119k views
0 votes
Find the input power and efficiency of a DC generator supplying a load of 30 A at an output voltage of 220 V.

The losses are as follows:
a. Friction 230 Watts
b. Iron Core 145 Watts
c. Field 220 Watts
d. Armature Copper Losses 480 Watts
e. Other stray losses 75 Watts

User OLP
by
7.7k points

1 Answer

0 votes

Final answer:

The input power of the DC generator is calculated to be 7750 W (7.75 kW) and the efficiency is found to be 85.16% by dividing the output power by the input power and multiplying the result by 100.

Step-by-step explanation:

To find the input power of a DC generator supplying a load, we start by calculating the output power which is given by multiplying the output current by the output voltage (Pout = IV). In this case, with a load of 30 A at an output voltage of 220 V, the output power is:

Pout = 30 A × 220 V = 6600 W (or 6.6 kW).

Next, we consider the losses to find the total input power, which is the sum of the output power and all the losses:

  • Friction loss: 230 W
  • Iron core loss: 145 W
  • Field loss: 220 W
  • Armature copper loss: 480 W
  • Other stray losses: 75 W

Total losses = 230 W + 145 W + 220 W + 480 W + 75 W = 1150 W.

Then, input power (Pin) is the sum of the output power and total losses:

Pin = Pout + Total losses = 6600 W + 1150 W = 7750 W (or 7.75 kW).

To calculate the efficiency of the DC generator, use the efficiency formula:

Efficiency = (Output power/Input power) × 100 = (6600 W / 7750 W) × 100

Efficiency = 0.8516 × 100 = 85.16%

User YGA
by
7.4k points