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Chapter 1: Problem 74

Most of the energy we consume as food is converted to thermal energy in the process of performing all our bodily functions and is ultimately lost as heat from our bodies. Consider a person who consumes \(2100 \mathrm{kcal}\) per day (note that what are commonly referred to as food calories are actually kilocalories), of which \(2000 \mathrm{kcal}\) is converted to thermal energy. (The remaining \(100 \mathrm{kcal}\) is used to do work on the environment.) The person has a surface area of \(1.8 \mathrm{~m}^{2}\) and is dressed in a bathing suit. (a) The person is in a room at \(20^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). At this air temperature, the person is not perspiring much. Estimate the person's average skin temperature. (b) If the temperature of the environment were \(33^{\circ} \mathrm{C}\), what rate of perspiration would be needed to maintain a comfortable skin temperature of \(33^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
To maintain a comfortable skin temperature of 33 掳C, the person would need an average skin temperature of approximately 28.1 掳C in a 20 掳C environment, thus having a rate of perspiration of about \(4.8\times10^{-4} \, \mathrm{kg/s}\) in a 33 掳C environment.

Step by step solution

01

(a) Calculate energy conversion to thermal

First, we will calculate the amount of energy converted to thermal energy per second. We are given that the person consumes 2100 kcal/day, and that 2000 kcal is converted to thermal energy. Since 1 kcal = 4184 J, we can convert this to Joules and divide by the number of seconds in a day to obtain the rate of energy conversion to thermal energy. \(2000 \, \mathrm{kcal/day} * \frac{4184 \, \mathrm{J}}{1 \, \mathrm{kcal}} * \frac{1 \, \mathrm{day}}{86400 \, \mathrm{s}} = E_t \)
02

(a) Calculate heat transfer rate

Now we need to calculate the rate of heat transfer between the person's skin and the environment. The heat transfer rate can be calculated using the formula: \(q = hA(T_s - T_r)\) Where q is the heat transfer rate, h is the convection heat transfer coefficient, A is the person's surface area, \(T_s\) is the skin temperature, and \(T_r\) is the room temperature. We need to find the skin temperature when the heat transfer rate equals the rate of energy conversion to thermal energy.
03

(a) Find the skin temperature

Set the heat transfer rate equal to the rate of energy conversion to thermal energy and solve for the skin temperature, \(T_s\): \(E_t = hA(T_s - T_r)\) Substitute the values and solve for \(T_s\): \(E_t = (3 \, \mathrm{W/m^2 \cdot K})(1.8 \, \mathrm{m^2})(T_s - 20^\circ \mathrm{C})\)
04

(b) Calculate heat transfer rate at 33 掳C

Now we need to calculate the heat transfer rate when the environment is at 33 掳C and the comfortable skin temperature is also 33 掳C. In this case, we will have no heat transfer due to convection since the skin and environment temperatures are the same: \(q = hA(T_s - T_r) = 0\)
05

(b) Calculate perspiration heat transfer rate

To maintain a comfortable skin temperature of 33 掳C, the excess heat generated in the body must be removed by perspiration. The heat transfer rate due to perspiration can be calculated using the latent heat of vaporization for water: \(q_p = m_p L_v\) Where \(q_p\) is the heat transfer rate due to perspiration, \(m_p\) is the rate of perspiration in kg/s, and \(L_v\) is the latent heat of vaporization of water (2.26x10^6 J/kg). We need to find the rate of perspiration needed to equal the rate of energy conversion to thermal energy.
06

(b) Find perspiration rate

Set the heat transfer rate from perspiration equal to the rate of energy conversion to thermal energy and solve for the perspiration rate, \(m_p\): \(E_t = m_p L_v\) Substitute the values and solve for \(m_p\): \(E_t = m_p (2.26\times10^6 \, \mathrm{J/kg})\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Convection
Convection is a fundamental concept of heat transfer, which occurs when thermal energy is transferred between a solid surface and a moving fluid or gas, such as air or water, that comes into contact with it. This process relies on the movement of the fluid, which can enhance the rate of heat exchange by carrying away or bringing thermal energy.
Convection plays a key role in regulating our body's temperature. When our skin is exposed to air, the heat from our body can be transferred to the surrounding air through convection. The rate of this transfer depends on several factors:
  • **Convection Heat Transfer Coefficient**: This determines how effectively heat is transferred between a solid surface and fluid. In the provided exercise, the coefficient is mentioned to be 3 W/m虏路K, and it influences how quickly heat can move from the skin to the surrounding air.
  • **Temperature Difference**: The difference between the skin temperature and the surrounding air temperature drives the heat transfer. Larger differences increase the rate of heat exchange.
  • **Surface Area**: Larger surface areas can exchange more energy, a notable factor since skin acts as the body's main heat exchange surface.
Convection ensures that excess body heat is dissipated efficiently into the environment, which is crucial especially in varying temperatures to maintain a stable internal temperature without causing discomfort or overheating.
Thermal Energy
Thermal energy refers to the internal energy present in a system due to its temperature. It represents the collective kinetic energy of particles within a substance as they move and vibrate. In the context of human metabolism and activity, thermal energy is the byproduct of converting food into usable energy for body functions.
When we consume food, our body metabolizes a significant portion into thermal energy, supporting processes such as:
  • **Muscle Activity**: Even small activities like breathing or maintaining posture generate thermal energy.
  • **Digestion and Cellular Processes**: Metabolic reactions produce heat as a byproduct.
In our exercise, the breakdown of caloric intake shows that 2000 of the 2100 kcal consumed daily is converted into thermal energy. This conversion is necessary for maintaining our constant body temperature and supporting metabolic functions. Managing the release of this heat is essential to prevent overheating.
Thermal energy balance ensures that our body doesn't absorb more heat than it can release. Through mechanisms like convection and perspiration, we control this energy to maintain optimal conditions, especially since excess thermal energy must be consistently transferred away from the body for thermal regulation.
Perspiration
Perspiration, commonly known as sweating, is the body's natural mechanism to regulate temperature and deal with excess thermal energy. Especially important in high temperatures, perspiration involves the secretion of fluid (mainly water) containing salts and other waste products, which evaporates to cool the body.
When the body heats up, sweat glands produce perspiration to cool down the skin once it evaporates. Here's how it helps regulate temperature:
  • **Evaporative Cooling**: The phase change from liquid sweat to water vapor requires energy. This energy is taken from the body in the form of heat, thus lowering the skin temperature.
  • **Latent Heat of Vaporization**: Sweating utilizes the high latent heat value of water, meaning it removes considerable amounts of heat as sweat evaporates. In the exercise, the latent heat of vaporization for water is cited as 2.26x10鈦� J/kg, indicating how effective this process is in heat transfer.
In scenarios where air temperatures are similar to body temperature, such as 33掳C, the body cannot rely on convection alone. Perspiration becomes critical to maintain a comfortable body temperature by actively removing excess thermal energy. By calculating the required perspiration rate, as shown in the exercise, we ensure the body can maintain thermal equilibrium even without the aid of convection. This sweat evaporative method enables effective temperature control under varying environmental conditions.

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Most popular questions from this chapter

The heat flux that is applied to one face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The opposite face is exposed to air at temperature \(30^{\circ} \mathrm{C}\), with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface temperature of the wall exposed to air is measured and found to be \(50^{\circ} \mathrm{C}\). Do steady-state conditions exist? If not, is the temperature of the wall increasing or decreasing with time?

An internally reversible refrigerator has a modified coefficient of performance accounting for realistic heat transfer processes of $$ \mathrm{COP}_{m}=\frac{q_{\text {in }}}{\dot{W}}=\frac{q_{\text {in }}}{q_{\text {out }}-q_{\text {in }}}=\frac{T_{c, i}}{T_{h, i}-T_{c, i}} $$ where \(q_{\text {in }}\) is the refrigerator cooling rate, \(q_{\text {out }}\) is the heat rejection rate, and \(\dot{W}\) is the power input. Show that \(\mathrm{COP}_{m}\) can be expressed in terms of the reservoir temperatures \(T_{c}\) and \(T_{h}\), the cold and hot thermal resistances \(R_{L, c}\) and \(R_{t, h}\), and \(q_{\text {in }}\), as $$ \mathrm{COP}_{m}=\frac{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}}{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{tot}}} $$ where \(R_{\mathrm{tot}}=R_{t, c}+R_{t, h}\). Also, show that the power input may be expressed as $$ \dot{W}=q_{\mathrm{in}} \frac{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{id \textrm {t }}}}{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}} $$

Consider a carton of milk that is refrigerated at a temperature of \(T_{m \mathrm{r}}=5^{\circ} \mathrm{C}\). The kitchen temperature on a hot summer day is \(T_{\infty}=30^{\circ} \mathrm{C}\). If the four sides of the carton are of height and width \(L=200 \mathrm{~mm}\) and \(w=100 \mathrm{~mm}\), respectively, determine the heat transferred to the milk carton as it sits on the kitchen counter for durations of \(t=10 \mathrm{~s}, 60 \mathrm{~s}\), and \(300 \mathrm{~s}\) before it is returned to the refrigerator. The convection coefficient associated with natural convection on the sides of the carton is \(h=10\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface emissivity is \(0.90\). Assume the milk carton temperature remains at \(5^{\circ} \mathrm{C}\) during the process. Your parents have taught you the importance of refrigerating certain foods from the food safety perspective. Comment on the importance of quickly returning the milk carton to the refrigerator from an energy conservation point of view.

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

The free convection heat transfer coefficient on a thin hot vertical plate suspended in still air can be determined from observations of the change in plate temperature with time as it cools. Assuming the plate is isothermal and radiation exchange with its surroundings is negligible, evaluate the convection coefficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d t)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\circ} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).
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