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Chapter 3: Problem 181

One wall of a refrigerated warehouse is \(10.0\)-m-high and \(5.0\)-m-wide. The wall is made of three layers: \(1.0\)-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fibreglass \((k=\) \(0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0-\mathrm{cm}\) thick gypsum board \((k=\) \(0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts \((k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), each \(2.0 \mathrm{~cm}\) in diameter and \(12.0 \mathrm{~cm}\) long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

Short Answer

Expert verified
Answer: The percent change in the rate of heat transfer across the wall due to metal bolts is 60.2%.

Step by step solution

01

Calculate Heat Transfer Across Each Layer Without Metal Bolts

To find the rate of heat transfer across each wall layer, first calculate the cross-sectional area and temperature difference for each layer, then apply the formula for heat transfer q for each layer. Area \(A\) for every layer: \(A = 10.0\textrm{ m} \times 5.0\textrm{ m} = 50.0\textrm{ m}^2\) Temperature difference \(\Delta T = 20^{\circ}\mathrm{C} - (-10^{\circ}\mathrm{C}) = 30^{\circ}\mathrm{C}\) Now calculate \(q\) for each layer: Aluminum: \(q_{Al} = \frac{200\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2 \times 30\textrm{ K}}{0.01\textrm{ m}} = 300000\textrm{ W}\) Fiberglass: \(q_{fib} = \frac{0.038\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2 \times 30\textrm{ K}}{0.08\textrm{ m}} = 713.5\textrm{ W}\) Gypsum: \(q_{gypsum} = \frac{0.48\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2 \times 30\textrm{ K}}{0.03\textrm{ m}} = 24000\textrm{ W}\)
02

Calculate the Total Heat Transfer Across Layered Wall Without Metal Bolts

To find the total heat transfer for the wall without metal bolts, first find the total thermal resistance without metal bolts, \(R_{total} = \frac{\Delta T}{q}\), and then apply the formula \(\Delta T=qR_{total}\): \(R_{total} = R_{Al} + R_{fib} + R_{gypsum} = \frac{0.01\textrm{ m}}{200\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2} + \frac{0.08\textrm{ m}}{0.038\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2} + \frac{0.03\textrm{ m}}{0.48\textrm{ W/m}\cdot\textrm{K} \times 50\textrm{ m}^2} = 1.3417\textrm{ K/W} \) Now, calculate the total heat transfer for the unbolted layered wall: \(q_{total-unbolted} = \frac{\Delta T}{R_{total}} = \frac{30\textrm{ K}}{1.3417\textrm{ K/W}} = 2235.6\textrm{ W}\)
03

Calculate Heat Transfer Across Metal Bolts

To calculate heat transfer across metal bolts, first calculate the cross-sectional area of one bolt and then apply the formula for heat transfer q for the bolts: Area of one bolt \(A_{bolt} = \pi(\frac{0.02\textrm{ m}}{2})^2 = 3.14\times10^{-4}\textrm{ m}^2\) \(q_{bolt} = \frac{43\textrm{ W/m}\cdot\textrm{K} \times 3.14\times10^{-4}\textrm{ m}^2 \times 30\textrm{ K}}{0.12\textrm{ m}} = 3.36\textrm{ W}\) Now, calculate the total heat transfer for 400 bolts: \(q_{bolts-total} = 400 \times 3.36\textrm{ W} = 1344\textrm{ W}\)
04

Calculate the Total Heat Transfer Across the Bolted Wall

To find the total heat transfer for the bolted wall, add the total heat transfer for the unbolted layered wall from Step 2 to the total heat transfer for the 400 metal bolts from Step 3: \(q_{total-bolted} = q_{total-unbolted} + q_{bolts-total} = 2235.6\textrm{ W} + 1344\textrm{ W} = 3579.6\textrm{ W}\)
05

Calculate the Percent Change in the Rate of Heat Transfer

To find the percent change in the rate of heat transfer across the wall due to metal bolts, use the following formula: Percent change \(= \frac{q_{total-bolted} - q_{total-unbolted}}{q_{total-unbolted}} \times 100\) Percent change \(= \frac{3579.6\textrm{ W} - 2235.6\textrm{ W}}{2235.6\textrm{ W}} \times 100 = 60.2\%\) Thus, the percent change in the rate of heat transfer across the wall due to metal bolts is 60.2%.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material property that measures a material's ability to conduct heat. Represented by the symbol 'k', it's quantified as the amount of heat, in watts, that passes through a meter-thick sample of the material when there's a temperature difference of one degree Celsius across the material. The higher the 'k' value, the better the material is at transferring heat. In the context of our problem, we see materials with a wide range of thermal conductivity; aluminum with a high 'k' rapidly transfers heat, while fiberglass with a low 'k' is much slower.
Steady State Heat Transfer
Steady state heat transfer refers to a condition where the temperature within each layer of the wall doesn't change with time, even though heat is constantly being transferred through the material. This implies that the rate at which heat enters a layer is equal to the rate at which it leaves, leading to a consistent temperature profile over time. In our problem, this principle enables us to calculate the heat transfer rate across different layers of the wall as if they were steady over time, simplifying the mathematical modeling of the process.
Thermal Resistance
Thermal resistance is the measure of a material's resistance to heat flow and is the reciprocal of thermal conductivity. High thermal resistance means the material is a good insulator, and it's measured in units of 'K/W' (Kelvin per Watt). In the given exercise, we add up the thermal resistances of each layer to find the total resistance, which then helps in determining the overall rate of heat transfer across the wall. Calculation of thermal resistance is key to designing insulating systems and for understanding how various construction materials will impact heat loss.
Conductance of Materials
Conductance, often denoted by 'C', is the reciprocal of resistance and is a measure of how easily heat flows through a material. Higher conductance values indicate that the material conducts heat more readily. In the context of our problem, the conductance of each material layer, and that added by the metal bolts, influences the overall rate of heat transfer. The bolts, acting as thermal bridges, considerably increase the total conductance and thus the rate of heat transfer, leading to the observed change in the insulation properties of the wall.

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Most popular questions from this chapter

What is a conduction shape factor? How is it related to the thermal resistance?

A turbine blade made of a metal alloy \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a length of \(5.3 \mathrm{~cm}\), a perimeter of \(11 \mathrm{~cm}\), and a cross-sectional area of \(5.13 \mathrm{~cm}^{2}\). The turbine blade is exposed to hot gas from the combustion chamber at \(973^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(538 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The base of the turbine blade maintains a constant temperature of \(450^{\circ} \mathrm{C}\) and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.

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Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.

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: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$
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