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-BARSIC- [3]
2 years ago
15

The two pond system is fed by a stream with flow rate 1.0 MGD (million gallons per day) and BOD (nonconservative pollutant) conc

entration of 20 mg/L. The rate of decay of BOD is 0.3/day. The volume of the first pond is 5 million gallons and the second is 3 million. Assuming complete mixing within the pond, find the BOD concentration leaving each pond.
Engineering
1 answer:
svlad2 [7]2 years ago
3 0

Answer: First pond= 8.0 mg/L

second pond = 4.2 mg/L

Explanation:

Cn/Co = [ 1/ [ 1 + (k*t/n)]]

for the first pond

where Co into the first pond is 20mg/L,

Cn = 20*[ 1/ [ 1+ ((1 x 5)/0.3)]]

Cn = 8 mg/L

into the second pond we calculate the BOD leaving the pond of volume 3million liters

we have Cn = 4.2 mg/L

You might be interested in
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge
ale4655 [162]

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = \pi d^2/4

A = \frac{3.142 * 70^2}{4} = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = v^2/2 = \frac{10^2}{2} = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

7 0
3 years ago
A square power screw has a mean diameter of 30 mm and a pitch of 4 mm with single thread. The collar diameter can be assumed to
poizon [28]

Answer:

i) The torque required to raise the load is 15.85 N*m

ii) The torque required to lower the load is 6.91 N*m

iii) The minimum coefficient of friction is -0.016

Explanation:

Given:

dm = mean diameter = 0.03 m

p = pitch = 0.004 m

n = number of starts = 1

The lead is:

L = n * p = 1 * 0.004 = 0.004 m

F = load = 7000 N

dc = collar diameter = 0.035 m

u = 0.05

i) The helix angle is:

tan\alpha =\frac{L}{\pi *d_{m} } =\frac{0.004}{\pi *0.03} \\\alpha =2.43

The torque is:

T=F\frac{d_{m} }{2} (\frac{\pi *u*d_{m}+L }{\pi *d_{m}-uL } )+(u_{c} F+\frac{d_{2} }{2} )=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03+0.004}{\pi *0.03-0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=15.85Nm

ii) The torque to lowering the load is:

T=7000*\frac{0.03}{2} (\frac{\pi *0.05*0.03-0.004}{\pi *0.03+0.05*0.004} )+(0.05*7000*\frac{0.035}{2} )=6.91Nm

iii)

T=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\  0=F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )+u_{c} *F*\frac{d_{c}}{2}\\F\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *F*\frac{d_{c}}{2}\\\frac{d_{m} }{2} (\frac{u*\pi *d_{m}-L}{\pi *d_{m}+uL} )=-u_{c} *\frac{d_{c}}{2}\\\\\frac{0.03}{2} (\frac{u*\pi *0.03-0.004}{\pi *0.03+u*0.004} )=-0.05*\frac{0.035}{2}

Clearing u:

u = -0.016

5 0
3 years ago
Carbon dioxide at a temperature of 0oC and a pressure of 600 kPa (abs) flows through a horizontal 40-mm- diameter pipe with an a
rusak2 [61]

Answer:

f = 0.04042

Explanation:

temperature = 0°C = 273k

p = 600 Kpa

d = 40 millemeter

e = 10 m

change in  P = 235 N/m²

μ = 2m/s

R = 188.9 Nm/kgk

we solve this using this formula;

P = ρcos*R*T

we put in the values into this equation

600x10³ = ρcos * 188.9 * 273

600000 = ρcos51569.7

ρcos = 600000/51569.7

=11.63

from here we find the head loss due to friction

Δp/pg = feμ²/2D

235/11.63 = f*10*4/2*40x10⁻³

20.21 = 40f/0.08

20.21*0.08 = 40f

1.6168 = 40f

divide through by 40

f = 0.04042

5 0
3 years ago
The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.
Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL_{max} / D = 57.915

we substitute

(4×0.005×L_{max}) / 4  = 57.915

0.005L_{max} = 57.915

L_{max} = 57.915 / 0.005

L_{max}  = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0
2 years ago
A power company desires to use groundwater from a hot spring to power a heat engine. If the groundwater is at 95 deg C and the a
prisoha [69]

Answer:

W  = 12.8 KW

Explanation:

given data:

mass flow rate = 0.2 kg/s

Engine recieve heat from ground water at 95 degree ( 368 K)  and reject that heat to atmosphere  at 20 degree (293K)

we know that maximum possible efficiency is given as

\eta = 1- \frac{T_L}{T_H}

\eta = 1 - \frac{ 293}{368}

\eta = 0.2038

rate of heat transfer is given as

Q_H = \dot m C_p \Delta T

Q_H = 0.2 * 4.18 8(95 - 20)

Q_H = 62.7 kW

Maximuim power is given as

W = \eta Q_H

W = 0.2038 * 62.7

W  = 12.8 KW

3 0
2 years ago
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