Question

An average person generates heat at a rate of \(84 \mathrm{~W}\) while resting. Assuming one-quarter of this heat is lost from the head and disregarding radiation, determine the average surface temperature of the head when it is not covered and is subjected to winds at \(10^{\circ} \mathrm{C}\) and \(25 \mathrm{~km} / \mathrm{h}\). The head can be approximated as a 30 -cm-diameter sphere. Assume a surface temperature of \(15^{\circ} \mathrm{C}\) for evaluation of \(\mu_{s}\). Is this a good assumption? Answer: \(13.2^{\circ} \mathrm{C}\)

Short answer

Based on the given conditions and assumptions, calculate the average surface temperature of the head. Solution: 1. Calculate the heat generated by the head: 21 W 2. Calculate the surface area of the head (assuming a 30 cm sphere): approximately 2827.43 cm² 3. Calculate the heat transfer coefficient: approximately 2.88 W/m²°C 4. Calculate the average surface temperature of the head: approximately 10.00257°C. Final Answer: The average surface temperature of the head under the given conditions is approximately 10.00257°C.

Step-by-step solution

Calculate the heat generated by the head

Since only one-quarter of the heat generated by an average person is lost by the head, we need to find out the heat generated by the head. We can find this by multiplying the total heat generated by 1/4. Total heat generated by the person: \(84 \mathrm{~W}\) Heat generated by the head: \(\frac{1}{4} \times 84 \mathrm{~W} = 21 \mathrm{~W}\)

Calculate the radiant heat transfer

To calculate the average surface temperature of the head, we'll need to find the radiant heat transfer by using the Lumped System Analysis equation: \(Q=hA\Delta T\) where: - \(Q = 21 \mathrm{~W}\) (Heat generated by the head) - \(h\) is the heat transfer coefficient - \(A\) is the surface area of the sphere - \(\Delta T\) is the temperature difference between the surface temperature and the air temperature

Calculate head surface area

Since we're assuming the head can be considered as a sphere with a 30 cm diameter, we can find the surface area of the head using the formula for the surface area of a sphere: \(A = 4\pi r^2\) where \(r\) is the radius of the sphere (half the diameter). Diameter of the sphere = 30 cm Radius: \(r = \frac{30}{2} = 15\) cm Now, we can find the surface area of the sphere: \(A = 4\pi (15)^2 = 4\pi (225) \approx 2827.43 \mathrm{cm^2}\)

Find the heat transfer coefficient, h

Now, we need to determine the heat transfer coefficient (\(h\)) accounting for the wind. \(h = \frac{10.45 - v + 10 \sqrt{v}}{t_{s}+ 273 - 273 / t_{s}}\) where: - \(v\) is the wind speed - \(t_{s}\) is the assumed surface temperature First, let's convert wind speed to m/s: Wind speed = \(25\mathrm{~km}/\mathrm{h} \times \frac{1000\mathrm{~m}}{1 \mathrm{~km}} \times \frac{1\mathrm{~h}}{3600\mathrm{s}} \approx 6.944\mathrm{~m}/\mathrm{s}\) With the wind speed and assumed surface temperature, we can find the heat transfer coefficient: \(h = \frac{10.45 - 6.944 + 10\sqrt{6.944}}{(15+273)-273/15} \approx 2.88 \mathrm{W}/\mathrm{m^2 °C}\)

Calculate average surface temperature of the head

Now we have everything to calculate the average surface temperature of the head. Rearranging the heat transfer equation: \(\Delta T = \frac{Q}{hA}\) and \(T_{\text{surface}} = T_{\text{air}} + \Delta T\) Calculating the temperature difference between the surface temperature and the air temperature: \(\Delta T = \frac{21\mathrm{~W}}{(2827.43\mathrm{cm^2} * 2.88\mathrm{W}/\mathrm{m^2 °C})} \approx 0.00257^{\circ} \mathrm{C}\) Now, we can find the average surface temperature of the head: \(T_{\text{surface}} = 10^{\circ}\mathrm{C} + 0.00257^{\circ}\mathrm{C} \approx 10.00257^{\circ}\mathrm{C}\) The head's surface temperature is approximately \(10.00257^{\circ}\mathrm{C}\), which is lower than the initial assumed temperature of \(15^{\circ}\mathrm{C}\). The surface temperature of the head when subjected to winds of \(10^{\circ}\mathrm{C}\) and \(25\mathrm{~km}/\mathrm{h}\) is in fact quite low compared to our assumption.

Key concepts

Lumped System Analysis

Lumped System Analysis is a method used in heat transfer to simplify complex problems. It assumes that the temperature within an object changes uniformly over time, which is true when the heat conduction within the object is much faster than the heat transfer across the boundary. This method is particularly applicable when an object has a small Biot number (Bi < 0.1), suggesting that the internal resistance to heat transfer is much less than the external resistance.

When applying the Lumped System Analysis to the human body, or specifically the head in our exercise, it allows us to use the formula for heat transfer, \(Q=hA\Delta T\), with the assumption that the entire surface of the head is at a uniform temperature. This greatly simplifies the calculation of the heat loss from the head to its surrounding environment.

Heat Transfer Coefficient

The heat transfer coefficient, \(h\), is a measure of the heat transfer rate per unit area per unit temperature difference. It is dependent on the properties of the fluid (air) around the object, the flow characteristics, and the nature of the surface. In our exercise, we calculate \(h\) considering the wind speed and temperature to determine how effectively the head loses heat to the surrounding air.

Understanding the heat transfer coefficient is crucial as it directly affects the rate of cooling or heating of the head. In reality, this coefficient can vary depending on several factors, such as the relative motion between the head and the air (wind speed), as well as the surface properties of the head.

Surface Temperature Calculation

Calculating the surface temperature of an object, especially a human body part like the head, involves determining the equilibrium temperature at which the heat generated by the body is equal to the heat lost to the surrounding environment. In the context of the provided exercise, several steps are taken to approximate the head's surface temperature when exposed to wind.

The surface temperature is critical for understanding thermal comfort, potential for frostbite, or hyperthermia conditions. However, it's important to note that our skin temperature can differ from our internal body temperature, and a range of environmental and physiological factors influence it.

Radiant Heat Transfer

Radiant heat transfer involves the transfer of heat via electromagnetic waves, without the need for a medium like air or water. This type of heat transfer is particularly important when considering the human body since we naturally emit infrared radiation. While our exercise discounts radiation to simplify the problem, in a more comprehensive analysis, accounting for radiant heat loss would be important for measuring the total heat loss from the body.

In many outdoor scenarios, especially in colder environments, radiant heat transfer can significantly affect a person's thermal comfort. Radiative heat can be lost to the surroundings if the environment's temperature is lower than the body's surface temperature, which is often the case in cold climate conditions.