Respuesta :
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[tex]\frac{100*(10^9)*(10^3)}{8760}[/tex]
                 = [tex]1.142*10^{10}[/tex]W
1000 MW Â Â Â Â Â Â Â Â Â = [tex]1000*10^{6} = 10^{9}W[/tex]
no. of plants = [tex]\frac{1.14155*10^{10} }{10^9}[/tex] = 11.4
So 12 plants required    Â
(b)
savings = unit price*total units
       = [tex]0.1 * 1.142*10^{10}( \frac{1}{1000*3600} )[/tex]
       = 3170.9 =3171 $
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
