Civil Engineering - Hydraulics - Discussion

Discussion Forum : Hydraulics - Section 2 (Q.No. 25)
25.
The height of water level in a tank above the centre of a circular hole 2.5 cm in diameter is 50 m. The velocity of water flowing through the hole, is
31.1 m/sec
31.2 m/sec
31.3 m/sec
31.4 m/sec.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Pradeep biswas said:   4 years ago
d = 2.5.
h = 50m.

V=√2gh.

V=√2*9.81*50,

V=31.32m/sec.
(2)

Surafel said:   5 years ago
Thank you all for explaining.

Jinish Patel said:   6 years ago
We can directly find by v = squrare root of 2*g*h.
So g is 9.82 m/sec.

H is height of the hole above = 50 m.
V = 31.3 m/sec note that here we only consider height above hole.

But we can also find it by velocity = height above hole (50 m) * height or dia of hole (2.5 cm) / 4 (noted dia of hole always take in cm and height above hole take always in meter)
= 31.3 m/sec.

But first equation is priority v = square root of 2gh.

Sai said:   6 years ago
Please Explain the answer. Clearly.

Saurabh said:   6 years ago
Here, the velocity is independent of dia of the hole. Right?

Nitai Maity said:   7 years ago
Velocity(v)=√(2gh).
Data given that,height of water(h)=50 m & diameter of hole =2.5 cm.
So,v=Sqrt(2*9.81*50)=31.3 m/s.

Gurjar said:   8 years ago
Explain it clearly.

Dinesh prasad sao said:   9 years ago
√2gh = √2 x 9.8 x 50.

Sreesha said:   9 years ago
@Ramkisan you are wrong m, cm you have to convert it into same units.

Jatin said:   9 years ago
V^2 = 2gh.
V = (2*9.81*50)^1/2.
V = 31.32.


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