Question

1. A sludge containing 4% solid (by weight) needs to be thickened by centrifugation to ≥15% solid for disposal. The centrifuge has a high constant speed and produces a dewatered sludge containing 20% solid. To increase the processing capacity while still meeting the disposal requirement, you decide to bypass (i.e., not dewater) some of the feed sludge and blend the untreated sludge with the dewatered (20% solid) sludge, so as to produce a final sludge with a 15% solid content. Assume all sludge streams have the density of water.

(a) Give the centrate production rate (in L/min) and the percent solid (of the total coming in) discharged via the centrate?

(b) What % of the feed sludge is bypassed (i.e., not centrifuged)?

2. Eutrophication is a natural aging process all lakes undergo as nutrients and organic matter accumulate from their drainage basins. Over time the accumulated nutrients enable increased aquatic growth and stimulate biological activities. The increase in phytoplankton causes the lake to become murky, while decaying organic matter depletes the available dissolved O2. Further accumulation of silt, nutrients, and organic matter causes the eutrophic ("well-fed") lake to become shallower and warmer with more plants growing in the shallow edges. This process transforms lakes into bogs and marshes over time. Natural eutrophication is usually quite slow, often measured in thousands of years before major changes occur. What human activity has done is rapidly accelerate this process (called "cultural eutrophication") by introducing large quantities of nutrients. Sources of nutrients include municipal wastewater, industrial wastes, and runoff from fertilized agricultural land. The introduction of superfluous nutrients disturbs the natural balance of the system, causing excessive algae growth and water quality deterioration. Algal blooms are the unsightly and often malodorous clumps of rotting debris along the shoreline and thick mats of dead organic matter in the lake. A nutrient of major importance that enter fresh water bodies such as lakes from the aforementioned sources is phosphorous. C = 20% C = 15% Q = 100 L/min C = 4% bypass centrifuge C = 0.1% ("centrate") 2 Consider a phosphorus-limited lake with a surface area equal to 8 x 107 m 2 that is being fed by a stream with a flow rate of 15 m3 /s and a phosphorus concentration of 0.01 mg/L. An outfall from a wastewater treatment plant also adds 1 g/s of phosphorus to the lake, but its volumetric flow rate is small and can be ignored. The outflow rate from the lake is the same as the inflow (15 m3 /s). Because phosphorus can attach to suspended particles in lake water, the phosphorus in the lake can also be removed through particle settling (to sediment), which occurs at a rate of 10 m/yr. For the questions below, assume the lake is a cylindrical CSTR.

(a) Draw a diagram to represent the processes involved in phosphorus mass balance. List all sources and sinks. Estimate the steady-state phosphorus concentration in the lake.

(b) An acceptable level of phosphorus in the lake is 0.01 mg/L, above which eutrophication is considered likely. By what percent must the wastewater treatment plant reduce its phosphorous discharge into the lake to achieve this concentration?

3. A textile finishing process involves drying fabric that has been treated with a volatile solvent. The drying process involves evaporation of solvent and removal of solvent vapor by air. The wet fabric entering the dryer contains 50% solvent by weight. Air enters the dryer at a rate of 8 kg per kg of solvent-free fabric. The drying process is 92% efficient; i.e., 92% of entering solvent is removed from fabric and carried out by air. What is the weight % of solvent in the dried fabric? What is the mole % of solvent vapor in the exhaust? The molecular weight of the solvent is 46 g/mol.(Note: Dried fabric still contains some solvent; i.e., dried fabric is not solvent-free).

4. A process heater burns 4.8 kg of natural gas per minute, and the air supply rate is 20% higher than the stoichiometric rate. Exhaust from the heater is sent to a heat exchanger, where water vapor is condensed and removed from the exhaust. Assume natural gas is all methane (CH4), air contains 21% O2 and 79% N2, and the combustion of CH4 is complete, what is the mole % of CO2 in the cooled exhaust?

Answer 1

Basis: 1kg of sludge feed per minute
Fraction of sludge bypassed = x
Sludge to centrifuge, w1t = (1-x)kg
Solids feed w1s= 0.04*(1-x)kg
Solids from centrifuge, w2s = 0.04(1-x)kg
Total weight of sludge from centrifuge = w2t
% of solid from centrifuge 20=0.04(1-x)/w2t*100
Weight of sludge from centrifuge, w2t=0.04(1-x)*100/20=0.2*(1-x)kg
Liquid from centrifuge, w2l = w2t - w2s =0.2(1-x)-0.04*(1-x)=0.16*(1-x)
weight of sludge bypassed = x kg
Solids in bypass w3s = 0.04*x
Liquid in bypass, w3l =x-0.04*x=0.96*x
Total solids in product, w4s = w2s+w3s = 0.04*(1-x)+0.04*x=0.04kg
Total liquid in product, w4l = w2l+w3l=0.16*(1-x)+0.96*x=0.16+0.8*x
Total product weight, w4t = w4s+w4l=0.04+0.16+0.8*x=0.2+0.8*x
% of solid in product,15=100*w4s/w4t=0.04*100/(0.2+0.8*x)
Fraction of bypass, x = 0.083
(a)
production rate of concentrate from centrifuge, kg/min = w2l =0.16*(1-x)=0.14kg/min
Assume density of liquid = 1kg/l
Production rate of concentrate = 0.14L/min
Percent of solid from centrifuge =20%
(b)
% of bypass =x*100=8.3%