Pipes and Cistern Concepts

Understanding the Concepts of Pipes and Cistern

Pipes and Cistern is a subtopic of Time & Work - individual efficiency concept with the inclusion of negative work notion.

In this case, time spent filling the Cistern equals time spent doing the work, the volume of the Cistern equals total work, and the speed of filling the Cistern equals work efficiency.

When a pipe is linked to a cistern to replenish it, it is referred to as an inlet.

This is an example of good work. On the other hand, when another pipe is attached to a cistern to draw down the water level, it must be addressed as an outlet or a 'leak.' This denotes a negative type of work done .

Pipes & Cistern terms can be properly represented using the diagram below.

These types of problems can be solved using 2 methods:

1. Fraction Method or Unitary Work Method
2. LCM Method

1. Fraction Method or Unitary Work Method

In this method, we will assume the total work to be 1 unit, calculate the individual efficiency in fractions, and solve the problem.

Let's solve the problem above using the fraction method to understand it better.

Example Problem

Pipes A and B can fill a tank in 30 and 40 hours respectively. Pipe C can empty it in 20 hours. If all the three pipes are opened together, then the tank will be filled in:

Step 1: Assume the total work to be 1.

Step 2: Calculate the individual efficiencies of the given pipes

Here,

• Pipe A will fill the tank in 30 hours. So in 1 hour, it will fill 1/30th of the tank.
• Pipe B will fill the tank in 40 hours. So in 1 hour, it will fill 1/40th of the tank.
• Pipe C will empty the tank in 20 hours. So in 1 hour, it will empty 1/20th of the tank.

Step 3: Calculate the combined efficiency of all the pipes in an hour

Here it is,

Combined work done in 1 hr = ((1/30) + (1/40) - (1/20)), Solving these we get ,

Combined work done in 1 hr = 1/120

Step 4: Taking the inverse of work done in 1hr will give us the total time taken to complete the task.

So, the tank will be filled in 120 hours.

2. LCM Method

In this method, we will assume the total work to be the LCM of the given numbers, calculate the individual efficiency in terms of integers, and solve the problem.

This will help us to solve the problem faster. Let's solve the problem above using the fraction method to understand it better.

Example Problem

Pipes A and B can fill a tank in 30 and 40 hours respectively. Pipe C can empty it in 20 hours. If all the three pipes are opened together, then the tank will be filled in:

Step 1: Assume the total work to be LCM of 30, 40 and 20

Here LCM = 120.

Step 2: Calculate the individual efficiencies of the given pipes with LCM as total work.

Here,

• Pipe A will fill the tank (120 work) in 30 hours. So in 1 hour, it will fill 120/30 of the tank= 4 work/hour.

• Pipe B will fill the tank (120 work) in 40 hours. So in 1 hour, it will fill 120/40 of the tank= 3 work/hour.

• Pipe C will empty the tank (120 work) in 20 hours. So in 1 hour, it will empty 120/20 of the tank=6 work/hour.

Step 3: Calculate the combined efficiency of all the pipes in an hour

Here it is,

Combined work done in 1 hr = 4+3-6. Solving these, we get

Combined work done in 1 hr = 1 work.

Step 4: Divide the total work by the work done in 1hr ⇒ 120/1 ⇒ 120

So, the tank will be filled in 120 hours.