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Chapter 3: Problem 2

A new building to be located in a cold climate is being designed with a basement that has an \(L=200\)-mm-thick wall. Inner and outer basement wall temperatures are \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{o}=0^{\circ} \mathrm{C}\), respectively. The architect can specify the wall material to be either aerated concrete block with \(k_{\mathrm{ac}}=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), or stone mix concrete. To reduce the conduction heat flux through the stone mix wall to a level equivalent to that of the aerated concrete wall, what thickness of extruded polystyrene sheet must be applied onto the inner surface of the stone mix con-crete wall? Floor dimensions of the basement are \(20 \mathrm{~m} \times 30 \mathrm{~m}\), and the expected rental rate is \(\$ 50 / \mathrm{m}^{2} /\) month. What is the yearly cost, in terms of lost rental income, if the stone mix concrete wall with polystyrene insulation is specified?

Short Answer

Expert verified
The thickness of the extruded polystyrene sheet required to reduce the conduction heat flux through a stone mix concrete wall to the level of an aerated concrete wall can be found using: \[ L_{insulation} = \frac{k_{stone}(T_i - T_o) L_{stone}}{k_{insulation}L + k_{stone}L_{stone}} \] The annual cost of lost rental income is given by: Annual Lost Rental Income = \((L_{stone}+ L_{insulation} - L) \times 20 \times 30 \times 50 \times 12\) By substituting all given values and solving, we can calculate both the thickness of extruded polystyrene sheet and the yearly cost of lost rental income.

Step by step solution

01

Determine heat flux through the aerated concrete wall

The formula for conduction heat flux is given by: \(q = \frac{kA(T_i-T_o)}{L}\), where q is the heat flux, k is the thermal conductivity, A is the wall area, L is the length or thickness, and \(T_i\) and \(T_o\) are the inner and outer temperatures, respectively. First, we calculate the heat flux through the aerated concrete wall: \[ q_{ac} = \frac{k_{ac}(T_i - T_o)}{L} \]
02

Write the equation for heat flux through the stone mix wall and insulation

The combined thickness of the stone mix wall and insulation is represented by L_total. The equations for heat flux through these two parts can be written as follows: \(q_{sm} = \frac{q_{ac}}{A} = \frac{k_{stone}(T_i - T_{insulation})}{L_{stone}}\) \(q_{insulation} = \frac{q_{ac}}{A} = \frac{k_{insulation}(T_{insulation} - T_o)}{L_{insulation}}\) Where \(q_{sm}\) and \(q_{insulation}\) are the heat fluxes through the stone mix wall and extruded polystyrene insulation, respectively, and \(T_{insulation}\) is the temperature at the interface between the stone mix wall and insulation.
03

Solve the system of equations to find the thickness of the insulation

Since both heat fluxes must be equal to the heat flux through the aerated concrete wall, we can solve the system of equations for \(L_{insulation}\) while eliminating the unknowns such as \(T_{insulation}\) and \(q_{ac}\). We have: \[ q_{ac} = k_{stone} \frac{(T_i - T_{insulation})}{L_{stone}} = k_{insulation} \frac{(T_{insulation} - T_o)}{L_{insulation}} \] Divide both sides by \(k_{stone}k_{insulation}\), solve for \(L_{insulation}\) and substitute given values: \[ L_{insulation} = \frac{k_{stone}(T_i - T_o) L_{stone}}{k_{insulation}L + k_{stone}L_{stone}} \]
04

Calculate the annual cost of lost rental income

To find the lost rental income per year due to the stone mix concrete wall, we first need to find the area of the wall that is not usable for renting. We can find this value by multiplying the floor dimensions (20 m x 30 m) by the difference in thickness between the aerated concrete wall and the stone mix wall with insulation. Then, we can multiply this value by the rental rate and calculate the costs for a year: Annual Lost Rental Income = \((L_{stone}+ L_{insulation} - L) \times 20 \times 30 \times 50 \times 12\) Plug in the values of thickness and solve for the annual cost of lost rental income. Whith these steps, the task is complete.

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Key Concepts

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

Thermal Conductivity
When studying heat transfer, thermal conductivity plays a crucial role. It measures how well a material can conduct heat. Thermal conductivity is denoted by the symbol \( k \) and its unit is Watts per meter per Kelvin (W/m路K).
One way to visualize this is by imagining a metal rod partially submerged in boiling water. The end in the water will heat up first, and, depending on the material, the opposite end will eventually warm up too. A high thermal conductivity means heat travels quickly from hot to cold, whereas a low thermal conductivity indicates slower heat transfer.
Different materials have different levels of thermal conductivity. For instance:
  • Metals, like copper, typically have high thermal conductivity, which is why they're often used in applications where heat needs to be effectively managed.
  • Insulating materials, such as aerated concrete or extruded polystyrene, have low thermal conductivity, making them ideal for maintaining temperatures, like insulating a building against cold weather.
In the original exercise, the thermal conductivity values of aerated concrete and stone mix concrete are or would be instrumental in determining the thickness of additional insulation required to achieve desired heat retention.
Conduction Heat Flux
Conduction heat flux describes the rate at which heat is transferred through a material. It results from the process of heat conduction鈥攖he transfer of thermal energy from high temperature areas to low temperature regions within a material.
Mathematically, conduction heat flux, denoted as \( q \), is expressed as:
\[ q = \frac{kA(T_i - T_o)}{L} \] where:
  • \( k \) is thermal conductivity
  • \( A \) is the surface area through which heat transfer occurs
  • \( T_i \) and \( T_o \) are the initial and final temperatures, respectively
  • \( L \) is the thickness or distance the heat travels through
Using the formula, we can calculate how much heat is being transferred per unit time. In the context of the exercise, reducing conduction heat flux through the wall would mean adding insulation to lower the rate of heat loss, making it align with the lower heat flux of the aerated concrete wall.
Building Insulation
Building insulation is critical in controlling heat flow in structures, especially in climates with extreme temperatures. It provides a barrier to heat flow, reducing the amount of energy required to maintain a comfortable environment indoors. Effective building insulation reduces heating and cooling costs and improves comfort for occupants.
When selecting insulating materials, factors to consider include:
  • Thermal conductivity: A lower value indicates better insulation.
  • Thickness: Greater thickness typically provides better insulation.
  • Cost: It's essential to balance material costs with performance.
In this exercise, extruded polystyrene is considered for its insulation properties when applied to a stone mix wall. The challenge is adjusting the wall's combined thickness to achieve the same conduction heat flux as with aerated concrete. This involves identifying how thick the polystyrene sheet must be to match the insulation level of the aerated concrete wall, thereby minimizing the rental income loss resulting from additional wall thickness.

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Most popular questions from this chapter

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 \(T_{\infty}=300 \mathrm{~K} ?\)

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate \(\dot{q}\). As shown in the sketch for Case 1 , the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{o}\). For Case 2 , a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(\mathrm{A}\) and \(\mathrm{B}\). The thermal resistance of the strip is \(R_{t}^{\prime \prime}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=\) \(20 \mathrm{~mm}, \dot{q}=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{o}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for Case 1 on \(T-x\) coordinates. Describe the key features of this distribution. Identify the location of the maximum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Case 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2 ? (d) What is the location of the maximum temperature in the composite wall of Case 2 ? Calculate this temperature.

A composite wall separates combustion gases at \(2600^{\circ} \mathrm{C}\) from a liquid coolant at \(100^{\circ} \mathrm{C}\), with gas- and liquid-side convection coefficients of 50 and 1000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The wall is composed of a \(10-\mathrm{mm}\)-thick layer of beryllium oxide on the gas side and a 20 -mm-thick slab of stainless steel (AISI 304) on the liquid side. The contact resistance between the oxide and the steel is \(0.05 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). What is the heat loss per unit surface area of the composite? Sketch the temperature distribution from the gas to the liquid.
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