need help for 6.8 and 6.9 ,10 68 mixed if 25 acres of a receives 600 ft problem.

Question

need help for 6.8 and 6.9

,10 68 mixed if 25 acres of a receives 600 ft problem. of steady state 8.5 lblyd and the week at a deten- Draw mate- the (completely per the landfillin substrate 5 ft influent days of The to life 1000 0.5 desired. solid expected h 43,560 l. waste a removal compacted acre the is k of waste removal estimate assuming value municipal a (S is (1 first-order available on still 10 solid the 90% assuming feet, balance rate-constant the recycle) to cell If solve chemostat, are the is space lbyds. for each hours landfill volume 500 diagram is without of 150 in the a depth materials of te rials-balance years average A a density sanitary calculate concentration 69 The volumetric of an ideal plug flow reactor for the following scenario. converted flow rate is 6500 m lday and Species A is according to a fin tordar

Explanation / Answer

6.8 question

SW input= (600ft^3/d) (5d/wk)(52wk/yr)(I yd^3/27ft^3)(500 lb/yd^3)=2.89*10^6 lb/yr

SWIN à

2.89x10^6 lb/y

2.89x10^3 yd3 /y

= 1000 lb/yd^3

10 ft per lift

--------------à

SWOUT = 0

à

SW input 2.89* 10^6 lb/y (1yd^3 / 1000 lb)= 2889 yd^3/ yr

Area =25 ac(43,560 ft^2/ ac) =1.089 x 10^6 ft^2

SW input =2889 *[27/yd^3] =7.8 10^4ft^3/ yr

Volume of one 10 ft cell = 1.089×10^6 ft 10ft = 1.089×10^7 ft^3

Expected life =1.089x10^7 ft^3/ 7.8 x10^4 ft^3/yr =140 yr

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6.9 question

S = (1 0.90 ) 150 mg/l =15mg /L

V = QS -QS + r V

r = kS

[ dS/dt]

V= QS- QS+ k S V

At steady state, the accumulation term goes to zero; and the above equation reduces to the following equation

Q( So S )= k S V

=V/Q =( So S )/ k S = (150 -15) mg/L /0.5h^-1 *(15 mg/l) =18 h

SWIN à

2.89x10^6 lb/y

2.89x10^3 yd3 /y

= 1000 lb/yd^3

10 ft per lift

--------------à

SWOUT = 0

à

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