a) Power = Incident power flux x Area b) if the panel is 10% efficient, then the actual power will be P actual = Pideal X efficiency

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1 10 Renewable Energy Technology 10.1 A solar cell whose active surface is 100 mm by 100 mm is exposed to a light intensity of 1000 walts per square metre. (a) Calculate the power incident on the cell. (b) When the cell is operating at 10% efficiency, what electrical power is it supplying? (c) If the cell maintains a voltage of 0.5 V, what current is it delivering? Ans [1 OW, 1 W, 2 Amps] a) Power = Incident power flux x Area = 1000 x (0.100 x 0.100) = 10W b) if the panel is 10% efficient, then the actual power will be P actual = Pideal X efficiency = 10xO.1 = 1 W c) Electrical Power = voltage x current hence current = Power / voltage = 1 /0.5 = 2 Amps.

2 10.2 A site receives 600 walts per square meter of solar radiation in July. Assume that the solar panels are 10% efficient and that the average sunny day is 6 hours. a. How many square meters would be required to generate 10 kwh of electricity. b. If a typical solar panel is 1.2 m 2, how many panels will be needed? ANS( 27.8 m 2, 24 units) Solution: a) Energy output = incident energy x panel efficiency Hence Each square meter gives you 600 xo.1 = 60 W/m2 in 6 hours you would get 6 x 60 = 360 walt-hours per square meter In order to get 10 kwh Area needed = = 27.8 square meters of collecting area b) The number of panels needed = = panels., say 24 panels.

3 10.3 Calculate the power delivered by the wind, in walts per square metre, for wind speed over a range from 1 to 20 metres per second. (Assume the density of air at normal pressure to be 1.23 kg per cubic metre.) Ans [1.6 kw/m2 at 20 m/s 1 The actual power output is calculated using p = C p * 1/2 * p * A * V 3 Assume that Cp=O.4 A = 1 m 2 1"]gb = 1 1"]gen = 1 * * 1"]gb 1"]gen Hence P = 0.2 X V 3 The power output (in W/m 2 ) for each speed is given in the table below Wind Speed Turbine Output P = 0.2 V' (m/sec) (W/m2)

4 10.4 The rotor diameter of a two-bladed HAWT is 7.0 metres. At its rated wind speed of 12.1 metres per second, the power output of the turbine is 15 I<W. (a) Find the efficiency of the turbine at this wind speed. (The density of air is 1.23 kg per cubic metre.) (b) Calculate the tip-speed ratio, if the blades are rotating at 240 rpm. Ans [0.357, 7.27] a) The actual power output is given by: P = C p * 1/2 * p * A* V 3 *1"]gb *1"]gen A = (1t/4) X 0 2 = m 2 Assume that both gear box and generator are 100% efficient, 1"]gb = 1 1"]gen = 1 Hence Cp = PI ( 112 * p * A * 1"]gb * 1"]gen X V 3 ) = 15 I ( 0.5 x 1.23 x x 1 x 1 X ) = b) to calculate the Tip speed ration, first convert the rotor speed into meters per second units. Vr = 2 1t r. N = 2 1t (7/2) x (240/60) = 88 mls Hence Tip Speed ratio = (Vr I V) = = 7.27

5 10.5 A two-bladed horizontal axis wind generator has a rotor diameter of 20 metres. (a) Operating at a wind speed of 8.1 metres per second, the rotor extracts 35% of the energy of the wind. If the efficiency of the generator which it drives is 85%, find the electrical power output in kilowatts. (The density of air is 1.23 kg per cubic metre.) (b) The output power of a turbine is equal to the torque exerted by the blades multiplied by the angular velocity (which is approximately 0.1 times the rpm). If the above turbine is rotating at 60 rpm, what is the torque? Ans [ 30.5 I<W, 4.8 knm 1 a) The actual power output is given by: P = C p * 1/2 * p * A* v 3 *1"]gb *1"]gen A = (1t/4) X 0 2 = m 2 The efficiency of the Turbine includes 1"]gb,1"]gen and Cp = 0.35 Hence the output power is calculated: P = 0.35*0.85 * 112 * 1.23 * * (8.1)3 = 30.5 kw b) P = torque x angular speed N = 60 rpm Nx = 0.1 x N = 6 rpm Hence Vr = 2 1t r. N = 2 1t (20/2) x (6160)= mls And Torque = Power 1 speed = W mls = 4.8 knm.

6 10.6 The Dinorwig pumped storage system in Wales can store 7.2 million cubic metres of water at an average height of 500 metres above its lower reservoir. (a) Calculate the maximum energy stored. (b) The water is pumped up overnight by six turbines operating as pumps, each consuming 281 MW. If the energy conversion efficiency of the system when pumping is 94%, estimate the time required to pump the full 7.2 million cubic metres of water. (c) The maximum output of each of the six generators of the system is 306 MW. If this output is produced when the height difference between the upper and lower reservoirs is 510 metres and the total flow rate to all six turbines is 390 cubic metres a second, what is the overall energy conversion efficiency? (d) How many kilowatt-hours are 'lost' in a complete cycle of storage and generation, if the above results represent the average efficiencies of pumping and generating? (e) The Dinorwig turbines rotate at 500 rpm. Calculate the specific speed under the above generating conditions. What type of turbine would you expect them to be? Ans: (9.8x1 0 6 MWh, 5.5h, 94%, 0.589x1 06MWh, 112 mixed turbine) a) PE = m g h = (1000x7.2 X106) x 9.81 x 500 = X J divide the above by 3.6x1 0 6 to get PE = 9.81 x 10 6 MWh b) P = p Q g H 1"] hence Q1 = P 1 1 P g H 1"] = (281 x 10 6 ) I ( 1000 x 9.81 x 500 x 0.94) = m 3 Is To empty the whole 7.2 m 3 between 6 such turbines it will take a time of T = (7.2 x1 06)/( 6 x ) = s = 5.5 hours. c) P=pQgH1"] 1"] =P/pQgH hence = (306x1 0 6 ) I [10 3 x (390/6) x 9.81 x 510] = 0.94 or 94 %

7 d) energy lost = 6% of (a) = 0.06 x 9.8 x1 0 6 =0.589 x1 0 6 MWh e) Ns = N = N JP = 500,/281x1000 = 112 'H The most suitable turbine is Mixed turbine. Radial Francis l\iixe,i.:'.. xial N s fol" yarious types (If tml'lnes mmllhllnps I I I I I I Q m3 is N revimill H m

8 10.7 (a) Find the volume flow rate required to deliver a power of 2.25 MW with each of the following effective heads: 50 m, 250 m and 1000 m. (b) Assuming no energy losses, calculate the speed at which water emerges from a jet after falling through each of the above heads. (c) Use the results of (a) and (b) to find the jet diameter required to deliver the 2.25 MW from each head. Ans[ a( 4.587,0.917,0.229 m 3 /s), b(31.3,70.0, 140 m/s), 0.432, 0.129, 0.046m)] a) Power Output = Efficiency x Water input power hence Q = P = 2.25xlO' = m 3 1 s IhxpxgxH lxlooox9.81x50 Repeat for other heads At 250m Q = m 3 /s At 1 OOOm Q=0.229 m 3 /s b) V = sqrt ( 2xg xh) at 50m, V = 31.3 m/s at 250m, V = 70 m/s at 1000m, V = 140 m/s c) Area = Q 1 V is used to calculate the jet diameter dj ( from Aj = (1t/4)x df At 50 m, Aj = 4.587/31.3 = m 2, hence dj = 0.432m At 250m, Aj = 0.917/70 = m 2, hence dj = 0.129m At 250m, Aj = 0.229/140 = m 2, hence dj = 0.046m

Wind Turbine Power Calculations

Wind Turbine Power Calculations RWE npower renewables Mechanical and Electrical Engineering Power Industry INTRODUCTION RWE npower is a leading integrated UK energy company and is part of the RWE Group,