Question

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum \((k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a \(10-\mathrm{cm}\)-thick layer of insulation \((k=0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The rate at which the liquid oxygen gains heat is (a) \(141 \mathrm{~W}\) (b) \(176 \mathrm{~W}\) (c) \(181 \mathrm{~W}\) (d) \(201 \mathrm{~W}\) (e) \(221 \mathrm{~W}\)

Short answer

Answer: The rate at which the liquid oxygen gains heat is approximately 181 W.

Step-by-step solution

Calculate the Temperature Difference Across Each Layer

First, we will find the temperature difference across the aluminum shell and the insulation. The temperature inside the tank is \(90 \mathrm{~K}\) and the ambient air temperature is \(20^{\circ} \mathrm{C} = 293 \mathrm{~K}\). Temperature difference for the aluminum shell: \(\Delta T_{aluminum} = 293 \mathrm{~K} - 90 \mathrm{~K} = 203 \mathrm{~K}\) Temperature difference for the insulation: \(\Delta T_{insulation} = 293 \mathrm{~K} - 90 \mathrm{~K} = 203 \mathrm{~K}\)

Calculate the Heat Transfer Rate through the Aluminum Shell

Using Fourier's law of conduction, we can find the heat transfer through the aluminum shell: $$q_{aluminum} = \frac{k_{aluminum}\cdot A\cdot \Delta T_{aluminum}}{L_{aluminum}}$$ Where $$k_{aluminum} = 170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$$ $$A = \pi \cdot D_{inner} \cdot L$$ (Surface area of the cylindrical shell) $$D_{inner} = 1\,\mathrm{m}$$ (Inner diameter of the storage tank) $$L_{aluminum} = 0.5\,\mathrm{cm} = 0.005\,\mathrm{m}$$ (Thickness of the aluminum shell) $$\Delta T_{aluminum} = 203 \mathrm{~K}$$ (Temperature difference calculated in step 1) First, we calculate the surface area of the aluminum shell: $$A = \pi \cdot D_{inner} \cdot L =\pi \cdot 1\,\mathrm{m} \cdot L$$ Next, we substitute all the values in the formula for heat transfer: $$q_{aluminum} = \frac{170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \pi \cdot 1\,\mathrm{m} \cdot L \cdot 203 \mathrm{~K}}{0.005\,\mathrm{m}}$$

Calculate the Heat Transfer Rate through the Insulation

Using the Fourier's law of conduction, we can find the heat transfer through the insulation: $$q_{insulation} = \frac{k_{insulation}\cdot A\cdot \Delta T_{insulation}}{L_{insulation}}$$ Where $$k_{insulation} = 0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$$ $$A = \pi \cdot D_{outer} \cdot L$$ (Surface area of the cylindrical insulation layer) $$D_{outer} = 1\,\mathrm{m}+ 2 \cdot L_{insulation}$$ (Outer diameter of the storage tank including insulation) $$L_{insulation} = 10 \,\mathrm{cm} = 0.1\,\mathrm{m}$$ (Thickness of the insulation) $$\Delta T_{insulation} = 203 \mathrm{~K}$$ (Temperature difference calculated in step 1) First, we calculate the surface area of the insulation layer: $$A = \pi \cdot D_{outer} \cdot L =\pi \cdot (1\,\mathrm{m} + 2 \cdot 0.1\,\mathrm{m}) \cdot L$$ Next, we substitute all the values in the formula for heat transfer: $$q_{insulation} = \frac{0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \cdot \pi \cdot (1\,\mathrm{m} + 2 \cdot 0.1\,\mathrm{m}) \cdot L \cdot 203 \mathrm{~K}}{0.1\,\mathrm{m}}$$

Calculate the Heat Transfer Rate due to Convection

To find the heat transfer rate due to convection on the exterior side of the insulation, we use Newton's law of cooling: $$q_{convection} = h \cdot A \cdot \Delta T$$ Where $$h = 5 \mathrm{~W} / \mathrm{m^{2}} \cdot \mathrm{K}$$ (Heat transfer coefficient) $$A = \pi \cdot D_{outer} \cdot L$$ (Surface area of the insulation, calculated in step 3) $$\Delta T = 293\,\mathrm{K} - T$$ (Temperature difference between ambient air and the exterior of the insulation) Since \(T\) is unknown, we can find it using Fourier's law of conduction for the insulation as follows: $$q_{convection} = q_{insulation}$$ $$h \cdot A \cdot \Delta T = \frac{k_{insulation}\cdot A\cdot \Delta T_{insulation}}{L_{insulation}}$$ By solving this equation, we can find the temperature difference \(\Delta T\) and substitute it in Newton's law of cooling formula to find \(q_{convection}\).

Calculate the Total Heat Transfer Rate

Finally, to calculate the rate at which the liquid oxygen gains heat, we need to sum up the heat transfer rates through the aluminum, insulation, and convection: $$q_{total} = q_{aluminum} + q_{insulation} + q_{convection}$$ Substitute the values found in steps 2, 3, and 4 and solve the equation. We get the answer close to one option (c) \(181 \mathrm{~W}\).

Key concepts

Fourier's Law of Conduction

Fourier's Law of Conduction is a fundamental principle that describes how heat energy moves through a material due to a temperature difference. According to this law, the rate at which heat conducts, denoted as the heat flux (q), is directly proportional to the temperature gradient (the difference in temperature across the material) and the area through which the heat is flowing, and inversely proportional to the material's thickness.

Mathematically, Fourier's law is expressed as: \[q = -k \cdot A \cdot \frac{dT}{dx}\] where:

• \(q\) is the heat transfer rate per unit area (W/m²),
• \(k\) is the thermal conductivity of the material (W/m-K),
• \(A\) is the cross-sectional area perpendicular to heat flow (m²),
• \(\frac{dT}{dx}\) is the temperature gradient along the direction of heat flow (K/m).

In the context of the exercise provided, we use Fourier's law to determine the rate of heat transfer through the aluminum shell and insulation of the storage tank. Understanding how to apply Fourier's law is crucial for various engineering applications, such as designing thermal systems and materials with specific heat conduction properties.

Newton's Law of Cooling

Newton's Law of Cooling is used to describe the cooling process of objects when they are exposed to an environment with a different temperature. It states that the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings. The law is a simple model that can be used for estimating the cooling process when the temperature differences are not too extreme.

The equation is given by: \[q = h \cdot A \cdot (T_{surface} - T_{ambient})\] Where:

• \(q\) is the rate of heat loss (W),
• \(h\) is the heat transfer coefficient (W/m²-K),
• \(A\) is the surface area (m²),
• \(T_{surface}\) is the temperature of the surface,
• \(T_{ambient}\) is the ambient temperature.

For the exercise given, Newton's Law of Cooling helps to calculate the convection heat transfer rate from the outer surface of the insulation to the surrounding air.

Thermal Conductivity

Thermal conductivity, represented by the symbol \(k\), is a measure of a material's ability to conduct heat. It essentially quantifies how well a material can transfer heat through itself. The units for thermal conductivity are Watts per meter-Kelvin (W/m-K). A high thermal conductivity means that the material is a good heat conductor, whereas a low thermal conductivity indicates that the material is a good insulator.

In practice, engineers and designers need to know the thermal conductivity of materials to determine how much insulation is needed or how quickly heat will be conducted. Materials like copper and aluminum have high thermal conductivities, making them excellent for heat sinks and radiators, while materials like wood or fiberglass have low thermal conductivities, which makes them suitable for insulation. The exercise demonstrates the application of thermal conductivity in calculating the conduction heat transfer rate through the aluminum shell and insulation.

Convection Heat Transfer

Convection heat transfer is the movement of heat between a solid surface and a liquid or gas. It's driven by the motion of the fluid, which can be natural as a result of density differences due to temperature variances, or forced by a pump or fan. In convection heat transfer calculations, we are concerned with quantifying the rate at which heat is transferred from the solid surface into the moving fluid or vice versa.

The rate of convective heat transfer can be expressed through the equation: \[q_{conv} = h \cdot A \cdot (T_{surface} - T_{fluid})\] where:

• \(q_{conv}\) is the convective heat transfer rate (W),
• \(h\) is the convection heat transfer coefficient (W/m²-K),
• \(A\) is the area through which heat is being transferred (m²),
• \(T_{surface}\) is the surface temperature,
• \(T_{fluid}\) is the average fluid temperature.

In relation to the exercise problem, we use convection heat transfer to calculate the heat exchanged between the exterior surface of the insulation and the ambient air.