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3 years ago

9

A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b

iodegradable contaminant. At steady state, the effluent from the lagoon must not exceed 20 mg/L of biodegradable contaminant. Assuming the lagoon is well mixed and that there are no losses or gains of water in the lagoon other than the sewage input, what biodegradation reaction rate coefficient (d^-1) must be achieved for a first-order reaction?

1 answer:

Marysya12 [62]3 years ago

6 0

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= $\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}$

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= $\frac{8640*100-8640*20}{20*10^5}$

k= 0.35 per day

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Explanation:

Given that

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We know that ρ of water is 1000 $\dfrac{kg}{m^3}$

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What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its

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<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil ( $N_{p}$ ) to the number of turns in its secondary coil ( $N_{s}$ ) is equal to the ratio of the input voltage ( $V_{p}$ ) to the output voltage ( $V_{s}$ ) of the transformer. i.e

$\frac{N_p}{N_s}$ = $\frac{V_p}{V_s}$ ----------------(i)

<em>From the question;</em>

$N_{p}$ = number of turns in the primary coil = 8 turns

$N_{s}$ = number of turns in the secondary coil = 515 turns

$V_{p}$ = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

$\frac{8}{515}$ = $\frac{117}{V_s}$

<em>Solve for </em> $V_{s}$ <em>;</em>

$V_{s}$ = 117 x 515 / 8

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3 years ago

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7.4 A pretimed four-timing-stage signal has critical lane group flow rates for the first three timing stages of 200, 187, and 21

Irina18 [472]

Answer:

16 seconds

Explanation:

Given:

C = 60

L = 4 seconds each = 4*4 =16

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200, 187, and 210 veh/h.

We are to find the estimated effective green time of the fourth timing stage. The formula for the estimated effective green time is:

$g = (\frac{v}{s}) (\frac{C}{X})$

Let's first find the fourth stage critical lane group ratio $\frac{v}{s}$ , using the formula:

$C = \frac{1.5L +5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{1.5*16 + 5}{1 - ( \frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800}) + ( \frac{v}{s})}$

$60 = \frac{24+5}{1 - (0.332 + ( \frac{v}{s}))}$

Solving for $(\frac{v}{s})$ , we have:

$(\frac{v}{s}) = 0.185$

Let's also calculate the volume capacity ratio X,

$X = (\frac{200}{1800} + \frac{187}{1800} + \frac{210}{1800} + 0.185)(\frac{60}{60-16}$

X = 0.704

For the the estimated effective green time of the fourth timing stage, we have:

$g_4 = (\frac{v}{s}) (\frac{C}{X})$

Substituting figures in the equation, we now have:

$g_4 = (0.185) (\frac{60}{0.704})$

$g_4 = 15.78 seconds$

15.78 ≈ 16 seconds

The estimated effective green time of the fourth timing stage is 16 seconds

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