Calculating Tank Capacity: A Practical Example of Inlet and Outlet Rates

Understanding how to calculate tank capacity when dealing with both inlet and outlet rates is crucial for various engineering and management applications. This article provides a detailed yet straightforward solution to a common problem: determining the capacity of a tank given the rates of a leak and an inlet pipe.

Problem Statement and Solution Outline

The problem revolves around a tank that can be emptied by a leak in 6 hours and can be filled by an inlet pipe at a rate of 4 liters per minute. When both the inlet and the leak are active, the tank takes 8 hours to empty. We need to determine the capacity of the tank.

Step 1: Determining the Leak Rate

We start by calculating the leak rate. If the leak can empty the full tank in 6 hours, its rate, denoted as ( L ), is:

$$ L frac{1 text{ tank}}{6 text{ hours}} frac{1}{6} text{ tank/hour} $$

Converting this to liters per minute, we use the fact that there are 60 minutes in an hour:

$$ L frac{1}{6} text{ tank/hour} times frac{1 text{ hour}}{60 text{ minutes}} frac{1}{360} text{ tank/min} $$

Step 2: Defining the Tank Capacity

Let the capacity of the tank be ( C ) liters. Therefore, the leak rate in liters per minute is:

$$ text{Leak rate} frac{C}{360} text{ L/min} $$

Step 3: Determining the Inlet Rate

The inlet pipe fills water at the rate of 4 liters per minute. So the inlet rate, ( Q_{text{in}} ), is:

$$ Q_{text{in}} 4 text{ L/min} $$

Step 4: Effect of Both Inlet and Outlet

When both the inlet and the leak are active, the effective rate at which the water is leaving the tank is:

$$ text{Net rate} Q_{text{in}} - text{Leak rate} 4 - frac{C}{360} text{ L/min} $$

The problem states that the tank empties in 8 hours or 480 minutes with both the inlet and the leak open. Thus, the effective rate of emptying the tank is:

$$ text{Net rate} frac{C}{480} text{ L/min} $$

Step 5: Setting Up the Equation

Equating the net rate to the effective rate of emptying, we get:

$$ 4 - frac{C}{360} -frac{C}{480} $$

Rearranging this equation:

$$ 4 frac{C}{360} - frac{C}{480} $$

Finding a common denominator (1440) and simplifying:

$$ 4 frac{4C}{1440} - frac{3C}{1440} $$

$$ 4 frac{C}{1440} $$

Multiplying both sides by 1440:

$$ C 4 times 1440 5760 text{ liters} $$

Therefore, the capacity of the tank is 5760 liters.

Alternative Solution

Another approach involves using the given steps. If a leak drains 12/4X in 12 hours, and an inlet fills 12 x 60 x 10 liters, then the equation becomes 12 x 60 x 10 - X 3X. Solving this gives:

$$ 12 x 60 x 10 - X 3X $$

$$ 12 x 60 x 10 4X $$

$$ X frac{12 x 60 x 10}{4} $$

$$ X 7200 div 2 $$

$$ X 3600 text{ liters} $$

This confirms the capacity of the tank as 3600 liters.

Conclusion

In conclusion, solving for the capacity of a tank given the rates of a leak and an inlet pipe involves carefully considering the net effect of both rates. By setting up the appropriate equations and solving them, we can accurately determine the tank's capacity, which is vital for various engineering applications. Whether the method involves balancing rates or using alternative equations, the solution provides a practical approach to solving such problems.