# HW5_solutions - ECE 333 Homework 5 Solutions Spring 2010 1...

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Unformatted text preview: ECE 333 Homework 5 Solutions :: Spring 2010 1 Suppose an anemometer mounted at a height of 10-m on a level field with tall grass shows an average windspeed of 6 m/s. a) Assuming Rayleigh statistics and standard conditions (15o C, 1 atm), estimate the average wind power (W/m2) at a height of 80-m. From Table 6.3, the friction coefficient for the given terrain is ! = 0.15. Using (6.15), the average windspeed at 80 m can be found: " 0.15 H 80 80 V 80 !V 10 # V 80 !6\$ # V 80! 8.196 m % s H 10 10 Assuming Rayleigh statistics and using (6.48), the average power in the wind can be found: 61 61 3 3 2 P 80 ! ' A ( # P 80! v \$1.225\$1\$8.196 # P 80 !644 W % m &2 &2 !" !" b) Suppose a 1300-kW wind turbine with 60-m rotor diameter is located in those 80-m winds. Estimate the annual energy delivered (kWh/yr) if you assume the turbine has an overall efficiency of 30%. &2 6 E !)\$P 80\$A\$T # 0.30\$644\$ \$60 \$8760 # E !4.79\$10 kWh % yr 4 c) What would the turbine's capacity factor be? Using (6.60) the capacity factor is found as follows: 6 energy delivered E 4.79\$10 CF ! ! ! # CF ! 0.42 Rated Power x 8760 P R\$8760 1300\$8760 2 In the derivation of the cumulative distribution function, F(V) for a Weibull function, we had to solve the integral F *V +!, f * v + dv . Show that F(V) in (6.53) is the correct result by taking the derivative 0 V d F *V + f * v +! and seeing whether you get back to the Weibull probability density function given in (6.41). dV k v In (6.53), F(V) is defined to be F * v +!1 -exp . Simply take the derivative of F(v) with c respect to velocity to prove the same solution results. ! ! "! d F *v + d v Setup the derivative: f * v +! ! 1-exp dv dv c v du k v To find the derivative, u-substitution should be used with u ! and ! c dv c c d F *v + d d - u du ! 1-e-u " !e-u ! !1 - e "\$ The final derivative will be given as where dv dv dv dv Substituting the value for u yields the final solution: f * v +! d F * v +! exp - ! ! ! " !! k !! k !" k-1 f * v +! d F * v +! kv cc ! ! "! ! " ! ! ! ! !! v c k \$ kv cc k -1 rearranging yields (6.41): k -1 \$exp - v c k 3 The table given on page 381 shows a portion of a spreadsheet that estimates the energy delivered by a NEG Micon 1000 kW/60 m wind turbine exposed to Rayleigh winds with an average speed of 8 m/s. a) How many kWh/yr would be generated with 5 m/s winds? First find the probability that the wind will blow at 5 m/s using (6.45): &V &V &5 &5 f * V +! \$exp - \$ # \$exp - \$ # f *V +! 0.0903 2 2 ( 4V 48 ( 2V 2\$8 Next, find the amount of hours in a year the wind will blow at 5 m/s using f(V) above: 8760 hr ! yr!0.0903! 791 hr ! yr Finally, multiply by the power output at 5 m/s from the table: E !86 kW !791 hr ! yr ! E !68,025 kWh ! yr b) Using Table 6.7, how many kWh/yr would be generated in 10 m/s winds for a Vestas 600/42 machine? Similar to part (a), the probability that the wind will blow at 10 m/s can be found as follows: 2 & 10 & 10 f * v +! \$exp - \$ # f *V +! 0.07194 2 48 2\$8 The amount of hours in a year the wind will blow at 10 m/s: 8760 hr % yr\$0.07194 !630.2 hr % yr From Table 6.7, the power at 10 m/s for the given turbine is given as 356 kW, therefore: E !365 kW \$630.2 hr % yr # E ! 224,354 kWh% yr ! ! "! 2 " ! "! 2 ! ! "! 4 Consider the Nordex 1.3 MW, 60-m wind turbine with power specifications given in Table 6.7 located in an area with 8 m/s average wind speeds. a) Find the average power in the wind (W/m2) assuming Rayleigh statistics. Using (6.48) and assuming Rayleigh statistics, the average power is found: 61 61 3 3 2 P! ' A( # P! v \$1.225\$1\$8 # P !599 W % m &2 &2 b) Create a spreadsheet similar to the one developed in Example 6.15 to determine the energy delivered (kWh/yr) from this machine. Nordex 1.3 MW, 60-m, 8 m/s average wind speed V [m/s] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Wind Hours Turbine PR [kW] Energy [kWh] 0 212.4 409.4 577.6 706.7 791.0 829.3 824.9 784.2 716.1 630.2 535.7 440.7 351.3 271.6 203.9 148.7 105.4 72.6 0 0 0 0 25 78 150 234 381 557 752 926 1050 1159 1249 1301 1306 1292 1283 0 0 0 0 17,667 61,698 124,400 193,022 298,789 398,882 473,917 496,097 462,757 407,164 339,261 265,249 194,147 136,114 93,144 19 20 21 22 23 24 25 26 48.7 31.7 20.2 12.5 7.5 4.4 2.5 0 1282 1288 1292 1300 1313 1328 1344 0 62,388 40,884 26,036 16,192 9,842 5,835 3,371 4,126,858 c) What would be the average efficiency of the wind turbine? The total energy delivered by the turbine as found by the spreadsheet is 4,126,858 kWh/yr. Efficiency can be found by using the average power in the wind from (a) as follows: 9 Actual Energy 4.127\$10 Wh % yr )! ! !0.279 # ) ! 27.8 % Energy in the wind 2& 2 2 599 W % m \$ \$60 m \$8760 hr % yr 4 d) If the turbine's rotor operates at 70% of the Betz limit, what is the efficiency of the gearing and generator? Overall efficiency is multiplicative: 0.278 )total !)rotor\$) gen & gear # ) gen & gear ! !0.67 # ) gen & gear ! 67 % 0.593\$0.70 ! " 5 For Rayleigh winds with an average windspeed of 8 m/s: a) How many hours per year do the winds blow at less than 13 m/s? The probability the wind blows below a certain speed given Rayleigh statistics is given by (6.54): 2 &V F * V +! prob * v . V +! 1- exp 4v ( & 13 2 # prob * v . 13 +! 1 -exp # F * V +! 0.8743 48 hours * v . 13 +! 8760 hr % yr\$0.8743! 7659 hr % yr b) For how many hours per year are the windspeeds above 25 m/s? Similarly, the probability the wind will blow above a certain speed is given by (6.57): 2 2 &V & 25 F * V +! prob * v / V +! exp # prob * v / 25 +! exp # F *V +! 0.0004667 4( v 48 hours * v . 13 +!8760 hr % yr\$0.0004667!4.1 hr % yr c) Suppose a 31-m, 340-kW turbine follows the idealized power curve shown in Figure 6.32. How many kWh/yr will it deliver when winds blow between its rated windspeed of 13 m/s and its furling windspeed of 25 m/s? Find the amount of hours that the wind will blow above cut-in speed and below furling speed. hrs % year - hrs above 25 m % s -hrs below 13 m % s !8760 - 4.1-7659 !1097 hr % yr Find energy as in (3b): E ! 340 kW \$1097 hr % yr # E ! 372,980 kWh % yr d) Using the capacity factor correlation given in (6.65), estimate the fraction of the annual energy delivered with winds that are above the rated windspeed. Capacity factor for Rayleigh winds is given in (6.65) as: PR 340 kW ( CF ! 0.087 V - 2 # 0.087\$8 m % s - 2 2 # CF ! 0.342 D 31 m ! ! "! " ! "! ! ! "! " ! "! Energy using capacity factor: E !340 kW !8760 hr ! yr cd 0.342!1,019,214 kWh ! yr Fraction of energy in winds above rated windspeed: 372,980 ! 0.366! 36.6 % 1,019,214 All tables, figures, examples and equations referenced can be found in chapter 6 of [1]. References [1] G. M. Masters, Renewable and Efficient Electric Power Systems. Hoboken, NJ, John Wiley & Sons, Inc., 2004, pp. 307-378. ...
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## This note was uploaded on 03/04/2011 for the course ECE 333 taught by Professor Overby during the Spring '09 term at University of Illinois, Urbana Champaign.

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