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evablogger [386]
3 years ago
15

Twenty six million gallons per day of wastewater with a DO of 1.00 mg/L is discharged into a river with a DO of 6.00 mg/L. If th

e flow rate of the river is 165 x 106 gal/d and saturation value of dissolved oxygen is 9.17 mg/L, what is the oxygen deficit he two flows? (0 Points)
Engineering
1 answer:
laiz [17]3 years ago
8 0

Answer:

oxygen deficit = 3.851 mg/L

Explanation:

given data

flow rate of the river=  165 × 10^{6} gal/d

saturation value of dissolved oxygen = 9.17 mg/L

to find out

oxygen deficit he two flows

solution

we will apply here formula for dissolved oxygen content after dilution is

Do mix = \frac{Qw*(Do)w +Qr*(Do)r}{Qw+Qr}     ..........................1

here Qw is rate of flow of waste water  i.e 26 ×10^{6} gal/d

(Do)w is Do of waste waster i.e 1 mg/L

Qr is aret of flow of river i.e 165 ×10^{6} gal/d

(Do)r is do of river water i.e 6 mg/L

so put all value in equation 1 we get

Do mix = \frac{26*10^6*1 +165*10^6*6}{(26+165)10^6}

solve we get

Do mix = 5.319 mg/L

so

oxygen deficit =  saturation oxygen - (Do) mix     ..............2

oxygen deficit = 9.17 - 5.319

oxygen deficit = 3.851 mg/L

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alukav5142 [94]

Answer:

Work done = 125π J

Explanation:

Given:

P = P_i * ( 1 - (x/d)^2 / 25)

d = 5.0 cm

x = 5 * d cm = 25d

Pi = 12 bar

Work done = integral ( F . dx )

F (x) = P(x) * A

F (x) =  (πd^2 / 4) * P_i * (1 - (x/d)^2 / 25)

Work done = integral ((πd^2 / 4) * P_i * (1 - (x/d)^2 / 25) ) . dx

For Limits 0 < x < 5d

Work done = (πd^2 / 4) * P_i  integral ( (1 - (x/d)^2) / 25)) . dx

Integrate the function wrt x

Work done = (πd^2 / 4) * P_i * ( x - d*(x/d)^3 / 75 )  

Evaluate Limits 0 < x < 5d :

Work done = (πd^2 / 4) * P_i * (5d - 5d / 3)

Work done = (πd^2 / 4) * P_i * (10*d / 3)

Work done = (5 π / 6)d^3 * P_i

Input the values:

Work done = (5 π / 6)(0.05)^3 * (1.2*10^6)

Work done = 125π J

5 0
3 years ago
Silicon chips are used primarily in ?
VMariaS [17]

Answer:

4th generation computers

5 0
2 years ago
A wooden cylinder (0 02 x 0 02 x 0 1m) floats vertically in water with one-third of ts length immersed. a)-Determine the density
Anuta_ua [19.1K]

Answer:

a)- the density of wood is 333.33 Kg/m³

b)-unstable condition

c)-unstable condition

Explanation:

Given data

wooden cylinder = 0.02m  x 0.02m x 0.1m

floating = 1/3 × Length

to find out

density of wood,  is it stable condition and wood would float stably in alcohol with density 700 kg/m3

Solution

First we find out density of wood

we know density of water is 1000 kg/m³

and we know wood float 1/3 of length so fraction of density will be

density of wood/ density of water = 1/3

density of wood = 1/3 ×  density of water

density of wood  = 1/3 × 1000 = 333.33 Kg/m³

Now in 2nd part we know for stable condition in partially submerged of body the metacentric height is greater than the centroid of body

we check these condition now,

metacentric height (GM)= I/v  

I = ( 0.02 × 0.02³ / 12 )  

v = ( 0.02 × 0.02 × 0.1 )

(GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

and we know centroid of body (BM) =  0.05 - 0.033 = 0.017

we know height is 0.1m so G act at 0.05 and B act at (0.1 × 0.3 ) = 0.033

we can see that now metacentric height is less than centroid of body so our body is unstable condition

Now in 3rd part we use alcohol so we calculate ratio of density of wood and density of alcohol i.e. = 333.33 / 700 = 0.48

so now our new G will be 0.05 and B will be (0.1 × 0.48 ) = 0.048

metacentric height (GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

centroid of body (BM) =  0.05 - 0.048 = 0.002

we can see that now metacentric height is less than centroid of body so our body is unstable condition

5 0
3 years ago
An article gave a scatter plot along with the least squares line of x = rainfall volume (m3) and y = runoff volume (m3) for a pa
kompoz [17]

Answer:

y = 0.834X - 1.58015

Slope = 0.8340 ; Intercept = - 1.5802

y = 40.9539

19.93

0.9765

Explanation:

X: Rainfall volume

6

12

14

16

23

30

40

52

55

67

72

81

96

112

127

Y : Runoff

4

10

13

14

15

25

27

48

38

46

53

72

82

99

100

The scatterplot shows a reasonable linear trend between the Rainfall volume and run off.

The estimated regression equation obtained using a linear regression calculator is :

y = 0.834X - 1.58015

y = Runoff ; x = Rainfall volume

Slope = 0.8340 ; Intercept = - 1.5802

Point estimate for Runoff, when, x = 51

y = 0.834X - 1.58015

y = 0.834(51) - 1.58015

y = 40.95385

y = 40.9539

d.)

Point estimate for standard deviation :

s = 5.145

σ = s * √n

σ = √15 * 5.145

= 19.93

e.)

r² = Coefficient of determination gives the proportion of explained variance in Runoff due to the regression line. From the model output, the r² value = 0.9765. Which means That about 97.65% Runoff is due to Rainfall volume.

7 0
3 years ago
Ignition for heavy fuel oil?
ipn [44]

Answer:

What heavy fuel oil?

Heavy Fuel Oil (HFO) is a category of fuel oils of a tar-like consistency. Also known as bunker fuel, or residual fuel oil, HFO is the result or remnant from the distillation and cracking process of petroleum.

Explanation:

7 0
3 years ago
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