Solving Leaky Tank Dilemmas: A Comprehensive Guide

When dealing with fluid dynamics and leaky tanks, understanding the relationship between the rate of water flow and the rate of water leakage can be crucial. This article will guide you through the steps to solve a classic problem involving a leaky tank, providing a detailed and step-by-step solution.

Case Study: The Leaky Tank Problem

The problem statement: A tank has a leak that would empty the tank in 20 minutes. A tap is turned on which admits 1 liter a minute into the cistern and it is emptied in 24 minutes. How many liters does the tank hold?

Solution Steps

To solve this problem, we can assume the tank's capacity is X liters. Let#39;s approach this in a structured manner, as detailed below.

Step 1: Determine the Leak Rate

Initially, we know the leak empties the tank in 20 minutes. Therefore, the rate of the leak can be calculated as:

Leak rate Tank capacity/20 minute
Leak rate X/20 (liters per minute)

Step 2: Account for Water Inlet

A tap admits 1 liter of water per minute. Given that the tank is now emptied in 24 minutes, the combined rate of the leak and the tap needs to be considered.

Let#39;s assume in 24 minutes, Y liters are added by the tap and subsequently removed by the leak. Hence, the effective drainage rate becomes:

Effective drainage rate Leak rate - Tap rate
Effective drainage rate X/20 - 1 (liters per minute)

Since the tank is emptied in 24 minutes, we can say:

Tank capacity/24 minute X/20 - 1 (liters per minute)

Step 3: Putting It All Together

Now, we set up the equation:

X/20 - 1 X/24 1 (since 24 minutes are required for the tank to be emptied with the tap now admitted)

Let#39;s solve this step-by-step:

X/20 - 1 X/24 1

X/20 - X/24 2

12X - 1 240 (multiplying through by 120 to clear the fractions)

2X 240

X 120 / 2 14400

Thus, the tank accommodates 14,400 liters of water.

Alternative Approaches

Various approaches can be employed to solve this problem. We present two additional methods for solving the given problem:

Method 1: Using Siproportional Rates

Let the capacity of the tank be V liters.

When the tap is off, the rate of leaking is V/10 liters per hour.

When the tap is on, the net rate of draining is (V - 2880)/12 liters per hour, since it takes 12 hours to drain 2880 liters.

The net rate of draining is equal to the initial draining rate:

V/10 (V - 2880)/12

Solving this equation yields the same result: V 14,400 liters.

Method 2: Direct Substitution

Assuming the tank holds X liters of water:

The tap admits 6 liters of water in 1 minute, thus 360 liters in 1 hour. Hence, X liters of water can be filled in X/360 hours.

The leak alone can empty the tank in 10 hours, so in one hour, 1/10 of the tank becomes empty.

With both the tap and leak working simultaneously, the tank is emptied in 10 hours, implying:

1/10 - 1/360 1/12

36 - 1 30

X 360 * 40 14400

Thus, the tank's capacity is 14,400 liters.

Conclusion

By understanding the dynamics of leak rates and inflow rates, you can effectively determine the capacity of a tank in the face of leak problems. This knowledge is not only useful in practical situations but also in theoretical fluid mechanics and engineering.