Question

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken off. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature?

Short answer

Answer: When the insulation with an outer radius less than the critical radius is removed, the rate of heat transfer from the pipe increases due to direct exposure to convective heat transfer.

Step-by-step solution

Understanding Critical Radius

The critical radius of insulation is the radius at which the addition or removal of insulation does not have an impact on the rate of heat transfer. Mathematically, it can be determined by considering that the sum of conduction and convection resistance is at its minimum. The critical radius (r_c) can be calculated using the formula: r_c = \frac{k}{h} where k is the thermal conductivity of the insulation material and h is the convection heat transfer coefficient.

Analyzing the initial given situation

We are given that the outer radius of the insulation is less than the critical radius (r_i < r_c). Therefore, the insulation is not effective in its current state, and the heat transfer rate (assuming steady state) can be calculated using the formula: q = 2\pi L k \frac{T_1-T_2}{\ln(\frac{r_2}{r_1})} where L is the pipe length, T_1 is the pipe surface temperature, T_2 is the temperature at the outer surface of the insulation, r_1 is the radius of the pipe, and r_2 is the outer radius of the insulation.

Analyzing the situation without insulation

Once the insulation is removed, the heat transfer rate will be determined by the convection heat transfer. The convection heat transfer rate (q') can be calculated using the formula: q' = 2\pi r_1 L h (T_1 - T_\infty) where T_\infty is the temperature of the surrounding fluid.

Comparing the heat transfer rates

Now, we need to compare the heat transfer rates with and without insulation (q and q', respectively). Since the outer radius of the insulation was less than the critical radius, the insulation was not efficient in reducing heat transfer. Removing the insulation will expose the pipe directly to convective heat transfer, so the rate of heat transfer from the pipe is expected to increase. Without insulation, the pipe will lose heat more quickly through convection. In conclusion, after removing the insulation, the rate of heat transfer from the pipe will increase due to direct exposure to convective heat transfer.

Key concepts

Heat Transfer Rate

The heat transfer rate is a fundamental concept in thermodynamics that describes how quickly heat energy flows from one body to another. Generally, this rate depends on various factors such as temperature difference, surface area, and thermal conductivity of the material. When dealing with pipes and insulation, the heat transfer rate is heavily influenced by whether or not insulation is present and its effectiveness.

In the context of the problem, we compare the heat transfer rate in two scenarios—one with insulation and one without. If the insulation's outer radius is less than the critical radius, it does not effectively reduce the heat transfer rate. This is because the insulation presents additional thermal resistance without significantly impacting convective heat loss.

When you remove the insulation, the heat transfer rate is determined primarily by the convective heat transfer from the pipe surface to the surrounding environment, potentially increasing the heat transfer rate from the pipe.

Thermal Conductivity

Thermal conductivity (k) is a measure of a material's ability to conduct heat. It is defined as the heat transfer rate through a material with a unit thickness and unit area due to a unit temperature difference. A material with high thermal conductivity efficiently conducts heat, while a material with low thermal conductivity does the opposite.

In insulating materials, low thermal conductivity is desirable because it minimizes heat transfer, helping to conserve energy. However, the effectiveness of insulation is not determined solely by thermal conductivity. The relationship between thermal conductivity and convection coefficient is crucial in calculating the critical radius of insulation.

The critical radius (r_c) can be calculated with: \( r_c = \frac{k}{h} \)where k is the thermal conductivity of the insulation material and h is the convection heat transfer coefficient. Understanding this relationship helps explain why, in certain cases, adding insulation can increase heat transfer when the outer radius is less than the critical radius.

Convection Heat Transfer

Convection heat transfer involves the transfer of heat through a fluid (which can be either a liquid or gas) in contact with a solid surface. It occurs when the fluid carries heat away from the surface, influenced by factors like fluid velocity, temperature difference, and surface area.

The convection heat transfer coefficient (h) is a crucial parameter, indicating the heat transfer capability between the surface and the fluid. It determines how effectively heat is transferred through convection.

When insulation is present and its radius is below the critical radius, the convective heat transfer plays a more significant role than the conductive resistance provided by the insulation. Once the insulation is removed, as in the exercise, convection becomes the primary mode of heat transfer. The equation for convection heat transfer rate is:\[ q' = 2\pi r_1 L h (T_1 - T_\infty) \] where r_1 is the radius of the pipe, L is the pipe length, h is the convection coefficient, T_1 is the pipe surface temperature, and T_\infty is the surrounding fluid temperature. Removing poorly performing insulation exposes the pipe directly to convective heat, increasing the heat transfer rate due to enhanced exposure.