A waste treatment plant operating under steady state conditions a waste liquor c

Question

A waste treatment plant operating under steady state conditions a waste liquor containing 500 parts per million (ppm )of a noxious impurity . The impurity is removed with practically no loss of the carrier fluid . The treating process can remove the impurity down to a level of 10 ppm . By local ordinance , a maximum of 100 ppm is allowed in the discharge of the processed effluent to a nearby river . What fraction of the waste liquor must be sent through the waste treatment unit, and what fraction may be bypassed?

Explanation / Answer

basis : 1 miilon =106 gm of water waste water. It contains 500 gm impurity

let x= gm of water sent to trreatment plant then 106-x= gm of water sent without treatment

the treated water contains 10 gm per 106 gm of water

106 gm of water contains 10gm

x gms contains x*10/106

rest of 106-x gm contains 500 ppm

106 gm contains 500 gm

106-x gm containsn (106-x)*500/106 gm of waste

the treated water contains 100ppm

since the basis is 106 waste , waste contains 100 gm per 106 gms

waste in untreated water- waste in treated water= waste in the resulting water

(106-x)*500/106 -x *10/106 = 100/106

(106-x)*500- 10x = 100

106*500-100= 510x

x= 9.83092*105

fraction sent = 9.83092*105/106 =0.983092

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