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
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.
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.
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.
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.
V = (2*9.81*50)^1/2.
V = 31.32.
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