Question

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

Short answer

Based on the given information, the required thickness of glass wool insulation to decrease the temperature rise of the water in the pipe is approximately 2.96 cm.

Step-by-step solution

Calculate the heat transfer rate without insulation

First, we need to find the rate of heat transfer in the original, non-insulated pipe. This can be found using the following formula: $$q = m \times C_p \times \Delta T$$ where \(q\) is the heat transfer rate (W), \(m\) is the mass flow rate of the water (\(0.98 kg/s\)), \(C_p\) is the specific heat capacity of water (\(4.18 kJ/kg \cdot K\), or \(4180 J/kg \cdot K\)), and \(\Delta T\) is the temperature rise of the water (\(8^{\circ}C - 7^{\circ}C = 1^{\circ}C\)). Now, let's calculate the heat transfer rate q: $$q = 0.98 kg/s \times 4180 J/kg \cdot K \times 1 K = 4096.4 W$$

Calculate the heat transfer rate with insulation

Next, we need to find the rate of heat transfer when the temperature rise of the water is decreased to \(0.25^{\circ}C\). The formula is the same as in Step 1: $$q_{insulated} = m \times C_p \times \Delta T_{insulated}$$ with \(\Delta T_{insulated} = 0.25^{\circ}C\). Let's calculate the heat transfer rate \(q_{insulated}\): $$q_{insulated} = 0.98 kg/s \times 4180 J/kg \cdot K \times 0.25 K = 1024.1 W$$

Calculate the insulation resistance

To decrease the heat transfer rate, we need to add insulation resistance to the system. The resistance to heat transfer for the insulated system can be calculated using the following formula: $$R_{insulated} = \frac{T_{air} - T_{out}}{q_{insulated}} - \frac{T_{in} - T_{out}}{q}$$ where \(T_{air}\) is the ambient air temperature (\(30^{\circ}C\)), \(T_{in}\) is the inlet water temperature (\(7^{\circ}C\)), and \(T_{out}\) is the outlet water temperature with insulation (\(7^{\circ}C + 0.25^{\circ}C = 7.25^{\circ}C\)). Now, let's find the insulation resistance \(R_{insulated}\): $$R_{insulated} = \frac{30^{\circ}C - 7.25^{\circ}C}{1024.1 W} - \frac{7^{\circ}C - 7.25^{\circ}C}{4096.4 W} = 0.0222 ~m^2 \cdot K / W$$

Calculate the required insulation thickness

Finally, we can calculate the required insulation thickness (\(\delta\)) using the formula: $$\delta = \frac{R_{insulated} \cdot k}{2 \cdot \pi \cdot L}$$ where \(k\) is the thermal conductivity of the insulation (\(0.05 W / m \cdot K\)), \(L\) is the length of the pipe (\(150 m\)), and \(2 \cdot \pi \cdot L\) is the total surface area of the cylindrical pipe. Let's calculate the required insulation thickness \(\delta\): $$\delta = \frac{0.0222 ~m^2 \cdot K / W \cdot 0.05 W / m \cdot K}{2 \cdot \pi \cdot 150 m} = 0.0296 m$$ So, the required thickness of the insulation is approximately \(2.96 cm\).

Key concepts

Thermal Insulation

To fully understand thermal insulation and its significance in the context of heat transfer in pipes, let's start by picturing a pipe transporting chilled water that's exposed to warmer air. The difference in temperature between the water and the air naturally drives heat transfer from the air to the water, raising the water's temperature.

Thermal insulation is a material or method used to reduce this unwanted heat exchange. In our exercise, glass wool insulation, which has a low thermal conductivity (\(k = 0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\)), is used to wrap the pipe and slow down the heat transfer.

Why is Thermal Conductivity Important?

Thermal conductivity is a measure of a material's ability to conduct heat. Materials with low thermal conductivity are good insulators because they do not allow heat to easily pass through them. Conversely, materials with high thermal conductivity are good conductors and can quickly transfer heat.To achieve effective insulation, it's critical to determine the required thickness of the insulation layer. In the exercise, this was done by calculating the drop in heat transfer rate that resulted from applying the insulation. An optimal thickness ensures energy efficiency by keeping the temperature rise within the desired limit, thereby saving on cooling costs and maintaining system performance.

Mass Flow Rate

When we talk about mass flow rate in pipes, we refer to the amount of mass moving through a pipe per unit time. In the context of our exercise, the mass flow rate is given as \(0.98 \mathrm{~kg} / \mathrm{s}\), signifying that 0.98 kilograms of water pass through the pipe each second.

Mass flow rate is a pivotal factor in determining both the initial and the insulated pipe's heat transfer rates. It helps us to quantify the total energy that needs to be handled by the cooling system to prevent excessive temperature rise.

Role of Mass Flow Rate in Heat Transfer Calculations

In systems involving heat transfer, altering the mass flow rate affects how much heat is transferred over time. The higher the mass flow rate, the greater the potential for heat absorption or release, and vice versa. This relationship is utilitarian in controlling the temperature of a fluid by changing how much of it flows through the system.

Specific Heat Capacity

Specific heat capacity is a property that measures the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). In the given exercise, water's specific heat capacity is \(4.18 \mathrm{~kJ/kg} \cdot \mathrm{K}\), or \(4180 \mathrm{~J/kg} \cdot \mathrm{K}\).

That means for each kilogram of water, 4180 joules of energy will raise its temperature by one Kelvin. This quantity is intrinsic to calculating the heat transfer rate before and after insulation because it directly relates to the energy absorbed or released by the water as its temperature changes.

Impact on Energy Calculations

Specific heat capacity enables us to compute the thermal energy change as water flows through the pipe. It's a fundamental factor in the formula used to calculate heat transfer rate (\(q = m \times C_p \times \Delta T)\), the product of mass flow rate, specific heat capacity, and temperature change. By understanding specific heat capacity, we can anticipate how much water's temperature will rise or fall when subjected to heat transfer, which is essential for designing thermal systems and ensuring their efficiency.