수질오염개론수질환경기사 · 2015년05월31일 · 14/100
14.아래와 같은 폐수의 생물학적으로 분해가 불가능한 불용성 COD는? (단, BODU/BOD5=1.5, COD=1583mg/L, SCOD=948mg/L, BOD5=659mg/L, SBOD5=484mg/L 이다.)
1
816.5 mg/L
2
574.5 mg/L
3
372.5 mg/L정답
4
235.5 mg/L
해설
1. BDCOD(BOUu)= BOD5 x 1.5 = 988.5mg/L 2. NBDCOD COD = BDCOD(BODu) + NBDCOD 1583mg/L = 988.5mg/L + NBDCOD = 594.5mg/L 3. ICOD COD = ICOD + SCOD 1583mg/L = ICOD + 948mg/L =635mg/L 4. BDSCOD = SBOD5 x 1.5 = 484 x 1.5 = 726mg/L 5. NBDCOD SCOD = BDCOD + NBDCOD 948 = 726 + NBDCOD NBDCOD = 222mg/L 6. NBDCOD = NBDICOD + NBDSCOD 594.5 = NDBICOD + 222 NBDCOD = 372.5mg/L [추가 해설] COD= BDCOD + NBDCOD = (BDSCOD + BDICOD) + (NBDSCOD+NBDICOD)= BODu + (NBDSCOD+NBDICOD) 문제에서 물어보는 건 NBDICOD 따라서 NBDICOD = COD - BODu - NBDSCOD를 구하면 됨 1. BODu = 1.5 * 659 = 988.5 2. NBDSCOD = SCOD - BDSCOD = SCOD - SBODu = 948 - (1.5*484) = 222 즉, NBDICOD = 1583 - 988.5 - 222 = 372.5