Question

A U.S. Department of Energy report estimates that over 100 billion kWh/year can be saved in the United States by various energy-conservation techniques applied to the pump-driven systems. Calculate (a) how many 1000-MW generating plants running constantly supply this wasted energy and (b) the annual savings in dollars if the cost of electricity is 0.10 $/kWh.

Answer 1

Answer:

a) Number of generating plants N = 11.42

That means N > 11

N = 12

b) annual savings S = $1×10^10

S = $10 billion

Explanation:

Given;

Amount of energy to be saved A=100 billion kWh/year

Capacity of each generating plant C= 1000 MW

Rate in dollars of cost of electricity R= $0.10/kWh

The number N of generating plants with capacity C that can supply the Amount A of of energy cam be given as;

N = A/C ......1

And also the Annual savings S in dollars if the rate of electricity cost R is used and amount of energy A is saved is:

S = AR .....2

But we need to derive the value of A in Watts

A = 100 billion kWh/year

There are 8760hours in a year,

A = 1×10^14 ÷ 8760 W

A = 11415525114.1W or 11415525.1141kW

C = 1000MW = 1× 10^9 W

a) Using equation 1,

N = 11415525114.1/(1×10^9)

N = 11.42

That means N > 11

N = 12

b) using equation 2

S = 1×10^11 kWh × $0.10/kWh

S = $1×10^10

S = $10 billion

Answer 2

Answer:

(a). 12 plants

(b). 3171 $

Explanation:

(a)first convert units of 100 billion kWh/year into Watts(W)

also convert the units of 1000 MW into Watts(W)

1 billion = 10^9

1 year = 365*24 = 8760 hrs

so

100 billion kWh/year = 1$\frac{100*(10^9)*(10^3)}{8760}$

                                  = $1.142*10^{10}$W

1000 MW                  = $1000*10^{6} = 10^{9}W$

no. of plants = $\frac{1.14155*10^{10} }{10^9}$ = 11.4

So 12 plants required        

(b)

savings = unit price*total units

             = $0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )$

             = 3170.9 =3171 $