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A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3 of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.
Correct answer is '0.23-0.24'. Can you explain this answer?
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Verified Answer
A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250...
Concept:
Temporal Mean Velocity Gradient (G’)
μ = dynamic viscosity of water, 
V = Volume of the raw water to be treated, m3
P = Power dissipated, watts
Calculation:
Given:
G’ = 106/second
V = 2500 m3
νν = 8.9 × 10-7 m2/s
∴ dynamic viscosity (μ) = ρν
μ = 1000 kg/m3 × 8.9 × 10-7 m2/s
⇒ P = μVG’2
⇒ P = 8.9 × 10-4 × 2500 × 106 × 10-3
⇒ P = 0.23585 kW
Most Upvoted Answer
A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250...
°C is 0.89 x 10^-6 m^2/s, estimate the power requirement for paddles.
To estimate the power requirement for the paddles, we will use the power number (Np) correlation for paddle mixers given by the equation:
Np = P / (ρ * n^3 * D^5)
Where:
P = Power required for mixing (W)
ρ = Density of the fluid (kg/m^3)
n = rotational speed of paddles (rev/s)
D = Impeller diameter (m)
First, let's calculate the impeller diameter (D). The paddles are 18m long and 0.45m wide. Since they are rectangular in shape, we can assume the diameter to be equal to the width:
D = 0.45m
Next, let's calculate the rotational speed (n) of the paddles. The average velocity gradient (G) is given as 106/second. The relationship between G and n is given by:
G = n^2 * D^2 / ν
where:
ν = kinematic viscosity of water (m^2/s)
Solving for n:
n^2 = G * (ν / D^2)
n = √(G * (ν / D^2))
Substituting the given values:
G = 106/second
ν = 0.89 x 10^-6 m^2/s
D = 0.45m
n = √(106/second * (0.89 x 10^-6 m^2/s / (0.45m)^2))
n ≈ 0.0032 rev/s
Now, let's calculate the power number (Np) using the given formula:
Np = P / (ρ * n^3 * D^5)
Rearranging the formula to solve for power (P):
P = Np * ρ * n^3 * D^5
The density of water (ρ) can be assumed to be 1000 kg/m^3.
Substituting the given values:
Np = unknown
ρ = 1000 kg/m^3
n = 0.0032 rev/s
D = 0.45m
Now, we need to find the power number (Np) correlation for paddle mixers. This correlation is typically obtained from experimental data or from published literature. Without this information, it is not possible to estimate the power requirement accurately.
To estimate the power requirement for the paddles, we will use the power number (Np) correlation for paddle mixers given by the equation:
Np = P / (ρ * n^3 * D^5)
Where:
P = Power required for mixing (W)
ρ = Density of the fluid (kg/m^3)
n = rotational speed of paddles (rev/s)
D = Impeller diameter (m)
First, let's calculate the impeller diameter (D). The paddles are 18m long and 0.45m wide. Since they are rectangular in shape, we can assume the diameter to be equal to the width:
D = 0.45m
Next, let's calculate the rotational speed (n) of the paddles. The average velocity gradient (G) is given as 106/second. The relationship between G and n is given by:
G = n^2 * D^2 / ν
where:
ν = kinematic viscosity of water (m^2/s)
Solving for n:
n^2 = G * (ν / D^2)
n = √(G * (ν / D^2))
Substituting the given values:
G = 106/second
ν = 0.89 x 10^-6 m^2/s
D = 0.45m
n = √(106/second * (0.89 x 10^-6 m^2/s / (0.45m)^2))
n ≈ 0.0032 rev/s
Now, let's calculate the power number (Np) using the given formula:
Np = P / (ρ * n^3 * D^5)
Rearranging the formula to solve for power (P):
P = Np * ρ * n^3 * D^5
The density of water (ρ) can be assumed to be 1000 kg/m^3.
Substituting the given values:
Np = unknown
ρ = 1000 kg/m^3
n = 0.0032 rev/s
D = 0.45m
Now, we need to find the power number (Np) correlation for paddle mixers. This correlation is typically obtained from experimental data or from published literature. Without this information, it is not possible to estimate the power requirement accurately.
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A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer?
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A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer? 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 A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer?.
A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer? 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 A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A flocculation chamber 50m long, 15 m wide and 5m deep is to treat 250 MLD of water. It is equipped with 18m long and 0.45m wide paddles. The tank contains 2500m3of water which is mixed using paddles at an average velocity gradient of 106/second. If the kinematic viscosity of water at 28° C temperature is 8.9 × 10-7m2/s, then the power required (in kW, up to two decimal place) to achieve adequate mixing will be ________.Correct answer is '0.23-0.24'. Can you explain this answer?.
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