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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

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

A storage tank consists of a cylindrical section that has a length and inner diameter of \(L=2 \mathrm{~m}\) and \(D_{i}=1 \mathrm{~m}\), respectively, and two hemispherical end sections. The tank is constructed from 20-mm-thick glass (Pyrex) and is exposed to ambient air for which the temperature is \(300 \mathrm{~K}\) and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The tank is used to store heated oil, which maintains the inner surface at a temperature of \(400 \mathrm{~K}\). Determine the electrical power that must be supplied to a heater submerged in the oil if the prescribed conditions are to be maintained. Radiation effects may be neglected, and the Pyrex may be assumed to have a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

A thin electrical heater is wrapped around the outer surface of a long cylindrical tube whose inner surface is maintained at a temperature of \(5^{\circ} \mathrm{C}\). The tube wall has inner and outer radii of 25 and \(75 \mathrm{~mm}\), respectively, and a thermal conductivity of \(10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal contact resistance between the heater and the outer surface of the tube (per unit length of the tube) is \(R_{t, c}^{\prime}=\) \(0.01 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\). The outer surface of the heater is exposed to a fluid with \(T_{\infty}=-10^{\circ} \mathrm{C}\) and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the heater power per unit length of tube required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C} .\)

Consider a power transistor encapsulated in an aluminum case that is attached at its base to a square aluminum plate of thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(L=6 \mathrm{~mm}\), and width \(W=20 \mathrm{~mm}\). The case is joined to the plate by screws that maintain a contact pressure of 1 bar, and the back surface of the plate transfers heat by natural convection and radiation to ambient air and large surroundings at \(T_{\infty}=T_{\text {sur }}=\) \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(\varepsilon=0.9\), and the convection coefficient is \(h=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The case is completely enclosed such that heat transfer may be assumed to occur exclusively through the base plate. (a) If the air-filled aluminum-to-aluminum interface is characterized by an area of \(A_{c}=2 \times 10^{-4} \mathrm{~m}^{2}\) and a roughness of \(10 \mu \mathrm{m}\), what is the maximum allowable power dissipation if the surface temperature of the case, \(T_{s, c}\), is not to exceed \(85^{\circ} \mathrm{C}\) ? (b) The convection coefficient may be increased by subjecting the plate surface to a forced flow of air. Explore the effect of increasing the coefficient over the range \(4 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A stainless steel (AISI 304) tube used to transport a chilled pharmaceutical has an inner diameter of \(36 \mathrm{~mm}\) and a wall thickness of \(2 \mathrm{~mm}\). The pharmaceutical and ambient air are at temperatures of \(6^{\circ} \mathrm{C}\) and \(23^{\circ} \mathrm{C}\), respectively, while the corresponding inner and outer convection coefficients are \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) What is the heat gain per unit tube length? (b) What is the heat gain per unit length if a \(10-\mathrm{mm}\) thick layer of calcium silicate insulation \(\left(k_{\text {ins }}=\right.\) \(0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tube?
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