Chemical Engineering - Fluid Mechanics - Discussion
Discussion Forum : Fluid Mechanics - Section 4 (Q.No. 28)
28.
A pipe of I.D. 4 m is bifurcated into two pipes of I.D. 2 m each. If the average velocity of water flowing through the main pipe is 5 m/sec, the average velocity through the bifurcated pipes is
Discussion:
3 comments Page 1 of 1.
Alok Manna said:
2 months ago
AV1 = 2AV2.
5(π/4) × (4)^2 = 2(π/4) × 2^2 × V2,
V2 = 10m/s.
5(π/4) × (4)^2 = 2(π/4) × 2^2 × V2,
V2 = 10m/s.
Rana said:
5 years ago
In my opinion:
A1 x V1 = A2 x V2.
By using this relation V2 = 20m/s.
A1 x V1 = A2 x V2.
By using this relation V2 = 20m/s.
Sumaya liwa said:
7 years ago
Q1=u1*A1=5* (π/4)*(4)^2=20 π
Q2=u2*A2=u2*(π/4)*(2)^2=π*u2
Q2/Q1 = π*u2 / 20 π ----> 0.5 Q1 /Q1 = u2/20.
u2 = 10 m/s.
Q2=u2*A2=u2*(π/4)*(2)^2=π*u2
Q2/Q1 = π*u2 / 20 π ----> 0.5 Q1 /Q1 = u2/20.
u2 = 10 m/s.
(4)
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