Respuesta :
Answer:
(a) Losses 11760 watt, Efficiency = 94.06 %
(b) Shunt current 10.21 A
(c) Armature resistance = 0.0084 ohm
Armature voltage = 222.97 volt
Explanation:
Output of the dc motor 2500 hp
As 1 hp = 746 watt
So output power
[tex]P=250\times 746=186500W[/tex]
Full load current = 862 A
(A) Input of the motor is equal to
[tex]=VI=230\times 862=198260W[/tex]
Therefore losses = input - output
= 192860-186500 = 11760 watt
Efficiency = [tex]\eta =\frac{186500}{192860}=0.9406[/tex] = 94.06 %
(b) Power of shunt field is equal to
[tex]P_{shunt}=11760\times \frac{20}{100}=2352watt[/tex]
So shunt current [tex]I_{sh}=\frac{2352}{230}=10.21A[/tex]
(C) It is given that 50% of the loss is due to armature current it means 50% loss will be due to shunt current
So loss due to shunt current [tex]=11760\times 0.5=5880W[/tex]
Shunt current is equal to [tex]I_{sh}=\frac{5880}{230}=25.56A[/tex]
Armature current [tex]I_a=862-25.56=836.43A[/tex]
Power loss due to armature = 11760-5880 = 5880 W
Therefore [tex]I_a^2R_a=5880[/tex]
[tex]836.43^2\times R_a=5880[/tex]
[tex]R_a=0.0084 ohm[/tex]
Emf is equal to [tex]E=V-I_aR_a[/tex]
[tex]E=230-836.43\times 0.0084=222.97V[/tex]
