Question

Two concentric cylinders of diameters \(D_{i}=30 \mathrm{~cm}\) and \(D_{o}=40 \mathrm{~cm}\) and length \(L=5 \mathrm{~m}\) are separated by air at \(1 \mathrm{~atm}\) pressure. Heat is generated within the inner cylinder uniformly at a rate of \(1100 \mathrm{~W} / \mathrm{m}^{3}\), and the inner surface temperature of the outer cylinder is \(300 \mathrm{~K}\). The steady-state outer surface temperature of the inner cylinder is (a) \(402 \mathrm{~K}\) (b) \(415 \mathrm{~K}\) (c) \(429 \mathrm{~K}\) (d) \(442 \mathrm{~K}\) (e) \(456 \mathrm{~K}\) (For air, use \(k=0.03095 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7111, v=\) \(\left.2.306 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\)

Short answer

Answer: The steady-state outer surface temperature of the inner cylinder is approximately 429 K.

Step-by-step solution

Understand the given data

We are given the following information: Inner cylinder diameter: \(D_i = 30\,\mathrm{cm}\) Outer cylinder diameter: \(D_o = 40\,\mathrm{cm}\) Length: \(L = 5\,\mathrm{m}\) Heat generation: \(q_g = 1100\,\mathrm{W}/\mathrm{m}^3\) Outer cylinder inner surface temperature: \(T_o = 300\,\mathrm{K}\) For air: \(k = 0.03095\,\mathrm{W}/\mathrm{m}\cdot \mathrm{K}\), \(\operatorname{Pr} = 0.7111\), \(\nu = 2.306\times10^{-5}\,\mathrm{m}^2/\mathrm{s}\) Convert the cylinder diameters to meters: \(D_i = 0.3\,\mathrm{m}\) \(D_o = 0.4\,\mathrm{m}\)

Calculate the radius of the cylinders

Calculate the radius of each cylinder: $$r_i = \frac{D_i}{2} = 0.15\,\mathrm{m}$$ $$r_o = \frac{D_o}{2} = 0.20\, \mathrm{m}$$

Calculate the total heat energy generated in the inner cylinder

We are given the heat generation rate per unit volume. Multiply this by the volume of the inner cylinder to find the total heat generated: $$Q_g = q_g \cdot V = q_g \cdot (\pi r_i^2 L) = 1100\,\frac{\mathrm{W}}{\mathrm{m}^3} \cdot (\pi \cdot (0.15\,\mathrm{m})^2 \cdot 5\,\mathrm{m}) = 1169.21\,\mathrm{W}$$

Calculate the heat transfer coefficient

For heat transfer through a cylindrical air gap, we will use the following equation to calculate the effective heat transfer coefficient \(h_\mathrm{eff}\): $$h_\mathrm{eff} = \frac{2k}{r_i+r_o}$$ Substitute the values of \(r_i\), \(r_o\), and \(k\): $$h_\mathrm{eff} = \frac{2 \times 0.03095\, \mathrm{W/m}\cdot\mathrm{K}}{0.15\, \mathrm{m} + 0.20\, \mathrm{m}} = 0.2063\, \mathrm{W/m}^2\cdot\mathrm{K}$$

Calculate the steady-state outer surface temperature of the inner cylinder

Use the energy balance equation to determine the temperature of the outer surface, \(T_i\): $$Q_g = h_\mathrm{eff}\cdot A \cdot (T_i - T_o)$$ Where \(A\) is the outer surface area of the inner cylinder. $$A = 2\pi r_i L = 2\pi(0.15\,\mathrm{m})(5\,\mathrm{m}) = 4.71\,\mathrm{m}^2$$ Solve for the outer surface temperature of the inner cylinder, \(T_i\): $$T_i = T_o + \frac{Q_g}{h_\mathrm{eff}\cdot A}$$ Substitute the given values and the calculated values for \(Q_g\), \(h_\mathrm{eff}\), and \(A\): $$T_i = 300\,\mathrm{K} + \frac{1169.21\,\mathrm{W}}{0.2063\,\mathrm{W/m}^2\cdot\mathrm{K}\cdot 4.71\,\mathrm{m}^2} = 429.01\,\mathrm{K}$$ Thus, the steady-state outer surface temperature of the inner cylinder is approximately \(429\,\mathrm{K}\). So the correct option is (c) \(429\,\mathrm{K}\).

Key concepts

Steady-State Temperature

In heat transfer analysis, steady-state temperature refers to a condition where the temperature in the system does not change with time, implying a balance between energy entering and leaving the system. In the context of concentric cylinders, thermal energy is generated in the inner cylinder and must pass through the cylindrical air gap to the outer cylinder. When steady-state is reached, the temperature at any point within the air gap remains constant over time.

For students looking to grasp this concept, it's important to understand that reaching steady-state can take time after an initial disturbance or change in conditions. However, once achieved, the steady-state temperature makes calculations much simpler as time-variant factors are no longer relevant. This also means that in such a scenario, if we know the rate at which heat is generated and the properties of the materials involved, we can predict temperatures at the boundaries quite accurately.

Thermal Conductivity

The thermal conductivity (\( k \)) of a material is a measure of its ability to conduct heat. It appears in many of the fundamental equations of heat transfer and is a crucial property for analyzing how heat moves through solids or fluids. For air in the given problem, thermal conductivity directly affects the rate at which heat is transferred from the inner to the outer cylinder.

When helping students with problems involving thermal conductivity, it's useful to illustrate that materials with high thermal conductivity, like metals, are good heat conductors, whereas materials with low thermal conductivity, like air or insulation materials, are good insulators. The value of thermal conductivity is often provided in textbooks or reference materials and is a key parameter in determining the effectiveness of heat transfer across different mediums.

Cylindrical Air Gap Insulation

Insulation plays a significant role in controlling the rate of heat transfer. In our exercise, the air gap between the concentric cylinders acts as an insulator. Cylindrical air gap insulation refers to this kind of setup, where an air gap is intentionally included to reduce heat transfer due to its low thermal conductivity compared to solid materials.

Understanding how this air gap reduces thermal energy transfer is essential. This setup exploits the natural insulating properties of air to minimize the undesired loss or gain of heat. In real-world applications, proper insulation can help save energy and maintain desired temperatures. For students dealing with heat transfer problems, it's helpful to visualize the air gap as a barrier that elongates the heat path between the cylinders thereby reducing the flow due to the air's thermal resistance.

Heat Generation Rate

The heat generation rate (\( q_g \)) in a material, such as the inner cylinder in this problem, is the amount of heat produced per unit volume per unit time. This is usually given in Watts per cubic meter (\( \text{W/m}^3 \)). In applications such as electrical devices, radioactive decay, or chemical reactions, internal heat generation can be significant.

For the exercise in question, the heat generation rate is crucial to calculating the resultant temperature difference across the air gap. When conveying this to students, emphasize that the heat generation within the inner cylinder causes the temperature to increase, which in turn induces a heat flow to the outer cylinder. The higher the heat generation rate, the greater the resultant temperature of the inner cylinder will be, assuming all other factors remain constant.