Calculating the Rate at Which Water Level in a Cylindrical Tank Rises

Water is poured into a cylindrical tank 5 meters in diameter at a rate of 200 liters per minute. This article explores how to find the rate at which the water level in the tank rises using dimensional analysis and the formulas for the volume of a cylinder. We will walk through the process step-by-step to ensure clarity and understanding.

Dimensional Analysis and the Problem Setup

Given a cylindrical tank with a diameter of 5 meters, we need to determine the rate at which the water level in the tank will rise. We know that water is being added at a rate of 200 liters per minute. The first step is to convert the diameter into the radius and then calculate the cross-sectional area of the tank.

Convert the Diameter to Radius

The radius (r) of the cylindrical tank can be calculated as follows:

Radius r frac{Diameter}{2} frac{5 , text{m}}{2} 2.5 , text{m}

Calculate the Cross-Sectional Area of the Tank

The cross-sectional area (A) of a cylinder is given by the formula:

A pi r^2

Substituting the radius:

A pi (2.5 , text{m})^2 pi times 6.25 , text{m}^2 approx 19.6349 , text{m}^2

Convert the Flow Rate from Litres to Cubic Meters

Since 1 liter 0.001 cubic meters, the flow rate can be converted as:

200 , text{litres/min} 200 times 0.001 , text{m}^3/text{min} 0.2 , text{m}^3/text{min}

Find the Rate of Rise of the Water Level

The volume of water (V) added to the tank can be expressed in terms of the height (h) of the water:

V A cdot h

Using calculus, we can find the rate of change of volume with respect to time:

frac{dV}{dt} A cdot frac{dh}{dt}

Rearranging to find (frac{dh}{dt}):

frac{dh}{dt} frac{1}{A} cdot frac{dV}{dt}

Substituting the known values:

frac{dh}{dt} frac{1}{19.6349 , text{m}^2} cdot 0.2 , text{m}^3/text{min} approx 0.0102 , text{m/min}

Therefore, the level of the water in the tank rises at approximately 0.0102 meters per minute or about 10.2 mm per minute.

Dimensional Analysis for a Simplified Approach

We can also solve this problem using dimensional analysis. The volume of a cylinder is given by:

V A cdot h pi frac{D^2}{4} cdot h

Dividing both sides of the equation by time (t):

frac{V}{t} pi frac{D^2}{4} cdot frac{h}{t}

The left-hand side of the equation is (frac{V}{t}), which is the volume flow rate. And (frac{h}{t}) is the rate at which the level is changing.

Given the flow rate in liters per minute:

200 , text{litres/min} times frac{text{m}^3}{1000 , text{litres}} 0.2 , text{m}^3/text{min}

Using the diameter of the tank:

0.2 , text{m}^3/text{min} pi frac{5^2}{4} cdot frac{h}{t}

Solving for (frac{h}{t}):

frac{h}{t} 0.0102 , text{m/min} 10.2 , text{mm/min}