# Solution_HW6 - Solution HW 6 Wnet = 535 MW Thermal efficiency = thermal = 0.40 Heat in = QH = Wnet thermal = 535 MW 0.4 = 1337.5 MW or MJ/s Heat

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Solution HW 6 W net = 535 MW Thermal efficiency = η thermal = 0.40 Heat in = Q H = W net / η thermal = 535 MW / 0.4 = 1337.5 MW or MJ/s Î Heat out = Q L = Q H – W net = 1337.5 – 535 = 802.5 MJ/s W net = 175,000 MW Option 1: η thermal = 0.32 Cost at \$1,150/kW = (1.75 x 10 8 kW)(\$1,150/kW) = \$ 2.0125 x 10 11 Q H = W net / η thermal = 175,000 MW / 0.4 = 546875 MW Coal required in 5 years = (5.46875 x 10 8 kJ/s) (5x365x24x3600 s) / (2.8 x 10 7 kJ/T) = 3.0979 x 10 9 T Option 2: η thermal = 0.47 Cost at \$1, 520/kW = (1.75 x 10 8 kW)(\$1, 250/kW) = \$ 2.66 x 10 11 Q H = W net / η thermal = 175,000 MW / 0.47 = 372340.4 MW Coal required in 5 years = (3.723404 x 10 8 kJ/s) (5x365x24x3600 s) / (2.8 x 10 7 kJ/T) = 2.0968 x 10 9 T Excess cost of more efficient plants = (\$2.66 - \$2.0125) x 10 11 = \$6.475 x 10 10 Savings in coal over 5 years = 3.0979 x 10 9 - 2.0968 x 10 9 = 9.8288 x 10 8 T

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If the extra cost is to be recovered from fuel savings in 5 years, the cost of coal should be \$6.475 x 10 10 / 9.8288 x 10 8 T = \$ 65.88 per ton Q L = 7.2 kW COP = 1.21 COP R = Q L / W Î W = Q L / COP = 7.2 / 1.21 kW = 5.95 kW Heat pump condenser Fluid R-134a Flow rate = m = 0.045 kg/s Inlet P = 800 kPa, T = 35 0 C = 308.15 K From table A-12 T sat = 31.31 0 C --> superheated state h in = 267.29 + (276.45-267.29) (35 – 31.31)/(40 – 31.31) kJ/kg = 271.18 kJ/kg Outlet P = 800 kPa Saturated liquid
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## This note was uploaded on 11/25/2009 for the course ENGR thermo taught by Professor Staff during the Spring '09 term at University of Louisiana at Lafayette.

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Solution_HW6 - Solution HW 6 Wnet = 535 MW Thermal efficiency = thermal = 0.40 Heat in = QH = Wnet thermal = 535 MW 0.4 = 1337.5 MW or MJ/s Heat

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