Question

The plumbing system of a house involves a \(0.5-\mathrm{m}\) section of a plastic pipe \((k=0.16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of inner diameter \(2 \mathrm{~cm}\) and outer diameter \(2.4 \mathrm{~cm}\) exposed to the ambient air. During a cold and windy night, the ambient air temperature remains at about \(-5^{\circ} \mathrm{C}\) for a period of \(14 \mathrm{~h}\). The combined convection and radiation heat transfer coefficient on the outer surface of the pipe is estimated to be \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\). Assuming the pipe to contain stationary water initially at \(0^{\circ} \mathrm{C}\), determine if the water in that section of the pipe will completely freeze that night.

Short answer

Answer: Yes, the water in the pipe section will completely freeze during the 14-hour period, as the heat loss due to convection and radiation is greater than the heat required to completely freeze the water.

Step-by-step solution

Calculate the surface area of the exposed pipe section

The surface area of a cylinder can be calculated using the formula \(A = 2\pi L (R_1 + R_2)\), where \(L\) is the length of the cylinder, \(R_1\) is the inner radius, and \(R_2\) is the outer radius. In our case, \(L = 0.5\ \text{m}\), \(R_1 = 1\ \text{cm}\), and \(R_2 = 1.2\ \text{cm}\). First, we convert radii from centimeters to meters: \(R_1 = 0.01\ \text{m}\) and \(R_2 = 0.012\ \text{m}\) Then, we calculate the surface area: \(A = 2\pi (0.5\ \text{m})(0.01\ \text{m} + 0.012\ \text{m}) = 0.207\ \text{m}^2\).

Calculate the heat transfer from the pipe's surface to the ambient air

The heat transfer from the pipe's surface to the ambient air due to convection and radiation can be calculated using the formula \(Q = hA(T_s-T_\infty)\), where \(h\) is the combined convection and radiation heat transfer coefficient, \(A\) is the surface area of the pipe, \(T_s\) is the surface temperature of the pipe, and \(T_\infty\) is the ambient air temperature. In our case, \(h = 40\ \text{W/m}^2 \cdot\text{K}\), \(A = 0.207\ \text{m}^2\), \(T_s = 0^\circ\text{C}\), and \(T_\infty = -5^\circ\text{C}\). Using these values, we can calculate the heat transfer: \(Q = (40\ \text{W/m}^2 \cdot\text{K})(0.207\ \text{m}^2)(0^\circ\text{C} - (-5^\circ\text{C})) = 41.4\ \text{W}\).

Calculate the amount of heat loss during the 14-hour period

Now we need to calculate the total amount of heat loss during the 14-hour period. This can be done using the formula \(Q_\text{total} = Qt\), where \(Q\) is the heat transfer rate, and \(t\) is the time. In our case, \(Q = 41.4\ \text{W}\) and \(t = 14\ \text{h}\). First, we convert the time from hours to seconds: \(t = 14\ \text{h} \times 3600\ \text{s/h} = 50,\!400\ \text{s}\) Then, we calculate the total amount of heat loss: \(Q_\text{total} = (41.4\ \text{W})(50,\!400\ \text{s}) = 2,\!085,\!760\ \text{J}\).

Calculate the volume of water in the pipe section

We can find the volume of water in the pipe section by finding the volume of the cylinder with radius \(R_1\) and length \(L\) and then subtracting the volume of the cylinder with radius \(R_2 \) and length \(L\). The volume of a cylinder can be calculated using the formula \(V = \pi R^2 L\). In our case, \(R_1 = 0.01\ \text{m}\), \(R_2 = 0.012\ \text{m}\), and \(L = 0.5\ \text{m}\). So, the volume of water in the pipe section is: \(V_\text{water} = \pi (0.012^2 - 0.01^2)(0.5\ \text{m}) = 0.0004712\ \text{m}^3\).

Calculate the mass of water in the pipe section

Now we need to calculate the mass of water in the pipe section using the formula \(m = \rho V\), where \(\rho\) is the density of water and \(V\) is the volume of water. In our case, \(\rho = 1000\ \text{kg/m}^3\) and \(V = 0.0004712\ \text{m}^3\). So, the mass of water in the pipe section is: \(m = (1000\ \text{kg/m}^3)(0.0004712\ \text{m}^3) = 0.4712\ \text{kg}\).

Calculate the amount of heat that needs to be removed from the water to completely freeze it

We can find the amount of heat that needs to be removed from the water to completely freeze it using the formula \(Q_\text{freeze} = mL_f\), where \(m\) is the mass of water and \(L_f\) is the heat of fusion of water. In our case, \(m = 0.4712\ \text{kg}\) and \(L_f = 333.7\ \text{kJ/kg} = 333,\!700\ \text{J/kg}\). So, the amount of heat that needs to be removed from the water to completely freeze it is: \(Q_\text{freeze} = (0.4712\ \text{kg})(333,\!700\ \text{J/kg}) = 157,\!218\ \text{J}\).

Compare the heat loss via convection and radiation to the heat required to freeze the water

Now we compare the total amount of heat loss due to convection and radiation, \(Q_\text{total}\), to the amount of heat required to completely freeze the water, \(Q_\text{freeze}\): \(Q_\text{total} = 2,\!085,\!760\ \text{J}\) \(Q_\text{freeze} = 157,\!218\ \text{J}\) Since \(Q_\text{total} > Q_\text{freeze}\), this means that the water in the pipe section will completely freeze during the 14-hour period.

Key concepts

Convection and Radiation Heat Transfer

Understanding how heat transfers through the environment is crucial for various applications, including the freezing of water in pipes as seen in the exercise. Convection and radiation are two of the three modes of heat transfer, with conduction being the third.

Convection is the transfer of heat by the movement of fluids, which can be either liquids or gases. It occurs when a fluid near a hot surface gains heat, becomes less dense, and rises, while cooler fluid moves in to replace it, forming a circulation pattern. In the context of the exercise, we have a cold ambient air temperature that causes heat to be transferred away from the water in the pipes through the air.

Radiation refers to the transfer of heat in the form of electromagnetic waves. All objects emit radiation, and the intensity depends on their temperature. In the exercise, the pipe's surfaces radiate heat to the cooler surroundings.

The combined effect of convection and radiation is represented by a heat transfer coefficient in the exercise. This coefficient is used to calculate the heat loss from the pipe's surface to the surrounding air. It's important to understand that the heat transfer rate here depends not just on the heat transfer coefficient, but also on the surface area of the pipe and the temperature difference between the pipe surface and the ambient air.

Thermal Conductivity

The rate at which heat is transferred through a material is defined by its thermal conductivity, usually represented by the symbol 'k'. Thermal conductivity is a measure of a material's ability to conduct heat; it appears as a constant in the exercise for the plastic pipes. The units for thermal conductivity are \( \mathrm{W/m}\cdot\mathrm{K} \), entailing the amount of heat \( \mathrm{W} \) that can pass through a meter \( \mathrm{m} \) of the material with a thickness of one meter for every degree of temperature difference \( \mathrm{K} \).

In the context of our exercise, the relatively low thermal conductivity of plastic (0.16 \( \mathrm{W/m}\cdot\mathrm{K} \) in this case) means that heat is not efficiently conducted through the pipe's material, which contributes to the cooling and potential freezing of the water within. High thermal conductivity materials would transfer heat more quickly, thus less prone to allow the water inside to freeze under similar conditions.

Heat of Fusion of Water

The heat of fusion is the amount of energy needed to change a substance from the solid phase to the liquid phase at its melting point without changing its temperature. For water, this is particularly important as it describes the energy required to melt ice or conversely, the energy released when water freezes. This parameter is represented in the exercise by \( L_f \) and for water, it is typically 333.7 \( \mathrm{kJ/kg} \).

To determine whether the water in the pipes will freeze, the mass of the water (calculated in the exercise steps) is multiplied by the heat of fusion. The resulting value corresponds to the total amount of heat that needs to be extracted for the water to become ice. By comparing this with the total heat loss over the 14-hour period due to convection and radiation, we can predict whether the water will completely freeze, which is the final step of the exercise.