Question

Suppose the total power is 7000W and used continuously for 10 hours. Now, you plan to cut 30% of usage, how many pounds of coal can you save, considering the overall efficiency for the electricity generation being 34%? A. 8.6 B. 14.2 C. 19 D. 26.6

Part 2:, how much money will be saved when you cut down 30%, assuming the cost of electricity is 10 cents per kWh? A. $2.11 B. $1.18 C. $0.83 D. $0.21

Answer 1

here,

P = 7000 W

time taken , t = 10 hours = 10 * 3600 s

a)

when we cut 30 % of usage

the electric energy saved , E1 = P * t * 0.3

the input energy used , Ei = P * t * 0.3 /0.34

the energy density of coal , E = 24 MJ /kg = 24 * 10^6 J/kg

the mass of coal saved , m = Ei /E

m = ( 7000 * 3600 * 10 * 0.3 /0.34) /(24 * 10^6)

m = 9.26 kg = 19 pounds

b)

the energy saved , Es = P * t * 0.3

Es = 7000 W * ( 10 h) * 0.3 = 21 KWh

the money saved , M = 10 * Es cents

M = 210 cents = 2.11 dollars