solution

This is a separately excited motor problem:

There 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, I A = 0 and therefore the terminal voltage equals the armature voltage: E A = V T 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. Knowing how the data in the question is useful is a significant part of the solution process for DC machine questions. a. This question requires you to read data from the table provided. Under no-load

a. This question requires you to read data from the table provided. Under no-load conditions, V T = 130V occurs when I F = 0.7A b. Armature voltage is given by E A = kϕω

i.e. voltage is a function of kϕ 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: VF = I F RF

gives I F = 0.4A From the table in the question, when: nm = 1000rpm and I F = 0.4A then E A = 102V . We need to find E A for the case when nm = 1200rpm and I F = 0.4A

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: 1200 E 1200 = E 1000

1000

Giving E A = 122.4V ii. Using the armature voltage equation, find kϕ : kϕ∣ ∣

IF =0.4

=

EA ω 2π

ω= n m

kϕ∣ ∣

IF =0.4

60

100π =

3

= 0.974

Now, at 1200 rpm armature voltage can be found directly from the armature voltage equation: EA ∣ ∣

1200

= kϕω 2π

ω= 1200

Giving E A

60

= 122.4V

c. To find the torque, there are three possible approaches. Two of the approaches require the calculation of kϕ 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 VT = E A + I A RA

IA

= 38.0A

i. Using power equations:

P conv = τ ω = E A I A

P conv = τ ω = E A I A

τ=

122.4 × I A × 30 1200π

τ = 37N m

ii. Using the torque equation directly: τ = kϕI A . This approach requires the calculation of kϕ , if not already done in the armature voltage calculation. iii. Rearranging the torque-speed equation: ω

=

VT kϕ

−

RA (kϕ)

2

τ

d. The final part of the question includes mechanical losses. In this case P conv = P out + P + rotational =4000 + 160 = 4160W

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: VT I A = E A I A + I ∴ I

2 A

2 A

RA

R A − V T I A + P conv = 0

Solving the quadratic for armature current results in two values, I A = 616A or I A = 33.75A . The correct answer will be the one that results in the lowest power loss, I A = 33.75A . The armature voltage can be found by solving the armature circuit equation VT = E A + I A RA E A = 123.25A

At this point, the armature losses can be found, but this is a separately excited machine and it is important to remember to account for power flow in the field circuit. To find the field current that gives E A = 123.25A 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. EA ∣ ∣

gives E A ∣∣ 1000

= 85.0V

1000

= EA ∣ ∣

1000 1450

1450

which, from the table, corresponds to a field current

I F = 0.3A

Finally, efficiency can be found from P out η=

P out + P losses

P losses = I

2 A

4000 η=

giving η

= 91.0%

Series Motor Example

2

R A + I F R F + P rotational = 395W

4395

Series Motor Example

A series DC motor has combined armature and field resistance of R A + R S = 1.2Ω . When connected to a supply of V T = 48V at standstill, the motor develops a torque of 1500Nm. a. Calculate the armature current at standstill and combined motor constant kc 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) Solution

This is a series motor problem:

a. At standstill, the angular velocity is zero, therefore the armature voltage is zero (E A = kcI A ω = \0) . The full terminal voltage is dropped across the winding resistances and V T = I A + (R A + R S ) gives I A = 40A Using the torque equation for a series excited DC machine, τ motor constant can be found: kc

2

= kcI A

= 1500/1600 = 0.9375N mA

, the combined

−2

b. To find the torque it is possible to either substitute directly into the series motor torquespeed equation, or to first find the armature currents: i. VT ω=

1

− − √ kc

ω= n m

−

RA + RS kc

√τ

2π 60

50π =

3

Re-arranging the torque speed equation and submitting for kc and speed gives τ = 0.854N m

ii. Alternately, substitute E A = kcI A ω into in the armature loop equation to obtain V T = I A (kcω + R A + R S ) and solve for armature current. 2 I A = 0.955A and use the torque equation τ = kcI . A c. To find efficiency neglecting rotational losses: η =

eta = 97.0%

© Andy Knight (mailto:[email protected])

P conv P in

=

EA IA VT I A

=

EA VT