Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion


A tank of uniform cross-sectional area (A) containing liquid upto height (H1) has an orifice of cross-sectional area (a) at its bottom. The time required to empty the tank completely will be


Answer: Option A


No answer description available for this question.

Nihal said: (Jan 16, 2015)  
It should be.


= AH/(Cd*a*(2gH)^.5).

= (A*(H)^.5)/(Cd*a*(2g)^.5).

Why is there an extra 2 in the numerator?

Heera Lal Choudhary said: (Jun 22, 2016)  
Why is there extra 2?

Ashutosh Gupta said: (Jul 19, 2016)  
Because the above formula is applicable for a particular height say H1.

As the water level decreases speed of flow will also decrease from max to minimum (zero) hence the average.

Niladri Shree Chandan said: (Aug 26, 2016)  
Q = a * Velocity = A * (dh/dt).
=> a * (2gh)^0.5 = A * (dh/dt).
=> a * (2g)^0.5 * dt = A * (dh/(h^0.5)).

Integrating above gives the result.

Dn Bhaskar said: (Dec 8, 2017)  
Right @Ashutosh.

Ayush said: (Oct 5, 2019)  
Let at time t height of water in container is h and after time dt, dh height of water flows out. At this postition dQ/dt = sqrt(2gh)*a.


Integrating this with lower limit 0 and higher limit H we will get the answer.

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