# Hemispherical

Time of emptying a hemispherical tank through an orifice at its bottom

**Contents**

## Overview

Consider a hemispherical tank, containing some liquid and having an orifice at its bottom as shown in figure. Let,- = Radius of the tank
- = Initial height of the liquid
- = Final height of the liquid
- a = Area of the orifice

At some instant, let the height of the liquid be *h* above the orifice.

*r*be the radius of the liquid surface. Then the surface area of the liquid, After a small interval of time

*dt*, let the liquid level fall down by the amount

*dh*. Therefore volume of the liquid that has passed in time

*dt*,

The value of

We know that the volume of liquid that has passed through the orifice in time *dh*is taken as negative, as its value will decrease with the increase in discharge.*dt*, = Coefficient of discharge Area Theoretical velocity Time Equating equations (1) and (2) From the geometry of the tank, we find that, Substituting this value of in equation (3) Now the total time

*T*required to bring the liquid level from to may be found out by integrating the equation (4) between the limits to i.e., If the vessel is to be completely emptied, then putting in this equation, If the vessel was full at the time of the commencement and is to be completely emptied, then putting in the above equation,

Example:

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##### Example - Time of emptying a hemispherical tank

Problem

A hemispherical tank of 2 meters radius contains water up to a depth 1meter. Find the time taken to discharge the tank completely through an orifice of 1000 mm

^{2}provided at the bottom. Take = 0.62.Workings

Given,

- = 2m
- = 1m
- = 1000 mm
^{2} - = 0.62

Solution

Time taken to discharge the tank completely = 43 min 13 sec