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Romashka [77]
3 years ago
14

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)

with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
Required:

a. Find the initial DO deficit (Do) after mixing.
b. What is the reaeration constant after mixing?
c. What is the critical time for this system?
d. What is the downstream distance fitting the critical time?
Engineering
1 answer:
4vir4ik [10]3 years ago
8 0

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

Your Welcome

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Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
3 years ago
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied
lyudmila [28]

Answer:

I=0.3636

Explanation:

See the attached picture for explanation.

4 0
3 years ago
If he wants to keep the height the same, what could the other dimensions be for him to get the volume he wants?
Fiesta28 [93]

tbm queria saber essa pergunta

8 0
3 years ago
A circuit contains a resistor, an inductor, and a capacitor. When an AC voltage is applied across the circuit, the impedance of
katen-ka-za [31]

Answer:

The impedance of  the circuit depends on the angular frequency of the voltage source.

Explanation:

  • In a electric circuit, the magnitude of  the impedance, is given by the following expression:

       Z = \sqrt{R^{2} + (Xl-Xc)^{2} (1)

        where R = Resistance

                   Xl = Inductive reactance = ω*L

                   Xc = Capacitive Reactance = 1/ωC

        and  ω = angular frequency of the voltage source.

  • So, it can be seen that the impedance depends on the value of the constants R,L and C, and on the angular frequency ω.

3 0
2 years ago
Astronauts who landed on the moon during the Apollo 15, 16, and 17 missions brought back a large collection of rocks to the eart
stiks02 [169]

Answer: a) W(earth) = 935.62 lbs

b) Mass of rocks in slugs = 29.06 slugs

Explanation:

a) From Newton's law, W = mg. Whether on the moon or on earth. Although, the mass of the rocks everywhere is the same, that is, mass of rocks on the moon = mass of rocks on earth.

W(moon) = mg(moon)

W(moon) = 154 lbs

g(moon) = 5.30 ft/s2

m = W(moon)/g(moon) = 154/5.3 = 29.06 lb.s2/ft

W(earth) = m g(earth)

g(earth) = 32.2 ft/s2

W(earth) = 29.06 × 32.2 = 935.62 lbs.

b) A slug = 1 lb.s2/ft, therefore the mass of the rocks in slugs is 29.06 slugs.

QED!

8 0
3 years ago
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