Consider a pipe at a constant temperature whose radius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid?

Short Answer

Expert verified
Answer: No, the rate of heat loss will not increase in this situation. In fact, it will likely decrease.

Step by step solution

01

Identifying the formula and definition for heat loss and critical radius

To analyze the situation, we will use the following formula for heat loss through a cylinder (pipe in this case): Q = 2 * pi * k * L * (T1 - T2) / (ln(r2/r1)), where Q is the heat loss, k is the thermal conductivity, L is the length of the pipe, T1 and T2 are the inner and outer surface temperatures, and r1 and r2 are the inner and outer radius. The critical radius of insulation (rc) for a cylinder is defined as: rc = k / h, where h is the convective heat transfer coefficient.
02

Identifying the relationship between heat loss and critical radius

When the radius is greater than the critical radius of insulation (r1 > rc), adding insulation (increasing r2) should decrease the rate of heat loss (Q), as the increased resistance due to the added insulation outweighs the increase in surface area.
03

Calculate the initial heat loss

Let's represent the initial conditions with r1 > rc. We can calculate the initial heat loss (Q1) using the formula mentioned before: Q1 = 2 * pi * k * L * (T1 - T2) / (ln(r2_initial/r1))
04

Add insulation to the pipe and calculate the new heat loss

Now, let's add insulation to the pipe, which means increasing the outer radius: r2_new > r2_initial. With the new radius, we can calculate the new heat loss (Q2) using the same formula: Q2 = 2 * pi * k * L * (T1 - T2) / (ln(r2_new/r1))
05

Compare initial and new heat loss

Now, we need to compare the initial heat loss (Q1) and the new heat loss (Q2) to see if the claim is valid. If Q2 > Q1, then the claim is valid; otherwise, it's not. Notice that in the formula for heat loss, the terms (2 * pi * k * L * (T1 - T2)) are constant, and the only variable is the ratio of r2 to r1. When r2 increases, the denominator of the fraction (ln(r2/r1)) also increases, which means the overall value of heat loss (Q) should decrease.
06

Conclusion

Since the rate of heat loss decreases when the pipe radius is greater than the critical radius of insulation and we add more insulation, the claim that the rate of heat loss would increase under these circumstances is not valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Loss
Understanding heat loss is essential when discussing thermal systems. Heat loss refers to the transfer of thermal energy from an object to its surroundings. This transfer occurs due to the temperature difference between the object and its environment, following the second law of thermodynamics, which states that heat moves from warmer to cooler areas.

Going back to our exercise, heat loss (\( Q \) ) from a pipe can be expressed mathematically using the formula \( Q = 2 \pi k L (T1 - T2) / ln(r2/r1) \) where \( k \) is the thermal conductivity, \( L \) is the length of the pipe, \( T1 \) and \( T2 \) are the temperatures on the inner and outer surfaces of the pipe respectively, and \( r1 \) and \( r2 \) are the inner and outer radii. This formula explains that as the radius changes, so does the heat loss, demonstrating the importance of insulation thickness in managing heat transfer.
Thermal Conductivity
Thermal conductivity (\( k \) ) is a measure of a material's ability to conduct heat. It quantifies how easily thermal energy moves through a material due to a temperature gradient. Certain materials, like metals, have high thermal conductivity, meaning they transfer heat quickly; others, like wood or fiberglass, have low thermal conductivity, acting as good insulators.

In our pipe scenario, the material's thermal conductivity dictates how much energy is lost through the pipe walls. The value of \( k \) directly impacts the heat loss: the higher the \( k \) , the more heat is lost. Thus, choosing the right insulation material with a lower thermal conductivity is pivotal in reducing heat loss from the pipe, which is why it is crucial to understand this property when calculating and comparing heat loss before and after adding insulation.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient (\( h \) ) characterizes the convective heat transfer occurring between a surface and a fluid (liquid or gas) in motion. A higher \( h \) means a fluid is more effective at removing heat from a surface, leading to potentially higher rates of heat loss.

When you look at the critical radius of insulation (\( rc \) = \( k/h \) ), it represents the insulation thickness at which heat loss due to conduction through insulation equates to the heat lost because of convection at the surface. If the pipe's radius is greater than \( rc \) , as mentioned in the exercise, adding insulation actually reduces the rate of heat transfer because it increases the thermal resistance to heat flow. Hence, the convective heat transfer coefficient is a fundamental concept in determining the effectiveness of insulation and optimizing thermal performance of systems.

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Most popular questions from this chapter

An 8-m-internal-diameter spherical tank made of \(1.5\)-cm-thick stainless steel \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located in a room whose temperature is \(25^{\circ} \mathrm{C}\). The walls of the room are also at \(25^{\circ} \mathrm{C}\). The outer surface of the tank is black (emissivity \(\varepsilon=1\) ), and heat transfer between the outer surface of the tank and the surroundings is by natural convection and radiation. The convection heat transfer coefficients at the inner and the outer surfaces of the tank are \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -h period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\).

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of \(0^{\circ} \mathrm{C}\). The vessel is covered with a \(5.0\)-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

A 0.3-cm-thick, 12-cm-high, and 18-cm-long circuit board houses 80 closely spaced logic chips on one side, each dissipating \(0.04 \mathrm{~W}\). The board is impregnated with copper fillings and has an effective thermal conductivity of \(30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the chips is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(40^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the temperatures on the two sides of the circuit board. (b) Now a \(0.2\)-cm-thick, 12-cm-high, and 18-cmlong aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with 864 2-cm-long aluminum pin fins of diameter \(0.25 \mathrm{~cm}\) is attached to the back side of the circuit board with a \(0.02\)-cm-thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

Exposure to high concentration of gaseous ammonia can cause lung damage. To prevent gaseous ammonia from leaking out, ammonia is transported in its liquid state through a pipe \(\left(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}\right.\), \(D_{o}=4 \mathrm{~cm}\), and \(L=10 \mathrm{~m}\) ). Since liquid ammonia has a normal boiling point of \(-33.3^{\circ} \mathrm{C}\), the pipe needs to be properly insulated to prevent the surrounding heat from causing the ammonia to boil. The pipe is situated in a laboratory, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\). The convection heat transfer coefficients of the liquid hydrogen and the ambient air are \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Determine the insulation thickness for the pipe using a material with \(k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to keep the liquid ammonia flowing at an average temperature of \(-35^{\circ} \mathrm{C}\), while maintaining the insulated pipe outer surface temperature at \(10^{\circ} \mathrm{C}\).

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