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Chapter 1: Problem 168

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

Short Answer

Expert verified
Question: Describe the relationship between air temperature, wind speed, and the risk of frostbite according to the Wind Chill Temperature Index (WCTI). Answer: The Wind Chill Temperature Index (WCTI) indicates that as air temperature decreases and wind speed increases, there is a more significant cooling effect on the human body, increasing the risk of frostbite. It is essential to take precautions such as wearing layered clothing, covering exposed skin, and minimizing time spent outdoors in cold and windy conditions to mitigate this risk.

Step by step solution

01

Convert the WCTI equation from metric units to English units

First, we need to convert the temperature. Fahrenheit to Celsius conversion formula is: \(T_c = \frac{5}{9}(T_f - 32)\) Next, we have to convert the wind speed from \(\mathrm{km/h}\) to \(\mathrm{mph}\). The conversion factor is: \(1 \; \mathrm{mph} = 1.609 \; \mathrm{km/h}\) So, \(V_{\mathrm{mph}} = \frac{V_{\mathrm{km/h}}}{1.609}\) Now, we have to substitute the new expressions for \(T_c\) and \(V_{\mathrm{mph}}\) into the original WCTI equation to derive the equivalent equation in English units.
02

Prepare a table for WCTI

We will now prepare a table for WCTI for air temperatures ranging from \(10^{\circ} \mathrm{C}\) to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from \(10 \mathrm{~km} / \mathrm{h}\) to \(80 \mathrm{~km} / \mathrm{h}\) using the WCTI equation in English units: WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\)
03

Comment on the cooling effect and the danger of frostbite

The table generated in step 2 shows how the wind chill temperature index (WCTI) decreases as the air temperature decreases and the wind speed increases. This means that colder air temperatures combined with higher wind speeds result in a more significant cooling effect on the human body. High wind speeds and low temperatures increase the risk of frostbite, especially on exposed skin. It is essential to take appropriate precautions, such as wearing layered clothing, covering exposed skin, and minimizing time spent outdoors in such conditions.

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Most popular questions from this chapter

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick \(2-\mathrm{m} \times 2-\mathrm{m}\) window glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were \(1 \mathrm{~cm}\) thick?

The heat generated in the circuitry on the surface of a silicon chip \((k=130 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is conducted to the ceramic substrate to which it is attached. The chip is \(6 \mathrm{~mm} \times 6 \mathrm{~mm}\) in size and \(0.5 \mathrm{~mm}\) thick and dissipates \(5 \mathrm{~W}\) of power. Disregarding any heat transfer through the \(0.5-\mathrm{mm}\) high side surfaces, determine the temperature difference between the front and back surfaces of the chip in steady operation.

Consider a flat-plate solar collector placed on the roof of a house. The temperatures at the inner and outer surfaces of the glass cover are measured to be \(33^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The glass cover has a surface area of \(2.5 \mathrm{~m}^{2}\), a thickness of \(0.6 \mathrm{~cm}\), and a thermal conductivity of \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is lost from the outer surface of the cover by convection and radiation with a convection heat transfer coefficient of \(10 \mathrm{~W} /\) \(\mathrm{m}^{2} \cdot \mathrm{K}\) and an ambient temperature of \(15^{\circ} \mathrm{C}\). Determine the fraction of heat lost from the glass cover by radiation.

A person's head can be approximated as a 25-cm diameter sphere at \(35^{\circ} \mathrm{C}\) with an emissivity of \(0.95\). Heat is lost from the head to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and by radiation to the surrounding surfaces at \(10^{\circ} \mathrm{C}\). Disregarding the neck, determine the total rate of heat loss from the head. (a) \(22 \mathrm{~W}\) (b) \(27 \mathrm{~W}\) (c) \(49 \mathrm{~W}\) (d) \(172 \mathrm{~W}\) (e) \(249 \mathrm{~W}\)

While driving down a highway early in the evening, the air flow over an automobile establishes an overall heat transfer coefficient of \(18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The passenger cabin of this automobile exposes \(9 \mathrm{~m}^{2}\) of surface to the moving ambient air. On a day when the ambient temperature is \(33^{\circ} \mathrm{C}\), how much cooling must the air conditioning system supply to maintain a temperature of \(20^{\circ} \mathrm{C}\) in the passenger cabin? (a) \(670 \mathrm{~W}\) (b) \(1284 \mathrm{~W}\) (c) \(2106 \mathrm{~W}\) (d) \(2565 \mathrm{~W}\) (e) \(3210 \mathrm{~W}\)
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