ELEN280 Midterm Fall 2023
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Santa Clara University *
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Course
280
Subject
Mechanical Engineering
Date
Jan 9, 2024
Type
docx
Pages
12
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Name
:
ELEN280 / MECH287
MIDTERM FALL 2023
True/False:
__F__
1. On average, buildings in the US account for less than a quarter of our energy
use.
__F__ 2. The Carbon Bubble is an alternative term for the large quantity of CO
2
in
the Earth’s atmosphere.
___T___ 3. Over 90% of the energy consumed within the transportation sector is oil
based.
___F___ 4. There is a scientific consensus that global climate change forced by human-
induced greenhouse gas emissions is occurring and the vast majority of
these emissions are from methane released into the atmosphere.
___F___ 5. Without the Greenhouse Gas Effect the Earth would be at an average
temperature of 15
o
C.
___T___ 6. The energy per unit volume for hydrogen is low while the energy per unit
of mass is very high.
___T___ 7. Embodied energy can be decreased by using locally derived materials.
___F___ 8. On a per molecule basis, carbon dioxide is the most potent of the
greenhouse gases.
___F___ 9. A heat-pump that uses external air for extracting and dumping heat has a
coefficient of performance (COP) that is independent of the external air
temperature.
___F___ 10. Heat driven air-conditioning (e.g., absorption chillers) and refrigeration
are the latest technological breakthrough in AC and refrigeration.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
1
Name
:
Multiple Choice:
___C____ 1. The number 0.0070 has how many significant figures?
a. 5
b. 4
c. 2
d.1
___B___ 2. Long distance electric power transmission is most efficient when using
a. high-voltage AC
b. high-voltage DC
c. AC and DC are the same
d. None of
the above
___B
_
__ 3. The spectrum of the Sun’s light that impinges on the top of the Earth’s
atmosphere approximates a
____________ K blackbody.
a. 6500
b. 5800
c. 4300
d. 1000
___A_
_
_ 4.
In order to be able to optimize passive solar performance of a house, the
long direction of the house should be oriented:
a. East/West
b. North/South
c.
parallel to the street
d. perpendicular to the
street
___A___ 5.
If an aluminum product was made from recycled aluminum instead of
freshly mined aluminum its embodied energy would:
a. decrease
b. be the same
c. increase slightly
d. increase a lot
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
2
Name
:
Fill in the Blank:
1.
GIGO is an acronym for Garbage in, Garbage out
.
2.
The heat absorbed or released at a phase transition is known as the latent
heat.
3.
Trying to make something seem green that really is not green is called
Greenwashing
.
4.
Net Metering
refers to the case when the customer can buy and sell power to and
from the utility for the same price (retail).
5.
When making measurements on a construction site for solar panel installation it
is more important to measure the effects of shade
rather than the sun’s power
density in that location.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
3
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Name
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1. Externalities
In California vehicles designated “zero-emission” are allowed to use express lanes
for free even when they have only one passenger.
a.
In what way are these cars zero emission?
Cars are zero emission if they produce no direct exhaust or tailpipe emissions. In
other words, they do not produce carbon dioxide emissions from their local energy
source. Cars that are electric or hydrogen powered are considered ‘zero emission’.
b.
In what way are they not zero-emission?
Although electric and hydrogen powered vehicles may not emit any pollutants
locally from their tailpipe during operation, the production of their energy sources
are by no means ‘zero-emission’. The production of electricity itself, unless done
sustainably, often involves burning of fossil fuels, which inherently emits pollutants
into the atmosphere. A prime example is car charging stations. The electricity source
for these vehicles is an externality, which requires emissions at the power plant to
generate the electricity required for the vehicle to function.
c.
Is there a general name for things like these non-zero emissions?
The non-zero emissions that occur as a result are called externalities
.
d.
Are the emissions associated with the same electric car the same everywhere
in the US?
Very briefly explain why or why not?
The emissions associated with electric cars can vary based on the externalities
where the electricity is sourced for. The answer is
no
because the electricity
generation mix can differ based on the source of the electric car charging source. For
example, some electric car charging stations could be based on sustainable energy
sources such as solar / wind power. Another electric car charging station could
source its energy from powerplants where coal or fossil fuels are utilized to generate
electricity.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
4
Name
:
2. Embodied Energy
Embodied energy in a product is the energy that is used to get it from the cradle to
the current state it is in.
The example of an aluminum marker-board tray was used
in class to explain steps that might have been taken in order for the tray to be made
and mounted on the wall in the classroom.
a.
Give a list of steps that might be involved in going from bauxite, the
aluminum ore mined for aluminum processing, through the forming of the
tray, to the installation of the tray on the wall.
List the steps in which
embodied energy is added to the final product and specify if it requires a lot
or small amount of energy.
1.
Bauxite mining (a lot of energy)
2.
Transport of mined bauxite to refinement location (small amount of energy)
3.
Bauxite refinement/conversion into Aluminum Oxide (a lot of energy)
4.
Electrolysis of Aluminum Oxide into Aluminum (a lot of energy)
5.
Aluminum extrusion / compression / heat refinement (a lot of energy)
6.
Assembly of marker-broad tray (small amount of energy)
7.
Packaging (small amount of energy)
8.
Transport of final product to installation location (a lot of energy)
9.
Installation (small amount of energy)
b.
Is it possible to change the embodied energy of the aluminum tray?
If not,
why not?
If so, what can be done?
Yes indeed. As opposed to mining for bauxite, the aluminum can be utilized to skip
the high energy steps of the mining, refinement/conversion, and electrolysis. In
addition, mining can be done locally to reduce transportation energies. This would
greatly reduce the embodied energy of the aluminum tray.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
5
Name
:
3
.
Coefficient of Performance (COP)
For a heat pump running as a heater, the COP is the ratio of the amount of energy
delivered
per unit of work done
by the heat pump while when running as an air
conditioner the COP is the heat removed
per unit of work done
by the heat pump.
a.
Assume the efficiency of the electricity production is 33% and transmission
of electricity is 96%.
1 BTU is
0.293071 Wh.
For a heat pump operating as an
air conditioner with a COP of 2.4, how many BTUs per hour of heat can be
removed when the power consumption of the heat pump is 1 kW?
COP
=
Q
c
W
=
2.4
; Powerconsumption
=
P
=
1
kW
=
1000
W
Total Power
=
P
∗
(
COP
)
=
1000
W
(
2.4
)
(
1
BTU
0.293071
Wh
)
=
8189.14
BTU
/
h
Therefore, the total amount of heat that can be removed is 8190 BTU per hour
.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
6
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Name
:
b.
Later in the day the temperature has increased by 10
o
F.
What is the value of
the COP at the higher outside temperature?
OriginalCOP
=
T
c
T
H , original
−
T
c
=
2.4
=
Q
c
W
=
7575.76
3156.56
AssumeT
H , original
=
70
℉
=
21.1
℃
T
c
(
W
)=
Q
c
(
T
H ,original
−
T
c
)=
Q
c
(
T
H ,original
)
−
Q
c
(
T
c
)
T
c
(
W
)
+
Q
c
(
T
c
)
=
Q
c
(
T
H, original
)
=
T
c
(
W
+
Q
c
)
T
c
=
Q
c
(
T
H ,original
)
W
+
Q
c
=
7575.76
(
21.1
℃
)
3156.56
+
7575.76
=
14.89
℃
=
58.8
℉
Now ,T
H , new
=
T
H , original
+
10
℉
=
70
℃
+
10
℃
=
80
℃
=
26.67
℃
NewCOP
=
T
c
T
H ,new
−
T
c
=
14.89
℃
26.67
℃
−
14.89
℃
=
14.89
℃
11.78
℃
=
1.26
Therefore, if the external temperature increases by 10
o
F, the COP will decrease.
If we assume starting external temperature is
70
℉
, then the COP decreases to
1.26.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
7
Name
:
4. Shadows and Shading
You are doing a shade analysis of a site for a solar panel installatio
n.
Using a SunEye
210 you obtain the picture shown in the figure.
The long curves identify the paths of
the sun at different times of the year that are broken up into the hours of the day by
cross bars/lines.
a.
What do the long curves A and B correspond to?
Identify both A and B.
A: summer solstice
B: winter solstice
b.
Estimate the largest number of hours of sunshine and the smallest number of
hours of sunshine hitting your panels and identify on what day each will
occur.
The largest number of hours of sunshine will be about 11 ±1 hours, on the summer
solstice of any given year. The smallest number of hours of sunshine will be about 4
±1 hours, on the winter solstice of any given year.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
A
B
8
Name
:
5.
Impacts of
Energy Use Growth Rate
The annual growth rate of energy utilization in the world was 3.5% per year in the period
from 1950 to 1973.
a. Assume that the global energy resources available at the moment are sufficient to
sustain the world for 1000 years at the current utilization rate. How long would it take to
consume all available resources if the consumption growth rate of 3.5% per year is
maintained?
Utilizingageometric progression formula
:
years
=
Y
=
a
(
r
+
1
)
t
−
1
(
r
+
1
)
−
1
rate
=
r
=
3.5%
=
0.035
;Y
=
1000
;a
=
1
;solve for t
Y
=
a
(
r
+
1
)
t
−
1
(
r
+
1
)
−
1
=
1
(
0.035
+
1
)
t
−
1
(
0.035
+
1
)
−
1
=
1.035
t
−
1
0.035
=
1000
1.035
t
−
1
=
1000
(
0.035
)
=
35
1.035
t
=
35
+
1
=
36
t
=
log
(
36
)
log
(
1.035
)
=
1.556
0.0149
=
104
years
Therefore, it will take 104 years
to consume all available resources given the statement.
b. Assume that the global energy resources at the moment are sufficient to sustain the
world for 10,000 years at the current utilization rate. How long would it take to consume
all available resources if the consumption growth rate of 3.5% per year is maintained?
Same problem, but now Y = 10000 years
10000
=
1.035
t
−
1
0.035
1.035
t
−
1
=
10000
(
0.035
)
=
350
1.035
t
=
350
+
1
=
351
t
=
log
(
351
)
log
(
1.035
)
=
2.545
0.0149
=
171
years
Therefore, it will take 171 years
to consume all available resources given the statement.
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SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
10
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:
6.
Using Present Value Dollars
You are trying to decide if you should build a factory or simply leave the money in
the bank collecting interest.
You will use a present-value method to make this
evaluation.
Assume a discount rate of 4%.
It will cost $10M to build the factory and two years of time.
Assume that all of the
building costs are spent on day-one of construction.
Assume the factory will run for
25 years after its completion.
In the first five years of operation, it generates $1.5M
per year in profit each year in future dollars.
In the next five years it generates $2M
per year.
In the next five years it generates $2.5M per year in profits.
In the next five
years it generates $1.5M per year in profits.
And in the last five years it generates
$1m per year in profits.
(All of the profits here are given in future dollars.)
Should you build the factory or leave the money in the bank at a 3% interest rate?
Show your work to get credit.
Use additional page if you need it.
To understand if we should build a factory or leave the money in the bank, we will
first calculate how much we would make by leaving the money in the bank:
Bank:
Principal
=
P
=
$
10
7
;
discount rate
=
d
=
4%
=
0.04
;interest rate
=
i
=
3%
=
0.03
years
=
n
=
27
years ;Future value
=
F
F
i
=
P
(
1
+
i
)
n
=
10
7
∗
(
1
+
0.03
)
27
=
10
7
∗
2.22
=
$
2.22
¿
10
7
(future w/ interest)
P
d
=
F
(
1
+
d
)
n
=
2.22
¿
10
7
(
1
+
0.04
)
27
=
$
7.704
¿
10
6
(money w/ interest and discount rate in bank)
Next, we will calculate how much our factory would make over the same time frame:
(Factory calculations and final answer on next page)
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
11
Name
:
Factory:
Initial 2 years of investment: -
$
10
7
Profit:
1
st
5 years:
∑
3
7
1.5
∗
10
6
1.04
n
=
$
6.17
∗
10
6
2
nd
5 years:
∑
8
12
2.0
∗
10
6
1.04
n
=
$
6.77
∗
10
6
3
rd
5 years:
∑
13
17
2.5
∗
10
6
1.04
n
=
$
6.95
∗
10
6
4
th
5 years:
∑
18
22
1.5
∗
10
6
1.04
n
=
$
3.43
∗
10
6
5
th
5 years:
∑
23
27
1.0
∗
10
6
1.04
n
=
$
1.88
∗
10
6
Total =
$
6.17
∗
10
6
+
$
6.77
∗
10
6
+
$
6.95
∗
10
6
+
$
3.43
∗
10
6
+
$
1.88
∗
10
6
−
$
10
7
Total =
$
2.52
∗
10
7
−
¿
$
1
∗
10
7
Total =
$
1.52
∗
10
7
Investing in the bank would yield you $7.7 million after 27 years, and investing in
building and operating the factory would yield you $15.2 million after 27 years.
Therefore, it is clear to invest in the factory
rather than in the bank.
SCU ELEN280 / MECH287
MIDTERM
FALL 2023
PAGE
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