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matrenka [14]
3 years ago
10

A wastewater is to be disinfected using ultraviolet (UV) light. Batch experiments find that the bacterial concentration decays u

nder UV light with first-order kinetics, with a half of the initial bacteria killed after 6.2 minutes of illumination. You need to design a reactor than can kill 99% of bacteria in a stream having a flow rate of 140, 000 gal/d. Assume that regardless of the reactor type it will be possible to have UV illumination conditions identical to those of the batch experiment. a. Based on the batch experiments, what is the first-order rate constant k for bacterial inactivation
Engineering
1 answer:
Ksivusya [100]3 years ago
7 0

Answer:

k = 0.1118 per min

Explanation:

Assume;

Initial number of bacteria = N0

Number of bacteria IN 'T' time = Nt

So,

Nt=N0e^{-kt}\\\\in\ 6.2 min\\\\\\frac{N0}{2}= N0e^{-k(6.2)}\\\\ln\frac{1}{2} = -k[6.2]

k = 0.1118 per min

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Administrative controls can be broken down into two categories; programs such as
Anastaziya [24]

Administrative controls can be broken down into two categories; programs such as  HAZCOM and activities such as housekeeping is a TRUE statement.

Explanation:

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3 0
4 years ago
A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the
Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + \frac{u^2}{30(\frac{a}{g}\pm G)}     ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5  + \frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}

solve it we get

SSD = 644 ft  

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1  - cos (\frac{28.65 SSD}{Rv}) )        .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1  - cos (\frac{28.65 \times 664}{2294})  )  

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R  = \frac{u^2}{15(e+f)}  ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = \frac{65^2}{15(e+0.10)}

solve it we get

e = 2%

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