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**Q. 1 The ultimate BOD (L _{0}) a wastewater sample is estimated as 87% of COD. The COD of this wastewater is 300 mg/L. Considering first order BOD reaction rate constant k (use natural log) = 0.23 per day and temperature coefficient Î¸ = 1.047, the BOD value (in mg/L, up to one decimal place) after three days of incubation at 2 7 Â°C for the w astew ater will b e _______ [2018 : 2 Marks, Set-I]**

Ultimate BOD = 0.87

COD = 0.87 x 300 = 261 mg/l

For municipal sewage, at standard temperature, value of k((base e) = 0.23 per day. Thus, value 0.23 per day given is w.r.t. the standard temperature of 20Â°C.

= 0.23 (1,047)^{27-20}

= 0.317 days^{-1}

BOD_{3} = 261 (1 - e^{-0.317 x 3})

= 160.226 mg/l

**Q. 2 For a given water sample, the ratio between BOD _{5-day}, 2oÂ°c and the ultimate BOD is 0.68. The value of the reaction rate constant k (on base e) (in day^{-1}, up to two decimal places ) is [2017 : 2 Marks, Set-II]**

**Q. 3: Two wastewater streams A and B, having an indentical ultimate BOD are getting mixed to form the stream C. The temperature of the stream /A is 20Â°C and the temperature of the stream C is 10Â°C. It is given that **

**the 5-day BOD of the stream A measured at 20Â°C = 50 mg/l****BOD rate constant (base 10) at 20Â°C = 0.115 per day****temperature coefficient = 1.135**

**The 5-day BOD (in mg/l, up to one decimal place) of the stream C, calculated at 10Â°C , is [2017 : 2 Marks, Set-I]**

**Ans:** 21.21mg/l

For stream A,

For stream C,

**Q. 4 For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of is estimated as 200 mg/l. Taking the value of the first order BOD reaction rate constant as 0.22 day ^{-}^{1}, the five-day BOD (expressed in mg/l) of the wastewater at incubation temperature of would be______ . **

From equation (i) and (ii), we get,

**Q. 5 : The 2-day and 4-day BOD values of a sewage sample are 100 mg/l and 155 mg/l, respectively. The value of BOD rate constant (expressed in per day) is________ . [2016 : 2 Marks, Set-I]****Ans: **0.299

1.55 - 1,55x = 1 - x^{2}

x^{2} - 1.55x + 0.55= 0

**Q. 6: Ultimate BOD of a river water sample is 20 mg/L. BOD rate constant (natural log) is 0.15 day ^{-1}. The respective values of BOD (in %) exerted and remaining after 7 days are: [2015 : 2 Marks, Set-II]**

BOD remaining after 7 days

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