Problem 9.83 Waste Incinerator The wastewater treatment plant at the Ossabow Pap
ID: 701855 • Letter: P
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Problem 9.83 Waste Incinerator The wastewater treatment plant at the Ossabow Paper Company paper mill generates about 24.0 tonnes of sludge per day The consistency of the sludge is 35.0%, meaning that the sludge contains 35.0 wt% solids and the balance liquids. The mill currently spends $40.0/tonne to dispose of the sludge in a landfil to generate useful energy and to eliminate the enviromental problems associated with landfill disposal. A flowchart for a preliminary design of the proposed sludge-treatment process follows to 75.0%, the sludge could be incinerated (burned) Water vapor satd 1 atm Sludge 24.0 350% consistency A Concentrated sludge -75.0%consistency 75.0 % c INCINERA Hot waste T = 300 gas Preheated air T=170.0 °C (saturated) a, a, Saturated Steam P 5.00 bar Natural gas 90.0 mole% CH 10.0 mole% T= 25.0 Boiler feedwater BOILER (62.0 % efficient) 7: 25.0 T=20.0 °C Hot combustion Q products ó, (kJld) preheat boiler furnace air | Q1 (kJ/d) generate steam in boiler | | Q2 (kJ/d) evaporate sludge water ! |03 (kld) heat loss from dryer 04 (kJ/d)--preheat incinerator air #6 fuel oil T=65.0 °C Preheated air |T= 125cc 7-25.0 BOILER AIR a, PREHEATERExplanation / Answer
In dryer,
Inlet 35% sludge flowrate =24 tonne/day
=>Total sludge quantity = .35*24 = 8.4 tonne/day
=>Total water quantity = (24-8.4) = 15.6 tonne/day
Outlet 75% sludge flowrate = F
Since, Mass of solid sludge in Inlet = Mass of solid sludge in outet
=>.35*24=F*.75
=>F=11.2 tonne/hr
=> Total water boiled in dryer = 24-11.2 = 12.8 tonne/hr
Specific heat of sludge = 2.5 KJ/Kg/C
Specific heat of water = 4.18 KJ/Kg/C
Drying occurs at 1 atm
Boiling Point of water at 1 atm = 100 C
=> 1st whole 35% slude mass temperature will rise to 100 C and then required water will evaporate to make 75% slude
Dryer inlet Temperature = 30C
Heat required to rise temperature = Mass flowrate*Specific heat*Change in Temperature
Heat require to boil water = Boiler water flowrate*Latent Heat
Latent heat of water at 1 atm = 2250 KJ/kg
=> Total Heat required for Dryer Process,Q2= 8.4*10^3*2.5*(100-30)+15.6*10^3*4.18*(100-30)+12.8*10^3*2250 =34834560 KJ/day
Efficiency of steam used = 55%
Using steam table, latent heat of stream at 5 bar = 2114 KJ/kg
Therefore, Steam required B, = 34834560 /.55*2114 = 29.96 ton/day
Heat lost to surrounding,Q3 = 2114*29.96*10^3-34834560=28500880 KJ/day
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