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Chapter 1: Problem 21

An electric resistance heater is embedded in a long cylinder of diameter \(30 \mathrm{~mm}\). When water with a temperature of \(25^{\circ} \mathrm{C}\) and velocity of \(1 \mathrm{~m} / \mathrm{s}\) flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of \(90^{\circ} \mathrm{C}\) is \(28 \mathrm{~kW} / \mathrm{m}\). When air, also at \(25^{\circ} \mathrm{C}\), but with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) is flowing, the power per unit length required to maintain the same surface temperature is \(400 \mathrm{~W} / \mathrm{m}\). Calculate and compare the convection coefficients for the flows of water and air.

Short Answer

Expert verified
The convection coefficients for water and air flowing over the heated cylinder are calculated as follows: For water: \(h_w = 4307.3244 \mathrm{~W/m^2\cdot K}\) For air: \(h_a = 68.7164 \mathrm{~W/m^2\cdot K}\) The convection coefficient for water is much higher than that for air, which means that heat is transferred more efficiently from the surface of the cylinder to water than to air.

Step by step solution

01

Calculate the surface area per unit length for the cylinder

The surface area per unit length (A) for a cylinder can be calculated using the following formula: A = 蟺d where d is the diameter of the cylinder. Given the diameter (d) = 30mm = 0.03 m, we can calculate the surface area per unit length (A): A = 蟺 * 0.03 m = 0.09424778 m虏/m
02

Calculate the temperature difference between the surface and the fluid for water and air

The temperature difference 螖T is calculated as follows: 螖T = T_s - T_f where T_s is the surface temperature and T_f is the fluid temperature. For water: 螖T_w = 90掳C - 25掳C = 65 K For air: 螖T_a = 90掳C - 25掳C = 65 K
03

Calculate the convection coefficients for water and air

Now, we can calculate the convection coefficients (h) for both water and air using the equation for heat transfer rate: Q = h * A * 螖T => h = Q / (A * 螖T) For water: h_w = (28,000 W/m) / (0.09424778 m虏/m * 65 K) = 4307.3244 W/m虏路K For air: h_a = (400 W/m) / (0.09424778 m虏/m * 65 K) = 68.7164 W/m虏路K
04

Compare the convection coefficients

Now that we have calculated the convection coefficients for both water and air, we can compare them: h_w = 4307.3244 W/m虏路K h_a = 68.7164 W/m虏路K The convection coefficient for water is much higher than that for air, which means that heat is transferred more efficiently from the surface of the cylinder to water than to air. This is because water has a higher thermal conductivity and specific heat capacity than air, allowing for more efficient heat transfer through convection.

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Key Concepts

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

Understanding the Convection Coefficient
The convection coefficient, denoted as \( h \), is a measure of how effectively heat is transferred between a solid surface and a fluid (like air or water) flowing over it. This coefficient is essential in scenarios where objects undergo heat exchange with their surroundings. The higher the convection coefficient, the more effective the heat transfer. It's influenced by various factors, such as the fluid's properties, flow velocity, and temperature difference between the surface and the fluid.

For example, in the exercise, two different fluids, water and air, flow across a heated cylinder at different velocities. The task is to calculate the convection coefficients for both cases. Given the higher convection coefficient for water compared to air, it鈥檚 clear that water is more effective at transferring heat. This is partly because water has a higher density and specific heat capacity than air, which facilitates better energy absorption and transfer.

To find the convection coefficient, you use the equation:
  • \( h = \frac{Q}{A \cdot \Delta T} \)
  • Where \( Q \) is the power per unit length, \( A \) is the surface area per unit length, and \( \Delta T \) is the temperature difference between the surface and the fluid.
This equation shows how \( h \) is directly related to how much heat energy is being transferred.
The Fundamentals of Cylinder Heat Transfer
Heat transfer around a cylinder is a common practical problem, especially in designing equipment like heaters, radiators, and pipes that must maintain certain temperatures when surrounded by fluids. The cylinder in the exercise is a good example of this, where heat is supplied to maintain its surface at a specified temperature while fluids (water and air) pass over it.

The heat transfer process involves both conduction and convection. Conduction happens within the cylinder materials, while convection occurs between the surface and the fluid. The magnitude of heat transfer is influenced by the cylinder鈥檚 surface area, which is given by:
  • Surface area per unit length, \( A = \pi d \)
  • Where \( d \) is the cylinder's diameter.
By calculating the cylinder's surface area, one can determine how much of the surface is available for heat exchange with the fluid. Larger surface areas typically result in more significant heat transfer, assuming the same convection environment.

In practical applications, understanding how heat is transferred from a cylindrical surface helps in improving energy efficiency and heating system performance.
Significance of Thermal Conductivity
Thermal conductivity is a physical property of a material that describes its ability to conduct heat. It is an intrinsic property, meaning it doesn't change with the size or shape of the material. Materials with high thermal conductivity, like metals, are excellent heat conductors, whereas those with low thermal conductivity, such as rubber, are poor conductors and serve as insulators.

In the context of the exercise, water has a higher thermal conductivity than air. This means it can transfer heat more effectively, which explains the higher convection coefficient for water compared to air. Higher thermal conductivity implies less resistance to the flow of heat, allowing energy to be rapidly transported through the medium.

Factors affecting thermal conductivity include material composition, temperature, and state of the material (solid, liquid, gas). For instance, metals generally have high thermal conductivity, making them suitable for applications requiring efficient heat dissipation. Understanding thermal conductivity helps in selecting the right materials for different thermal management challenges.

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Most popular questions from this chapter

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.

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 \(33^{\circ} \mathrm{C}\) ?

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 glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{g}=\) \(1.4 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining panes are separated by an air space. If the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(1 \mathrm{~m} \times 2 \mathrm{~m}\) window? The themal conductivity of air is \(k_{a}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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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