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Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s?
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Design a rapid mixing unit to treat a flow of 10 MLD with detention ti...
**Problem Statement**
Design a rapid mixing unit to treat a flow of 10 MLD (million liters per day) with a detention time of 1 min. Determine the diameter of the tank if the depth is 1 m. Also, find the velocity gradient if the mixing unit imparts a power of 1000 Watts. Assume the viscosity of the fluid is 10^-3 kg/m-s.
**Designing a Rapid Mixing Unit**
1. **Detention Time Calculation:**
- Given flow rate = 10 MLD
- Detention time = 1 min
- Detention time (T) can be calculated using the formula:
T = Volume of tank / Flow rate
- Rearranging the formula, we get:
Volume of tank = Flow rate * Detention time
- Substituting the values, we have:
Volume of tank = 10 MLD * (1/1440) days (since 1 day = 24 hours = 1440 minutes)
Volume of tank = 6944.44 liters
2. **Tank Diameter Calculation:**
- Given tank depth = 1 m
- The volume of a cylindrical tank can be calculated using the formula:
Volume of tank = π * (D/2)^2 * H
- Rearranging the formula, we get:
D = √(4 * Volume of tank / (π * H))
- Substituting the values, we have:
D = √(4 * 6944.44 / (π * 1))
D ≈ 2.97 m
3. **Velocity Gradient Calculation:**
- Given power input (P) = 1000 Watts
- The velocity gradient (G) can be calculated using the formula:
G = (P * μ) / (ρ * V^3)
where μ is the viscosity, ρ is the density, and V is the velocity
- Since the flow rate is given in MLD, we need to convert it to m^3/s:
Flow rate (Q) = 10 MLD * (1/24) hours * (1/60) minutes * (1/60) seconds
Flow rate (Q) = 0.00463 m^3/s
- The velocity (V) can be calculated using the formula:
V = Q / (π * (D/2)^2)
- Substituting the values, we have:
V = 0.00463 / (π * (2.97/2)^2)
V ≈ 0.166 m/s
- Substituting the values into the velocity gradient formula, we have:
G = (1000 * 10^-3) / (1000 * 0.166^3)
G ≈ 379.3/s
**Conclusion**
The diameter of the tank for the rapid mixing unit is approximately 2.97 meters, and the velocity gradient is approximately 379.3/s. Therefore, the correct option is (1) D = 2.97 and G = 379.3/s.
Design a rapid mixing unit to treat a flow of 10 MLD (million liters per day) with a detention time of 1 min. Determine the diameter of the tank if the depth is 1 m. Also, find the velocity gradient if the mixing unit imparts a power of 1000 Watts. Assume the viscosity of the fluid is 10^-3 kg/m-s.
**Designing a Rapid Mixing Unit**
1. **Detention Time Calculation:**
- Given flow rate = 10 MLD
- Detention time = 1 min
- Detention time (T) can be calculated using the formula:
T = Volume of tank / Flow rate
- Rearranging the formula, we get:
Volume of tank = Flow rate * Detention time
- Substituting the values, we have:
Volume of tank = 10 MLD * (1/1440) days (since 1 day = 24 hours = 1440 minutes)
Volume of tank = 6944.44 liters
2. **Tank Diameter Calculation:**
- Given tank depth = 1 m
- The volume of a cylindrical tank can be calculated using the formula:
Volume of tank = π * (D/2)^2 * H
- Rearranging the formula, we get:
D = √(4 * Volume of tank / (π * H))
- Substituting the values, we have:
D = √(4 * 6944.44 / (π * 1))
D ≈ 2.97 m
3. **Velocity Gradient Calculation:**
- Given power input (P) = 1000 Watts
- The velocity gradient (G) can be calculated using the formula:
G = (P * μ) / (ρ * V^3)
where μ is the viscosity, ρ is the density, and V is the velocity
- Since the flow rate is given in MLD, we need to convert it to m^3/s:
Flow rate (Q) = 10 MLD * (1/24) hours * (1/60) minutes * (1/60) seconds
Flow rate (Q) = 0.00463 m^3/s
- The velocity (V) can be calculated using the formula:
V = Q / (π * (D/2)^2)
- Substituting the values, we have:
V = 0.00463 / (π * (2.97/2)^2)
V ≈ 0.166 m/s
- Substituting the values into the velocity gradient formula, we have:
G = (1000 * 10^-3) / (1000 * 0.166^3)
G ≈ 379.3/s
**Conclusion**
The diameter of the tank for the rapid mixing unit is approximately 2.97 meters, and the velocity gradient is approximately 379.3/s. Therefore, the correct option is (1) D = 2.97 and G = 379.3/s.
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Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s?
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Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s?.
Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Design a rapid mixing unit to treat a flow of 10 MLD with detention time 1 min. Find its diameter if the depth of the tank is 1m. Also find velocity gradient if mixing unit imparts a power of 1000 Watts. Assume viscosity = 10^-3 kg/m-s (1) D=2.97 and G= 379.3/s (2) D=3.28 and G= 412.4/s (3) D=1.54 and G= 216.7/s (4) D = 4.54 and G = 186.9/s?.
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