Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 31)
31.
A pipe of 0.1 m2 cross sectional area suddenly enlarges to 0.3 m2 cross-sectional area. If the discharge of the pipe is 0.3 m3 /sec, the head loss is
Discussion:
23 comments Page 1 of 3.
Taba Tallum said:
1 decade ago
h(en) = ( v1-v2)^2 / 2g.
= (Q/A1 - Q/A2)^2 / 2g.
= (0.3/0.1 - 0.3/0.3)^2 / 2g.
= (3-1)^2 / 2g.
= 2/g m of water.
= (Q/A1 - Q/A2)^2 / 2g.
= (0.3/0.1 - 0.3/0.3)^2 / 2g.
= (3-1)^2 / 2g.
= 2/g m of water.
(2)
Abhijit said:
9 years ago
Nice explanation @Taba Tallum.
Sai said:
9 years ago
Thanks @Taba Tallum.
Dar said:
9 years ago
Thanks @Taba Tallum.
Anil said:
9 years ago
Thank you for explaining the solution @Taba.
Srinu said:
9 years ago
Clear explanation, Thank you @Taba Tallum.
Tinku said:
9 years ago
Thank you @Taba.
SAKTHIMUNIVEl said:
9 years ago
Thanks for the explanation @Taba.
Dineah prasad sao said:
9 years ago
V1 = .3/.1 = 3
V2 = .3/.3 = 1
Head loss = v^2/2g.
= (3-1)^2/2g,
= 2/g.
V2 = .3/.3 = 1
Head loss = v^2/2g.
= (3-1)^2/2g,
= 2/g.
(1)
Hitesh said:
8 years ago
Good, Thanks @Dineah.
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