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A 5-m-wide, 4-m-high, and 40-m-long kiln used to cure concrete pipes is made of 20 -cm-thick concrete walls and ceiling \((k=0.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The kiln is maintained at \(40^{\circ} \mathrm{C}\) by injecting hot steam into it. The two ends of the kiln, \(4 \mathrm{~m} \times 5 \mathrm{~m}\) in size, are made of a 3 -mm-thick sheet metal covered with 2 -cm-thick Styrofoam \((k=0.033 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The convection heat transfer coefficients on the inner and the outer surfaces of the kiln are \(3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Disregarding any heat loss through the floor, determine the rate of heat loss from the kiln when the ambient air is at \(-4^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Answer: The rate of heat loss from the kiln is approximately 1684.22 W.

Step by step solution

01

Calculate the Surface Areas

For this step, we should calculate the surface area of the walls, ceiling, and the ends of the kiln. This is necessary for calculating the heat transfer through each section. Walls: \(2 * (5 \mathrm{m} * 4 \mathrm{m}) = 40 \mathrm{m}^2\) Ceiling: \(5 \mathrm{m} * 40 \mathrm{m} = 200 \mathrm{m}^2\) Kiln Ends: \(2 * (4 \mathrm{m} * 5 \mathrm{m}) = 40 \mathrm{m}^2\)
02

Calculate the Resistance of Each Layer

For each layer's resistance, we will use the formula: \[R = \frac{L}{kA}\], where L is the thickness, k is the thermal conductivity, and A is the surface area. For concrete walls and ceiling (we will calculate this in one step because they have the same thermal conductivity): \[R_{concrete} = \frac{0.2 \mathrm{m}}{(0.9 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 9.26\times 10^{-4} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] For sheet metal and Styrofoam ends: \[R_{metal} = \frac{3 \times 10^{-3} \mathrm{m}}{(45 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(40 \mathrm{m}^2)} = 1.67\times 10^{-5} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] \[R_{Styrofoam} = \frac{0.02 \mathrm{m}}{(0.033 \frac{\mathrm{W}}{\mathrm{m} \cdot \mathrm{K}})(40 \mathrm{m}^2)} = 1.5152\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]
03

Calculate the Resistance of Convection

Now, we will calculate the convection resistance on both the inner and outer surfaces of the kiln. We use the formula: \[R_{conv} = \frac{1}{hA}\], where h is the convection heat transfer coefficient and A is the surface area. For the inner surface: \[R_{conv\_in} = \frac{1}{(3000 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 1.39\times 10^{-4} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\] For the outer surface: \[R_{conv\_out} = \frac{1}{(25 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}})(240 \mathrm{m}^2)} = 1.67\times 10^{-3} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]
04

Calculate the Total Resistance

Now we will add the resistances together: Total resistance: \[R_{total} = R_{conv\_in} + R_{concrete} + R_{conv\_out} + R_{metal} + R_{Styrofoam} = 1.39\times 10^{-4} + 9.26\times 10^{-4} + 1.67\times 10^{-3} + 1.67\times 10^{-5} + 1.5152\times 10^{-2} = 2.6167\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}\]
05

Calculate the Rate of Heat Loss

Now that we have the total resistance, we can calculate the rate of heat loss using the formula: \[q = \frac{\Delta T}{R_{total}}\], where q is the rate of heat loss, and \(\Delta T\) is the temperature difference between the inside and outside of the kiln. \[q = \frac{40^{\circ} \mathrm{C} - (-4^{\circ} \mathrm{C})}{2.6167\times 10^{-2} \frac{\mathrm{K} \cdot \mathrm{m}^2 }{\mathrm{W}}} = 1684.22 \mathrm{W}\] Therefore, the rate of heat loss from the kiln is approximately \(1684.22 \mathrm{W}\).

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Key Concepts

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

Thermal Resistance
Thermal resistance is a key concept in understanding how heat moves through different materials. It represents how well a material can resist heat flow. The higher the thermal resistance, the better the material is at insulating. This is calculated using the formula: \[ R = \frac{L}{kA} \]where:
  • \(L\) is the thickness of the material,
  • \(k\) is the thermal conductivity, and
  • \(A\) is the surface area.

For instance, in calculating the thermal resistance of concrete walls in the kiln, we're finding how effectively the concrete can reduce the flow of heat out of the kiln. In this exercise, different layers like metal, Styrofoam, and concrete each contribute their own resistances. Adding these resistances gives us the total thermal resistance for the entire structure. This total helps us eventually find out how quickly heat escapes from the kiln. Remember, effectively managing thermal resistance is crucial in design to ensure minimal heat loss and efficient energy usage.
Convection Heat Transfer
Convection heat transfer is another way that heat moves, especially through fluids like air and water. It happens when molecules near a hot surface gain energy, move away, and are replaced by cooler molecules. This affects surfaces such as the inside and outside of our kiln.For calculating convection resistance, we use the formula:\[ R_{conv} = \frac{1}{hA} \]where:
  • \(h\) is the convection heat transfer coefficient,
  • \(A\) is the surface area.

In this study of the kiln, convection adds a layer of complexity by impacting the heat flow on the surfaces. High convection coefficients mean more heat can be transferred quickly. The coefficient for the inner surface of the kiln is significantly higher, at 3000 W/m²·K, compared to 25 W/m²·K for the outer surface, showcasing the ability of the inner surface to transfer heat much more efficiently. Recognizing convection's role is essential to understand total heat loss and designing for thermal efficiency.
Thermal Conductivity
Thermal conductivity refers to a material's ability to conduct heat, represented by the symbol \(k\). It measures how quickly heat can pass through a material under a temperature gradient. Materials with high thermal conductivity are good conductors of heat, such as metals, whereas materials with low thermal conductivity, like Styrofoam and concrete, are better insulators.In this exercise:
  • The concrete has a conductivity of 0.9 W/m·K,
  • Styrofoam has a much lower conductivity of 0.033 W/m·K, making it a better insulator.

Understanding the thermal conductivity of building materials helps us design structures that minimize unwanted heat loss or gain. In the kiln, choosing a material with the right thermal conductivity is crucial for maintaining the desired internal temperature without excessive energy input. It's a balancing act between material choices to maintain energy efficiency and effectiveness.
Concrete Structures
Concrete structures play a vital role in the construction of kilns, acting both as a supportive frame and as an insulator due to concrete's specific properties. Concrete has a relatively low thermal conductivity, which makes it moderately effective in insulating against both heat loss and gain. The given structure has 20-cm-thick concrete walls. Interestingly, while not the best insulator compared to other materials used in construction, concrete's robustness and availability make it a popular choice in engineering and architecture. In our kiln, the concrete not only aids in retaining the heat necessary for curing but also adds physical strength to the overall structure. Understanding how concrete's characteristics impact thermal behavior helps in optimizing the kiln’s performance and energy usage. This reflects the importance of thoughtful material selection in achieving desired thermal performance and structural integrity in various projects.

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Most popular questions from this chapter

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

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: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

A 25 -cm-diameter, 2.4-m-long vertical cylinder containing ice at \(0^{\circ} \mathrm{C}\) is buried right under the ground. The cylinder is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are \(18^{\circ} \mathrm{C}\) and \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) respectively. The rate of heat transfer to the cylinder is (a) \(37.2 \mathrm{~W}\) (b) \(63.2 \mathrm{~W}\) (c) \(158 \mathrm{~W}\) (d) \(480 \mathrm{~W}\) (e) \(1210 \mathrm{~W}\)

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

A 3-cm-long, 2-mm \(\times 2-\mathrm{mm}\) rectangular crosssection aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the fin efficiency is 65 percent, the effectiveness of this single fin is (a) 39 (b) 30 (c) 24 (d) \(18 \quad(e) 7\)

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