# Answer to Question #14448 in Electromagnetism for John

Question #14448

what is the final delivered electricity in kWh by burning one pound of coal (13,100 BTU) with a boiler loss of 20%, generator loss of 50%, and transmission loss of 2%?

Expert's answer

Q0 = 13,100 BTU = 13,821,231.7 J.

1 kWh = 3.6 MJ, so Q0 = 3.839 kWh.

The

final delivered electricity Qf = Q0 * (100 - Lb - Lg - Lt) / 100 =

3.839 *

(100 - 20 - 50 - 2) / 100 = 1.075 kWh.

1 kWh = 3.6 MJ, so Q0 = 3.839 kWh.

The

final delivered electricity Qf = Q0 * (100 - Lb - Lg - Lt) / 100 =

3.839 *

(100 - 20 - 50 - 2) / 100 = 1.075 kWh.

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