Quiz 1 (at the end of Lecture Notes, Week 2). Solution:
The molar mass of C2H6O is 2×12 + 6×1 + 16 = 46.
Burning of one molecule of C2H6O will produce two molecules of CO2 . The
molar mass of each is 12 + 2×16 = 44; the combined molar mass of the
two is then 88. So, we can set up a proportion:
1 kg 𝑥
=
46
88
By solving, we get that 𝑥 =
88
46
× 1 kg = 1.913 kg.
The burning will emit 29.7 MJ of heat, so that answer to (b) is
1.913 kg/29.7 MJ = 0.0644 kg/MJ. One GJ is 1000 times larger than 1 MJ,
so we get 64.4 kg/GJ. And 1 kW-h = 3.6 MJ, so by multiplying 0.0644 by
3.6 we get 0.232 kg/kW-h.
Quiz 2 (also at the end of Lecture Notes, Week 2). Solution:
One US Gallon is 3.785 liters, and the mass one liter of gasoline is about
0.75 kg, so that the mass of 1 Gallon is 2.839 kg. The heat of combustion
of gasoline is 47.3 MJ/kg (also given in Lecture Notes, week 2, a bit
earlier), so that burning one Gallon of gasoline releases 2.839 kg×47.3
MJ/kg = 134.28 MJ, which is equivalent to 134.28 MJ/(3.6 MJ/kW-h) = 37.3
kW-h. Now, one mile costs 0.25 kW-h/mile, so that, if the efficiency of a
gasoline engine were 100%, the energy from burning one Gallon of
gasoline would propel the car by 37.3 kW-h/(0.25 kWh/mile) = 149.2 miles.
Many more miles than a real car makes per one Gallon of fuel. An
“economy car” usually makes up to 30 miles/Gallon; it corresponds to the
efficiency of 30/149.2 = 0.2, or 20%.
Quiz 3 (end of Lecture Notes, Week 3):
The mass of 852 m3 of water is 852 000 kg. The water flows down 385
meters, so its potential energy relative to the power house level is
852000kg×385m×9.81m/s2 = 3.218×109 kg-m2/s2 = 3.218×109 J = 3.218
GJ. It’s the energy released in one second, so it corresponds to the power
of 3.218 GW, or 3218 MW. So, since the powerhouse generates 3001 MW
of power, its efficiency is 3001/3218 = 0.933 = 93.3%.
The capacity of the upper pool is 14 million m3, and the flow is 852 m3 per
second, so the flow can last 14000000/852 = 16 432 s = 4.56 hours.
The total energy generated is 3001 MW × 16432 s = 4.931 ×107 MJ
= 4.931×104 GJ.
...
.
'
. '
..·.
fl' .
:)
,
-!.-.,
Ph313, Spring 2017 – Classroom Quiz #5, April 28, 2017.
The rotor diameter (diameter -- not radius!) of the Vestas 80 three-blade wind
turbine is 80 m. Estimate how much electric power can be generated by this
turbine if the wind speed is 20 mph. Express the result in the units of MW.
Take the air density as 1.25 kg/m3;
1 mile = 1609 meters; 1 hour = 3600 s.
For a wind speed in the 15-20 mph range the conversion efficiency of the Vestas
80 turbine is the highest, reaching about 70% of the Betz Limit efficiency.
(
UJ
~:r \ ·
·_ ·
I ~i I<,,) 3/J1io ~ ) ='B' .c1-4 wi /~.
M'1lD \ ( /60'1
\ \\·r
r~ AOvV'I,
J
W'l
\(
j
A= If rz.-=-- 7T (4o)
2 -;;
so2t> .ssVI)'
::: \. 32 MW
~·
Ph313, Spring 2017 – Classroom Quiz #6, May 03, 2017.
A true story:
Nevada Solar One is a parabolic trough plant owned by the Acciona energy company is rated at
64MW. It does not contain any heat storage devices. It is listed as a project site of 1.6 km 2. The
solar collectors are designed, as claimed by Solar Electric Power Association, to produce 134
GWh/year, which would equate to a reasonable capacity factor for a plant without storage.
1.
Calculate the average output power density of the plant (in the units of W/m2).
2. The insolation of that area of Nevada is 225 W/m2. Based on the result you have
obtained above, find the Nevada Solar One's overall efficiency:
Ph313, Spring 2017 – Classroom Quiz #7, May 03, 2017.
Potassium nitrate KNO3 is not-so-bad salt for using it in thermal
storage, because of its not-so-bad latent heat for solidifying of 100
kJ/kg.
Find how much molten KNO3 is needed for delivering thermal
power of 1 MW over the period of 1 hour.
Now, suppose that the latent heat from solidifying KNO3 is used
for producing steam, which is sent to a steam turbine operating an
electric generator. Suppose that the overall efficiency of the
turbine-generator system is 25%. So, how much solidifying KNO3
is needed for obtaining electric power of 1 MW over the period o 1
hour?
Q. v \ 7- . -#
·~
I'
.
/
-- ,...o-
-.J
o.)
•
; =.~1_.,
-
l -...
fi_.'J--/t__,···
. I
f
i
-~-
~)
6.\.-f . . -=~
I
!
l;
!..
\.,
})
b
~~"·
""""
~-.