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.
Amare Yihunie said:
7 years ago
Q1 = A1 * V1,
V1 = Q1/A1,
Q = 0.3,
A1 = 0.1,
V = 0.3/0.1 = 3.
Q2 = Q1 = 0.3.
= A2 * V2,
V2 = Q2/A2, 0.3/0.1=1.
HL = (V1-V2)^2/2g,(3-1)^2/2g.
=2^2/2g,4/2g,
=2/g.
V1 = Q1/A1,
Q = 0.3,
A1 = 0.1,
V = 0.3/0.1 = 3.
Q2 = Q1 = 0.3.
= A2 * V2,
V2 = Q2/A2, 0.3/0.1=1.
HL = (V1-V2)^2/2g,(3-1)^2/2g.
=2^2/2g,4/2g,
=2/g.
(20)
Arpan Dey said:
3 years ago
@All.
If pressure head and static heads are neglected then the Bernoulli equation becomes
[(V1)^2]/2.g = [(V2)^2 ]/2g+ head loss.
The answer should be 4/g mtr.
If pressure head and static heads are neglected then the Bernoulli equation becomes
[(V1)^2]/2.g = [(V2)^2 ]/2g+ head loss.
The answer should be 4/g mtr.
(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)
Rajesh said:
7 years ago
Thank you somuch.
(2)
Ipsa said:
5 years ago
Thanks for explaining @Taba Tallbum.
(1)
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)
Vinod said:
8 years ago
Thanks for the given solution.
Reborn said:
4 years ago
Thanks all for explaining the answer.
Ravi said:
4 years ago
Nice explanation, Thanks.
Venkatesh said:
5 years ago
Thanks all for explaining the answer clearly.
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