16. A portion of wastewater sample was subjected to standard BOD test (5 days, 20°C), yielding a value of 180 mg/1. The reaction rate constant (to the base 'e') at 20°C was taken as 0.08 per day. The reaction rate constant at other temperature may be estimated by k_r = (k 20 (1.047)^T-20. The temperature at which the other portion of the sample should be tested, to exert the same BOD in 2.5 days, is

1. 1. 4.9°C

   
   
   

   2. 24.9°C

   
   
   

   3. 31.7°C

   
   
   

   4. 35.0°C

   
   
   

---

1. View Hint View Answer Discuss in Forum

   BOD₅²⁰ = 180 mg/L = (L₅²⁰)
   BOD₅²⁰ = BOD_2.5^T = 180
   L₅²⁰ = L₀ [1 – e^–kt]
   

   L0 =	L520	=	L2.5T
   	1 – e–kt		1 – e–kt

   
   180 = L₀ (1 – e^{–k₂₀×5}) = L₀(1 – e^–k_T×2.8)
   Upon simplifying, we get,
   –k₂₀ × 5 = – k_T × 2.5
   k_t = k₂₀ (1.04T)^T–20
   ∴ 0.18 × 5 = 0.18 × (1047)_T-20 × 2.5
   ∴ T = 35°C

   Correct Option: D

   BOD₅²⁰ = 180 mg/L = (L₅²⁰)
   BOD₅²⁰ = BOD_2.5^T = 180
   L₅²⁰ = L₀ [1 – e^–kt]
   

   L0 =	L520	=	L2.5T
   	1 – e–kt		1 – e–kt

   
   180 = L₀ (1 – e^{–k₂₀×5}) = L₀(1 – e^–k_T×2.8)
   Upon simplifying, we get,
   –k₂₀ × 5 = – k_T × 2.5
   k_t = k₂₀ (1.04T)^T–20
   ∴ 0.18 × 5 = 0.18 × (1047)_T-20 × 2.5
   ∴ T = 35°C

---

17. The following data are given for a channel-type grit chamber of length 7.5 m.
    (i) flow through velocity = 0.3 m/s
    (ii) the depth of wastewater at peak flow in the channel = 0.9 m
    (iii) specific gravity of inorganic particles = 2.5
    (iv) g = 9.80 m/s², m = 1.002 × 10^–3 N/s/m² at 20°C, pw 1000 kg/m³ Assuming that the stokes is valid, the largest diameter particle that would be removed with 100 parent efficiency is

1. 1. 0.01 mm

   
   
   

   2. 0.21 mm

   
   
   

   3. 1.92 mm

   
   
   

   4. 6.64 mm

   
   
   

---

1. View Hint View Answer Discuss in Forum

   v =	μ	=	1002 × 10-3
   	l		1000

   
   = 1.002 × 10^-6 m²/s
   

   Vs =	g	(s-1)d²
   	18v

   
   

   0.036 =	9.81	(2.5-1) ×	d²
   	180		1.002 × 10-6

   
   ∴ d = 0.21 mm

   Correct Option: B

   v =	μ	=	1002 × 10-3
   	l		1000

   
   = 1.002 × 10^-6 m²/s
   

   Vs =	g	(s-1)d²
   	18v

   
   

   0.036 =	9.81	(2.5-1) ×	d²
   	180		1.002 × 10-6

   
   ∴ d = 0.21 mm

---

---

18. Bulking sludge refers to having

1. 1. F/M < 0.3 /d

   
   
   

   2. 0.3/ d < F/M < 0.6/d

   
   
   

   3. F/ M = zero

   
   
   

   4. F/ M > 0.6/d

   
   
   

---

1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   NA

---

19. In a certain situation, wastewater discharged into a river,mixes with the river water instantaneously and completely. Following is the data available:
    Wastewater: DO = 2.00 mg/L
    Discharge rate = 1.10 m³/s
    River water DO = 8.3 mg/L
    Flow rate = 8.70 m³/s
    Temperature = 20°C
    Initial amount of DO in the mixture of waste and river shall be

1. 1. 5.3 mg/L

   
   
   

   2. 6.5 mg/L

   
   
   

   3. 7.6 mg/L

   
   
   

   4. 8.4 mg/L

   
   
   

---

1. View Hint View Answer Discuss in Forum

   DO of mixture is given by
   

   DOmix =	As × DOs + AR × DOR
   	As + AR

   
   

   =	(1.1 × 2) + (8.7 × 8.3)
   	1.1 + 8.7

   
   = 7.6 mg/l

   Correct Option: C

   DO of mixture is given by
   

   DOmix =	As × DOs + AR × DOR
   	As + AR

   
   

   =	(1.1 × 2) + (8.7 × 8.3)
   	1.1 + 8.7

   
   = 7.6 mg/l

---

---

20. A circular primary clarifier processes an average flow of 5005 m³/d of municipal wastewater. The overflow rate is 35 m³/d. The diameter of clarifier shall be

1. 1. 10.5 m

   
   
   

   2. 11.5 m

   
   
   

   3. 12.5 m

   
   
   

   4. 13.5 m

   
   
   

---

1. View Hint View Answer Discuss in Forum

   Area =	Q	=	5005	= 143 m²
   	OFR		35

   
   

   	π	× D² = 143m²
   	4

   
   ∴ D = 13.5 m

   Correct Option: D

   Area =	Q	=	5005	= 143 m²
   	OFR		35

   
   

   	π	× D² = 143m²
   	4

   
   ∴ D = 13.5 m

---