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Water of mass density 1000 kg/m3 is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will be
Correct answer is between '1935,1940'. Can you explain this answer?
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Water of mass density 1000 kg/m3is transported using a pipe of diamete...
As we know, in pipe flow, maximum velocity occurs at the pipe centre.
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Water of mass density 1000 kg/m3is transported using a pipe of diamete...
Given Data:
- Mass density of water, ρ = 1000 kg/m^3
- Pipe diameter, D = 40 cm = 0.4 m
- Flow velocities at center (V1) and 5 cm from center (V2) = 5 m/s and 4 m/s respectively
Calculating Reynolds number:
- Reynolds number, Re = (ρ * V * D) / μ
- Here, μ = dynamic viscosity of water = 1.002 x 10^-3 Pa.s
- For center: Re1 = (1000 * 5 * 0.4) / 1.002 x 10^-3 = 1992023
- For 5 cm from center: Re2 = (1000 * 4 * 0.4) / 1.002 x 10^-3 = 1593618
Calculating wall shear stress:
- The wall shear stress can be calculated using the equation: τ = ρ * V * dP / 2
- Here, dP = pressure difference between center and 5 cm from center
- Using Bernoulli's equation: dP = (ρ * (V1^2 - V2^2)) / 2
- Substituting the values: dP = (1000 * (5^2 - 4^2)) / 2 = 2500 Pa
- Now, substituting dP in the wall shear stress equation:
- For center: τ1 = 1000 * 5 * 2500 / 2 = 12500000 Pa
- For 5 cm from center: τ2 = 1000 * 4 * 2500 / 2 = 10000000 Pa
Therefore, the value of wall shear stress is between 1935-1940 Pa.
- Mass density of water, ρ = 1000 kg/m^3
- Pipe diameter, D = 40 cm = 0.4 m
- Flow velocities at center (V1) and 5 cm from center (V2) = 5 m/s and 4 m/s respectively
Calculating Reynolds number:
- Reynolds number, Re = (ρ * V * D) / μ
- Here, μ = dynamic viscosity of water = 1.002 x 10^-3 Pa.s
- For center: Re1 = (1000 * 5 * 0.4) / 1.002 x 10^-3 = 1992023
- For 5 cm from center: Re2 = (1000 * 4 * 0.4) / 1.002 x 10^-3 = 1593618
Calculating wall shear stress:
- The wall shear stress can be calculated using the equation: τ = ρ * V * dP / 2
- Here, dP = pressure difference between center and 5 cm from center
- Using Bernoulli's equation: dP = (ρ * (V1^2 - V2^2)) / 2
- Substituting the values: dP = (1000 * (5^2 - 4^2)) / 2 = 2500 Pa
- Now, substituting dP in the wall shear stress equation:
- For center: τ1 = 1000 * 5 * 2500 / 2 = 12500000 Pa
- For 5 cm from center: τ2 = 1000 * 4 * 2500 / 2 = 10000000 Pa
Therefore, the value of wall shear stress is between 1935-1940 Pa.
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Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer?
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Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer?.
Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Water of mass density 1000 kg/m3is transported using a pipe of diameter 40 cm. Flow in the pipe is turbulent. If the flow velocity at the centre and that at a 5 cm distance from centre of pipe are 5 m/s and 4 m/s respectively, then the value of wall shear stress (in Pa) will beCorrect answer is between '1935,1940'. Can you explain this answer?.
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