Question

\(80^{\circ} \mathrm{C}\). Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1-133 It is well known that wind makes the cold air feel much colder as a result of the wind chill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The wind chill effect is usually expressed in terms of the wind chill temperature (WCT), which is the apparent temperature felt by exposed skin. For outdoor air temperature of \(0^{\circ} \mathrm{C}\), for example, the wind chill temperature is \(-5^{\circ} \mathrm{C}\) at \(20 \mathrm{~km} / \mathrm{h}\) winds and \(-9^{\circ} \mathrm{C}\) at \(60 \mathrm{~km} / \mathrm{h}\) winds. That is, a person exposed to \(0^{\circ} \mathrm{C}\) windy air at \(20 \mathrm{~km} / \mathrm{h}\) will feel as cold as a person exposed to \(-5^{\circ} \mathrm{C}\) calm air (air motion under \(5 \mathrm{~km} / \mathrm{h}\) ). For heat transfer purposes, a standing man can be modeled as a 30 -cm- diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C}\). For a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from this man by convection in still air at \(20^{\circ} \mathrm{C}\). What would your answer be if the convection heat transfer coefficient is increased to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as a result of winds? What is the wind chill temperature in this case?

Short answer

Question: Determine the rate of heat loss for a standing man in still air and windy conditions, and calculate the wind chill temperature, given the following parameters: the man is modeled as a vertical cylinder with a diameter of 0.3m and height of 1.7m, the convection heat transfer coefficient in still air is 15 W/m²K, the convection heat transfer coefficient in windy conditions is 30 W/m²K, and the ambient temperature is 20°C. Answer: The rate of heat loss in still air is approximately 565.49 W, whereas the rate of heat loss in windy conditions is approximately 1130.97 W. The wind chill temperature is approximately 14°C.

Step-by-step solution

Find exposed surface area of the man

The standing man is modeled as a vertical cylinder with a diameter of \(d = 0.3~\mathrm{m}\) and length (height) of \(h = 1.7~\mathrm{m}\). Since the top and bottom surfaces are insulated, we only need to determine the curved surface area of the cylinder. The exposed surface area for the curved surface is given by the formula \(A = 2 \pi r h\), where \(r\) is the radius of the cylinder. The radius of the cylinder is half its diameter: \(r = 0.15~\mathrm{m}\). Now, let's calculate the surface area \(A\): $$A = 2 \pi(0.15)(1.7)$$

Calculate the rate of heat loss in still air

To find the rate of heat loss in still air, we will use the convection heat transfer equation \(Q = hA(T_{s}-T_{\infty})\), where \(h\) is the convection heat transfer coefficient, \(A\) is the surface area, \(T_s\) is the surface temperature, and \(T_\infty\) is the ambient temperature. \(h = 15~\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (still air) \(T_s = 34^\circ \mathrm{C}\) (surface temperature of the man) \(T_\infty = 20^\circ \mathrm{C}\) (ambient temperature) Using the surface area calculated in Step 1, we can now find the rate of heat loss in still air: $$Q = (15)(2\pi(0.15)(1.7))(34-20)$$

Calculate the rate of heat loss in windy conditions

To find the rate of heat loss in windy conditions, we will again use the convection heat transfer equation \(Q = hA(T_{s}-T_{\infty})\), with the updated convection heat transfer coefficient due to the wind: \(h = 30~\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (windy conditions) The surface area, surface temperature, and ambient temperature remain the same. We can now find the rate of heat loss in windy conditions: $$Q = (30)(2\pi(0.15)(1.7))(34-20)$$

Calculate the wind chill temperature

To find the wind chill temperature, we can set the rate of heat loss in windy conditions equal to the rate of heat loss in still air at the wind chill temperature (\(T_{wct}\)). We can use the convection heat transfer equation: $$(30)(2\pi(0.15)(1.7))(34-T_{wct}) = (15)(2\pi(0.15)(1.7))(34-20)$$ Now, we can solve for the wind chill temperature \(T_{wct}\). By following the steps above and solving the equations, the rate of heat loss for still air and windy conditions, as well as the wind chill temperature, can be determined.

Key concepts

Heat Loss Calculation

Understanding how to calculate heat loss is pivotal in various scientific and engineering tasks. It is a key concept in thermodynamics and can be necessary for optimizing thermal comfort in buildings or understanding the thermal balance of living organisms. The fundamental formula for calculating heat loss due to convection is given by the equation ewline
ewline\( Q = hA(T_s - T_{\text{ambient}}) \).ewline
ewlineHere, \(Q\) stands for the rate of heat loss, \(h\) is the convection heat transfer coefficient, \(A\) denotes the exposed surface area, \(T_s\) is the surface temperature, and \(T_{\text{ambient}}\) is the ambient temperature. For a person or an object losing heat, it's critical to determine the surface area exposed to the cooler air, as it directly affects the total heat loss. In our example, we are considering a cylindrical model to represent a man. This model simplifies a complex human shape to a more straightforward geometry for which the surface area can be easily calculated. By plugging in the specific values into the formula, we can deduce the amount of heat being lost over a given time period, a fundamental step in assessing the body's thermal comfort or energy efficiency of a system.

Wind Chill Temperature

The wind chill temperature (WCT) is an index designed to reflect how cold the environment feels on human skin due to wind speed. It is not an actual air temperature but a perceived equivalent temperature. For instance, even when the thermometer reads \(0^{\text{circ}} \text{C}\), a brisk wind can make it feel much colder. Wind increases the rate of heat loss from the body through convective heat transfer, making us feel colder than we would in still air.ewline
ewlineThe concept of wind chill is crucial when discussing outdoor comfort or safety during cold weather. Wind chill is calculated by using temperature and wind speed to determine how the combination of these affects human sensation. This calculation can help in making safety guidelines for outdoor activities, determining potential frostbite times, or guiding proper clothing choices. In the exercise example, a wind chill temperature is calculated by equating the rate of heat loss from the skin at two different conditions – one with still air and one with wind – to find the perceived temperature or the wind chill.

Convection Heat Transfer Coefficient

The convection heat transfer coefficient (h) is a measure of the convective heat transfer rate per unit area and temperature difference. It primarily depends on the properties of the fluid (in our case, air), the flow conditions, and the nature of the surface. This coefficient is typically measured in watts per square meter per Kelvin (\( \text{W/m}^2\text{K} \)). In practical terms, it signifies how effective the convective process is at removing heat from a surface.ewline
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Impact of Wind on Heat Transfer Coefficient

The exercise demonstrates that the heat transfer coefficient can vary significantly due to environmental conditions like wind speed. Wind effectively enhances the convection process by replacing the warmed-up air next to the surface (which naturally has lower convection potential) with cooler air continuously, thus maintaining a higher temperature gradient and, consequently, increasing the value of \(h\). This increase is why we feel much colder in windy conditions, and it's also the reason the rate of heat loss from an object or person accelerates, as illustrated when the coefficient doubles from \(15 \text{W/m}^2\text{K}\) in still air to \(30 \text{W/m}^2\text{K}\) in windy conditions. Understanding and measuring this coefficient is important for a wide range of applications, including weather forecasting, designing clothing, and calculating heat loss in buildings.