Understanding Water Tank Emptying Time and Pipe Diameter

The time it takes to empty a water tank using a pipe depends directly on the flow rate through the pipe. The flow rate, in turn, is proportional to the cross-sectional area of the pipe. A larger area allows more water to flow per unit time, thus reducing the time needed to empty the tank.

Calculating Pipe Cross-Sectional Area

The cross-sectional area of a circular pipe is given by the formula for the area of a circle:

\(A = \pi r^2\)

where \(r\) is the radius of the pipe. Since the diameter \(d\) is twice the radius (\(d = 2r\)), we can also write the area in terms of diameter:

\(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\)

Comparing Areas for Different Diameters

Let's compare the cross-sectional area for the two pipes mentioned in the question:

1. For a pipe with diameter \(d\), the area is \(A_1 = \frac{\pi d^2}{4}\).
2. For a pipe with diameter \(2d\), the diameter is doubled. Let's call the new diameter \(d_2 = 2d\). The new area \(A_2\) is:

\(A_2 = \frac{\pi (d_2)^2}{4} = \frac{\pi (2d)^2}{4} = \frac{\pi (4d^2)}{4} = \pi d^2\)

Now, let's find the ratio of the new area \(A_2\) to the original area \(A_1\):

\(\frac{A_2}{A_1} = \frac{\pi d^2}{\frac{\pi d^2}{4}} = \frac{\pi d^2}{1} \times \frac{4}{\pi d^2} = 4\)

This shows that doubling the pipe diameter increases the cross-sectional area by a factor of 4.

Relating Flow Rate and Emptying Time

The flow rate (volume of water flowing per unit time) is directly proportional to the cross-sectional area of the pipe. If the area increases by a factor of 4, the flow rate increases by a factor of 4 (assuming the water pressure or head remains constant or changes predictably, which is implied in typical introductory problems like this).

The total volume of water in the tank is constant. The relationship between volume, flow rate, and time is:

\(\text{Volume} = \text{Flow Rate} \times \text{Time}\)

Therefore, for a constant volume, the time taken is inversely proportional to the flow rate:

\(\text{Time} = \frac{\text{Volume}}{\text{Flow Rate}}\)

If the flow rate increases by a factor of 4, the time taken to empty the same volume will decrease by a factor of 4.

Calculating the New Emptying Time

We are given that the time taken to empty the tank with a pipe of diameter \(d\) (and area \(A_1\)) is 40 minutes.

\(T_1 = 40 \text{ minutes}\)

For the pipe with diameter \(2d\), the area is \(A_2 = 4 A_1\), which means the flow rate is 4 times the original flow rate.

Let \(T_2\) be the time taken with the larger pipe. Since time is inversely proportional to flow rate (and thus area):

\(\frac{T_2}{T_1} = \frac{\text{Flow Rate}_1}{\text{Flow Rate}_2} = \frac{A_1}{A_2} = \frac{A_1}{4 A_1} = \frac{1}{4}\)

So,

\(T_2 = T_1 \times \frac{1}{4}\)

Substituting the given value for \(T_1\):

\(T_2 = 40 \text{ minutes} \times \frac{1}{4}\)

\(T_2 = 10 \text{ minutes}\)

Therefore, it will take 10 minutes for a 2d diameter pipe to empty the water tank.

Let's summarize the key relationship:

The final answer is 10 minutes.

Revision Table: Water Tank Emptying Calculations

Additional Information: Fluid Flow Principles

This problem demonstrates a basic principle in fluid dynamics related to flow rate through pipes. Here are a few related points:

• Flow Rate (Q): Defined as the volume of fluid passing through a point or region per unit of time. Units can be \(m^3/s\) or \(L/min\), etc.
• Continuity Equation: For an incompressible fluid (like water, approximately) flowing in a pipe, the flow rate is constant throughout the pipe, even if the diameter changes (\(A_1 v_1 = A_2 v_2\)), where \(v\) is the flow velocity. In our tank emptying problem, the velocity at the pipe exit is influenced by the height of water in the tank (pressure head).
• Torricelli's Law: This law, derived from Bernoulli's principle, states that the speed \(v\) of efflux of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth \(h\) is the same as the speed that a body would acquire in falling freely from a height \(h\), i.e., \(v = \sqrt{2gh}\). In reality, pipe flow is more complex due to viscosity and pipe shape, but the core idea that larger pipes allow greater flow holds true.
• Practical Considerations: In real-world scenarios, factors like friction within the pipe (viscosity and pipe surface roughness) and minor losses due to bends or valves also affect the flow rate. However, for simplified problems like this, we often assume ideal flow conditions where flow rate is primarily determined by the pipe's cross-sectional area and the driving pressure (like the height of water in the tank).