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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 air temperature equivalent 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 $\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 \mathrm{~V}^{0.16}+0.3965 T V^{0.16}$ 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
The wind-chill temperature index (WCTI) formula can be converted from metric units to English units by applying two conversion factors: one for temperature (from degrees Celsius to degrees Fahrenheit) using the formula \( T(°F) = \frac{9}{5} \times T(°C) + 32\), and one for wind speed (from kilometers per hour to miles per hour) using the formula \(V(mph) = V(km/h) \times 0.621371\). After substituting these conversions into the metric formula and simplifying, the English units formula becomes: WCTI\((^{\circ}F) = 35.74 + 0.6215T - 35.75V^{0.16} + 0.4275TV^{0.16}\).

Step by step solution

01

Identify relevant conversion factors

To convert the given formula into English units, we need two conversion factors: one to convert the temperature from degrees Celsius (°C) to degrees Fahrenheit (°F), and one to convert the wind speed from kilometers per hour (km/h) to miles per hour (mph). 1. Convert temperature from Celsius to Fahrenheit: \( T(°F) = \frac{9}{5} \times T(°C) + 32\) 2. Convert wind speed from km/h to mph: \(V(mph) = V(km/h) \times 0.621371\) Now that we have the conversion factors, we can proceed with converting the formula.
02

Convert the formula using conversion factors

To convert the formula, we will substitute \(T(°C)\) with \(\frac{5}{9}(T(°F) - 32)\) and \(V(km/h)\) with \(V(mph) \div 0.621371\) in the given metric formula. WCTI\((^{\circ}F) = 13.12 + 0.6215(\frac{9}{5} T - 32) - 11.37(\frac{V}{0.621371})^{0.16} + 0.3965 (\frac{9}{5} T - 32)(\frac{V}{0.621371})^{0.16}\) Simplify and rearrange the formula: WCTI\((^{\circ}F) = 35.74 + 0.6215T - 35.75V^{0.16} + 0.4275TV^{0.16}\)
03

Create a table for WCTI

Use the metric formula to calculate WCTI for given air temperatures ranging from \(10^{\circ}\)C to \(-60^{\circ}\)C and wind speeds ranging from \(10\) km/h to \(80\) km/h. For simplicity, the results are presented only for a sample of values. | Air Temperature (°C) | Wind Speed (km/h) | WCTI (°C) | | -------------------- | ----------------- | --------- | | 10 | 10 | 4.72 | | -10 | 10 | -25.08 | | -30 | 10 | -45.08 | | 10 | 40 | -10.14 | | -20 | 40 | -42.52 | | -40 | 40 | -62.52 | We could calculate more values if needed, or use a similar process using English units formula.
04

Comment on the cooling effect and danger of frostbite

The table shows that as the wind speed increases, the cooling effect (WCTI) also decreases, which indicates that wind has a significant cooling effect on the body. In extremely cold temperatures and high wind speeds, the danger of frostbite increases significantly. It is essential to take adequate precautions and protect exposed skin in such conditions to avoid frostbite and potentially life-threatening situations.

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Most popular questions from this chapter

A person's head can be approximated as a \(25-\mathrm{cm}\) 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) \(249 \mathrm{~W}\)

Heat is lost through a brick wall $(k=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(, which is \)4 \mathrm{~m}\( long, \)3 \mathrm{~m}\( wide, and \)25 \mathrm{~cm}\( thick at a rate of \)500 \mathrm{~W}$. If the inner surface of the wall is at \(22^{\circ} \mathrm{C}\), the temperature at the midplane of the wall is (a) \(0^{\circ} \mathrm{C}\) (b) \(7.5^{\circ} \mathrm{C}\) (c) \(11.0^{\circ} \mathrm{C}\) (d) \(14.8^{\circ} \mathrm{C}\) (e) \(22^{\circ} \mathrm{C}\)

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment in both summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

A person standing in a room loses heat to the air in the room by convection and to the surrounding surfaces by radiation. Both the air in the room and the surrounding surfaces are at \(20^{\circ} \mathrm{C}\). The exposed surface of the person is \(1.5 \mathrm{~m}^{2}\) and has an average temperature of \(32^{\circ} \mathrm{C}\) and an emissivity of \(0.90\). If the rates of heat transfer from the person by convection and by radiation are equal, the combined heat transfer coefficient is (a) \(0.008 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (b) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(5.5 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) (d) \(8.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(10.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

A \(3-\mathrm{m}^{2}\) black surface at \(140^{\circ} \mathrm{C}\) is losing heat to the surrounding air at \(35^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of \(16 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\) and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat loss from the surface is (a) \(5105 \mathrm{~W}\) (b) \(2940 \mathrm{~W}\) (c) \(3779 \mathrm{~W}\) (d) \(8819 \mathrm{~W}\) (e) \(5040 \mathrm{~W}\)

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