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For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is —10 Pa/m. The magnitude or axial velocity (in m/s) at a radial location of 0.2 cm is ___________ m/sec
- a)6.2
- b)6.9
- c)5.4
- d)5.9
Correct answer is option 'A'. Can you explain this answer?
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For a fully developed laminar flow of water (dynamic viscosity 0.001 P...
For incompressible laminar flows through pipes velocity distribution
= 6.24m/sec.
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For a fully developed laminar flow of water (dynamic viscosity 0.001 P...
The axial pressure gradient for a fully developed laminar flow of water through a pipe can be calculated using the Hagen-Poiseuille equation.
The Hagen-Poiseuille equation for laminar flow in a circular pipe is given by:
ΔP = (8μQ) / (πr^4)
Where:
ΔP is the pressure drop along the pipe
μ is the dynamic viscosity of the fluid (0.001 Pa-s for water)
Q is the volumetric flow rate
r is the radius of the pipe (5 cm = 0.05 m)
Since the flow is fully developed, the volumetric flow rate remains constant along the pipe.
Assuming a constant volumetric flow rate (Q), we can rearrange the equation to solve for the axial pressure gradient (ΔP):
ΔP = (8μQ) / (πr^4)
Substituting the given values, we have:
ΔP = (8 * 0.001 Pa-s * Q) / (π * (0.05 m)^4)
Simplifying,
ΔP = (8 * 0.001 Pa-s * Q) / (π * 0.000000625 m^4)
ΔP = (8000 Q) / (π * 0.000000625)
Therefore, the axial pressure gradient for a fully developed laminar flow of water through a pipe of radius 5 cm is (8000 Q) / (π * 0.000000625), where Q is the volumetric flow rate.
The Hagen-Poiseuille equation for laminar flow in a circular pipe is given by:
ΔP = (8μQ) / (πr^4)
Where:
ΔP is the pressure drop along the pipe
μ is the dynamic viscosity of the fluid (0.001 Pa-s for water)
Q is the volumetric flow rate
r is the radius of the pipe (5 cm = 0.05 m)
Since the flow is fully developed, the volumetric flow rate remains constant along the pipe.
Assuming a constant volumetric flow rate (Q), we can rearrange the equation to solve for the axial pressure gradient (ΔP):
ΔP = (8μQ) / (πr^4)
Substituting the given values, we have:
ΔP = (8 * 0.001 Pa-s * Q) / (π * (0.05 m)^4)
Simplifying,
ΔP = (8 * 0.001 Pa-s * Q) / (π * 0.000000625 m^4)
ΔP = (8000 Q) / (π * 0.000000625)
Therefore, the axial pressure gradient for a fully developed laminar flow of water through a pipe of radius 5 cm is (8000 Q) / (π * 0.000000625), where Q is the volumetric flow rate.
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For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is —10 Pa/m. The magnitude or axial velocity (in m/s) at a radial location of 0.2 cm is ___________ m/seca)6.2b)6.9c)5.4d)5.9Correct answer is option 'A'. Can you explain this answer?
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For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is —10 Pa/m. The magnitude or axial velocity (in m/s) at a radial location of 0.2 cm is ___________ m/seca)6.2b)6.9c)5.4d)5.9Correct answer is option 'A'. 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 For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is —10 Pa/m. The magnitude or axial velocity (in m/s) at a radial location of 0.2 cm is ___________ m/seca)6.2b)6.9c)5.4d)5.9Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For a fully developed laminar flow of water (dynamic viscosity 0.001 Pa-s) through a pipe of radius 5 cm. the axial pressure gradient is —10 Pa/m. The magnitude or axial velocity (in m/s) at a radial location of 0.2 cm is ___________ m/seca)6.2b)6.9c)5.4d)5.9Correct answer is option 'A'. Can you explain this answer?.
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