ECE370 POWER & ENERGY SYSTEMS Homework Set 3 Solutions

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1 ECE37 POWER & ENERGY SYSTEMS Homework Set 3 Solutions 3.1 a) A coal-fired power plant employs a condenser that extracts 16 MJ for each kg of burned fuel. A neighboring oil-fired power plant employs a condenser that extracts 17 MJ for each kg of burned fuel. Calculate the mechanical energy of their turbines and their ideal efficiencies for: i) Petroleum with TEC = 4 MJ/kg ii) Coal with TEC = 24 MJ/kg b) Which plant will produce cheaper electricity if petroleum costs 75 /kg and coal costs $25 per tonne if both generating units have the same efficiency (equal ηgen). a) i) For Petroleum: WPet = 4 17 = 23 MJ/kg. W Pet Q 1 23 η ideal = = = 57.5 %. 4 ii) For Coal: WCoal = = 8 MJ/kg. W Coal Q 1 8 η ideal = = = 33⅓ %. 24 b) 1 kg of petroleum produces 23 MJ and costs 75, so the energy cost is /MJ. Coal costs $25/tonne, which is 25 /kg. 1 kg of coal produces 8 MJ and costs 25, so the energy cost is /MJ. Coal is 95.8% the price of Oil. (Coal is 4 % cheaper.)

2 3.2 a) Calculate the mass of U 235 required to produce an average of 75 MW of electricity annually, if the plant is 33⅓% efficient. b) Compare this with the equivalent mass of coal for a 25% efficient plant. c) Determine the ratio of volume of coal : volume of uranium. Assume: ρur = 21 x 1 3 kg/m 3, ρcoal = 4¼ x 1 3 kg/m 3, TECcoal = 6. kwh/kg Since one joule requires 31 x 1 9 fission events, one watt will require 31 x 1 9 fission events per second. Then 75 MW requires 75 x 1 6 x (31 x 1 9 ) = x 1 18 fission events per second. 75 MW for 1 h requires 36 x (23.25 x 1 18 ) = 83.7 x 1 21 fission events. 75 MW for 1 year requires 876 x (83.7 x 1 21 ) = x 1 24 fission events. The number of annual fission events is 33.2 x 1 24 if the plant is 1% efficient. For 33⅓% efficiency this becomes 2.2 x Since 1 kg of U 235 can have 2.54 x 1 24 fission events, the fuel needed for the reactor annually is: x 1 a) Mass of U 235 annually = 24 = 866 kg 2.54 x 1 b) If 1 kg of coal contains 6. kwh of thermal energy and the plant is 25% efficient, we only get 1.5 kwh/kg as output. To produce 75 MW of thermal power for one hour, we need to burn 5 x 1 3 kg of coal. To produce 75 MW of thermal power annually, we need to burn: 876 x 5 x 1 3 = 4.38 x 1 9 kg of coal. c) Mass of coal/mass of uranium = 5.6 x 1 6 olume Coal olume Uranium = 5.6 x 1 6 x = 25 x 1 6 : 1.

3 3.3 A 25 ka, 36/24, single-phase transformer has the following test data: oltage () Current (A) Power (W) O/C Test S/C Test Use the approximate equivalent circuit to calculate: a) The voltage regulation and efficiency when the load takes 11 A at 22 and.6 lag pf. (NOTE: this is not rated load). b) The voltage regulation and efficiency at rated load conditions and.8 lag pf. From the OC Test: 4985 pf oc = =.359 lag then: 24 x θ oc = cos = 69 Ioc = = 2.77 j54 A R c = = Ω and: X m = = Ω These values are referred to the lv side. Referring to the hv side gives: Rc = 26 Ω and Xm = 1 Ω From the SC Test: 4823 pf sc = =.3714 lag then: 187 x θ sc = cos = 68.2 I 2 = A 187 Z = = Ω then convert to rectangular to get: R = 1. Ω and X = 2.5 Ω, which are referred to the hv side a) I2 = A I 2 = 73⅓ A 2 = 22 2 = 33 p = 33 (1. j2.5) x 73⅓ = I = ' p I2 R P c p jx m R= x1%= 5.79% = = A 26 j1 Pin = Re{ x } = kw and Pout = 22 x 11 x.6 = kw η = x1% = 93.5% 155.3

4 25 x 1 b) I2 = cos 1.8 = I2 A I 2 = a = A p = 36 (1. j2.5) x = R = x1%= 4.47% 36 I = ' p I2 R P c jx p m = = A 26 j1 Pin = Re{ x } = 21.3 kw and Pout = 25 x 1 3 x.8 = 2 kw 2 η = x1% = 95.1% 21.3

5 3.4 A hydroelectric development has two identical powerhouses each one has three penstocks and each penstock passes 2 m 3 /s of water with velocity 35 m/s when the average head behind the dam is m. The generators operate at.9185 lag pf and the electricity is transmitted at 345 k on three parallel transmission lines. a) Calculate the penstock efficiency. b) Assuming the coefficient of performance of the turbine is.6 and the generator efficiency is 95¼%, what is the total generated real electrical power for the development? c) Calculate the magnitude of the current in the individual transmission lines. a) Since: ηpen = 2 v 2gH 2 35 = = 75%. 2 x 9.81x b) Pelec = ηgen CP Pmech and mratev 2 x 1 x 35 Pmech = = = MW / Penstock 2 2 Then: Pelec =.9525 x.6 x = 7 MW for each penstock, resulting in: Ptotal = 6 x 7 = 42 MW. for the development. 42 c) The Apparent Power is given by: S = = MA.9185 The total current is: x 1 I = = A 3 3 x 345 x 1 A third of this flows in each circuit giving: 255 A/circuit.

6 3.5 You have a requirement for a three-phase transformer to handle 1 MA with a voltage ratio of 2.4 k: k k. You are provided with three single-phase transformers each rated MA with voltage ratios of k : 37.4 k. a) Draw a diagram showing how each transformer should be connected. Be sure to indicate the voltage applied to each winding. The tolerance on voltage is.25 k. b) By what percentage is the bank underloaded or overloaded? a) Given: a = = and required a '= = a Then = 3 a' Y required. The configuration is shown below. 6 1 x 1 b) For k I L = = 463 A 3 3 x x 1 Transformer primary rating x 1 I S = = A 37.4 x Load Ratio = = Transformers are 9.9% overloaded.

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