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Chapter 3: Problem 10

The wind chill, which is experienced on a cold, windy day, is related to increased heat transfer from exposed human skin to the surrounding atmosphere. Consider a layer of fatty tissue that is \(3 \mathrm{~mm}\) thick and whose interior surface is maintained at a temperature of \(36^{\circ} \mathrm{C}\). On a calm day the convection heat transfer coefficient at the outer surface is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), but with \(30 \mathrm{~km} / \mathrm{h}\) winds it reaches \(65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). In both cases the ambient air temperature is \(-15^{\circ} \mathrm{C}\). (a) What is the ratio of the heat loss per unit area from the skin for the calm day to that for the windy day? (b) What will be the skin outer surface temperature for the calm day? For the windy day? (c) What temperature would the air have to assume on the calm day to produce the same heat loss occurring with the air temperature at \(-15^{\circ} \mathrm{C}\) on the windy day?

Short Answer

Expert verified
The ratio of heat loss per unit area from the skin on a calm day to that on a windy day is \(\frac{q_c}{q_w} = \frac{h_c \cdot (T_{sc} - T_\infty)}{h_w \cdot (T_{sw} - T_\infty)}\). To find the outer surface temperature for calm and windy days, use the conduction heat transfer formula: \(T_c = T_h - \frac{q \cdot L}{k}\). To find the required air temperature on a calm day for the same heat loss as on a windy day, solve for \(T_\infty\): \(T_\infty = T_{sc} - \frac{q_w}{h_c}\).

Step by step solution

01

Calculate Heat Loss Per Unit Area on Calm and Windy Days

To find the ratio of heat loss, we need to first calculate the heat loss per unit area on both calm and windy days. We will use the following convection heat transfer formula: \(q = h \cdot A \cdot (T_s - T_\infty)\) Where: \(q\) - Heat loss per unit area \(h\) - Convection heat transfer coefficient \(A\) - Surface area \(T_s\) - Outer surface temperature \(T_\infty\) - Ambient air temperature For a calm day: \(q_c = h_c \cdot A \cdot (T_{sc} - T_\infty)\) For a windy day: \(q_w = h_w \cdot A \cdot (T_{sw} - T_\infty)\) Since we are asked to find the ratio, the surface area A will cancel out. So we can write it as: \(\frac{q_c}{q_w} = \frac{h_c \cdot (T_{sc} - T_\infty)}{h_w \cdot (T_{sw} - T_\infty)}\)
02

Find the Outer Surface Temperature for Calm and Windy Days

To find the outer surface temperature for calm and windy days, we need to make use of the conduction heat transfer formula: \(q = k \cdot \frac{T_h - T_c}{L}\) Where: \(k\) - Thermal conductivity of fatty tissue (W/m.K) \(T_h\) - Interior surface temperature \(T_c\) - Outer surface temperature \(L\) - Layer thickness Since we are given the interior surface temperature, we can rewrite the formula as: \(T_c = T_h - \frac{q \cdot L}{k}\) We'll need to perform this calculation for both calm and windy days, substituting the appropriate values for \(q\), \(h\), and \(T_\infty\).
03

Find the Required Air Temperature on Calm Day to Produce the Same Heat Loss as Windy Day

To find the required air temperature on a calm day that would produce the same heat loss as on a windy day, we need to solve for \(T_\infty\) in the convection heat transfer formula for the calm day: \(q_w = h_c \cdot (T_{sc} - T_\infty)\) Rearrange and solve for \(T_\infty\): \(T_\infty = T_{sc} - \frac{q_w}{h_c}\) Input the known values from step 1 and 2 to find the required air temperature on the calm day.

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Key Concepts

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

Convection Heat Transfer
Convection heat transfer is a mode of heat transfer that occurs when a fluid (such as air or water) moves over a surface and transfers heat between the surface and the fluid. This process is affected by the movement of the fluid, which can be natural, as in buoyancy, or forced, such as wind blowing over a surface. The rate of heat transfer is governed by the convection heat transfer coefficient, denoted as \( h \), which depends on various factors like fluid velocity and the nature of the fluid flow (laminar or turbulent).

In the given exercise, we see two scenarios: one with calm air conditions and the other with wind present. The higher the speed of the wind, the greater the convection heat transfer coefficient. This means more heat is lost from the surface as the wind blows faster, which can be observed through the increased coefficient value from \(25 \ ext{W/m}^2 \cdot \text{K}\) to \(65 \ ext{W/m}^2 \cdot \text{K}\).

Thus, knowing how to calculate the convection heat transfer coefficient is key to understanding and predicting how different environmental conditions affect heat loss from a surface.
Thermal Conductivity
Thermal conductivity is a material property indicating a material's ability to conduct heat. It is denoted by \( k \) and measured in watts per meter-kelvin (\( ext{W/m} \, \text{K} \)). High thermal conductivity means heat can pass through the material quickly, while low conductivity indicates that the material is a good insulator.

Consider the layer of fatty tissue mentioned in the exercise. This tissue acts as an insulator for the human body, preventing rapid heat loss. Its thermal conductivity value determines how much heat transfer occurs between the body's interior and the outer surface. Using the formula for conduction, \[ q = k \cdot \frac{T_h - T_c}{L} \] where \( T_h \) is the temperature on the inside of the tissue, \( T_c \) is the temperature at the surface, and \( L \) is the thickness of the tissue, one can find how temperature changes across the tissue layer.

When comparing calm and windy days, thermal conductivity continues to drive the conduction process, while the convection heat transfer coefficient changes, altering the outer temperature of the tissue due to environmental effects.
Wind Chill Effect
The wind chill effect is a phenomenon where the perceived temperature on a human or object is lower than the actual air temperature due to wind. This occurs because the wind increases the convective heat transfer from the skin to the environment, making it feel colder.

In the exercise, this effect is quantified by an increased heat loss rate as wind speed increases, illustrated by the change in the convection heat transfer coefficient. On a calm day, the body loses heat at a lower rate than on a windy day. As the wind might not actually change the temperature of the air, it changes the rate of heat transfer and thereby affects how cold it feels or how quickly heat loss from the body occurs.

Understanding how to calculate the change in temperature the body perceives due to wind (the wind chill temperature) involves not just the actual air temperature but also the wind speed and the thermodynamics of heat transfer. This insight helps one understand why, on windy days, one feels colder and requires better insulation to maintain body warmth.

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Most popular questions from this chapter

A nuclear reactor fuel element consists of a solid cylindrical pin of radius \(r_{1}\) and thermal conductivity \(k_{f}\). The fuel pin is in good contact with a cladding material of outer radius \(r_{2}\) and thermal conductivity \(k_{c^{*}}\). Consider steady-state conditions for which uniform heat generation occurs within the fuel at a volumetric rate \(\dot{q}\) and the outer surface of the cladding is exposed to a coolant that is characterized by a temperature \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain equations for the temperature distributions \(T_{f}(r)\) and \(T_{c}(r)\) in the fuel and cladding, respectively. Express your results exclusively in terms of the foregoing variables. (b) Consider a uranium oxide fuel pin for which \(k_{f}=2\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{1}=6 \mathrm{~mm}\) and cladding for which \(k_{c}=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(r_{2}=9 \mathrm{~mm}\). If \(\dot{q}=2 \times 10^{8}\) \(\mathrm{W} / \mathrm{m}^{3}, h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and \(T_{\infty}=300 \mathrm{~K}\), what is the maximum temperature in the fuel element? (c) Compute and plot the temperature distribution, \(T(r)\), for values of \(h=2000,5000\), and 10,000 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the operator wishes to maintain the centerline temperature of the fuel element below \(1000 \mathrm{~K}\), can she do so by adjusting the coolant flow and hence the value of \(h\) ?

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at \(5-\mathrm{mm}\) increments along the tube length, what is the heat loss per meter of tube length?

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.
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