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Chapter 1: Problem 16

A cartridge electrical henter is shaped as a cylinder of length \(L\). \(=200 \mathrm{~mm}\) and outer diameter \(D=20 \mathrm{~mm}\), Under normal operating conditions the heater dissipares \(2 \mathrm{~kW}\) while subencrged in a water flow that is at \(20 \mathrm{~F}^{\mathrm{C}}\) and provides a coevection heat transfer cuefficient of \(h=5000 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). Neglecting heat transfer from the ends of the heater, delermine its uurface temperature \(T_{w}\) If the water flow it inadveriently termanated while the healer continues to operate, the heuter sarface is exponed to air that is alwo at \(2 O T C\) but for which \(h=50\) Whan \({ }^{2}-\mathbb{K}\). What is the coercsponding surface temperature? What are the cunsequerces of such an event?

Short Answer

Expert verified
The heater's surface temperature is 324.8 K in water and 3520 K in air, risking material failure in air.

Step by step solution

01

Calculate Surface Area of the Cylinder

The surface area of a cylindrical heater can be calculated as \( A = \pi D L \), where \( D \) is the diameter and \( L \) is the length. Substituting the values, we have:\[A = \pi \times 0.02 \, \text{m} \times 0.2 \, \text{m} = 0.01256 \, \text{m}^2\]
02

Determine Surface Temperature in Water

To find the surface temperature \( T_w \) when the heater is in water, we use the formula: \( Q = h A (T_w - T_\infty) \), where \( Q \) is the power (2000 W), \( h \) is the heat transfer coefficient for water (5000 W/m虏路K), and \( T_\infty \) is the surrounding temperature (293 K, since \( 20掳C = 293 K \)). Rearrange to solve for \( T_w \):\[T_w = \frac{Q}{hA} + T_\infty = \frac{2000}{5000 \times 0.01256} + 293 = 324.8 \, \text{K}\]
03

Determine Surface Temperature in Air

When the heater is exposed to air, use the same heat transfer equation, but update \( h \) to the air value (50 W/m虏路K):\[T_w = \frac{Q}{hA} + T_\infty = \frac{2000}{50 \times 0.01256} + 293 = 3520 \, \text{K}\]
04

Analyze Consequences

The surface temperature of the heater when exposed to air is exceedingly high (3520 K), which surpasses practical material limits. This could lead to material failure, posing safety hazards due to excessive heating and potential damage to the heater and surrounding equipment.

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Key Concepts

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

Convection
Convection is a mechanism of heat transfer where heat is carried away by a fluid, such as water or air, over a surface. In the context of the heater problem, convection is crucial as it dictates how efficiently heat dissipates from the heater to its surroundings. In general, there are two types of convection:
  • Natural Convection: Occurs due to temperature differences in the fluid causing movement.
  • Forced Convection: Occurs when a fluid is forced to flow over a surface, such as a fan blowing air or water being pumped.
In this exercise, we deal with forced convection, as the water flow is controlled around the heater. The heat transfer coefficient, denoted as \( h \), represents the efficiency of this heat exchange process. With a high \( h \) value, more heat is transferred per unit area per degree of temperature difference. This is why the water significantly cools the heater compared to air.
Surface Temperature
The surface temperature of an object, such as the heater, is the temperature at which the heat loss and heat input are balanced. It depends on several factors: the power output of the heater, the surrounding fluid's properties, and the surface area available for heat exchange. In this problem, understanding surface temperature is essential for evaluating the heater's stability and safety.
The surface temperature differs dramatically between water and air due to the convection efficiency of each medium. When in contact with water, the heater maintains a safe temperature of 324.8 K. However, once exposed to air, which has a lower heat transfer coefficient, the temperature skyrockets to 3520 K. This is because the air fails to dissipate the heat effectively, highlighting the importance of the medium in controlling surface temperature.
Heat Transfer Coefficient
The heat transfer coefficient, \( h \), is a pivotal parameter in evaluating heat transfer via convection. It quantifies the convection efficiency between a surface and a fluid per unit area per unit temperature difference, measured in W/m虏路K. This coefficient varies according to the type of fluid and its flow characteristics.
In the examples provided, water has a high heat transfer coefficient of 5000 W/m虏路K due to its effective ability in conducting and transferring heat. This means that water can absorb more heat from the heater, maintaining a cooler surface temperature.
Conversely, air has a much lower coefficient, 50 W/m虏路K in this context, which makes it less effective at removing heat from the surface. Consequently, the surface temperature escalates dangerously. Knowing \( h \) values allows engineers to predict temperature changes and ensure safety in designs exposed to different environments.

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Most popular questions from this chapter

An electric resistance heater is embedded in a lony 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.

Consider a surface-mount type transistor on a circuit board whose temperature is muintained at \(35^{\circ} \mathrm{C}\). Air at \(207 \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm}\) by \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm}\) by \(0.25 \mathrm{~mm}\) and length 4 nim, conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting rudiation, eximate the case temperature uhen \(150 \mathrm{~mW}\) is dissipoted by the thansistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal condoctivitios of the wire leadk, air, and cunductive paste are 25, \(0.0263\), and \(0.12\) W/m - \(\mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case lemperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a lager convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m}+\mathrm{K}\), compute and plot the maximum allowable hean dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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 \(k_{e}=\) \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). 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 \(I \mathrm{~m} \times 2 \mathrm{~m}\) window? The thermal conductivity of air is \(k_{t}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

The free convection heat transfer cocfficient 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 cuoks. Ascuming the plate is iscthermal and radiation exchange with its surroundings is negligible, evaluate the convection cocfficient at the instant of time when the plate temperature is \(225^{\circ} \mathrm{C}\) and the change in plate temperature with time \((d T / d)\) is \(-0.022 \mathrm{~K} / \mathrm{s}\). The ambient air temperature is \(25^{\prime} \mathrm{C}\) and the plate measures \(0.3 \times 0.3 \mathrm{~m}\) with a mass of \(3.75 \mathrm{~kg}\) and a specific heat of \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{h}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). Air enters the computer console at a temperiture of \(T_{6}=20^{\circ} \mathrm{C}\), and flow is driven by a fin whese power consumption is \(P_{f}=25 \mathrm{~W}\). Iniet ain \(\forall_{1} T_{i}\) (a) If the tempensture rise of the air flow. \(\left(T_{e}-T_{i}\right)_{\text {is }}\) is ot to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\forall\) of the air? 'The density and specific heat of the air may be approximated is \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most suncejtible to thermal failure dissipates I W/cm \({ }^{2}\) of surface area. To minimive the potential for thermal failure, where should the component be installed on a PCB? What is its surface lemperature at this location?
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