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This question belongs to Mechanical Engineering Hydraulics And Fluid Mechanics In ME

Hydraulics and Fluid Mechanics in ME

?

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

A) $$\frac{{2{\text{A}}\sqrt {{{\text{H}}_1}} }}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$ B) $$\frac{{2{\text{A}}{{\text{H}}_1}}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$ C) $$\frac{{2{\text{AH}}_1^{\frac{3}{2}}}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$ D) $$\frac{{2{\text{AH}}_1^2}}{{{{\text{C}}_{\text{d}}} \times {\text{a}}\sqrt {2{\text{g}}} }}$$

Answer & Solution

Correct Answer: Option A

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Question posted by Examveda

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Madhavan Palmani 7 years ago

should be.

Volume/(Discharge).

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

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

Why is there an extra 2 in the numerator

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