# DC Motor help

Discussion in 'Homework Help' started by dan_man555, Sep 5, 2010.

1. ### dan_man555 Thread Starter New Member

Sep 1, 2010
18
0
I need help working this one out. Would really appreciate the help.

An 80-V DC motor has a constant field flux (separately excited) and a nameplate speed of 1150 rpm with 710 W output power. The nameplate armature current is 10 Amps and the no-load current it 0.5 A. Assume constant rotational losses.

a) Determine the armature resistance
b) Estimate the rotational loss
c) Find the machine constant in V/(rad/s)
d) What is the no-load speed in rpm
e) Find the effieciency at nameplate load, including field circuit losses of 30 W.

I know the formula for part a) is Ra=(V-E)/Ia, but how do you find E?

2. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,795
951
The voltage supplied to a DC motor MUST equal the voltage dropped by armature resistance + the back EMF.

3. ### dan_man555 Thread Starter New Member

Sep 1, 2010
18
0
Alright, so if when the motor is at rest, the induced voltage E = 0, and so the starting current is Ia= V/R
Ra=V/Ia
=80/10
=0.8 ohms

then for b)

P-losses = P-developed - P-out

E=V-Ia*Ra
=80-0.5*0.8
=79.6v

P-developed=E * Ia
=79.6*0.5
=39.8W

P-losses=39.8-710
=-670.2W

That doesn't seem right

Last edited: Sep 5, 2010
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I would approach it as follows

1. At rated load, the armature current is 10A at 80V DC supply.

So P_electrical input at rated mechanical power = .....?

2. The rated load mechanical power P_mechanical = 710W

3. What are the total losses at rated power? P_loss_total = ....?

5. The no load losses are the sum of the constant rotational losses and the no-load armature resistance losses (i.e. at Ia=0.5A).

6. The losses at rated load would be sum of the constant rotational losses and the rated armature resistance losses (i.e. at Ia=10A).

7. Set up two equations of the form

8. Solve for Ra and then P_loss_rotational.

And so on .......

5. ### Kermit2 AAC Fanatic!

Feb 5, 2010
3,795
951
torque is proportional to the current through the windings,

T = kI where T is the torque, I is the current, and k is a constant

The wire coils have both a resistance, R, and an inductance, L. When the motor is turning, the current is switching, causing a voltage,

V = L dI/dt

This voltage is known as the back-emf(electromotive force), e.

If the angular velocuty of the motor is w, then e = kw (like a generator)

This voltage, e, is working against the voltage we apply across the terminals, and so,

(V- kw) = IR where I = T/R

which implies (V-kw) = (T/k) R

The maximum or stall torque is the torque at which w = 0 or T = kV/R, and

The stall or starting current, I = V/R

The no load speed, w = V/k, is the maximum speed the motor can run.

Based on the information you have,you will need to make some substitutions for the missing information you have been given. Go back and try your first calculation again, but try to get to a ohm value for the armature instead. The value for a backEMF looks right. Take that and the 80 volts given and determine what the resistance would be that allows 10 amps to flow in the loaded motor.

6. ### dan_man555 Thread Starter New Member

Sep 1, 2010
18
0
Trying it your way t_n_k,

P_electrical input = 10*80
=800W

P_mechanical
= 710W

P_loss_rotational+0.5^2*Ra=800

P_loss_rotational+10^2*Ra=710

P_loss_rotational=800.23
Ra=-0.9023

How does that look?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
Ra is not 0.8Ω. As you indicated earlier you don't have the motor back emf. Check my approach.

I have Ra = 0.49875Ω.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
The no load loss will be source electrical input power under no load = 80*0.5=40W - not 800W.

The rated load loss = rated electrical input power - P_mechanical

Making it 800-710=90W

9. ### dan_man555 Thread Starter New Member

Sep 1, 2010
18
0
P-rot + Ra*(0.5^2)=40
P-rot + Ra*(10^2)=90
That gives me Ra=0.5013Ω, and P-rot losses=39.87W

For the machine constant I have the formula:
T-dev= K Ia

P-dev=Pout-P-rot losses
=710-39.87
=670.13W

T-dev= Wm*P-dev
=670.13 * (1150*2pi)/60
=670.13*120.43
=80,703N.m

K=Ia/T-dev
=0.5/80,703

or

E=K*Wm

E = V - Ia*Ra
=80-0.5*0.5013
=79.75W

K= 79.75 / (1150*2pi)/60
=79.75/120.43

Last edited: Sep 6, 2010
10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I agree with your Ra and rotational losses.

My earlier Ra estimate was incorrect - algebraic error on my part.

I have the machine constant as 0.62267 V/rad/sec.

Based on this reasoning......

At rated load the armature drop is Ra*Ia=.50125*10=5.0125V

Hence E_rated=80-5.0125=74.987V

Rated speed = 1150rpm=120.428 radians/sec

Machine constant= 74.987/120.420=0.62267 V/rad/sec

This would give (I believe) a no-load speed of 1223 rpm.

11. ### dan_man555 Thread Starter New Member

Sep 1, 2010
18
0
Your right, i put the wrong current.
Here's my attempt for d:

k=0.6227
Wm=79.749/0.6227
Nm=1223rpm

and e:

efficiency is usually P-out/P-in * 100%,
so (710-30)/800 * 100
=85%
I assume you can just subtract the 30W field loss directly from the output power.

12. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
For the efficiency I would calculate the total input power as the base and the mechanical output as the power yield.

Total rated input power = 830W (including field excitation)

The yield is 710W mechanical.

Efficiency = (710/830)*100=85.54% which is close to yours but slightly different.

dan_man555 likes this.