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Chapter 1: Problem 22

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 \(2770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
The heat transfer coefficient is approximately 12.63 W/m虏K.

Step by step solution

01

Understand the Problem

We need to evaluate the free convection heat transfer coefficient (h) for a thin vertical plate using the given data at the moment when the plate temperature is 225掳C. The calculation assumes negligible radiation exchange.
02

Use the Heat Transfer Equation

The primary equation relating the rate of change of temperature to heat transfer for a plate is given by the equation \[\frac{dT}{dt} = -\frac{h \cdot A \cdot (T_s - T_\infty)}{m \cdot c_p}\]Where:- \(dT/dt\) is the rate of change of temperature with time.- \(h\) is the convection heat transfer coefficient.- \(A\) is the surface area of the plate.- \(T_s\) is the surface temperature.- \(T_\infty\) is the ambient air temperature.- \(m\) is the mass of the plate.- \(c_p\) is the specific heat of the material.
03

Calculate the Plate's Surface Area

Calculate the surface area of the plate:\[A = 0.3 \times 0.3 = 0.09 \ \text{m}^2\]
04

Substitute Values into the Equation

Substitute the given values into the heat transfer equation:\[-0.022 = -\frac{h \cdot 0.09 \cdot (225 - 25)}{3.75 \cdot 2770}\]
05

Solve for the Heat Transfer Coefficient

Re-arrange the equation to isolate \(h\): \[h = \frac{3.75 \cdot 2770 \cdot 0.022}{0.09 \cdot 200}\]Simplify and solve for \(h\):\[h = \frac{227.25}{18} \approx 12.63 \, \text{W/m}^2\cdot\text{K}\]
06

Conclusion

The calculated free convection heat transfer coefficient at the instant the plate temperature is 225掳C is approximately 12.63 W/m虏K.

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Key Concepts

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

Convection Heat Transfer Coefficient
The concept of convection heat transfer coefficient, denoted as \( h \), is pivotal in understanding how heat is transferred between a solid surface and a fluid (like air or water) flowing past it. In the context of free convection, this transfer happens without any external influence like fans or pumps.

The convection heat transfer coefficient depends on several factors:
  • The properties of the fluid, such as its viscosity and thermal conductivity.
  • The temperature difference between the surface and the fluid.
  • The flow conditions of the fluid over the surface, although it is limited in free convection.
In this problem, the value of \( h \) is found using a specific equation that incorporates given parameters like the area of the plate, the mass, and specific heat of the plate material, alongside the temperature difference and rate of temperature change. This calculation allows us to estimate how effectively heat is being transferred by convection from the plate to the air at the specific instant when the plate temperature is 225掳C.
Isothermal Plate Assumption
The isothermal plate assumption simplifies the problem by presuming that the entire plate surface is at a uniform temperature at all times. This means we don't have to consider temperature gradients across the plate's surface.
This assumption is quite practical because it allows for straightforward calculations of heat transfer. By assuming that the whole plate is at the same temperature, the heat transfer equation focuses only on the differences between this uniform plate temperature and the ambient air temperature. This simplification is crucial in determining the convection heat transfer coefficient, \( h \), as it ensures that the surface temperature, \( T_s \), remains constant during the analysis.
Rate of Change of Temperature
The rate of change of temperature, denoted as \( \frac{dT}{dt} \), is a measure of how quickly or slowly the temperature of the plate is changing over time. In the given exercise, this rate is provided as \(-0.022 \, \text{K/s}\). This negative sign indicates that the plate is cooling down.

This rate is a critical component of the heat transfer equation used to calculate the convection heat transfer coefficient. It represents the dynamic aspect of heat transfer, combining with physical parameters to give us insight into how rapidly temperature changes affect heat transfer efficiency. Therefore, understanding and accurately calculating the rate of change is essential as it directly influences the estimation of the convection heat transfer coefficient \( h \).
In many real-world scenarios, knowing this rate helps engineers and scientists design and optimize systems for better thermal management.

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Most popular questions from this chapter

A vacuum system, as used in sputiering electrically conducting thin films on microcircuits, is comprised of a buseplate maintained by an electrical heater at \(300 \mathrm{~K}\) and a shroud within the enclosure maintained at \(77 \mathrm{~K}\) by a liquid-nitrogen coolant loop. The circular baseplate, insulated on the lower side, is \(0.3 \mathrm{~m}\) in diameter and has an emissivity of \(0.25\). (a) How much clectrical power must be provided to the baseplate heater? (b) At what rate must liquid nitrogen be supplied to the shiroud if its heat of vaporization is \(125 \mathrm{~kJ} / \mathrm{kg}\) ? (c) To reduce the liquid-nitrogen consumption, it is proposed to bond a thin sheet of aluminura foil \((\varepsilon=0.09)\) to the baseplate. Will this have the desfred effect?

A vertical slab of Woods metal is joined to a substrate on one surface and is melted as it is uniformly irradialed by a laser souree on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\). and the melt rans off by gravity as soon as it is formed. The abserptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{y}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantanecus rate of melting in \(\mathrm{kg} / \mathrm{s}-\mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\mathrm{w}}=20 r \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \(\left(\varepsilon=0.4, T_{\text {max }}=\right.\) \(20^{\circ} \mathrm{C}\), determine the instantaneous rate of melting during imadiation.

Electronic power devices are mounted to a heat sink having an exposed surface area of \(0.045 \mathrm{~m}^{2}\) and an emissivity of \(0.80\). When the devices dissipate a total power of \(20 \mathrm{~W}\) and the air and sarroundings are at \(27^{\circ} \mathrm{C}\), the average sink temperature is \(42^{\circ} \mathrm{C}\). What average temperature will the heat sink reach when the devices dissipate \(30 \mathrm{~W}\) for the same environmental condition?

The concrete slab of a basement is \(11 \mathrm{~m}\) long. \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basemient is heated by a gas furnace operating at an cfficiency of \(\eta_{r}=0.90\) and natural gas is priced at \(C_{r}=50.01 \mathrm{MJ}\), what is the daily cost of the heat loss?

The inner and outer surface temperatures of a glass window \(5 \mathrm{~mm}\) thick are 15 and \(5^{\circ} \mathrm{C}\). What is the heat loss through a window that is \(1 \mathrm{~m}\) by \(3 \mathrm{~m}\) on a side? The thermal conductivity of glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).
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