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Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)
Correct answer is '19.2'. Can you explain this answer?
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Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube ...
Calculation of Reynolds Number
- Reynolds number (Re) = (density × velocity × diameter) / viscosity
- Substituting the given values, Re = (1000 × 800 × 0.5 × 10^-3) / (p × 10^-4)
- Simplifying, Re = 80000 / p
Calculation of Friction Factor
- Friction factor (f) = (pressure drop × diameter) / (length × density × velocity^2 / 2)
- Substituting the given values, f = (2.0 × 10^6 × 0.5 × 10^-3) / (2 × 1000 × 800^2 / 2)
- Simplifying, f = 0.03125
Calculation of Colebrook-White Equation
- Colebrook-White equation: 1 / √f = -2.0 log10 [(2.51 / Re) √f + (k / 3.71)]
- Substituting the values of f and Re, 1 / √0.03125 = -2.0 log10 [(2.51 / 80000 / p) √0.03125 + (k / 3.71)]
- Simplifying and solving for p, p = 19.2
Therefore, the value of p is 19.2, which is the answer to the given question.
- Reynolds number (Re) = (density × velocity × diameter) / viscosity
- Substituting the given values, Re = (1000 × 800 × 0.5 × 10^-3) / (p × 10^-4)
- Simplifying, Re = 80000 / p
Calculation of Friction Factor
- Friction factor (f) = (pressure drop × diameter) / (length × density × velocity^2 / 2)
- Substituting the given values, f = (2.0 × 10^6 × 0.5 × 10^-3) / (2 × 1000 × 800^2 / 2)
- Simplifying, f = 0.03125
Calculation of Colebrook-White Equation
- Colebrook-White equation: 1 / √f = -2.0 log10 [(2.51 / Re) √f + (k / 3.71)]
- Substituting the values of f and Re, 1 / √0.03125 = -2.0 log10 [(2.51 / 80000 / p) √0.03125 + (k / 3.71)]
- Simplifying and solving for p, p = 19.2
Therefore, the value of p is 19.2, which is the answer to the given question.
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Community Answer
Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube ...
μ = 0.00192 Ns/m2
μ = 19.2 × 10-4 N-s/m2
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Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. Can you explain this answer?
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Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. 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 Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. 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 Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. Can you explain this answer?.
Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. 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 Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. 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 Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. Can you explain this answer?.
Solutions for Flow rate of a fluid (density = 1000 kg/m3) in a small diameter tube is 800 m3/s. The length and the diameter of the tube are 2 m and 0.5 mm respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is p × 10-4 N-s/m2. The value of p is (Answer up to one decimal place)Correct answer is '19.2'. Can you explain this answer? in English & in Hindi are available as part of our courses for Civil Engineering (CE).
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