Engineering Building Room 1333 Mail Code Phone: 818.677.6448 s1

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. E k  rotational,max  m max m 

We can eliminate the mass of the rotor in favor of the volume by using the equation, m = V; E E k  E rotational,max  rotational,max  m max  V  rotational,max m V  kmmax

Look at two cases: solid disk for which km = 1 and the thin rim configuration for which km = 0.5. Using the problem data for solid steel, we can find the required volume for the solid steel disc as follows.

3 Erotational,max 60 MJ 1 MPa m 3 Vsteel cyl    0.1091 m 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 Steel RX-2002 Configuration Solid Disc Thin Rim Solid Disc Thin Rim Volume 0.1091 m3 0.2182 m3 0.02857 m3 0.05714 m3 Mass 850.9 kg 1702 kg 222.9 kg 445.7 kg

2 The size of the disk is related to the rotational energy by the equation Erotational = I /2. The moment of inertia for the solid disc of radius R is mR2/2; the moment of inertia for the thin rim, 2 2 with radius Ri and Ro is m(Ro + Ri )/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 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062 1 kg m2 2 60x106 J I2 2E   2 E   I   J  s  844.3kg m2 2 2 2   3600 rev m 2     m 60 s rev  Using the formula that the moment of inertia for a solid cylinder is mR2/2 gives the required radius for the solid steel cylinder as follows.

mR2 2I 2 844.3kg  m2 I   R      1.409m 2 m 850.9kg

The disc thickness of the disc with this radius and the required volume of 0.1091 m3, is found as follows. V 0.1091m3 V  R2h  h    0.01750m R2  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 1 kg  m2 2 60x106 J 2E   2 I   J  s  6.839kg  m2 2 2   3   40x10 rev m 2     m 60 s rev  Using the formula that the moment of inertia for a solid cylinder is mR2/2 gives the required radius for the solid RX-2002 cylinder as follows.

mR2 2I 2 6.839kg  m2 I   R      0.2477m 2 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 0.02857 m3 V  R2h  h    0.01482m R2 0.2477 m2 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 RX- 2002 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

Engineering Building Room 1333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062 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  m2 1 ft 4.05x1013 J  E 2 0.3048 m E  mgz  m   J  s  4.979x109 kg 9.81065 m gz 8000 ft  5280 ft s2 Dividing this mass by the water density of 1000 kg/m3 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 km2/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 km2/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.

Engineering Building Room 1333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062 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.

Engineering Building Room 1333 Mail Code Phone: 818.677.6448 E-mail: [email protected] 8348 Fax: 818.677.7062