Renewable Energy Systems

T3 Solutions 4.3 a. Fuel saved by waste heat: b. Electricity savings: c. Annual cost of gas for the system: d. Annual savings: Savings = gas + electric savings – fuel cost – O&M Annual savings = $14,016 + $175,200 - $63,598 – 0.02 x $1.5x10 6 = $95,618/ yr e. Simple payback Boiler fuel savings = 300 , 000 Btu/hr 0 . 75 x $ 5 10 6 x 0 . 80 x 8760 hr/yr = $ 14 , 016 /yr Electricity savings = 250 kW x 0 . 8 x 8760 hr/yr x $ 0 . 10 /kWh = $ 175 , 200 /yr Gas cost = 250 kW x 7260 Btu/kWh x 0.80 x 8760 hr/yr x $5 10 6 Btu  $63, 598/yr Simple payback = $1.5 x10 6 $95,618/yr  15.6 yrs

ECON | 443 | Renewable Energy Systems | assessment | T3 Solutions 4.3 a. Fuel saved

T5 Solutions 6.1. a. Power from the turbine would be b. From Table 6.1, the density 10 o C is 1.247 kg/m 3 giving a K T ratio of 1.247/1.225 = 1.018 (the table rounds that to 1.02). Unfortunately Table 6.2 doesn’t provide a density correction for 2500 m so we need to use (6.13) giving an altitude correction factor of Using (6.14) c. Turbine power is proportional to air density, so P    1 2  Av 3  0.30  0.5  1.225   4 20 2  10 3  57, 727 W  57.73 kW P=P 0 exp  1.185 x 10  4 H    1.225exp  1.185 x 10  4  2500    0.9109 K A  0.9109 1.225  0.743   1.225 K T K A  1.225  1.018  0.743  0.927 kg / m 3 P  57.73 kW  0.927 1.225  43.7 kW 6.2. From Table 6.3, the friction coefficient  for ground with hedges, etc., is estimated to be 0.20. From the 15 o C, 1 atm conditions, the air density is  = 1.225 kg/m 3 . (a) Windspeed at the lowest point of the rotor (30 m) will be: and the specific power will be: v 30  5  30 10     0.20  6.229 m / s P 30  1 2  v 3  0 . 5 x 1 . 225 x 6 . 229 3  148 W / m 2 (b) At the highest point (90 m) the rotor will see: and the specific power will be: (c ) The ratio of power top-to-bottom, is: Using (6.17) gives: 6.3. a. For it to be a pdf, the area under the curve must equal 1: v 90  5  90 10     0.20  7.759 m / s P 90  1 2  v 3  0 . 5 x 1 . 225 x 7 . 759 3  286 W / m 2 Power Ratio  286 W / m 2 148 W / m 2  1.93 P P 0  H H 0       3   90 30     3 x 0.20  1.93 which agrees Area = 1 2 10 k  5 k  1  k= 0 . 2 so f(V) = 0.02V (0 ≤ V≤ 10 m/s) b. P AVG = 1/2  A (V 3 ) AVG P AVG = 0.5 x 1.225 x 1 x 400 = 245 W/m 2 6.4. a. What is the probability that the wind is blowing between 5 and 15 m/s? It is just the fraction of the total area that is between 5 and 15 m/s: ANS: 2 / 3 b. What is the annual energy that the wind turbine would generate? The only wind the turbine responds to is that between 5 and 15 m/s, during which time the power delivered is 1 kW: E = 2 / 3 x 8760 hr/yr x 1 kW = 5840 kWh/yr V 3  AVG  fV  0    V 3 dV  0.02 V  V 3 0 10  dV  0.02 V 5 5 0 10  0.02  10 5 5  400 c. What is the average power in the wind (W/m 2 )? WE NEED “k” FROM THE f(V) GRAPH: AREA UNDER CURVE = 1, SO k = 1/15 6.5. a. The wind power density (W/m 2 ) at a height of 80 m: From Table 6.3, the friction coefficient  is 0.15, so using (6.15) b. The annual energy delivered with an overall efficiency of 30%: E =  x P WIND x A x  t = 0.30 x 644 W/m 2 x (  /4 x 60 2 ) m 2 x 8760 hr/yr E = 4.79 x 10 9 W-hr/yr = 4.8 x 10 6 kWh/yr c. What would the turbine’s capacity factor be? P  1 2  AV 3  AVG V 3  AVG  V 3 fV  0   dV  V 3  1 15 5 20   dV  1 15  V 4 4 from 5 to 20 = 1 60 20 4  5 4    2656 m / s P  0.5 x 1.225 x 2656 = 1626 W/m 2 V V 0  H H 0         80 10     0.15  1.366 V 80  1.366 x 6 m/s = 8 . 196 m/s P 80  1.91  1 2  A V 3  1.91 x 0 . 5 x 1 . 225 x 1 x 8 . 196   3  644 W / m 2 CF  Energy delivered Rated kWh x 8760 hrs/yr  4.8 x 10 6 kWh/yr 1300 kW x 8760 hr/yr  0.42

ECON | 443 | Renewable Energy Systems | assessment | T5 Solutions 6.1. a. Power fro

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