Problem Analysis: Cistern Filling with Pipes

The question asks us to find the time it takes for a certain number of pipes with a different diameter to fill the same cistern, given the time it takes for a different number of pipes with the original diameter. The key factors are the number of pipes, their diameter, and the flow speed, which is stated to be the same.

The rate at which a pipe fills a cistern (its flow rate) depends on the cross-sectional area of the pipe and the speed of the fluid flow. Since the speed of flow is constant, the flow rate of a single pipe is proportional to its cross-sectional area. The cross-sectional area of a circular pipe is given by the formula for the area of a circle, which is $\pi r^2$ or $\pi (d/2)^2 = \frac{\pi d^2}{4}$, where $r$ is the radius and $d$ is the diameter.

This means the flow rate is proportional to the square of the pipe's diameter ($d^2$).

Understanding Pipe Flow Rate and Diameter

Let $d_1$ be the original diameter of the pipes.

The area of an original pipe is $A_1 = \frac{\pi d_1^2}{4}$.

The flow rate of one original pipe, $R_1$, is proportional to $A_1$. We can write $R_1 = k \cdot A_1 = k \cdot \frac{\pi d_1^2}{4}$ for some constant $k$ (related to flow speed and $\pi/4$). For simplicity in proportional calculations, we can just say $R_1 \propto d_1^2$. Let's assume $R_1$ is just proportional to $d_1^2$, so $R_1 = c \cdot d_1^2$ for some constant $c$.

Original Setup: 15 Pipes

Number of pipes = 15.

Diameter = $d_1$.

Flow rate per pipe = $R_1 = c \cdot d_1^2$.

Total flow rate for 15 pipes = $15 \times R_1 = 15 \times c \cdot d_1^2$.

Time taken to fill the cistern = 12 minutes.

Let $V$ be the volume of the cistern.

The relationship between volume, total flow rate, and time is: Volume = Total Flow Rate × Time.

So, $V = (15 \times c \cdot d_1^2) \times 12 = 180 \times c \cdot d_1^2$.

New Setup: 6 Pipes with Double Diameter

Number of pipes = 6.

New diameter, $d_2$, is double the original diameter, so $d_2 = 2d_1$.

The flow rate of one new pipe, $R_2$, is proportional to the square of the new diameter, $d_2^2$.

$R_2 = c \cdot d_2^2 = c \cdot (2d_1)^2 = c \cdot (4d_1^2) = 4 \times (c \cdot d_1^2) = 4 \times R_1$.

So, one new pipe has a flow rate 4 times that of an original pipe because its diameter is double (and area is proportional to diameter squared, $2^2=4$).

Total flow rate for 6 new pipes = $6 \times R_2 = 6 \times (4 \times R_1) = 24 \times R_1$.

Calculating Time for 6 Pipes

Let $T$ be the time taken for the 6 new pipes to fill the same cistern.

Using the formula Volume = Total Flow Rate × Time:

$V = (24 \times R_1) \times T$.

We know from the original setup that $V = 180 \times R_1$.

So, we can equate the two expressions for $V$:

$180 \times R_1 = 24 \times R_1 \times T$.

We can divide both sides by $R_1$ (since $R_1$ is a flow rate, it's not zero):

$180 = 24 \times T$.

Now, solve for $T$:

$T = \frac{180}{24}$.

Step-by-Step Calculation

1. Understand that the flow rate of a pipe is proportional to the square of its diameter. If diameter doubles, flow rate increases by $2^2 = 4$ times.
2. Calculate the effective total flow rate for the original setup: 15 pipes × Rate of one original pipe. Let the rate of one original pipe be $R$. Total original rate = $15R$.
3. Calculate the total volume of the cistern: Volume = Total original rate × Time = $15R \times 12 = 180R$.
4. Calculate the rate of one new pipe: Diameter is double ($2d$), so rate is $4 \times R = 4R$.
5. Calculate the effective total flow rate for the new setup: 6 pipes × Rate of one new pipe = $6 \times 4R = 24R$.
6. Calculate the time taken for the new pipes: Time = Volume / New total rate = $\frac{180R}{24R}$.
7. Simplify the expression to find the time: $\frac{180}{24} = \frac{30 \times 6}{4 \times 6} = \frac{30}{4} = \frac{15 \times 2}{2 \times 2} = \frac{15}{2} = 7.5$ minutes.

Detailed Calculation Steps

Original Number of pipes $N_1 = 15$. Original Diameter $D_1$. Time $T_1 = 12$ minutes.

Flow rate per original pipe $R_1 \propto D_1^2$.

Total flow rate $F_1 = N_1 \times R_1 = 15 \times R_1$.

Volume of cistern $V = F_1 \times T_1 = 15 \times R_1 \times 12 = 180 R_1$.

New Number of pipes $N_2 = 6$. New Diameter $D_2 = 2 D_1$.

Flow rate per new pipe $R_2 \propto D_2^2 = (2 D_1)^2 = 4 D_1^2$. So, $R_2 = 4 R_1$.

Total flow rate $F_2 = N_2 \times R_2 = 6 \times (4 R_1) = 24 R_1$.

Time taken $T_2 = \frac{V}{F_2} = \frac{180 R_1}{24 R_1}$.

$T_2 = \frac{180}{24} = \frac{15 \times 12}{2 \times 12} = \frac{15}{2} = 7.5$ minutes.

Conclusion on Cistern Filling Time

Based on the calculation, it will take 7.5 minutes for 6 pipes, each with double the original diameter, to fill the same cistern at the same speed of flow.

Revision Table: Pipes and Cistern Filling

Additional Information: Work Rate Problems Explained

Problems involving multiple entities working together to complete a task, like pipes filling a cistern or workers building a wall, are known as work rate problems. The fundamental principle is that the total amount of work done is equal to the rate of work multiplied by the time taken.

• Work = Rate × Time

In this specific problem:

• The "work" is filling the cistern, which has a fixed volume $V$.
• The "workers" are the pipes.
• The "rate" is the flow rate of the pipes. The total rate is the sum of the rates of all active pipes.

When dealing with pipes, the rate of a single pipe is determined by its capacity (cross-sectional area) and the speed of the fluid. Assuming constant speed, the rate is proportional to the area ($\pi r^2$ or $\pi d^2/4$). Thus, Rate $\propto$ Diameter$^2$. This square relationship is crucial in solving problems where pipe diameters change.

If you have $N$ identical pipes, the total rate is $N$ times the rate of a single pipe. If the pipes are not identical (like having different diameters), you must calculate the rate of each type of pipe based on its characteristics and then sum the rates of all active pipes to get the total rate.