Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 26 (Q.No. 6)
6.
The velocity distribution for flow over a plate is given by u = 0.5 y - y2 where u is the velocity in m/s at a distance y metre above the plate. If the dynamic viscosity of the fluid is 0.9 N/m2, then what is the shear stress at 0.20 m from the boundary?
Discussion:
2 comments Page 1 of 1.
Karthika.c said:
8 years ago
Shear stress=dynamic viscosity*(velocity gradient).
Velocity gradient=du/dy.
here u=0.5y-y^2,
du/dy=0.5-2y,
Hence Shear stress=0.9*[(0.5)-2 (0.2)].
Hence, the answer is 0.09N/m^2.
Velocity gradient=du/dy.
here u=0.5y-y^2,
du/dy=0.5-2y,
Hence Shear stress=0.9*[(0.5)-2 (0.2)].
Hence, the answer is 0.09N/m^2.
Anomie said:
4 years ago
Given:
u = 0.5y - y^2.
du/dy = 0.5-2y.
Shear stress = dynamic viscosity x du/dy.
= 0.9 x (0.5-2y),
= 0.9 x (0.5-2x0.20),
= 0.9 x (0.5-0.4),
= 0.9 x (0.1),
= 0.09 N/m^2.0
u = 0.5y - y^2.
du/dy = 0.5-2y.
Shear stress = dynamic viscosity x du/dy.
= 0.9 x (0.5-2y),
= 0.9 x (0.5-2x0.20),
= 0.9 x (0.5-0.4),
= 0.9 x (0.1),
= 0.09 N/m^2.0
(1)
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