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## User: pragya23121997

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#### Posts by pragya23121997

<prev • 71 results • page 1 of 8 • next >
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... **1)the siphonage**:- Due to siphonage action (due to pressure) the water seal of the trops gets broked up. It is called as siphonage. It happens when the water is sudenly discharged from any fixture on the upper floor. **2)Anti-siphonage**:- To provide protection and to remove difficult a separa ...
written 19 hours ago by pragya231219970
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... **Subject**:Environmental engineering-II **Topic**:House drainage & Environmental sanitation **Difficulty**:Medium ...
written 19 hours ago by pragya231219970
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... **given**:- 5 day 20°C BOD =350mg/l $K_{20}=0.1/d$ 1)To find organic matter present(L) at the start of BOD:- $Y_t=L[1-(10)^{-K_{20}*t}]$ Where $Y_t$:total amount of organic matter oxidised in t days =320mg/l L= organic matter present at the start of BOD=? $t=5 day K_{20}=0.1$ $350=L[1-(10)^{ ... written 4 days ago by pragya231219970 • updated 4 days ago by Sanket Shingote ♦♦ 220 1 answer 47 views 1 answer ... If 5 day BOD of waste water is 350mg/l what will be it's 7 day 25°C BOD?$K20=0.1/day$. Assume data if required: --- **Subject**: Environmental Engineering-II **Topic**: Secondary treatment method **Difficulty**: Medium ... written 5 days ago by pragya231219970 1 answer 42 views 1 answers ... **Given**:- population equivalent =7000 loading rate=0.09 kg/capita/day volatile solids in raw sludge =70% moisture content of raw sludge-=96% digestion period=25 days$P_2$= volatile solids resolution during digestion = 50%$P_2$=moisture content of digested sludge=92% storage period re ... written 5 days ago by pragya231219970 1 answer 42 views 1 answer ... **Subject**:-Environmental Engineering-II **Topic**:-Secondary treatment method **Difficulty**:High Find the volume of digester for population equivalent -7000, loading rate-0.09 kg/capita/day volatile,solids in raw sludge -70%, moisture content of raw sludge- 96%, digestion period-25 days, volat ... written 6 days ago by pragya231219970 1 answer 25 views 1 answers ... **Given**:- 1)sewage flow:4MLD 2)Recirculation ratio:1:5 3)$BOD_5$of raw sewage:280 mg/l 4)BOD removal in primary clarifier=25% 5)Final effluent$BOD_5$desired=20mg/l 1)To find efficiency of filter(n):- Total BOD present in raw sewage=4 ml *280 mg/lit=1120kg BOd removed in primary tank=25% ... written 6 days ago by pragya231219970 1 answer 25 views 1 answer ... **Subject**:-Environmental Engineering-II **Topic**:-Secondary treatment method **Difficulty**:High Determine the size of high rate trikling filter for following data 1)sewage flow:4MLD 2)Recirculation ratio:1:5 3)$BOD_5$of raw sewage:280 mg/l 4)BOD removal in PST=25% 5)Final effluent$BOD_ ...
written 6 days ago by pragya231219970
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... 1. It represents the concentration of the sludge in the system which helps in controlling rate of return of sludge. 2. Inorder to maintain the desired MLSS concentration corresponding to a particular $\frac{R}{M}$ ratio to achieve the requird degree of treatment. ![][1] 3. Sludge Volume ...
written 6 days ago by pragya231219970
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