Hi guys, this is a simple problem on DC motors and I would like to know if I am doing the problem right. Here's the problem
A 250W 24V 1800rpm DC machine is operated as a shunt connected motor and connected to a fan that draws 60% full load current through the armature.
1) Sketch the circuit
2) The field winding requires a current of 0.8A to produce a motor EMF of 24V. Determine the field resistance required to produce the field current if the armature resistance is 0.1 ohms.
3) Determine the input voltage and current required
4) Determine the efficiency if the rotational losses are 15W
Answers:
1) Sketch
2)
If the field current is 0.8A since that's the current that flows through the windings, then the field resistance would be Rf = E/IF = 24/0.8 = 30 ohms, Is that correct ? or does it depend on the terminal voltage ?
3)
By input voltage they mean terminal voltage, right ?, so the terminal voltage has to be slightly higher than the machines rated voltage in order to account for losses ?, so
First, armature current
Ia = Rated Power/Rated Voltage = 250/24 = 10.42A
So It(Terminal Current) = Ia+If
= 10.42+0.8 (Field Current)
= 11.22A
Vt (Terminal Voltage) = E + Ia*Ra
= 24 + (10.42)(0.1)
= 25.04V
4)
Input Power = Vt*It
= 25.04*11.22
= 280.95 W
They said the load draws 60% of the full load current, so
0.6*10.42
= 6.25A
I'm not sure if it will be 60% of the rated voltage also, will it ?, I did it in my calculation
0.6*24 = 14.4V
Pout = Pmech - Plosses
= (14.4)(6.25) - (6.25)2(0.1) This is the armature loss - 15 (Rotational loss)
= 71.09 W
So, efficiency = Pout/Pin
= 0.25*100
= 25%
A 250W 24V 1800rpm DC machine is operated as a shunt connected motor and connected to a fan that draws 60% full load current through the armature.
1) Sketch the circuit
2) The field winding requires a current of 0.8A to produce a motor EMF of 24V. Determine the field resistance required to produce the field current if the armature resistance is 0.1 ohms.
3) Determine the input voltage and current required
4) Determine the efficiency if the rotational losses are 15W
Answers:
1) Sketch
2)
If the field current is 0.8A since that's the current that flows through the windings, then the field resistance would be Rf = E/IF = 24/0.8 = 30 ohms, Is that correct ? or does it depend on the terminal voltage ?
3)
By input voltage they mean terminal voltage, right ?, so the terminal voltage has to be slightly higher than the machines rated voltage in order to account for losses ?, so
First, armature current
Ia = Rated Power/Rated Voltage = 250/24 = 10.42A
So It(Terminal Current) = Ia+If
= 10.42+0.8 (Field Current)
= 11.22A
Vt (Terminal Voltage) = E + Ia*Ra
= 24 + (10.42)(0.1)
= 25.04V
4)
Input Power = Vt*It
= 25.04*11.22
= 280.95 W
They said the load draws 60% of the full load current, so
0.6*10.42
= 6.25A
I'm not sure if it will be 60% of the rated voltage also, will it ?, I did it in my calculation
0.6*24 = 14.4V
Pout = Pmech - Plosses
= (14.4)(6.25) - (6.25)2(0.1) This is the armature loss - 15 (Rotational loss)
= 71.09 W
So, efficiency = Pout/Pin
= 0.25*100
= 25%
