Answer:
474.59 mg/L
Explanation:
Given that
BOD = 30 mg/L
Original BOD = 30 mg/L × dilution factor
Original BOD = 30 mg/L × 10 = 300 mg/L
![L_o = \frac{BOD}{1-e^{-5t}}](https://tex.z-dn.net/?f=L_o%20%3D%20%5Cfrac%7BBOD%7D%7B1-e%5E%7B-5t%7D%7D)
here
is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day
![L_o = \frac{300}{1-e^{-5(0.20)}}](https://tex.z-dn.net/?f=L_o%20%3D%20%5Cfrac%7B300%7D%7B1-e%5E%7B-5%280.20%29%7D%7D)
Answer:
The time required is 10.078 hours or 605 min
Explanation:
The formula to apply here is ;
K=(d²-d²₀ )/t
where t is time in hours
d is grain diameter to be achieved after heating in mm
d₀ is the grain diameter before heating in mm
Given
d=5.5 × 10^-2 mm
d₀=2.4 × 10^-2 mm
t₁= 500 min = 500/60 =25/3 hrs
t₂=?
n=2.2
First find K
K=(d²-d²₀ )/t₁
K={ (5.1 × 10^-2 mm)²-(2.4 × 10−2 mm)² }/ 25/3
K=(0.051²-0.024²) ÷25/2
K=0.000243 mm²/h
Re-arrange equation for K ,to get the equation for d as;
d=√(d₀²+ Kt) where now t=t₂
![d=\sqrt{0.024^2+0.000243*t} \\\\0.055=\sqrt{0.024^2+0.000243t} \\\\0.055^2=0.024^2+0.000243t\\\\0.055^2-0.024^2=0.000243t\\\\0.002449=0.000243t\\\\0.002449/0.000243=t\\\\10.078=t\\\\t=605min](https://tex.z-dn.net/?f=d%3D%5Csqrt%7B0.024%5E2%2B0.000243%2At%7D%20%5C%5C%5C%5C0.055%3D%5Csqrt%7B0.024%5E2%2B0.000243t%7D%20%5C%5C%5C%5C0.055%5E2%3D0.024%5E2%2B0.000243t%5C%5C%5C%5C0.055%5E2-0.024%5E2%3D0.000243t%5C%5C%5C%5C0.002449%3D0.000243t%5C%5C%5C%5C0.002449%2F0.000243%3Dt%5C%5C%5C%5C10.078%3Dt%5C%5C%5C%5Ct%3D605min)
no artical shoul be used here