Problem 32 - Physics

Question

Many hot-water heating systems have a reservoir tank connected directly to the pipeline, so as to allow for expansion when the water becomes hot. The heating system of a house has \(76 \mathrm{~m}\) of copper pipe whose inside radius is \(9.5 \times 10^{-3} \mathrm{~m}\). When the water and pipe are heated from 24 to \(78^{\circ} \mathrm{C}\), what must be the minimum volume of the reservoir tank to hold the overflow of water?

Step-by-Step Solution

Answer

The reservoir tank must hold at least 0.00024617 m³ of overflow water.

Step 1: Determine the Initial Volume of Water

First, we need to calculate the initial volume of water in the copper pipe before heating. This volume is a cylinder, which is calculated using the formula \(V = \pi r^2 h\), where \(r\) is the radius and \(h\) is the height (length) of the pipe. Here \(r = 9.5 \times 10^{-3} \text{ m}\) and \(h = 76 \text{ m}\).

Step 2: Calculate the Initial Volume

Plug the values into the formula: \(V = \pi (9.5 \times 10^{-3})^2 \times 76\). This results in:\[V = \pi \times 9.025 \times 10^{-5} \times 76 = 0.021643 \text{ m}^3\].

Step 3: Determine the Coefficient of Thermal Expansion for Water

The coefficient of volume expansion for water is approximately \(\beta = 210 \times 10^{-6} \text{ per } ^\circ\text{C}\). This will be used to calculate the change in volume due to temperature change.

Step 4: Calculate the Volume Expansion

The change in volume \( \Delta V \) can be calculated using the formula \( \Delta V = \beta V_0 \Delta T \), where \(V_0\) is the initial volume (0.021643 m³) and \(\Delta T\) is the temperature change. Here, \(\Delta T = 78 - 24 = 54 ^\circ \text{C}\).

Step 5: Compute the Change in Volume

Plug these values into the formula: \( \Delta V = 210 \times 10^{-6} \times 0.021643 \times 54 \). Calculating this gives \( \Delta V \approx 2.4617 \times 10^{-4} \text{ m}^3\).

Step 6: Conclusion - Minimum Volume of the Reservoir Tank

The reservoir tank must be able to accommodate the overflow, which is \( \Delta V \), the change in water volume due to heating. This means the tank should have a minimum capacity of 0.00024617 m³ to hold the overflow.

Key Concepts

Volume Calculation

When dealing with thermal expansion in heating systems, understanding volume calculation is crucial. Here, we need to find the initial volume of water inside a copper pipe before it is heated. The pipe is in the shape of a cylinder, and the formula for the volume of a cylinder is given by \[ V = \pi r^2 h \]where:

• \( V \) = volume,
• \( r \) = radius of the cylinder's base,
• and \( h \) = height (or length) of the cylinder.

For our copper pipe:

• the radius \( r = 9.5 \times 10^{-3} \text{ m} \)
• the height \( h = 76 \text{ m} \).

By substituting these values into the formula, we determine the initial volume of the water inside the pipe as approximately \( 0.021643 \text{ m}^3\).This volume calculation is the starting point for understanding how much space the water will require after thermal expansion when heated.

Copper Pipe

Copper pipes are commonly used in plumbing due to their excellent heat conduction and durability. With a copper pipe, it's important to consider not just the volume of the water inside it, but how the pipe itself might expand with temperature changes.Copper expands when heated, albeit less so than water. The inside radius of the copper pipe in this task, given as \(9.5 \times 10^{-3} \text{ m} \), tells us only about the space for the water flow. Any expansion due to heat will minimally affect the piping but can slightly impact the flow.When designing systems that involve heating fluids, it is essential to account for both the pipe and fluid expansion to ensure seamless performance.

Coefficient of Thermal Expansion

The coefficient of thermal expansion represents how much a material expands per unit temperature change. For water, this is symbolized by \( \beta \) and has a typical value of \( 210 \times 10^{-6} \text{ per } ^\circ\text{C} \).This coefficient is vital in calculating the volume change of water when it is heated. The formula used is:\[ \Delta V = \beta V_0 \Delta T \]where:

• \( \Delta V \) = change in volume,
• \( V_0 \) = initial volume,
• and \( \Delta T \) = temperature change.

In this exercise, the temperature change \( \Delta T = 54 ^\circ \text{C} \), calculated from \( (78 - 24) ^\circ \text{C} \). By using the coefficient of thermal expansion in our calculation, we find that the water's volume increase is about \( 2.4617 \times 10^{-4} \text{ m}^3 \).This tells us exactly how much more space the water will need as it expands with heat.

Reservoir Tank

A reservoir tank in heating systems is a critical component that holds the overflow of water as it expands due to heating. This process avoids damage to the system by preventing pressure build-up.To determine the size of the reservoir tank, you calculate the increased volume of the water when heated. This additional volume, roughly \( 0.00024617 \text{ m}^3 \) in this situation, is the capacity the reservoir tank must handle.The reservoir tank should be slightly larger than this value to account for any unexpected variations or measurement errors, ensuring the system operates safely and effectively. This allows the heating system to function smoothly without causing damage or leakage.