College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 694ALT
Alternative Energy
Spring 2008 Number: 18411 Instructor: Larry Caretto
February 24 Homework Solutions
1. Problem 16.1 text: You have been asked to design a rotating mechanical flywheel to
supply 1 MWe of power for one minute. You have two choices of materials – steel and a
proprietary composite (Rx-2002). The properties of both materials are given below. Key
parameters to specify in your design are the volume and area (radius), mass, and
maximum rotational speed. Discuss the factors that led to your final design. Are there
other factors that need to be considered before your flywheel is practical?
Density – steel = 7.8 g/cm3 and RX-2002 = 2.0 g/cm3
Maximum tensile strength – steel = 550 MPa and RX-2002 = 2,100 MPa
The energy stored in the flywheel is the product of the power and the time. Thus we have to
store E = (1 MW)(60 s) = 60 MW·s = 60 MJ. Equation 16-17 gives the specific energy density
(energy per unit mass) by the following equation.
Erotational , max
m

kmmax

We can eliminate the mass of the rotor in favor of the volume by using the equation, m = V;
Erotational , max
m

Erotational ,max
V

kmmax

 V
Erotational ,max
kmmax
Look at two cases: solid disk for which k m = 1 and the thin rim configuration for which k m = 0.5.
Using the problem data for solid steel, we can find the required volume for the solid steel disc as
follows.
Vsteel cyl 
Erotational,max
60 MJ
1 MPa  m3

 0.1091 m3
km ,cyl max, steel 1550 MPa 
MJ
Similar calculations for the other combinations of material and geometry and multiplying volume
by density to give mass produce the results below.
Material
Configuration
Volume
Mass
Steel
Solid Disc
0.1091 m3
850.9 kg
Thin Rim
0.2182 m3
1702 kg
RX-2002
Solid Disc
Thin Rim
0.02857 m3
0.05714 m3
222.9 kg
445.7 kg
The size of the disk is related to the rotational energy by the equation E rotational = I2/2. The
moment of inertia for the solid disc of radius R is mR2/2; the moment of inertia for the thin rim,
with radius Ri and Ro is m(Ro2 + Ri2)/2. If the rotational speed of the steel disc were 3,600 RPM,
the upper limit listed on page 663, the moment of inertia for the required rotational energy would
be
Engineering Building Room 1333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
February 24 homework solution
E
I2
2

ME 496ALT, L. S. Caretto, Spring 2009
I
2E

2

2 60 x106 J

1 kgJ  sm
Page 2
2
2
 3600 rev m 2 


m
60 s rev 

2
 844.3 kg  m 2
Using the formula that the moment of inertia for a solid cylinder is mR 2/2 gives the required radius
for the solid steel cylinder as follows.
I
mR2
2
 R


2I
2 844.3 kg  m2

 1.409 m
m
850.9 kg
The disc thickness of the disc with this radius and the required volume of 0.1091 m3, is found as
follows.
V  R 2 h  h 
V
0.1091m3

 0.01750 m
R 2  1.409 m2
For the RX-2002, assume that the material can be used to a higher RPM, say 40,000 RPM that is
typical of synthetic materials (see http://www.testdevices.com/flywheel_article.htm, accessed
March 1, 2008). In this case the required moment of inertia is
I
2E

2

6
2 60 x10 J


1 kg  m 2
J  s2
 40 x103 rev m 2 



m
60 s rev 

2
 6.839kg  m 2
Using the formula that the moment of inertia for a solid cylinder is mR 2/2 gives the required radius
for the solid RX-2002 cylinder as follows.
I
mR2
2

R
2I

m


2 6.839kg  m 2
 0.2477 m
222.9kg
The disc thickness of the disc with this radius and the required volume of 0.02857 m3, is found as
follows.
V  R 2h  h 
V
R 2

0.02857 m3
0.2477 m2
 0.01482 m
Choose the RX-2002 material because of its smaller size.
Other items to consider in the design include the power transfer system, the durability of the RX2002 in actual applications, the reliability of the material at the higher RPM, and the costs of the
material.
2. Problem 16.2 text. A large pumped hydropower energy storage system is being
considered for a site in Colorado near Denver where peak demand can be high at times.
The plan calls for 1,000 MWe of dispatchable power for up to six hours. The proposed
design would have two reservoirs, one located at Denver’s mile-high elevation of 5,280 ft.
and the other in the foothills for the front range of the Rocky Mountains at an elevation of
8,000 ft. Estimate the minimum required volume of water that would need to be stored
(state and justify all assumptions made). Assuming a margin of safety of 50% and an
average water depth of 20 ft in both reservoirs, how large an area does this system
February 24 homework solution
ME 496ALT, L. S. Caretto, Spring 2009
Page 3
impact? Are there other issues that may be important in assessing the sustainability
attributes of this proposal?
The required energy storage is the product of the energy rate and the time: E = (1000 MW)(6
h)(3600 s/h)(106 W/MW)(1 J/W·s) = 2.16x1013 J. The required storage would have to account for
the efficiency with which this stored energy could be converted into electric power. This is not the
overall efficiency of the energy storage process (energy out over energy in), but the only the
efficiency for the second step. Assume that this efficiency is 80%, then the required storage
would be (2.16x1013 J)/(0.8) = 2.7x1013 J. Providing a safety margin of 50% would require an
additional 1.35x1013 J of energy giving a total required storage of 4.05x1013 J. This storage is
provided by the potential energy difference between the two elevations. The required mass of
water is thus found as follows.


1 kg  m 2 1 ft
4.05 x10 J
E
J  s 2 0.3048 m  4.979 x109 kg
E  mgz  m 

9.81065 m
gz
8000 ft  5280 ft 
s2
13
Dividing this mass by the water density of 1000 kg/m 3 gives a volume requirement of 4.979x106
m3(1 ft/0.3048 m)3 = 1.758x108 ft3. If the average depth of the each storage reservoir is 20 ft,
each reservoir will have to have an area of (1.758x108 ft3) / (20 ft) = 8.792x106 ft2. Dividing by the
factor of (5280 ft/mi)2 gives a required area of 0.3154 mi2. Thus the area impacted by the two
storage reservoirs would be about 0.6 mi2 .
Other issues to consider include the problem of constructing and maintaining the higher reservoir
in an environmentally sensitive area. Access to the higher reservoir in winter could also be a
problem. One important aspect of energy storage is that is lowers the costs of peak electricity
production, but requires a process with an overall efficiency of 60-70% to do so. Thus, it
increases overall energy use.
3. Problem 17.2 text. Compare the land requirements given by Flavin in Table 17.7 with the
data presented in Table 9.2. Are the values reasonably consistent? If not, what do you
think the reasons are for discrepancies?
The data in Table 17.2 give the land requirements in km 2/exajoule·yr, whereas the data in Table
9.2 give the global energy flux in TW. Obviously one table has an area factor and the other does
not. The basis can be made the same by dividing the data in Table 9.2 by the land area of the
earth (148,939,100 km2) and converting to energy units. Note that 1 TW = 1 TJ/s = 10-6 EJ/s.
The energy in 1 TW·yr with (365 days/yr)(24 h/day)(3600 s/h) = 3.1536x107 s/yr is 31.536 EJ.
The total land flux of solar energy in Table 9.2 is given as 27,000 TW. This flux for one year
corresponds to (27,000 TW·yr)(31.536 EJ/TW·yr) = 8.5x105 EJ. Dividing this by the land area of
148,939,100 km2 gives a figure of 0.0057 EJ/km2. The reciprocal of this is 175 km 2/EJ which is
less than the areas in Table 17.7 of 1,700 – 3,300 km2/EJ for solar photovoltaic central stations
and 700 – 3,000 km2/EJ for thermal trough. The differences are due to the efficiency of
conversion. The flux in Table 9.2 represents the solar energy input; the energy in Table 17.7
represents the electrical power output. Note that the variation is due to the local solar flux. The
global average flux for an entire year is less than the flux in an area where there is a large
amount of sun.
I just noticed that I do not have a consistent set of units. According to the column heading in
Table 17.7, the second column represents the land area per exajoule·yr. I am not sure what this
unit means and have chosen to ignore the yr.
Here is another check: Table 9.2 lists the biomass flux as 30 TW. As noted above this flux for
one year corresponds to (30 TW·yr)(31.536 EJ/TW·yr) = 946 EJ. Dividing this by the land area of
148,939,100 km2 gives a figure of 6.35x10-6 EJ/km2. The reciprocal of this is 1.57x105 km2/EJ;
this is within the range of 1.25x105 km2/EJ to 2.5x105 km2/EJ in Table 17.7.
February 24 homework solution
ME 496ALT, L. S. Caretto, Spring 2009
Page 4
Table 9.2 lists the wind flux as 100 TW. For one year this is (100 TW·yr)(31.536 EJ/TW·yr) =
3.15x103 EJ. Dividing this by the land area of 148,939,100 km2 gives a figure of 2.12x10-7
EJ/km2. The reciprocal of this is 4.72x104 km2/EJ; this is above the range of 0.03x104 km2/EJ to
1.7x104 km2/EJ in Table 17.7. This discrepancy is possibly due to the use of only high wind sites
(rather than the global average) for producing electricity from wind.