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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

Hot water at an average temperature of \(53^{\circ} \mathrm{C}\) and an average velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) is flowing through a \(5-\mathrm{m}\) section of a thin-walled hot-water pipe that has an outer diameter of \(2.5 \mathrm{~cm}\). The pipe passes through the center of a \(14-\mathrm{cm}\)-thick wall filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the surfaces of the wall are at \(18^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the pipe to the air in the rooms and \((b)\) the temperature drop of the hot water as it flows through this 5 -m-long section of the wall. Answers: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) 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.

A room at \(20^{\circ} \mathrm{C}\) air temperature is loosing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4 -m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\) thick insulation with a conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)

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