A water treatment plant has 6 settling tanks
that operate in parallel (the flow gets split into
six equal flow streams), and each tank has a
volume of 600m3. If the flow to the plant is
10 mgd, what is the retention time in each
of the settling tanks? If, instead, the tanks
operated in series (the entire flow goes first
through one tank, then the second, and so
on), what would be the retention time in each
tank?
23
1.) parallel
Q1=Q2=Q3=Q4=Q5=Q6= (440L/s)/(6)
t=V/Q=((600m^3)/(440/g L/s))x (1000 L)/ (1m^3L = 8181.82 s
2.) series
Q1=Q2=Q3=Q4=Q5=Q6= 440L/s
t= (600 m^3)/(440 L/s) x (1000L/m^3) = 1363.64 s
Ah, the settling tanks are getting quite popular these days. Let me entertain you with some math humor to answer your question.
If we have 6 tanks operating in parallel, each with a volume of 600m3, and a flow rate of 10 mgd, then we need to find the retention time in each tank.
To calculate this, we divide the total volume of all tanks (6 tanks x 600m3) by the flow rate:
Total volume = 6 tanks x 600m3 = 3600m3
Flow rate = 10 mgd
Retention time in each tank = Total volume / Flow rate
= 3600m3 / 10 mgd
Now the tricky part - converting mgd to m3/s. I won't water it down for you, so here it goes:
1 mgd = 4.381 x 10^-3 m3/s
So, 10 mgd = 10 x 4.381 x 10^-3 m3/s
Now, let's plug in the values and find out the retention time:
Retention time in each tank = 3600m3 / (10 x 4.381 x 10^-3 m3/s)
*Takes out a calculator and starts typing*
Wait a second, let me calculate this for you...
*muffled sounds of calculator buttons being pressed*
Oh, here we go! The retention time in each tank operating in parallel is approximately...
*drumroll*
1.638 seconds!
Now, let's move on to the series scenario, where the flow goes through one tank at a time. Brace yourself for some hilarious mathematical calculations!
If the tanks are operating in series, the total volume remains the same (3600m3), but the flow rate remains constant (10 mgd). So, the retention time in each tank is simply:
Retention time in each tank (series) = Total volume / Flow rate
= 3600m3 / (10 x 4.381 x 10^-3 m3/s)
*Calculates with gusto*
And the answer is...
*drumroll gets louder*
6.756 seconds!
So, there you have it! In parallel, the retention time is approximately 1.638 seconds in each tank, whereas in series, it's a delightful 6.756 seconds in each tank. I hope my calculations made a splash with you!
To calculate the retention time in each settling tank, we will first convert the flow rate from million gallons per day (mgd) to cubic meters per day (m3/day).
Given:
Number of settling tanks = 6
Volume of each settling tank = 600 m3
Flow to the plant = 10 mgd
Conversion:
1 million gallons = 3,785.41 m3 (approximately)
1 day = 24 hours
Flow rate in m3/day = 10 mgd * 3,785.41 m3/1 mgd = 37,854.1 m3/day
Now, let's calculate the retention time for the parallel operation:
Total volume of the settling tanks = Volume of one tank * Number of tanks
Total volume = 600 m3 * 6 = 3,600 m3
Retention time = Total volume / Flow rate
Retention time = 3,600 m3 / 37,854.1 m3/day
Retention time ≈ 0.0952 days ≈ 2.29 hours
Therefore, the retention time in each of the settling tanks for the parallel operation is approximately 0.0952 days or 2.29 hours.
Let's calculate the retention time for the series operation:
For the series operation, the entire flow goes through each tank one by one.
Retention time = Volume of one tank / Flow rate
Retention time = 600 m3 / 37,854.1 m3/day
Retention time ≈ 0.0158 days ≈ 0.378 hours
Therefore, the retention time in each of the settling tanks for the series operation is approximately 0.0158 days or 0.378 hours.
To calculate the retention time in each of the settling tanks, we can use the formula:
Retention Time = Volume of tank / Flow rate
First, let's calculate the flow rate in cubic meters per day (m³/day), as we need to ensure the units are consistent. To convert million gallons per day (mgd) to cubic meters per day, we can use the conversion factor of 1 mgd = 4,546.09 m³/day.
Flow rate = 10 mgd * 4,546.09 m³/day / 1 mgd
Flow rate = 45,460.9 m³/day
Now we can calculate the retention time for the settling tanks when they are operating in parallel.
Retention Time (parallel) = Volume of tank / Flow rate
Retention Time (parallel) = 600 m³ / 45,460.9 m³/day
To simplify the calculation, we can convert m³/day to m³/hour by dividing by 24.
Retention Time (parallel) = 600 m³ / (45,460.9 m³/day / 24 hours/day)
Retention Time (parallel) = 600 m³ / 1,894.204 m³/hour
Therefore, the retention time in each of the settling tanks when operating in parallel is 0.316 hours (or approximately 19 minutes). Please note that this value is constant for each tank because the flow is split equally among all six tanks.
Now let's calculate the retention time for the settling tanks when they are operating in series.
In series operation, the entire flow goes through one tank before flowing to the next tank. Therefore, the flow rate in each subsequent tank will be the same as the initial flow rate.
For the first tank:
Retention Time (series, tank 1) = Volume of tank / Flow rate
Retention Time (series, tank 1) = 600 m³ / 45,460.9 m³/day
For the second tank:
Retention Time (series, tank 2) = Volume of tank / Flow rate
Retention Time (series, tank 2) = 600 m³ / 45,460.9 m³/day
This calculation will be the same for all six tanks in series operation.
Therefore, the retention time in each of the settling tanks when operating in series is also 0.316 hours (or approximately 19 minutes).