Energy Systems & Climate Change – week 3 Winter – Zita solns – Friday 27 Jan. 2012
Wolfson Ch.11: Electricity and Ch.9: Solar
Ch.11 Electricity #1, 5; Ch.9 #2, 3, 4, 6, 7, 8, 10, 13, 15
11. 1 For a given length of wire, resistance is inversely proportional to the wire’s crosssectional area. A utility is replacing a 115-kV transmission line with one using wire
whose diameter is half that of the line it’s replacing. At what voltage must the new line
operate if transmission losses are to stay the same?
.
Solution: Since area is proportional to diameter2, a wire with half the diameter has 4
1
1
times the resistance: Area  r 2  d 2 , and R   2 , so
A d

R  d0   d0
  
R0  d   d 0
 2
2
2

  22  4 .


2
P

We learned in Example 11.2 that Ploss  I R   source  R .
 V 
Since Psource does not change and we want to keep Ploss constant, we will have a constant
ratio of
R0
R
 2 . Therefore the new line must operate at a voltage of
2
V0 V
2
R
115kV 4  230 kV .
R0
If the resistance increases, the voltage must also increase to keep the current the same.
V  V0
11. 5. In Example 11.2, find the voltage between the two ends of either wire - that is, the
voltage between power plant and city across either wire, not the voltage between the two
wires at the generator output. Answer for both choices of transmission voltage.
Solution:
The output power of the plant is P  650 MW, and the resistance of each wire is R  2  .
(a) If the generator’s output voltage is  a  115kV, then the current in the line is
Ia 
Power
a

650 106 W
 5650 Amps, and the voltage across the line is
115 103 V
Va  I a R  5650 A  2   11.3kV , with a power loss of Pa  I a2 R  Va I a  64 MW.
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(b) If the generator’s output voltage is  b  345kV, then the current in the line is
Ib 
Power
b

650 106 W
 1880 Amps, and the voltage across the line is
345 103 V
Vb  I b R  1880 A  2   3.8 kV , with a power loss of Pb  I b2 R  Vb I b  7 MW.
Solar: Ch.9 #2, 3, 4, 6, 7, 8, 10, 13, 15
2. Use the values of Earth’s radius and the solar constant given in the text to verify the
figure 174 PW for the total solar power incident on Earth’s sunward side.
Solution: Earth intercepts solar radiation effectively as a flat disk of radius RE  6.37106
m.
2
Power
Power from Sun =
disk area = S   RE 2  1364 W 2   6.37 106 m   1.74 1017 W
m
Area
Power from Sun  174 PW
3. How long does it take the Sun to deliver to Earth the total amount of energy
humankind uses in a year?
Solution: Power  energy / time. First find how much energy people use in a year, if we
use power Pp = 16 TW.
J
Annual energy use = power  time = Pp 1 year   16 1012  107 s   5 1020 J
s
Now we can find the time it takes the Sun to deliver this much energy:
energy used by people
5.0 1020 J
min
time =

 2.9 103 s
 48 minutes
17 J
power from Sun
60 s
1.74 10
s
4. A passive solar house requires an average heating power of 8.4 kW in the winter. It
stores energy in 10 tonnes (10,000 kg) of Glauber salt, whose heat of fusion is 241 kJ/kg.
At the end of a sunny stretch, the Glauber salt is all liquid at its 32C melting point.
If the weather then turns cloudy for a prolonged period, how long will the Glauber salt
stay at 32C and thus keep the house comfortable?
Solution: The salt’s temperature won’t change as it solidifies, since freezing is a phase
transition. The heat energy required to solidify the salt is Q  mL, and the time this takes
can be found in the relation between power and energy:
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energy Q mL
 
, therefore
time
t
t
4
3 J
mL 10  24110 kg
hr
time =

 2.87 105 s
 80 hr = 3.3 days
3 J
P
3, 600s
8.4 10
s
power =
6. A household uses 100 gallons of hot water per day (1 gallon of water has a mass of
3.78 kg), and the water temperature must be raised from 10˚C to 50˚C. (a) Find the
average power needed to heat the water. (b) Then assume you have solar collectors with e
= 55% average efficiency at converting sunlight to thermal energy when tilted at your
location’s latitude. Use data from Table 9.1 to find the collector area needed to supply
this power in Albuquerque in June.
Solution: The energy needed to heat water is Q  mC T, where the specific heat of
3.78 kg
water c  4184 J/kgK, the mass of water is m  100 gallons 
 378 kg, and the
gallons
temperature change T  50C – 10C = 40C  40 K.
(a) The average power needed to heat the water
energy mc T
is power 


time
one day

378 kg  4184 J
kg  K
3, 600 s
24 hr
hr
  40 K  732 W
(b) The average horizontal insolation in Albuquerque in June is S = 296 W/m2 = power /
power
732 W

 4.5 m 2 .
area, so the required area of solar panels is: area 
eS
0.55  296 W 2
m
7. Repeat part (b) of the preceding problem for Minneapolis in January, assuming a
lower efficiency of e = 32% (because of higher losses at the lower temperature).
Solution: The average horizontal insolation in Minneapolis in January is S  146 W/m2,
power
732 W

 16 m 2 .
so the required area of solar panels is area 
e S
0.32 146 W 2
m
8. Find the average temperature in the passive solar house of Example 9.2 if it’s located
in (a) Fairbanks and (b) San Francisco.
Solution: First recall the analysis of Example 9.2:
Heat loss through walls and roof  QWR  120 T (W/C), where T  Tin  Tout
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Heat loss through windows (glazing)  QG 
A windows T
 34 T (W/C)
R
Total heat loss Qout  QWR QG 154 T (W/C)
Heat gain from Sun onto floor Qin  Afloor S  39 S (W), where S depends on latitude
Our strategy for each location is to look up S and the average outdoor temperature for a
given month, and calculate the temperature from the equilibrium condition:
Qin  Qout
39 S (W)  154 T (W/C)
39  S
T  C  
 Tin  Tout
154
(a) For Fairbanks in January, the average outdoor temperature is Tout  –23.4C and the
insolation (for solar panels tilted at latitude) is S  4 W/m2. Therefore,
39  4 156
T  C  

 1.01C and the average January temperature is:
154 154
Tin  Tout  T  23.4C  1C  22.4C (this is too cold!)
(b) For San Francisco in January, the average outdoor temperature is Tout 9.3C and
the
insolation (for solar panels tilted at latitude) is S  92 W/m2. Therefore,
39  92 3,590
T  C  

 23.3C and the average January temperature is:
154
154
Tin  Tout  T  9.3C  23.3C  32.6C (this is too warm!)
10. My college’s average electric power consumption is about 2 MW. What area of 15%
efficient PV modules would be needed to supply this power during a summer month
when insolation on the modules averages 225 W/m2? Compare with the area of the
college’s football field (55 m by 110 m overall).
Solution: One football field has an area of 55 m 110 m = 6 103 m 2 . The PV modules’
effective insolation = power / area, so
P
2 106 W

 6 104 m 2 or
the area required =
e S 0.15  225 W 2
m
60 thousand square meters, which is about 10 football fields.
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13. A solar hot-water system for a U.S. home costs about $6,000 after government
subsides. If the system reduces electrical energy consumption by 3,500 kWh of electrical
energy each year, and if electricity costs 14¢/kWh, how long will the system take to pay
for itself (neglect the economics of borrowed money and similar complications).
Solution: What is the rate of energy savings?
Energy savings 3500 kW  h $0.14 $490
Savings rate 




year
year
energy
year
kW  h
How long will it take to pay for the cost of the system?
Cost
$6000
Payback time 

 12 years .
savings rate $490
year
Plus, the system is preventing CO2 emissions for water heating supplied by coal plants or
other fossil fuel burning.
15. The controversial Cheviot open-pit coal mine in Alberta, Canada, occupies 7,455
hectares and produces 1.4 million tonnes (1 tonne = 1,000 kg) of coal per year. (a) Use
the energy content of coal from Table 3.3 to find the power in watts corresponding to this
rate of coal production, assuming it’s burned in power plants with 35% efficiency. (b)
Again using the assumptions of Example 9.4, find the area of a PV plant with this power
output, and compare that number with the area of the Cheviot mine.
Solution: To find how big a PV plant we would need to replace this coal mine, first let’s
find out how much power this mine produces. Recall (Table 3.3) that coal produces
dE
MJ
approximately
.
 29
dm
kg
(a) The power produced is:
6
3

dE
dE dm
J
6 J  1.4  10  10 kg
Pcoal  e
e

 0.35  29 10
 4.5 108  450 MW

7
dt
dm dt
kg 
s
 10 s

year
year
(b) The solar PV power produced is the efficiency  insolation  area: Psolar  e S A . In
order to have Psolar  Pcoal , the required PV area would be 2A:
e S A  Pcoal
2A 
2 Pcoal
2  4.5 108 W
ha

 2.36 107 m 2 4 2  2,360 ha
eS
10 m
0.15  254 W 2
m
The solar PV array requires only about
2,360
 32% the area of the Cheviot mine to
7, 455
generate the same amount of power.
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