im doing some revision and im stuck at this question.

A 10kW separately excited dc generator has an armature resistance of 0.4ohm. When delivering full load, the output voltage is 400v. Calculate the voltage on no load and on half load.

using this formula:
V = E - Ia Ra

at no load, Ia=0 thus V=E
therefore, voltage on no load is same as the output voltage which is 400v.

the problem now is how do i find the voltage on half load? thanks

 

Rather than just telling you the answer, I'll show you how to work it out for yourself.
a)
If the full load is 10kW (10,000W) and the output voltage is 400V, then the full load current must be:-
Ifl = 10000/400 + 25A (because W =V*I or I = W/V)
b)
Now the armature has an internal resistance of .4R, so at full load there will be an internal voltage drop of .4*25 = 10V (Ohms Law V=IR). So for 400V output, the EMF (ie. no-load voltage) must be 400+10 = 410V
c)
The half load current must be Ifl/2 = 25/2 = 12.5 Amps. So you can calculate the output voltage using the same technique as I showed you in b) above

Go for it!

 

Oops typo!

line 4 should read:-
Ifl = 10000/400 = 25A (because W =V*I or I = W/V)

 

The source of your misunderstanding is that your formula "V = E - Ia Ra" is not precise enough! Which value of V does it refer to?

This is a more precise version"Va = E - Ia*Ra"

 

This question may be a bit more complicated than I first thought.

If "half load" means 1/2 of "full load" current, then my solution c) is correct.

However, if "half load" means "half power" (ie. 5000W) then you need to first calculate the load resistance R using Ohms Law with Vfl and Ifl. Then calculate the required "half load" voltage Vh using the alternate equation for power (Wh = Vh*Vh/R)

note: from W=V*I you can use Ohms Law to derive alternatives W=I*I*R or W=V*V/R

 

The problem may be even more complicated!

I showed you how to calculate Vh for the case where the load R is constant and E is the variable. [solution d)]

There is another possible interpretation where E is constant and the load R is the variable. In this case:-
e) Wh=5000 = Vh*Ih and Vh = E -Ih*Ra. Eliminate Ih by substitution, then solve for Vh. (E and Ra are constants)

If you solve all 3 possible cases, you will have learned quite a lot! (and impressed your teacher!!!!)

 

ozjon69, thank you so much for your help mate

 

now there is another question that im quite confused with.

Q: A 20kW dc motor rotates at 1500rpm when fed from a 220 volt dc supply and delivering full rated torque. If the efficiency is 91% and the armature resistance is 0.15ohm, what is the back emf of the motor and the torque developed on the rotor?

It would be nice if someone can explain to me the steps and the formulas needed to solve this type of question. So far, i've managed to get the formula for the torque part which is:

T=(60*E*Ia)/(2*pie*n)
where
E is the back emf
Ia is the armature current
n is the speed

thanks

 

so is there anyone can help me with that question?

 

Hi kuridak,

I don't remember many formulas - I work things out from basics (first principles).
I'll show you how I do that (and hopefully improve your general understanding so that you can solve many future problems).

Some basics you need to memorise:-
Torque = Force * radius Tq = F * r
Work = Force * distance moved W = F * d
Power = Work/unit time P = W/t
Efficiency = Output/Input E = Po/Pi or Wo/Wi

You need to use a consistent set of units when working on particular cases.
If the problem is stated with mixed units you will need to apply some conversion factors when you calculate numerical answers.
For this problem, you need to know that 746Watts = 1HP and 1HP= 550 ft.lbs/sec (both well worth memorising)

Let's walk through the torque calculation:-
Torque is a rotational force F applied at a radius r. Tq = F * r
in 1 revolution of the motor, the force F will have moved distance d = 2*pi*r, so the output work done in one rev Wo=F*2*pi*r
In N revolutions Wo = N*F*2*pi*r = N*2*pi*Tq

Output power Po is the output work Wo done per unit time, so Po = Wo/t = N*2*pi*Tq/t ie. Tq = Po*t/(2*pi*N)

We will convert everything to ft.lb.sec units.
Converting Po (20kW) to ft,lb,sec units Po=20000/746 HP = 20000*550/746 ft.lb/sec and N/t = 1500RPM=1500/60 rev/sec.

Using these numbers will give you the torqueTq (in ft.lbs) ....... you do the arithmetic!

You should be able to do the electrical calculations using the methods I showed you previously together with the basic definition "Efficiency = Output/Input E = Po/Pi"

Best Wishes, John

 

My choice of symbol "E" for efficiency in my last post was unfortunate.
I suggest that you change it to "Eff" so as not to get confused with "E" for EMF in earlier posts.

 

It is usual to specify a motor by it's output power in HP (sometimes in kW).

I had assumed this to be the case with this question and I used 20kW as the output power to calculate torque.

If, however the questioner meant 20kW to be the input power, then you need to first adjust the output power by the efficiency to calculate torque.