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Chapter 1: Problem 24

A cartridge electrical heater is shaped as a cylinder of length \(L=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\). Under normal operating conditions, the heater dissipates \(2 \mathrm{~kW}\) while submerged in a water flow that is at \(20^{\circ} \mathrm{C}\) and provides a convection heat transfer coefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, determine its surface temperature \(T_{s}\). If the water flow is inadvertently terminated while the heater continues to operate, the heater surface is exposed to air that is also at \(20^{\circ} \mathrm{C}\) but for which \(h=50\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding surface temperature? What are the consequences of such an event?

Short Answer

Expert verified
Under normal water flow conditions, the surface temperature of the heater is approximately 53.97 \(^\circ \mathrm{C}\). However, when the water flow is inadvertently terminated, the heater's surface temperature increases drastically to 540 \(^\circ \mathrm{C}\). This sudden increase in temperature could cause the heater to overheat, potentially resulting in damage to the heater or its surroundings and posing safety hazards, such as fire risks.

Step by step solution

01

Identify given values

We are given the following information: - Heater length: \(L = 200 \,\mathrm{mm} = 0.2 \,\mathrm{m}\) - Outer diameter: \(D = 20 \,\mathrm{mm} = 0.02 \,\mathrm{m}\) - Heater power: \(P = 2 \,\mathrm{kW} = 2000 \,\mathrm{W}\) - Water temperature: \(T_\infty = 20^\circ \, \mathrm{C}\) - Water convection heat transfer coefficient: \(h_\mathrm{water} = 5000 \, \mathrm{W/m^2K}\) - Air convection heat transfer coefficient: \(h_\mathrm{air} = 50 \, \mathrm{W/m^2K}\)
02

Calculate the surface area of the heater cylinder

For this, we need to find the lateral surface area of the cylindrical heater. We can neglect the ends' heat transfer, as mentioned in the question. The lateral surface area formula for a cylinder is given by: \(A_s = 2 \pi r L\) where \(r\) is the radius of the cylinder, and \(\pi\) represents the mathematical constant (~3.14159). For this heater, we have: \(r = \frac{D}{2} = 0.01 \, \mathrm{m}\) Now, we can calculate the surface area: \(A_s = 2 \pi (0.01 \, \mathrm{m})(0.2 \, \mathrm{m}) \approx 0.0126 \, \mathrm{m^2}\)
03

Calculate the heater's surface temperature in water

To find the surface temperature, we will use the convection heat transfer equation, which is given by: \(P = h A_s (T_s - T_\infty)\) Solving for the surface temperature, we have: \(T_s = \frac{P}{h A_s} + T_\infty\) For the heater in water: \(T_{s,\mathrm{water}} = \frac{2000 \, \mathrm{W}}{5000 \, \mathrm{W/m^2K} \times 0.0126 \, \mathrm{m^2}} + 20^\circ \, \mathrm{C} \approx 53.97^\circ \, \mathrm{C}\) So the surface temperature of the heater in water is approximately 53.97 \(^\circ \mathrm{C}\).
04

Calculate the heater's surface temperature in air

Similarly, we can find the surface temperature of the heater in air by using the same equation, but now with the air convection heat transfer coefficient: \(T_{s,\mathrm{air}} = \frac{2000 \,\mathrm{W}}{50 \, \mathrm{W/m^2K} \times 0.0126 \,\mathrm{m^2}} + 20^\circ \,\mathrm{C} \approx 540^\circ \, \mathrm{C}\) So the surface temperature of the heater in air is approximately 540 \(^\circ \,\mathrm{C}\).
05

Discuss the consequences of the water flow termination

When the water flow is terminated, the heater is exposed to air instead, which has a much lower convection heat transfer coefficient. As a result, the heater's surface temperature increases drastically from 53.97 \(^\circ \,\mathrm{C}\) to 540 \(^\circ \,\mathrm{C}\). This sudden and significant increase in temperature could cause the heater to overheat and potentially cause damage to the heater itself or its surroundings. In addition, the high surface temperature in the air might pose a fire risk or present other safety hazards.

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Key Concepts

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

Cylindrical Heater
A cylindrical heater is a type of electrical heater that is designed in a cylindrical shape to efficiently transfer heat. These heaters are commonly used due to their ability to provide uniform heat distribution across their surface.

The key features of a cylindrical heater include:
  • A circular cross-section, which helps in even heat dispersal
  • Ability to be submerged in fluids for direct heating applications
  • Compact design that can fit into narrow spaces

Given the exercise problem, our heater is cylindrical with a length of 200 mm and an outer diameter of 20 mm. This shape naturally helps in maintaining a consistent heat transfer environment around the cylinder. It's especially useful in applications where efficient energy use and limited space are factors. The heater works through electrical resistance, converting electric energy into heat, and in this exercise, it dissipates 2 kW of power while operating.
Surface Temperature Calculation
Calculating the surface temperature of a cylindrical heater is crucial for understanding how efficiently it transfers heat to a surrounding medium, such as water or air. The surface temperature is where the heat transition occurs from the heater to its environment.

To calculate the surface temperature, we use the convection heat transfer equation:
\( P = h A_s (T_s - T_\infty) \)
where:
  • \( P \) is the power being dissipated (2000 W in this example)
  • \( h \) is the convection heat transfer coefficient
  • \( A_s \) is the surface area of the heater ([calculated as \( A_s = 2 \pi r L \)](https://www.example.com))
  • \( T_s \) is the surface temperature
  • \( T_\infty \) is the surrounding temperature

For instance, when in water, the heater's surface temperature is about 53.97 掳C, as computed by rearranging the equation to solve for \( T_s \). Meanwhile, in the air, this jumps markedly to around 540 掳C due to the much lower heat transfer efficiency of air compared to water.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient, denoted as \(h\), is a measure of the heat transfer rate between a surface and a fluid per unit surface area and temperature difference. It is a critical factor in determining how effectively heat is transferred.

Several factors influence the value of \(h\):
  • The type of fluid (e.g., air, water)
  • The velocity of the fluid (how fast it moves over the heater surface)
  • The surface properties (smoothness, material)
  • The temperature difference between the surface and the fluid

In the original problem, the heat transfer coefficient for water is 5000 W/m虏K, while for air, it dramatically drops to 50 W/m虏K. This substantial difference illustrates why the heater's surface temperature is much higher in air than in water. Water provides an efficient medium for cooling the heater since its high convection heat transfer coefficient allows more heat to be drawn away, maintaining a lower surface temperature.

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Most popular questions from this chapter

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.

Three electric resistance heaters of length \(L=250 \mathrm{~mm}\) and diameter \(D=25 \mathrm{~mm}\) are submerged in a 10 -gal tank of water, which is initially at \(295 \mathrm{~K}\). The water may be assumed to have a density and specific heat of \(\rho=990 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) If the heaters are activated, each dissipating \(q_{1}=500 \mathrm{~W}\), estimate the time required to bring the water to a temperature of \(335 \mathrm{~K}\). (b) If the natural convection coefficient is given by an expression of the form \(h=370\left(T_{s}-T\right)^{1 / 3}\), where \(T_{s}\) and \(T\) are temperatures of the heater surface and water, respectively, what is the temperature of each heater shortly after activation and just before deactivation? Units of \(h\) and \(\left(T_{s}-T\right)\) are \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(\mathrm{K}\), respectively. (c) If the heaters are inadvertently activated when the tank is empty, the natural convection coefficient associated with heat transfer to the ambient air at \(T_{\infty}=300 \mathrm{~K}\) may be approximated as \(h=0.70\) \(\left(T_{s}-T_{\infty}\right)^{1 / 3}\). If the temperature of the tank walls is also \(300 \mathrm{~K}\) and the emissivity of the heater surface is \(\varepsilon=0.85\), what is the surface temperature of each heater under steady-state conditions?

The curing process of Example \(1.9\) involves exposure of the plate to irradiation from an infrared lamp and attendant cooling by convection and radiation exchange with the surroundings. Alternatively, in lieu of the lamp, heating may be achieved by inserting the plate in an oven whose walls (the surroundings) are maintained at an elevated temperature. (a) Consider conditions for which the oven walls are at \(200^{\circ} \mathrm{C}\), airflow over the plate is characterized by \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the coating has an emissivity of \(\varepsilon=0.5\). What is the temperature of the plate? (b) For ambient air temperatures of 20,40 , and \(60^{\circ} \mathrm{C}\), determine the plate temperature as a function of the oven wall temperature over the range from 150 to \(250^{\circ} \mathrm{C}\). Plot your results, and identify conditions for which acceptable curing temperatures between 100 and \(110^{\circ} \mathrm{C}\) may be maintained.

The heat flux that is applied to the left face of a plane wall is \(q^{\prime \prime}=20 \mathrm{~W} / \mathrm{m}^{2}\). The wall is of thickness \(L=10\) \(\mathrm{mm}\) and of thermal conductivity \(k=12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the surface temperatures of the wall are measured to be \(50^{\circ} \mathrm{C}\) on the left side and \(30^{\circ} \mathrm{C}\) on the right side, do steady-state conditions exist?

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that must be applied to the top and side walls to ensure a heat load of less than \(500 \mathrm{~W}\), when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?
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