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§
§
§
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§
The exam is closed book, closed notes, and “closed neighbor.” There are three parts to this exam—
(1) short answers & calculations (~20 minutes)
(2) longer calculations (~70 minutes)
(3) essay (~30 points)
The exam is out of a total of 100 (UG)/120 points (GR) (~2 hours if you pace yourself at one
point/minute).
The most critical part of problem solving is setting out your method and ideas, so you must show your
work, and if you need to make assumptions, be sure to state what you assumed and why. Partial credit
will be given.
Read the questions carefully, and be sure that your answer is what the question asks for, is in the
correct units, and has the correct number of significant figures.
You are encouraged to use a calculator.
Extra pages have been provided at the end in case you run out of space. If you need more than this,
ask the GSIs for extra paper.
You can remove this page from your exam for easy reference, we won’t collect it.
Some useful equations and constants
Energy Contents & Chemical Formulas:
Natural gas
3.9 × 106 J/m3
CH4
Oil
48 × 106 J/kg
-(CH2)Coal
29.3 × 106 J/kg
C10H8O2
Dry Biomass 16 × 106 J/kg
C6H12O6
Atomic:
Carbon
12 g/mol
Oxygen
16 g/mol
Hydrogen 1 g/mol
Sulfur
32g/mol
Nitrogen 14 g/mol
Power:
Power = Energy/time
1 watt (W) = 1 J/sec = 3.6 kJ/hr = 31.5 MJ/year
1 horsepower (hp) = 0.764 kilowatts (kW)
1 kWh = 3.6 × 106 J = 3414 Btu
P = IV
V = IR Ploss=I2R
Learning curves:
b
C ⎛ V ⎞ and the Progress Ratio PR = 2b €
⎟
= ⎜
C o ⎜⎝ Vo ⎟⎠
Metric prefixes:
10-6 micro (µ)
109 giga (G)
101 deka(da)
-3
10 milli (m)
1012 tera (T)
2
-2
10
hecto(h)
10 centi (c)
1015 peta (P)
1018 exa (E)
103 kilo(k)
106 mega(M)
Other useful energy equations:
Pwind =
1
ρArotor v 3
2
ΔG = ΔH – TΔS
Efficiency Equations
ηI = Wnet/Qin
ηII = ηI / ηc
ηc = 1 – Tlow/Thigh
ηmax-fuel-cell = ΔG/ΔH
Growth equations:
Compound Interest: N(t) = N0(1 + r)t
Exponential: N(t) = N0ert
Mass, volume, distance, area:
1000 liters = 264.2 gal 1 kg = 2.205lbs
1 km = 0.6214 mi 1 metric ton = 103 kg
1 m = 3.281 ft
sphere surface area=4𝜋𝑟 !
Hubbert curves for non-renewable resources:
⎡−1⎛ t − t ⎞2 ⎤
m
P(t) = Pm exp⎢ ⎜
⎟ ⎥
⎝
⎠ ⎦
2
σ
⎣
⎛ P
t m = t 0 + 2 ∗ ln⎜⎜ m
⎝ Po
⎞
⎟⎟ and Q = σPm
⎠
Financial Equations:
&
#
r
&1 − (1 + r ) − n #
U = P$
P
=
U
−n !
$
!
r
%1 − (1 + r ) "
%
"
n
F = P(1 + r )
P=
F
(1 + r )n
Constants:
ρair = 1.2 kg/m3
ρwater = 1,000 kg/m3 = 1000 g/liter
g = 9.8 m/s2
S = 1370 W/m2
Temperature Conversions:
K = °C + 273.15
ER 100/200 GSPP184/284 Fall 2014
Final Review
Structure of final
The final is closed book and closed notes. The exam will cover material from the entire semester.
There will be quantitative and qualitative questions covering the material presented from
lectures, readings, and problems sets. The first section will be short answers, second section
more involved calculations, and the third section will be an essay.
The equation sheet will be provided and you aren’t expected to have memorized more
complicated equations. However, you should be able to balance a combustion equation. All
conversion factors, energy contents, and molecular weights will be given.
Tips
§ Bring a calculator, but not one that can communicate with the outside world (e.g., a smart
phone).
§ Scan through the whole exam before beginning.
§ Review your lecture notes, discussion notes, and homework sets.
§ Review the key readings (given below). Make notes on the main points and make sure you
understand them. Think about the theme for each week and how the readings contribute to your
understanding of the topic.
§ Most of the rest of the readings were technical overviews—review these for each concept you
do not understand.
Key readings
Lovins (1976)
Goldemberg (1996)
Kammen & Dove (1997)
Hirsh (1999)
Arrow et al (2013)
Sager et al (2011)
Pacala and Socolow (2004)
Hansen et al (2012)
Important Topics:
Unit Conversions
Significant Digits
Metric Prefixes
Definitions of Energy and Power
**Make sure you understand and can quickly answer the calculations from the first problem
set.
IPAT
Linear and Exponential Growth Models
Hard Path versus Soft Path
GDP and Poverty
Gender and Labor
Fuelwood Gap Theory
Household Energy Ladder
Exposure to Pollution
Charcoal Production
Biomass Energy and Development
Combustion Chemistry/Stoichiometry
Thermodynamic Laws
Efficiencies: 1st, 2nd, and Carnot
Carnot Cycle
Brayton Cycle
Rankine Cycle
Combined Cycle
Cogeneration
Emissions Control – technologies and practices
Petroleum and Hubbert Curves
Power Plants and Pollutants
Economic Analysis (NPV, Levelized Annual Cost (Uniform Payments), Future payments,
discount rates)
Negawatts
Cost of Conserved Energy
Levelized Cost of Electricity
Life Cycle Assessment
Evolution of the Electricity Industry
Technology Learning Curves
PURPA, QFs, Avoided Costs
Energy Regulation and Deregulation
Physics of electricity (voltage, current, Ohm’s Law, transmission losses)
Direct versus alternate current
Macro-level grid operation principles
Fracking process and potential consequences
Nuclear physics (getting energy from fission reactions, radioactive decay)
Chain reactions (critical state)
Nuclear fuel cycle (from mining to disposal)
Comparative advantages and disadvantages of nuclear power
Environmental justice (definition, relevance to energy)
Learning curves
Energy polices (e.g., feed-in tariff, production tax credits, renewable portfolio standards, feebate
programs)
Solar power alternatives (photovoltaics and solar thermal)
Solar photovoltaic cell design (basic principles)
Solar PV efficiency (rated insolation)
Comparative advantages and disadvantages of solar power
Wind power density calculations
Wind speed distribution, Betz limit, and turbine power curves
Comparative advantages and disadvantages of wind and water power
Hydrogen fuel cell chemical reaction and thermodynamics (enthalpy, entropy, Gibbs free
energy)
PEM fuel cell design
Comparative advantages and disadvantages of hydrogen fuel cells
Transportation and the reliance on oil
Oil reserves versus resources
Ways to reduce the carbon footprint of transportation
Comparative advantages and disadvantages of alternative to conventional personal vehicles
Climate change and the role of greenhouse gases
Present and forecast consequences of climate change
Climate feedbacks
Climate stabilization “wedges”
Carbon abatement/mitigation supply curves (“Vattenfall/McKinsey curves”)
Cost of Conserved Carbon
Energy R&D
Short Answer
1) Hairdryers are devices that dry hair.
a) If you run an 1800-Watt hairdryer for 5 minutes per day, every day, how much energy do you use, in
kWh per year? Report your answer with two significant digits. (2 points)
1.8 kW x 5 min/day x 365 day/yr x 1 hr/60 min = 55 kWh
b) What is the nameplate capacity of the power plant (in MWe) needed to generate the yearly energy for
1000 of these hair dryers? Assume the plant has 30% efficiency (1st Law), and 80% capacity factor.
Report your answer with two significant digits. (3 points)
54.75 kWh/hairdryer-year x 1000 dryers = 54,750 kWh/yr 54,750 kWh/yr
= Nameplate capacity x 8760 hr/yr x 0.80 Nameplate capacity = 7.8x10-3
MW
c) Define capacity factor and efficiency (1st law) with respect to a power plant. What is the distinction
between these terms? (3 points)
Capacity factor is the ratio of the actual output of a power plant over a
given period of time to its potential output if it were to operate at
maximum capacity over that same time interval. A power plant’s first-law
efficiency is the ratio of useful work produced (electricity, or electricity
and purposeful heating in the case of cogeneration) to the amount of
primary energy put in to the power plant to produce that useful work.
While both terms speak to the operation of a power plant, capacity factor
describes to what degree a power plant operates (how often or at what
output level) and efficiency describes how well a power plant converts
energy from one form into another.
2) Explain using words (and equations if desired) why high voltage is used for transmission of
electricity. (3 points)
Ploss = I2R, which means that line losses increase at the current over the line goes up. Thus, we’d like to reduce current to reduce line loses. Power = Voltage x Current, meaning that if power flow is fixed, there is an inverse relationship between voltage and current. If we want a smaller current value at a fixed power flow, then the voltage has to increase. In sum, high voltage means low current, which in turn means smaller transmission loses.
3) Which of the following produces a larger increase in the power from a wind-turbine: (a) tripling the
diameter of the blades, or (b) or operating the windmill at a site with twice the average wind velocity,
assuming Rayleigh statistics? Show why, using equations. (4 points)
Pturbine-­‐average = 0.5ρAv3η(1.91)
Because Area = pi(r2), there is a square relationship between rotor length (r) and Power. Tripling the diameter of the blades would lead to a 32 = factor of 9 increase in turbine power.
There is a cubic relationship between average wind speed and power output. Doubling the average wind speed would lead to a 23 = factor of 8 increase in turbine power. Accordingly, tripling the blade length would be more effective in increasing power output.
4) Name two potential benefits of using biofuels for transportation and two potential pitfalls. (4 points)
Example potential benefits: Lower lifecycle carbon emissions (cellulosic ethanol), reduced dependence on foreign oil (US context), liquid fuels can work with existing vehicles and fueling infrastructure
Example potential pitfalls: Competition between food and fuel (increased food prices), indirect land use change, depletion of soil nutrients or increased use of artificial fertilizers
5) List two factors that spurred the electricity industry to develop from Edison’s small decentralized
utility model to one of large centralized power stations. (2 points)
Scales of economy and improving efficiencies meant that electricity could be produced more cheaply by larger turbines.
The development of alternating current allowed electricity to be transmitted over longer distances with less loss.
6) Many U.S. policymakers advocate building a large number of new nuclear power plants over the next
several decades. What are two potential benefits of this plan? What are two arguments against this
plan? (4 pts)
Example potential benefits: Low lifecycle carbon emissions, a baseload generator (replacement for coal), domestic supply of uranium
Example arguments against: Environmental damage from mining uranium, uncertainties about cost, nuclear waste disposal (including nuclear weapons proliferation), concerns about safety
7) PG&E pays about $200 million per year to run its Diablo Canyon nuclear power plant (this includes
operations and maintenance, as well as a significant levelized capital cost payment). If PG&E found
out suddenly that 20 years down the road (i.e., in 2030) it was going to have to pay an additional
billion dollars toward decomissioning the plant as a result of consumer law suits, how would that
change the yearly cost of running the plant? Assume a 5% discount rate. (3 points)
Future value (FV) = $1x109 Present value = FV/(1+r)20 = $1x109/(1.05)20 =
$3.769x108
U = PV(CRF) CRF = .05/[1-(1.05)-20] = 0.0802/yr U =
$3.769x108(0.0802/yr) = $30x106/yr is the additional annual cost
8) You have an electric lawnmower that runs on a rechargeable Li-ion battery. If the lawnmower draws
1800 W while in operation, you use the lawnmower for 1.25 hours per week, and the total (roundtrip)
efficiency of charging and discharging the battery is 90.0%, how large (in area) would a solar panel
need to be to fully charge the lawnmower every week? Assume that the solar panel efficiency is 18%
and that the solar resource is 2.4 kWh/m2/day. (3 points)
1.8kW x 1.25 hr/wk = 2.25 kWh/wk, or 4.5 kWh/two weeks Since the
overall efficiency of charging is 90%, we need 4.5/0.90 = 5.0 kWh for two
weeks. 2.5 kWh/week = Area x 2.4 kWh/m2-day x 7 days/week x
0.18 Area = 0.83 m2
Long-form Questions
The Art of Efficiency
In 1974, the average refrigerator in the U.S. used 1800 kWh per year. Today—thanks to Art
Rosenfeld's pioneering energy efficiency work—a new, average-sized refrigerator uses about 475
kWh per year.
a. Approximately 150 million refrigerators are currently in use in the U.S. How much energy
will be saved this year as a result of energy efficiency improvements? (Assume that a typical
refrigerator today use 475 kWh per year and would have used 1800 kWh per year without
energy efficiency improvements).
150x106 x (1800 kWh/yr – 475 kWh/yr) = 2.0x1011 kWh/yr
b. How many tons of annual carbon emissions will be saved as a result of these energy
efficiency improvements for refrigerators in the U.S.? To solve this, use the information in
part (a) and assume that the carbon intensity of electricity generation in the U.S. is 160 g
C/kWh.
1.9875x1011 kWh/yr x 160 g C/kWh x 1 t/106 g = 3.2x107 t C/yr
c. Imagine that these energy efficiency improvements never occurred and that all of the 150
million refrigerators currently in the U.S. use 1800 kWh per year. Calculate the number of 1
GW nuclear power plants that would be required to run these refrigerators. The capacity
factor of these nuclear plants is 0.80 and the thermal efficiency is 30%.
150x106 x 1800 kWh/yr = 2.7x1011 kWh/yr = 2.7x105 GWh/yr 2.7x105 GWh/yr = Capacity x 8760 hr/yr x CF = Capacity x 8760 hr/yr x 0.80 Capacity = 38.52 GW, so 39 1-GW NPPs would have been needed
Oil Resources
The 2006 paper by Farrell and Brandt, “Scraping the Bottom of the Barrel: The Risks of Oil
Transition,” estimates that the reserves of conventional and non-conventional liquid hydrocarbons is
about 4 trillion barrels. In 2006, total world crude oil consumption was 31 Gbbl/yr, and in the years
leading up to the publication of that paper, world-wide consumption of crude oil was increasing at a
rate of 1.7%/yr.
a. Estimate how long (to the nearest year) that it would take to exhaust the 4 trillion barrel stock of
liquid hydrocarbon reserves given a continued growth in consumption of 1.7%/yr. Use the
following formula:
𝑒 !" 1
𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑜𝑣𝑒𝑟 𝑡 𝑦𝑒𝑎𝑟𝑠 = (𝐴𝑛𝑛𝑢𝑎𝑙 𝑐𝑜𝑛𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑖𝑛 𝑦𝑒𝑎𝑟 1)(
− )
𝑟
𝑟
where
r is the annual consumption growth rate, and
t is the number of years of consumption.
Total consumption = 4x1012 bbl = 31x109 bbl/yr(e0.017t/0.017 – 1/0.017)
129.032 yr = e0.017t/0.017 – 1/0.017
2.1935 yr = e0.017t – 1
3.1935 yr = e0.017t
t = 68 years
b. Let’s be optimistic and assume that the reserve of liquid hydrocarbons will prove to be double the
Farrell and Brandt estimate, or 8 trillion barrels. Estimate how long (to the nearest year) that it
would take to exhaust that resource base.
Same math as part a), but with 8x1012 bbls instead of 4x1012 t = 99 years
(only 30 extra years with twice the resource!)
c. Now assume that 4 trillion barrels is the extent of the liquid hydrocarbon resource. How long will
that resource last if growth in demand is cut by 1 percentage point? Report your answer to the
nearest year.
Same math as part a), but with r = 0.007 t = 92 years
d. What do your answers from parts a-c tell you about the sustainability of reducing growth in
resource consumption versus continued expansion of the resource base?
Both reducing demand and increasing supply can extend the amount of
time we can use a given resource. In this case, a relatively small decrease in
the rate of growth of consumption was nearly as impactful as the more
improbable scenario of doubling the resource base. Accelerating rates of
consumption mean that even very large resource stocks get consumed in
short order.
The Nuclear Option
For this question, you’ll calculate the size of part of the monetary transfer (from government to the
nuclear industry) that facilitates the construction of nuclear power plants.
Loan guarantees are a form of public policy support for energy technologies. They are essentially a
promise by the government to pay off the debt obligation of power plant project in the event that
there is a default with an investment in building and deploying a technology. Loan guarantees serve
to shift the investment risk to the public sector, thereby making investments safer for private capital.
The GAO estimates that about half of the loans guaranteed by the US Department of Energy go
into default.
In his proposal for the 2012 budget, President Obama included $36 billion for loan guarantees for
the construction of new nuclear power plants. Historically, these loan guarantees have averaged
about $9 billion per unit, so it can be estimated that this budget item would support the construction
of 4 1.0-GW power plants.
a. What is the value ($/kWh of produced nuclear energy) of the cost transfer for this policy? To
calculate, assume that half of the money allocated goes to paying for defaulted loans (two of the 4
plants do not get built) and that the other half is unspent. Assume also that, if built, a nuclear power
plant has a capacity factor of 90.% and that it will operate for 40 years. Use a discount rate of 4%,
and report your answer to the nearest tenth of a cent.
$/kWh = ($/yr)/(kWh/yr)
Total cost = $18x109 CRF = .04/[1-(1.04)-40] = 0.0505/yr U = PV x CRF =
$18x109(0.0505/yr) = $9.09x108/yr
Two 1.0-GW NPPs are built 2.0 GW x 8760 hr/yr x 0.90 x 106 kWh/GWh
= 1.5768x1010 kWh/yr Cost per kWh = ($9.09x108/yr)/(1.5768x1010
kWh/yr) = 5.8 cents/kWh
b. Imagine that the number of nuclear power plants in the United States doubled, going from a
count of 104 to 208. Would this doubling be of sufficient magnitude to comprehend a nuclear
“wedge”? Assume the operation characteristics from part b, and that one kWh of nuclear electricity
will offset 850 g CO2.
104 NPP: 104 GW x 8760 hr/yr x 0.90 x 106 kWh/GWh x 850 g
CO2/kWh x 12 g C/44 g CO2 x 1 Gt C/1015 g C = 0.19 Gt C/yr; since a
wedge needs to abate 1 Gt C/yr, this doubling of US nuclear capacity
would not be sufficient to make a full wedge
c. Using your answer from part c, calculate the NPV of the loan guarantee of this doubling of the
US nuclear fleet. Use the assumptions from previous parts of the question, and, for simplicity’s sake,
assume that all power plants are constructed at the same time.
From the problem statement, we know that we need a guarantee of $9
billion per NPP and that half of the proposed NPPs don’t get built. Thus,
we need to plan to build 208 NPPs
208 x $9 billion = $1.8 trillion
Energy in China
Coal is the world’s fastest growing fossil fuel, and the largest human sources of GHG emissions.
China relies on coal for 70% of its primary energy supply and about 80% of its electricity generation.
As such, China, especially China’s coal-fired power sector, simultaneously presents the most
challenging and critical test for energy efficiency and carbon emission control. By end of 2010,
China’s total power generation was 4,228 TWh (about 24% of global total), among which 3,414
TWh (80.7%) was from coal. The remaining was filled up by hydro, nuclear and increasingly
renewables, accounting 16.2%, 1.8% and 1.2% respectively. Using what you have learned from the
class to understand the challenges it presents to both China’s energy and climate future and global
energy and climate policy.
a) As the energy content of coal varies by coal quality, China’s coal power plant efficiency is
defined as the mass of “standard coal” (1 ton of raw coal is equivalent to 0.7143 ton of
standard coal), or “grams of coal equivalent” (gce) required to generate one kilowatt-hour
electricity. From China Electricity Council, the average coal consumption of coal power
supply in 2010 was 335gce/kWh. What’s the average efficiency (first law) of China’s coalfired power plant? (1𝑔𝑐𝑒 = 7000𝑐𝑎𝑙)
335 gce x 7000 cal/gce x 4.184 J/cal x 1 kWh/3.6x106 J = 2.725
kWh Efficiency = Work out/Energy in = 1 kWh/2.725 kWh = 37%
(1 sf)
b) Given China’s coal fired power generation was 3,414 TWh in 2010, how much raw coal was
consumed in 2010 for electricity generation? According to Chinese Coal Transportation and
Distribution Association, China produced 3.3 billion tons of raw coal in 2010, what portion
of this coal was burned to generate electricity?
3.414x1012 kWh x 335 gce/kWh x 1 g raw coal/0.7143 standard coal x 1
Gt/1015 g = 1.6 Gt raw coal
1.6 Gt/3.3 Gt = 49% of all Chinese raw coal consumption was used to
make electricity
c) If we assume all Chinese power plants are instantly upgraded to global best practice power
generation efficiency at 45% in 2010, how much raw coal they can save in one single year?
Comparing what you learn from Rosenfeld energy efficiency unit, calculating how many
Rosenfeld units of energy and emission it means? (1 Rosenfeld = 0.03 EJ or 3 MMT CO2)
1.6 Gt coal x 0.37/0.45 = 1.3 Gt coal, so 0.3 Gt raw coal saved
0.3 Gt raw coal x 0.7143x1015 g standard coal/1 Gt raw coal x 7000
cal/gce x 4.184 J/cal x 1 EJ/1018 J x 1 Rosenfeld/0.03 EJ = 200
Rosenfelds or 600 MMT CO2
d) China sees the opportunity for clean energy to be the next frontier of economic competition.
As observed, China has progressively developed its renewable energy and is leading the
global production of PV and installation of wind in 2010. Its wind capacity has grown from
1.26 GW of installed capacity in 2005 to 75 GW installed in 2012. What potential challenges
would be presented to China’s coal-based power systems by exponentially expanding
renewables? (Hint: think about the requirement of power supply system and the feature of
renewable energy)
The most prominent renewables (wind and solar) are intermittent power
sources, meaning that wind turbines and PV panels have variable output
depending on the amount of wind and sun available from moment to
moment. Coal, on the other hand, is a baseload source of electricity,
very constant, directly under human management, and ease to use when
trying to maintain constant electricity quality (stable voltage, frequency,
etc.). Also unlike coal plants, wind and solar installations are quite
sensitive to where they are sited. China may not have a enough good
locations to place renewable facilities. Even with an abundance of sites,
the transmission grid might have to be extend to these sites, which
presents an additional cost.
"Greening" Transportation
Imagine that you are the sustainability coordinator for Best Buy, and the CEO is interested in
figuring out how to "green" the company's U.S.-based transportation operations. In particular, he
wants to determine how to reduce emissions from the Geek Squad vehicle fleet, which makes house
calls to customers in need of in-person technical help. The CEO has emailed you the following
questions about possible Geek Squad fleet changes and included the following facts about the
current Geek Squad Fleet to help you answer his questions.
Geek Squad Fleet Facts (as of December 2011)
Number of Geek Squad Vehicles in the U.S. = 2,210
Average annual miles driven by a Geek Squad vehicle = 15,5000
Typical Geek Squad vehicle: 2009 VW New Beetle (average combined fuel economy = 23.1 mpg)
Number of Geek Squad house calls made per year in the U.S. = 1.89 million
*Note: In accordance with company policy, Geek Squad vehicles are ONLY used for official Geek
Squad business (technicians do not use the vehicles for personal travel).
a. What would be the reduction in annual CO2 emissions of the entire U.S. Geek Squad fleet if
all vehicles were replaced with electric vehicles that charge from the grid?
From personal research, you learn the following:
• Combustion of gasoline leads to a production of 9.254 kg of CO2-e/gallon.
• Average carbon intensity of the U.S. grid = 1.306 lbs CO2/kWh.
Current emissions: 2210 vehicles x 15,500 mi/vehicle-yr x 1 gal/23.1
mi x 9.254 kg CO2/gal = 1.3723x107 kg CO2/yr
EV emissions: 2210 vehicles x 15,500 mi/vehicle-yr x 1 kWh/4 mi x
1.306 lb CO2/kWh x 1 kg/2.205 lb = 5.0722x106 kg CO2/yr
Reduction = 1.3723x107 kg CO2/yr - 5.0722x106 kg CO2/yr = 9x106
kg CO2/yr
b. What would be the reduction in annual CO2 emissions if one-half of the U.S. Geek Squad
fleet were replaced with hydrogen fuel cell vehicles?
From personal research and reasoning, you conclude the following:
• Nationally, most hydrogen fuel for fuel cell vehicles is produced via steam methane
reforming of natural gas. This translates to per-mile CO2 emissions of 143 g CO2/mi.
• Geek Squad fuel cell vehicles are likely to travel 5% further per year than the
conventional Geek Squad vehicles in order to access sparse H2 fueling stations.
7
Half of current emissions = 0.5 x 1.3723x10 kg CO2/yr = 6.8615x10
kg CO2/yr
6
H2 emissions = 1105 vehicles x (1.05 x 15,500 mi/vehicle-yr) x 0.143
6
kg CO2/mi = 2.5717x10 kg CO2/yr
Reduction = 6.8615x106 kg CO2/yr - 2.5717x106 kg CO2/yr = 4x106
kg CO2/yr
c. Best Buy's CEO is also considering expanding its call center and developing a new app that
allows technicians to help customers with virtual video technology. Both of these options
would reduce the number of house calls that would need to be made by the Geek Squad
vehicles.
The CEO asks you: How many fewer house calls would need to be in order to reduce CO2
emissions from Geek Squad cars to the same level as would be achieved by switching to
electric cars (i.e., the result that you calculated in Part a above for the electric vehicle
scenario)?
.
.
7
Emissions per visit = (1.3723x10 kg CO2/yr)/(1.89x106 visits/yr) =
7.2608 kg CO2/visit
(8.65x106 kg CO2/yr)/( 7.2608 kg CO2/visit) = 1.2x106 visits/yr need to
be eliminated to see the same emissions reduction as using EVs
d. The CEO also asks you: If one-half of the U.S. Geek Squad fleet is replaced by 2012 Prius
hybrids, how many fewer house calls would need to be made to reduce CO2 emissions to the
electric vehicle scenario levels (i.e., the result that you calculated in Part a)?
From personal research, you learn the following:
• Average combined fuel economy for the 2012 Prius: 49.5 mi/gal.
VW emissions = Half of current emissions = 6.8615x106 kg CO2/yr
Prius emissions = 1105 vehicles x 15,500 mi/vehicle-yr x 1 gal/49.5
mi x 9.254 kg CO2/gal = 3.2020x106 kg CO2/yr
Total emissions = Emissions from VWs + emissions from Prius hybrids
= 6.8615x106 kg CO2/yr + 3.2020x106 kg CO2/yr = 1.0064x107 kg
CO2/yr
7
6
6
(1.0064x10 kg CO2/yr)/(1.89x10 visits/yr) = (5.0722x10 kg
5
CO2/yr)/(X visits/yr) X = 9.525x10 visits/yr
6
5
New number of visits = 1.89x10 visits/yr - 9.525x10 visits/yr =
9.4x105 visits/yr
To help him decide what, if any, changes the company should focus to "green" the Geek Squad
fleet, the CEO has also asked you to provide some information about the advantages and
disadvantages of each of the three fleet changes presented above. For EACH of the three (listed
below), identify and describe ONE disadvantage/drawback, ONE advantage, and ONE strategy
that the company could explore to reduce CO2 emissions from employing this technology even
more. These descriptions should be short, ~1 sentence per advantage, disadvantage, strategy listed
(for a total of 9 sentences).
(1) switching to electric vehicles
Disadvantages: more frequent need to maintenance (battery replacement),
limited range
Advantages: electricity is abundant, refueling can be done at Best Buy
stores Strategy: Use PV panels to charge cars to reduce carbon intensity of
fuel
(2) switching ½ of the fleet to hydrogen fuel cell vehicles
Disadvantages: H2 is not commonly available, vehicles currently expensive
Advantages: Greater range before refueling is necessary, Strategy: Use fuel
made from electricity and water rather than natural gas
(3) reducing the number of house calls made
Disadvantages: Possible reduced customer satisfaction Advantages: No
need to purchase or store vehicles, faster response time Strategy: Impose a
limit on the distance the Geek Squad will travel, cutting
out the very long trips