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Assignment 4-2013 soln

# Assignment 4-2013 soln - Assignment 4 Solution Capacity of...

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Assignment # 4 Solution: Capacity of plant = 2300 MW Thermal efficiency ( Heating value of Natural gas = 975 Input rate of gas = output rate of gas, hence Or, Amount of gas required = 20.66 MMSCf / Hr

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Heating Value of Oil = 18250 BTU / lb This gives, Amount of oil required = 500745.8334 Kg / Hr = 1102312.67 lb / Hr Hence Oil required in Gallons per hour = ( ) KGPH = 139.239 KGPH Now Heating value for Coal = 10000 Btu / lb This gives, = 012758.78 Kg / Hr = 1004 ton / Hr (Emissions in lb / Hr) Natural Gas Residual Oil Coal Particulate matter 103.3 1392.39 60240 1136.3 14620.095 21786.8 12.396 10930.2615 68673.6 VOC 28.924 10582.164 60.04
Solution: (a) Height (Ft) Height (m) Temperature (°F) Temperature °C Lapse rate Height (m) Temperature °C 0 0 60 15.56 1°C / 100 m 0 20 1200 365.76 69 20.56 5.4° F / 1000 Ft 365.76 16.3424 Now, ( ) ( ( The condition is strongly stable (F). (b) Height (Ft) Height (m) Temperature (°F) Temperature °C Lapse rate Height (m) Temperature °C 0 0 55 12.78 1°C / 100 m 0 20 900 274.32 51 10.56 5.4° F / 1000 Ft 274.72 17.2568 0 50 100 150 200 250 300 350 400 0 5 10 15 20 25 Adiabatic lapse rate Environmental lapse rate

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Now, ( ) ( ( The condition is stable (E or F).
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