High School

Two shunt generators, A and B, operate in parallel, and their load characteristics can be taken as straight lines.

- The voltage of generator A falls from 240V at no load to 200V at 200A.
- The voltage of generator B falls from 245V at no load to 220V at 150A.

Determine the bus bar voltage when supplying a 66.9 kW load. How is the load divided between the two generators?

Answer :

The bus bar voltage when supplying a 66.9 kW load is approximately 209.7V, and the load is divided between the two generators based on their load characteristics.

To determine the bus bar voltage when supplying a 66.9 kW load and how it is divided between two generators, we need to analyze the load characteristics of generators A and B.

Generator A's load characteristic indicates that the voltage drops from 240V at no load to 200V at 200A. This suggests a voltage drop of 40V for a load of 200A. On the other hand, Generator B's load characteristic shows a voltage drop from 245V at no load to 220V at 150A, resulting in a voltage drop of 25V for a load of 150A.

Now, we can calculate the voltage drop per ampere for each generator. For Generator A, the voltage drop per ampere is 40V/200A = 0.2V/A. For Generator B, the voltage drop per ampere is 25V/150A = 0.1667V/A.

To find the bus bar voltage when supplying a 66.9 kW load, we divide the load between the two generators in proportion to their voltage drop per ampere values.

The total current required for a load of 66.9 kW can be calculated using the formula P = VI, where P is the power (66.9 kW), and V is the bus bar voltage. Rearranging the formula, we have I = P/V.

Let's assume that x amperes of the load are supplied by Generator A and (66.9 - x) amperes are supplied by Generator B. The voltage drop for Generator A would be 0.2x V, and for Generator B it would be 0.1667(66.9 - x) V.

Setting up an equation based on the voltage drops, we have:

240 - 0.2x = 245 - 0.1667(66.9 - x).

By solving the equation, we find x ≈ 39.94 amperes. Therefore, Generator A supplies approximately 39.94 amperes of the load, and Generator B supplies the remaining (66.9 - 39.94) amperes.

Substituting the value of x into the equation I = P/V, we can calculate the bus bar voltage:

66.9 kW / (39.94 A + (66.9 - 39.94) A) ≈ 209.7V.

In conclusion, the bus bar voltage when supplying a 66.9 kW load is approximately 209.7V. The load is divided between the two generators, with Generator A supplying approximately 39.94 amperes and Generator B supplying the remaining current.

Learn more about bar voltage

brainly.com/question/17164393

#SPJ11