Solution Manual for Sustainable Energy SI Edition 1st Edition by Richard Dunlap

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Chapter 2
Past, Present and Future
World Energy Use
Problem 2.1 Consider an island with a current population density of 50 people/km2 (about equal to the present world average). If the annual growth rate is 2%, determine the year in which the population density be equal to 20,000 people/km2 (approximately the population density of Monaco, the world's most densely populated nation).
Solution As the problem deals with population density, the actual area of the island is not relevant. Solving equation (2.8) for a as a function of R gives
a = ln[1+R/100]
Substituting in the annual growth rate of 2% gives
a = ln[1+0.02] = 1.98×10-2 y-1
From equation (2.4) the quantity as a function of time (relative to the present value) is
N(t)/N0 = exp(at)
Solving for t gives
t = (1/a)×ln[N(t)/N0]
In this problem N is taken to be the population in 1 km2 of land area so
N(t)/N0 = 20,000/50 = 400.
Using the above value for a and solving for t gives
t = (ln 400)/(1.98×10-2 y-1) = 303 y
Relative to the year 2013, this population density will be reached in the year 2316.
Problem 2.2 A quantity has a doubling time of 110 years. Estimate the annual percent increase in the quantity.
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Solution From equation (2.9) the annual rate of increase R is given as
D t
R
100ln 2

where tD is the doubling time. If tD is 110 years then
   
0.63% per year
110 y
100 0.693


R 
This is much less than 10% so the approximation given in equation (2.9) is valid.
Problem 2.3 The population of a particular country has a doubling time of 45 years.
When will the population be three times its present value?
Solution From equation (2.7) the constant a can be determined from the doubling time as
a
n
t D
1 2

so
D t
n
a
1 2

For tD=45 years then
1 0.0154y
45
0.693  a  
From equation (2.4) the quantity of any time is given in terms of the initial value as
Nt  N expat  0 
so solving for t we get
 
 


 



0
ln
1
N
N t
a
t
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for N(t) = 3N0 then we get
ln3 71.3years
0.0154y
1
1   


 


  t
Problem 2.4 Assume that the historical growth rate of the human population was
constant at 1.6% per year. For a population of 7 billion in 2012, determine the time in the
past when the human population was 2.
Solution As the annual percentage growth rate is small then we can use the
approximation of equation (2.4) to get the doubling time from R so
   
43.31years
1.6
100ln 2 100 0.693


 




R
tD
from equation (2.7) the constant a can be found to be
1 0.016y
43.31y
ln 2 0.693     


 



D t
a
from equation (2.4) we start with an initial population of N0 = 2 at t = 0 then N(t) =
6.7×109
then from
Nt  N expat  0 
so
 
1374y
2
7 10
ln
0.016
1
ln
1 9
0


 
N
N t
a
t
in the past or at year 2012-1374 = 638 (obviously growth rate was not constant)
Problem 2.5 What is the current average human population density (i.e., people per
square kilometer) on earth?
Solution The radius of the Earth is 6378 km (assumed spherical). The total area
(including oceans) is       2 2 8 2 A  4r  4  3.14  6378km  5.110 km . The total
current population is 6.7×109, so the population density is
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2
8 2
9
13.1people/km
5.1 10 km
6.7 10



If only land area is included, the land area on Earth is from various values given on the
web range from 1.483×108 km2 to 1.533×108 km2. Using 1.5×108 km2 we find
2
8 2
9
44.7people/km
1.5 10 km
6.7 10



Problem 2.6 The total world population in 2012 was about 7 billion, and Figure 2.11
shows that at that time the actual world population growth rate was about 1% per year.
The figure also shows an anticipated roughly linear decrease in growth rate that
extrapolates to zero growth in about the year 2080. Assuming an average growth rate of
0.5% between 2012 and 2080, what would the world population be in 2080? How does
this compare with estimates discussed in the text for limits to human population?
Solution If R = 0.5% per year then the doubling time is found from equation (2.9) to be
   
138.6y
0.5
10ln 2 100 0.693


 
R
t D
using equation (2.7) to get the constant a
1 0.005y
138.6y
ln 2 0.693    
D t
a
then equation (2.4) gives
Nt N expat  0 
so from 9
0 N  710 people and t  2080  2012  68yearswe find
  7 10 exp0.005y  68y 9.8 10 people 9 1 9       N t
This is consistent with comments in the text which suggest that the limit to human
population cannot be much more than 10 billion.
Problem 2.7 The population of a state is 25,600 in the year 1800 and 218,900 in the year
1900. Calculate the expected population in the year 2000 if (a) the growth is linear and
(b) the growth is exponential.
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Solution If population growth is linear then for 100 years between 1800 and 1900 it
grows by   3 3 218.9  25.6 10 193.310 , so the population would grow by another
3 193.310 during the 100 years from 1900 to 2000 for a total of
218.9 193.3 10 412.2 10 people 3 3    
If the population growth is exponential then from equation (2.4) for 9
0 N  6.7 10 in
1800 then for t=100 years, N(t) is 3 218.910 . From this a can be found to be
  1
3
3
0
0.0215y
25.6 10
218.9 10
ln
100y
1
ln
1    


 




  


 


 
N
N t
t
a
Then using 3
0 N  218.910 in year 1900 the population at 100y (i.e. in year 2000) is
   3   1    6  218.910  exp 0.0215y  100y 1.8710  N t
about 4.5 times the value for linear growth.
Problem 2.8 The population of a country as a function of time is shown in the following
table. Is the growth exponential?
year population (millions)
1700 0.501
1720 0.677
1740 0.891
1760 1.202
1780 1.622
1800 2.163
1820 2.884
1840 3.890
1860 5.176
1880 6.761
1900 8.702
1920 10.23
1940 11.74
1960 13.18
1980 14.45
2000 15.49
Solution For exponential growth
     0 0 N t  N exp a t  t so
 
  0
0
ln a t t
N
N t
   


 


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and the ln of the related population should be linear in time. Calculating   0 N t / N from
the values above gives the tabulated values. They are plotted as a function of t - tD as
shown
year population (millions) year - 1700 ln[N(t)/N(1700)]
1700 0.501 0 0
1720 0.677 20 0.301065
1740 0.891 40 0.575738
1760 1.202 60 0.875136
1780 1.622 80 1.174809
1800 2.163 100 1.462645
1820 2.884 120 1.750327
1840 3.89 140 2.049558
1860 5.176 160 2.335182
1880 6.761 180 2.60232
1900 8.702 200 2.854702
1920 10.23 220 3.016474
1940 11.74 240 3.154151
1960 13.18 260 3.26985
1980 14.45 280 3.361844
2000 15.49 300 3.431344
The graph shows that the ln is linear and hence the population is exponential until ~ 1900
when the increase is less than exponential.
0
0.5
1
1.5
2
2.5
3
3.5
4
0 50 100 150 200 250 300 350
year-1700
ln[n(t)/n(1700)]
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Problem 2.9 Consider a solar photovoltaic system with a total rated output of 10 MWe
and a capacity factor of 29%. If the total installation cost is $35,000,000, calculate the
decrease in the cost of electricity per kilowatt-hour if the payback period is 25 years
instead of 15 years. Assume a constant interest rate of 5.8%.
Solution From Example 2.3 the contribution to the cost of electricity per kWh due to the
capital cost is
 
 
1  1
1
8760h/y
1
 

 T
T
i
i i
Rf
Using I = 35,000,000, i = 0.058, R = 104 kW, f =0.29, then for a payback period of 15
years the cost per kWh is
 
   1.378 0.102 $0.140/kWh
1.058 1
0.058 1.058
10 0.29 8760
3.5 10
15
15
4
7
  



 

For a payback period of 25 years the cost is
 
   1.378 0.0767 $0.106 / kWh
1.058 1
0.058 1.058
1.378 25
25
  



or a decrease of (0.140-0.106)=$0.034 per kWh.
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