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Chapter 1: Problem 29

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The person feels chilled in the winter and comfortable in the summer due to the difference in net heat loss or gain caused by radiation and natural convection. In winter, the person loses more heat due to radiation as the wall temperature is lower than in summer. This difference in heat loss between summer and winter, mainly due to radiation, accounts for the differing comfort levels experienced by the person.

Step by step solution

01

Identify the knowns and unknowns

We know: - Room air temperature: \(T_{air} = 20^\circ \mathrm{C}\) - Wall temperature in summer: \(T_{wall,s} = 27^\circ \mathrm{C}\) - Wall temperature in winter: \(T_{wall,w} = 14^\circ \mathrm{C}\) - Person's surface temperature: \(T_{body} = 32^\circ \mathrm{C}\) - Person's surface emissivity: \(\epsilon = 0.90\) - Heat transfer coefficient for natural convection: \(h = 2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Objective: - Analyze the heat loss or gain in summer and winter due to radiation and natural convection.
02

Heat exchange due to radiation

First, let's analyze the heat exchange due to radiation between the person's body and the walls. We use the Stefan-Boltzmann law. The formula for the rate of heat exchange due to radiation: \[Q_{rad} = A \epsilon \sigma (T_1^4 - T_2^4)\] where, - A: surface area of the person's body, - \(\epsilon\): emissivity of the person's body, - \(\sigma\): Stefan-Boltzmann constant \(\approx 5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\), - \(T_1\) is the temperature of the person's body, and - \(T_2\) is the temperature of the surrounding walls. In our case, the heat exchange due to radiation looks like: 1. For summer: \(Q_{rad,s} = A \epsilon \sigma (T_{body}^4 - T_{wall,s}^4)\) 2. For winter: \(Q_{rad,w} = A \epsilon \sigma (T_{body}^4 - T_{wall,w}^4)\) Let's calculate the values of \(Q_{rad,s}\) and \(Q_{rad,w}\).
03

Heat exchange due to natural convection

Now, let's analyze the heat exchange due to natural convection between the person's body and room air. We use the formula for the rate of heat exchange due to natural convection: \[Q_{conv} = A h (T_{body} - T_{air})\] In our case, the heat exchange due to natural convection looks like: \[Q_{conv} = A h (T_{body} - T_{air})\] Let's calculate the value of \(Q_{conv}\).
04

Net heat loss or gain

Finally, let's determine the net heat loss or gain in summer and winter. 1. For summer: \(Q_{net,s} = Q_{rad,s} + Q_{conv}\) 2. For winter: \(Q_{net,w} = Q_{rad,w} + Q_{conv}\) Compare the net heat loss or gain in summer and winter to understand why the person feels chilled in winter and comfortable in summer. By calculating these values and comparing the net heat loss or gain in summer and winter, the person feels chilled in the winter because they lose more heat in winter due to radiation and natural convection compared to summer. The difference in heat loss due to radiation caused by the differing wall temperatures in summer and winter mainly explains the situation.

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Key Concepts

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

Radiative Heat Exchange
When we talk about radiative heat exchange, we're discussing how heat energy moves in the form of electromagnetic waves, without needing a medium to travel through. This type of heat transfer can occur in a vacuum and is how the Sun's energy reaches the Earth.

Consider a person sitting in a room: their body, at a higher temperature, emits infrared radiation, which can be absorbed, reflected, or transmitted by the surrounding walls and objects. The amount of heat exchanged through this process depends on the temperature difference, the surface's emissivity (a measure of how well it emits infrared energy), and the Stefan-Boltzmann constant. In our exercise, the person's body has a constant emissivity, indicating its effectiveness at emitting thermal radiation relative to a perfect black body.

The feeling of being colder in winter despite a constant room air temperature is partly because the walls, at a lower winter temperature, absorb more of this radiative heat from the person compared to summer. The walls, being colder, also emit less radiative heat back to the person, resulting in a net heat loss from the person that is greater in winter.
Natural Convection
Natural convection is a type of heat transfer that occurs due to the movement of fluids (which can be gases or liquids) caused by temperature differences within the fluid itself. When part of a fluid is heated, it becomes less dense and rises, while cooler, denser fluid sinks, creating a natural circulation pattern.

In the context of our exercise, the person's body warms the air around it, causing it to rise and be replaced by cooler air from other parts of the room. This circulation leads to heat being transferred away from the body. The rate of heat transfer depends on the temperature difference between the body and the surrounding air and a value called the heat transfer coefficient, which considers how easily heat is transferred from the surface to the fluid. A higher heat transfer coefficient indicates more effective convection.

In both the summer and winter scenarios, despite the constant temperature of the room air, the overall sensation of warmth or chilliness can also be influenced by the rate at which this convective heat loss occurs.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is fundamental in understanding radiant heat exchange. It states that the energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature.

The law is represented mathematically as: \[ Q_{rad} = A \epsilon \sigma (T^4) \] where \( Q_{rad} \) is the radiative heat transfer, \( A \) is the area of the emitting surface, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature of the body in kelvins.

In the problem we're examining, the Stefan-Boltzmann Law helps explain why a person loses more heat by radiation during winter. The fourth power dependency on temperature means that as the wall temperature drops in winter, the radiation heat loss from the person to the walls increases significantly. This increased loss of heat makes the person feel chilled even though the air temperature remains the same.

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Most popular questions from this chapter

1.42 One method for growing thin silicon sheets for photovoltaic solar panels is to pass two thin strings of high melting temperature material upward through a bath of molten silicon. The silicon solidifies on the strings near the surface of the molten pool, and the solid silicon sheet is pulled slowly upward out of the pool. The silicon is replenished by supplying the molten pool with solid silicon powder. Consider a silicon sheet that is \(W_{\mathrm{si}}=85 \mathrm{~mm}\) wide and \(t_{\mathrm{si}}=150 \mu \mathrm{m}\) thick that is pulled at a velocity of \(V_{\mathrm{si}}=20 \mathrm{~mm} / \mathrm{min}\). The silicon is melted by supplying electric power to the cylindrical growth chamber of height \(H=350 \mathrm{~mm}\) and diameter \(D=300 \mathrm{~mm}\). The exposed surfaces of the growth chamber are at \(T_{s}=\) \(320 \mathrm{~K}\), the corresponding convection coefficient at the exposed surface is \(h=8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the surface is characterized by an emissivity of \(\varepsilon_{s}=0.9\). The solid silicon powder is at \(T_{\mathrm{s}, i}=298 \mathrm{~K}\), and the solid silicon sheet exits the chamber at \(T_{\text {si, } o}=420 \mathrm{~K}\). Both the surroundings and ambient temperatures are \(T_{\infty}=T_{\text {sur }}=298 \mathrm{~K}\). (a) Determine the electric power, \(P_{\text {elec }}\), needed to operate the system at steady state. (b) If the photovoltaic panel absorbs a time-averaged solar flux of \(q_{\text {sol }}^{\prime \prime}=180 \mathrm{~W} / \mathrm{m}^{2}\) and the panel has a conversion efficiency (the ratio of solar power absorbed to electric power produced) of \(\eta=0.20\), how long must the solar panel be operated to produce enough electric energy to offset the electric energy that was consumed in its manufacture?

A thermodynamic analysis of a proposed Brayton cycle gas turbine yields \(P=5 \mathrm{MW}\) of net power production. The compressor, at an average temperature of \(T_{c}=400^{\circ} \mathrm{C}\), is driven by the turbine at an average temperature of \(T_{h}=1000^{\circ} \mathrm{C}\) by way of an \(L=1\)-m-long, \(d=70-\mathrm{mm}-\) diameter shaft of thermal conductivity \(k=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Compare the steady-state conduction rate through the shaft connecting the hot turbine to the warm compressor to the net power predicted by the thermodynamics-based analysis. (b) A research team proposes to scale down the gas turbine of part (a), keeping all dimensions in the same proportions. The team assumes that the same hot and cold temperatures exist as in part (a) and that the net power output of the gas turbine is proportional to the overall volume of the device. Plot the ratio of the conduction through the shaft to the net power output of the turbine over the range \(0.005 \mathrm{~m} \leq L \leq 1 \mathrm{~m}\). Is a scaled-down device with \(L=0.005 \mathrm{~m}\) feasible?

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 $$ \dot{W}=q_{\mathrm{in}} \frac{T_{h}-T_{c}+q_{\mathrm{in}} R_{\mathrm{id \textrm {t }}}}{T_{c}-q_{\mathrm{in}} R_{\mathrm{tot}}} $$

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?
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