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Chapter 3: Problem 5

Consider steady heat transfer through the wall of a room in winter. The convection heat transfer coefficient at the outer surface of the wall is three times that of the inner surface as a result of the winds. On which surface of the wall do you think the temperature will be closer to the surrounding air temperature? Explain.

Short Answer

Expert verified
Answer: The outer surface of the wall will have a temperature closer to the surrounding air temperature compared to the inner surface. This is because the higher convection heat transfer coefficient caused by winds on the outer surface allows for more efficient heat transfer, thus maintaining a smaller temperature difference between the surface and the air.

Step by step solution

01

Understand the problem

We need to compare the temperatures at the inner and outer surfaces of the wall, taking into consideration the convection heat transfer coefficients at both surfaces.
02

Convection heat transfer coefficients

Let's denote the convection heat transfer coefficient at the outer surface as h_out and the one at the inner surface as h_in. According to the problem statement, we have h_out = 3 * h_in.
03

Heat conduction through the wall

We can express the heat conduction through the wall using Fourier's Law of heat conduction, given by: q = -k * (dT/dx) Where q is the heat flux, k is the thermal conductivity of the wall, dT/dx is the temperature gradient across the wall.
04

Heat convection at the surfaces

Heat convection at both surfaces can be expressed using Newton's Law of cooling, as follows: For the outer surface: q = h_out * (T_wall,out - T_air,out) For the inner surface: q = h_in * (T_wall,in - T_air,in)
05

Steady heat transfer condition

Given that heat transfer is steady, the heat flux through the wall must be equal to heat convection at both surfaces. Therefore, h_out * (T_wall,out - T_air,out) = h_in * (T_wall,in - T_air,in)
06

Solve for the temperature difference

From the previous step, we can rewrite the equation in terms of temperature difference: 3 * (T_wall,out - T_air,out) = (T_wall,in - T_air,in)
07

Determine which surface is closer to the air temperature

Since the outer surface has a higher convection heat transfer coefficient (h_out = 3 * h_in), the temperature difference between the outer surface and the surrounding air temperature will be smaller than the temperature difference between the inner surface and the surrounding air temperature. Therefore, the temperature at the outer surface of the wall will be closer to the surrounding air temperature compared to the inner surface. This is because the higher convection heat transfer coefficient caused by winds on the outer surface allows for more efficient heat transfer, thus maintaining a smaller temperature difference between the surface and the air.

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Key Concepts

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

Convection Heat Transfer Coefficient
Heat transfer between a solid surface and the adjacent fluid that is in motion is described by the convection heat transfer coefficient, represented by the symbol 'h'. This coefficient measures the rate at which heat is transferred per unit area and per degree of temperature difference between the surface and the fluid. Essentially, a high 'h' means heat can more efficiently move from the surface to the fluid or vice versa.

For example, when the wind increases outside the room mentioned in our original exercise, the convection heat transfer coefficient at the wall's outer surface increases. This leads to a higher rate of heat loss from the wall to the cold outside air, which conveys the idea that with a higher 'h_out', the wall is more effective at

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Most popular questions from this chapter

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. What statement below is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

A 20-cm-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\) and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

What is an infinitely long cylinder? When is it proper to treat an actual cylinder as being infinitely long, and when is it not?

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

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) \(37 \mathrm{~W}\)
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