Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 2 (Q.No. 12)
12.
The discharge through a 100 mm diameter external mouth piece fitted to the side of a large vessel is 0.05948 m3/s. The head over the mouth piece is
Discussion:
10 comments Page 1 of 1.
Masood khan said:
4 years ago
This loss of head, sometimes, reduces the coefficient of discharge by a small amount (up to 0.82). But, for all practical purposes, the value of coefficient of discharge is taken as 0.855.
Anonymous said:
4 years ago
For external mouthpiece Cd = 1.
So, the Ans must be 3 m.
So, the Ans must be 3 m.
Achumi kits said:
5 years ago
What does "A" signifies here? Please explain.
Rabin Das said:
5 years ago
Thanks all.
Zia said:
6 years ago
Option D is correct.
Rajesh Ata said:
6 years ago
Q = .855a * (2gH)^1/2.
.05948 =.855*(3.14*.1*.1/4) * (2*9.81*H)^1/2.
H = 3.998 m.
.05948 =.855*(3.14*.1*.1/4) * (2*9.81*H)^1/2.
H = 3.998 m.
(4)
Anurag said:
7 years ago
Yes, right @Abhishek Anand.
Abhishek Anand said:
8 years ago
Q=.855a * (2gH)^1/2.
.05948 =.855*(3.14*.1*.1/4) * (2*9.81*H)^1/2.
H = 3.998 m.
.05948 =.855*(3.14*.1*.1/4) * (2*9.81*H)^1/2.
H = 3.998 m.
(1)
Kali said:
9 years ago
@Ambily.
It is not .82 it is 0.855.
It is not .82 it is 0.855.
(1)
Ambily said:
1 decade ago
Q = 0.82 A (2gH)^0.5.
0.05948 = 0.82*(3.14*.1*.1)/4*(2*9.81*H)^0.5.
= 9.24 = (2*9.81*H)^0.5.
H = 4.35.
0.05948 = 0.82*(3.14*.1*.1)/4*(2*9.81*H)^0.5.
= 9.24 = (2*9.81*H)^0.5.
H = 4.35.
(1)
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