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Chapter 3: Problem 17

Consider the oven of Problem 1.54. The walls of the oven consist of \(L=30\)-mm- thick layers of insulation characterized by \(k_{\text {ins }}=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) that are sandwiched between two thin layers of sheet metal. The exterior surface of the oven is exposed to air at \(23^{\circ} \mathrm{C}\) with \(h_{\text {ext }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The interior oven air temperature is \(180^{\circ} \mathrm{C}\). Neglecting radiation heat transfer, determine the steady-state heat flux through the oven walls when the convection mode is disabled and the free convection coefficient at the inner oven surface is \(h_{\mathrm{fr}}=3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heat flux through the oven walls when the convection mode is activated, in which case the forced convection coefficient at the inner oven surface is \(h_{\mathrm{fo}}=27 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Does operation of the oven in its convection mode result in significantly increased heat losses from the oven to the kitchen? Would your conclusion change if radiation were included in your analysis?

Short Answer

Expert verified
The heat flux when the oven is operating in forced convection mode is about 2.8 times higher than during free convection mode, indicating a significant increase in heat loss from the oven to the kitchen. The inclusion of radiation heat transfer would likely not change this conclusion, since the heat losses are proportional to the temperature difference between the inner and outer surfaces of the oven and could be comparable for both scenarios.

Step by step solution

01

Calculate the Thermal Resistance of the Insulation

First, we need to determine the thermal resistance of the insulation layer. We know the insulation layer has a thickness of \( L = 30 \) mm and a thermal conductivity of \( k_\text{ins} = 0.03 \) W/m路K. The formula for calculating the thermal resistance of the insulation is given by: \[R_\text{ins} = \frac{L}{k_\text{ins} A} \] where \( A \) is the surface area of the insulation. As the exercise provided no dimensions for the oven, we will keep the thermal resistance in terms of \( A \) for now. Substituting the given values, we can calculate the thermal resistance: \[R_\text{ins} = \frac{0.03}{0.03 A} = \frac{1}{A}\]
02

Calculate the Heat Flux for Free Convection

Next, we will calculate the heat flux during the scenario when the oven is in free convection mode with the given heat transfer coefficient \(h_\text{fr} = 3 \) W/m虏路K. The total thermal resistance in this case consists of convection at both the inner and outer surfaces of the oven and the conductive resistance of the insulation. We can use the formula for calculating the total thermal resistance: \[R_\text{tot,fr} = \frac{1}{h_\text{fr} A} + \frac{1}{A} + \frac{1}{h_\text{ext} A}\] Using the given values, the total thermal resistance during free convection is: \[R_\text{tot,fr} = \frac{1}{3 A} + \frac{1}{A} + \frac{1}{2 A} = \frac{19}{6 A}\] Now we can calculate the heat flux using the temperature difference and the total thermal resistance: \[q_\text{fr} = \frac{T_\text{in} - T_\text{out}}{R_\text{tot,fr}}\] Substituting the given values, the heat flux during free convection is: \[q_\text{fr} = \frac{180 - 23}{\frac{19}{6 A}} = \frac{6 A(157)}{19} = \frac{942 A}{19}\]
03

Calculate the Heat Flux for Forced Convection

Now, we will calculate the heat flux during the scenario when the oven is in forced convection mode with the given heat transfer coefficient \(h_\text{fo} = 27 \) W/m虏路K. The total thermal resistance during forced convection is: \[R_\text{tot,fo} = \frac{1}{h_\text{fo} A} + \frac{1}{A} + \frac{1}{h_\text{ext} A}\] Using the given values, the total thermal resistance during forced convection is: \[R_\text{tot,fo} = \frac{1}{27 A} + \frac{1}{A} + \frac{1}{2 A} = \frac{61}{54 A}\] Now we can calculate the heat flux using the temperature difference and total thermal resistance: \[q_\text{fo} = \frac{T_\text{in} - T_\text{out}}{R_\text{tot,fo}}\] Substituting the given values, the heat flux during forced convection is: \[q_\text{fo} = \frac{180 - 23}{\frac{61}{54 A}} = \frac{54 A(157)}{61} = \frac{8466 A}{61}\]
04

Compare Heat Fluxes and Discuss the Effects of Radiation

Now, we can compare the heat fluxes to see if there is a significant difference between operating the oven in convection mode or not. Calculating the ratio of the heat fluxes, we get: \[\frac{q_\text{fo}}{q_\text{fr}} = \frac{\frac{8466 A}{61}}{\frac{942 A}{19}} = \frac{8466(19)}{942(61)} = \frac{160854}{57462} \approx 2.80\] This shows that the heat flux when the oven is operating in forced convection mode is about 2.8 times higher than during free convection mode. This is a significant increase in heat loss from the oven to the kitchen. As for the effects of radiation, the inclusion of radiation would likely increase the heat flux. However, heat losses due to radiation would be proportional to the difference in temperatures between the inner and outer surfaces of the oven and could be comparable for both free and forced convection scenarios. Since the main concern is the heat flux ratio between these scenarios, adding radiation heat transfer would likely not change our conclusion that operation in the convection mode increases heat losses from the oven to the kitchen.

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Key Concepts

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

Convection
Convection is a mode of heat transfer that occurs when fluid motion distributes thermal energy. In the context of the oven exercise, we deal with two types of convection: free convection and forced convection. Free convection happens when thermal energy causes air to move around naturally, without any external force. In the oven, this would mean the air circulates slowly due to the temperature difference between the oven's interior and its surroundings.

On the other hand, forced convection occurs when an external source, like a fan, causes air to circulate more vigorously. This enhances the rate of heat transfer because the motion increases the interaction between the air particles and the surfaces they're in contact with.
  • Free Convection: Naturally occurring due to temperature differences.
  • Forced Convection: Externally induced to enhance heat transfer.
When the oven's convection mode is activated, a fan forces air to move faster, which can significantly increase the rate of heat loss from the oven to the kitchen. This is because forced convection improves the efficiency of energy transfer by reducing the thermal resistance that air provides. Hence, switching between these modes can notably change how much heat is lost.
Heat Flux
Heat Flux is the rate of heat energy transfer through a given surface area. In our oven problem, we express heat flux in terms of power per unit area, typically in W/m虏.

Heat flux is influenced by the temperature difference between the interior and exterior surfaces and the thermal resistance of the materials in between. A higher temperature difference or lower resistance leads to a greater heat flux.
  • More significant temperature differences drive a higher heat flux.
  • Lower thermal resistance allows more rapid energy transfer.
Calculating heat flux involves dividing the temperature difference by the total thermal resistance of the oven walls. The equations show how changing modes from free to forced convection significantly increases the heat flux. This is because the total thermal resistance is lower in forced convection, allowing more energy to pass through the oven's walls, thereby increasing heat loss despite the same overall temperature gradient.
Thermal Conductivity
Thermal Conductivity, denoted by the symbol "\( k \)," is a material property that describes a material's ability to conduct heat. In the context of the oven problem, the insulating material used has a thermal conductivity of \( k_{\text{ins}} = 0.03 \) W/m路K.

Materials with high thermal conductivity efficiently transfer heat, whereas those with low thermal conductivity act as insulators. This characteristic is crucial for managing heat flow in thermal systems like ovens. In the oven example, the low thermal conductivity of the insulation helps minimize heat loss.
  • Low \( k \) values indicate good insulating properties.
  • High \( k \) values suggest effective heat distribution.
The concept of thermal resistance, used in calculating heat flux, directly derives from thermal conductivity. The resistance depends on the material's conductivity and its thickness. For oven walls, choosing materials with low thermal conductivity is essential for effective insulation. This ensures the oven maintains its interior temperature with minimal heat loss, which becomes apparent when comparing the effects of convection modes.

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Most popular questions from this chapter

A motor draws electric power \(P_{\text {elec }}\) from a supply line and delivers mechanical power \(P_{\text {mech }}\) to a pump through a rotating copper shaft of thermal conductivity \(k_{s}\), length \(L\), and diameter \(D\). The motor is mounted on a square pad of width \(W\), thickness \(t\), and thermal conductivity \(k_{p}\). The surface of the housing exposed to ambient air at \(T_{\infty}\) is of area \(A_{h}\), and the corresponding convection coefficient is \(h_{h}\). Opposite ends of the shaft are at temperatures of \(T_{h}\) and \(T_{\infty}\), and heat transfer from the shaft to the ambient air is characterized by the convection coefficient \(h_{s}\). The base of the pad is at \(T_{\infty}\).

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)

The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material, as shown in the schematic. For wires of diameter \(D=125 \mu \mathrm{m}\) and a convection coefficient of \(h=700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the minimum separation distance between the two legs of the sting, \(L=L_{1}+L_{2}\), to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and alumel wires. Evaluate the thermal conductivity of copper and constantan at \(T=300 \mathrm{~K}\). Use \(k_{\mathrm{Ch}}=19 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(k_{\mathrm{Al}}=29 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the thermal conductivities of the chromel and alumel wires, respectively.

Consider two long, slender rods of the same diameter but different materials. One end of each rod is attached to a base surface maintained at \(100^{\circ} \mathrm{C}\), while the surfaces of the rods are exposed to ambient air at \(20^{\circ} \mathrm{C}\). By traversing the length of each rod with a thermocouple, it was observed that the temperatures of the rods were equal at the positions \(x_{\mathrm{A}}=0.15 \mathrm{~m}\) and \(x_{\mathrm{B}}=0.075 \mathrm{~m}\), where \(x\) is measured from the base surface. If the thermal conductivity of rod \(\mathrm{A}\) is known to be \(k_{\mathrm{A}}=70 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the value of \(k_{\mathrm{B}}\) for rod B.

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?
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