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EastWind [94]
3 years ago
6

g Four primary sedimentation basins are designed for total average flow of 1.2 m3/s. The TSS concentration in the primary treate

d effluent is 150 mg/L, and TSS removal is 63 percent. The sludge has an average solids concentration of 6 percent and specific gravity of 1.045. What is the capacity of the sludge pump in m3/min per basin if the pump is used on a 15-min pumping cycle per h

Engineering
1 answer:
fiasKO [112]3 years ago
5 0

Answer:

Capacity for the sludge pump = 0.217 m³/min

Explanation:

Detailed explanation is given in the attached document.

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allochka39001 [22]

Answer:

Any engineering job would be good YOU should be the one choosing which job.

Explanation:

Engineering is a great outlet for the imagination, and the perfect field for independent thinkers.

7 0
2 years ago
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12.50 An air conditioner operating at steady state takes in moist air at 28°C, 1 bar, and 70% relative humidity. The moist air f
Mandarinka [93]

Answer:

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Explanation:

8 0
3 years ago
water flows in a horizontal constant-area pipe; the pipe diameter is 75 mm and the average flow speed is 5 m/s. At the pipe inle
Veronika [31]

Answer:

Head loss = 28.03 m

Explanation:

According to Bernoulli's theorem for fluids  we have

\frac{P}{\gamma _{w}}+\frac{V^{2}}{2g}+z=Constant

Applying this between the 2 given points we have

\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}=\frac{P_{2}}{\gamma _{w}}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{l}

Here h_{l} is the head loss that occurs

\therefore h_{l}=\frac{P_{1}}{\gamma _{w}}+\frac{V_{1}^{2}}{2g}+z_{1}-\frac{P_{2}}{\gamma _{w}}-\frac{V_{2}^{2}}{2g}-z_{2}

Since the pipe is horizantal we have z_{1}-z_{2}=0

Applying contunity equation between the 2 sections we get

A_{1}V_{1}=A_{2}V_{2}\\\\\therefore V_{1}=V_{2}(\because A_{1}=A_{2})

Since the cross sectional area of the both the sections is same thus the speed

is also same

Using this information in the above equation of head loss we obtain

h_{l}=\frac{1}{\gamma _{w}}(P_{1}-P_{2})

Applying values we get

h_{l}=\frac{1}{9810}\times (275\times 10^{3})m\\\\\therefore h_{l}=28.03m

3 0
3 years ago
Write a program with total change amount as an integer input, and
AURORKA [14]

Answer:

amount = int(input())

#Check if input is less than 1

if amount<=0:

    print("No change")

else: #If otherwise

    #Convert amount to various coins

    dollar = int(amount/100) #Convert to dollar

    amount = amount % 100 #Get remainder after conversion

    quarter = int(amount/25) #Convert to quarters

    amount = amount % 25 #Get remainder after conversion

    dime = int(amount/10) #Convert to dimes

    amount = amount % 10 #Get remainder after conversion

    nickel = int(amount/5) #Convert to nickel

    penny = amount % 5 #Get remainder after conversion

    #Print results

    if dollar >= 1:

          if dollar == 1:

                print(str(dollar)+" Dollar")

          else:

                print(str(dollar)+" Dollars")

    if quarter >= 1:

          if quarter == 1:

                print(str(quarter)+" Quarter")

          else:

                print(str(quarter)+" Quarters")

    if dime >= 1:

          if dime == 1:

                print(str(dime)+" Dime")

          else:

                print(str(dime)+" Dimes")

    if nickel >= 1:

          if nickel == 1:

                print(str(nickel)+" Nickel")

          else:

                print(str(nickel)+" Nickels")

    if penny >= 1:

          if penny == 1:

                print(str(penny)+" Penny")

          else:

                print(str(penny)+" Pennies")

Explanation:

8 0
3 years ago
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Akimi4 [234]

Answer:

OMG GOD I HAVE WONDERED THAT

Explanation:

4 0
3 years ago
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