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Chapter 6: Problem 12

Air at a free stream temperature of \(T_{a}=20^{\circ} \mathrm{C}\) is in parallel flow over a flat plate of length \(L=5 \mathrm{~m}\) and temperature \(T_{s}=90^{\circ} \mathrm{C}\). However, obstacles placed in the flow intensify mixing with increasing distance \(x\) from the leading edge, and the spatial variation of temperatures measured in the boundary layer is correlated by an expression of the form \(T\left({ }^{\circ} \mathrm{C}\right)=20+70\) \(\exp (-600 x y)\), where \(x\) and \(y\) are in meters. Determine and plot the manner in which the local convection coefficient \(h\) varies with \(x\). Evaluate the average convection coefficient \(\bar{h}\) for the plate.

Short Answer

Expert verified
The local convection coefficient \(h(x)\) is determined as \(h(x) = \frac{-k\cdot{(-42000\exp(-600x))}}{90 - (20 + 70\exp(-600x))}\), where \(k\) is the thermal conductivity of air at the given temperature. By plotting this expression, we can visualize how \(h(x)\) varies with \(x\). To find the average convection coefficient \(\bar{h}\) for the plate, we integrate \(h(x)\) over the plate length and divide by the length, i.e., \(\bar{h} = \frac{1}{L}\int_{0}^{L} h(x) dx\). Assuming a value of \(k = 0.0263\, W/mK\), we can use numerical integration to calculate \(\bar{h}\).

Step by step solution

01

Define the temperature correlation expression

We are given the temperature correlation expression in the boundary layer: \(T({ }^{\circ}\mathrm{C}) = 20 + 70\exp(-600xy)\) This will be used to find the local convection coefficient \(h\).
02

Calculate the local heat flux

To find the local convection coefficient \(h\), we first need to calculate the local heat flux, which is given by the following formula: \(q" = -k\frac{dT}{dy}\) Where \(k\) is the thermal conductivity of air at the given temperature and \(\frac{dT}{dy}\) is the temperature gradient with respect to y. We can find the derivative of the temperature expression with respect to y: \(\frac{dT}{dy} = -42000\exp(-600xy)\) Now, we can plug this expression into the heat flux formula: \(q" = -k(-42000\exp(-600xy))\)
03

Newton's Law of Cooling

According to Newton's law of cooling, we have the equation: \(q" = h(T_s - T)\) Where \(T_s\) is the surface temperature of the flat plate, \(T\) is the local temperature, and \(h\) is the local convection coefficient. We need to find \(h\). Substitute the heat flux calculated in step 2 into the above equation and solve for \(h\): \(h = \frac{-k\cdot{(-42000\exp(-600xy))}}{T_s - T}\)
04

Obtain expression of h in terms of x

We know the given temperatures of the air and the surface, which are \(T_a = 20^\circ{}C\) and \(T_s = 90^\circ{}C\), respectively. Using the given expression for temperature, we can find the temperature \(T\) in terms of \(x\): \(T = 20 + 70\exp(-600x)\) Now, we can plug this expression of \(T\) in the equation we obtained in Step 3 to find \(h(x)\): \(h(x) = \frac{-k\cdot{(-42000\exp(-600x))}}{90 - (20 + 70\exp(-600x))}\) After obtaining the expression of h in terms of x, we can plot it.
05

Calculate the average convection coefficient for the plate

We need to evaluate the average convection coefficient \(\bar{h}\) for the entire plate. To do that, we can integrate the expression of \(h(x)\) over the length of the plate and divide by the plate length: \(\bar{h} = \frac{1}{L}\int_{0}^{L} h(x) dx\) Since h depends on the thermal conductivity k, we can either calculate k by finding a value for the specific temperature range or look it up in a table. Let's assume the value of k as 0.0263 W/mK. We can integrate the expression over the range of x from 0 to 5 meters. Finally, we can use a numerical integration method or software to calculate the average convection coefficient for the flat plate \(\bar{h}\).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics concerning the movement of thermal energy from one object or substance to another. It can occur through three primary mechanisms: conduction, convection, and radiation. Conduction involves heat transfer through direct molecular contact, convection includes the bulk movement of a fluid transferring heat, and radiation is the transfer of energy through electromagnetic waves.

In the exercise above, convection plays a significant role as air moves over a flat plate, carrying away heat from the surface. Specifically, we are interested in understanding how heat is convected away from a heated flat plate into the surrounding air flow. The ability of the air to take away this heat is quantified by the convection coefficient, which varies along the length of the plate due to the induced mixing caused by obstacles.

The Boundary Layer

As air flows over the surface of the plate, a boundary layer forms where the behavior of the air in terms of velocity and temperature differs from the free stream. The thickness of the boundary layer can affect the efficiency of heat transfer, and analyzing this region is crucial in predicting heat transfer rates.
Boundary Layer Analysis
The boundary layer is a thin zone at the surface interface between a fluid, such as air, and a solid where the effects of the fluid's viscosity are significant. Analysis of this layer is crucial in understanding heat transfer and fluid dynamics. It helps us predict how changes in flow velocity and temperature occur, which directly affects the heat transfer between the fluid and the surface.

In our exercise, the temperature distribution within the boundary layer is defined by a mathematical expression based on the position coordinates. Through boundary layer analysis, we deduce how the convection coefficient, a measure of the convective heat transfer capability of the air, varies with respect to the plate's length. The temperature gradient found by differentiating the temperature expression with respect to the vertical coordinate, 'y', is vital for calculating the local heat flux and therefore the local convection coefficient.
Newton's Law of Cooling
Newton's Law of Cooling is pivotal in the context of heat transfer problems, including the one described in the exercise. This law states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surrounding environment.

Applied to our exercise, it helps establish a relationship between the local heat flux and the convection coefficient. In a practical scenario, this allows for the determination of the convection coefficient 'h' at any point along the plate by knowing the temperatures and the local heat flux. The equation derived from Newton's Law of Cooling is useful for both the theoretical understanding and practical calculation of heat transfer characteristics in systems where convection is the primary mode of heat removal.
Thermal Conductivity
Thermal conductivity is a property of a material that indicates its ability to conduct heat. Represented by 'k' in scientific formulas, it plays an integral role in quantifying the rate at which heat is transferred through materials. The higher a material's thermal conductivity, the more effectively it transfers heat.

In our exercise, thermal conductivity of the air is a required parameter to calculate the local heat flux and, consequently, the local convection coefficient. Given that thermal conductivity can vary with temperature, it's important to use a value that corresponds to the average temperature experienced by the air in the boundary layer. Correct application of this property is essential to ensure accurate modeling of heat transfer rates for engineering applications and theoretical studies alike.

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Most popular questions from this chapter

An expression for the actual water vapor partial pressure in terms of wet-bulb and dry-bulb temperatures, referred to as the Carrier equation, is given as $$ p_{v}=p_{g w}-\frac{\left(p-p_{g w}\right)\left(T_{d b}-T_{\mathrm{wb}}\right)}{1810-T_{\mathrm{wb}}} $$ where \(p_{v}, p_{g w}\) and \(p\) are the actual partial pressure, the saturation pressure at the wet-bulb temperature, and the total pressure (all in bars), while \(T_{\mathrm{db}}\) and \(T_{\mathrm{wb}}\) are the dry- and wet-bulb temperatures in kelvins. Consider air at \(1 \mathrm{~atm}\) and \(37.8^{\circ} \mathrm{C}\) flowing over a wet-bulb thermometer that indicates \(21.1^{\circ} \mathrm{C}\). (a) Using Carrier's equation, calculate the partial pressure of the water vapor in the free stream. What is the relative humidity? (b) Refer to a psychrometric chart and obtain the relative humidity directly for the conditions indicated. Compare the result with part (a). (c) Use Equation \(6.65\) to determine the relative humidity. Compare the result to parts (a) and (b).

It is known that on clear nights the air temperature need not drop below \(0^{\circ} \mathrm{C}\) before a thin layer of water on the ground will freeze. Consider such a layer of water on a clear night for which the effective sky temperature is \(-30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient due to wind motion is \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The water may be assumed to have an emissivity of \(1.0\) and to be insulated from the ground as far as conduction is concerned. (a) Neglecting evaporation, determine the lowest temperature the air can have without the water freezing. (b) For the conditions given, estimate the mass transfer coefficient for water evaporation \(h_{\mathrm{m}}(\mathrm{m} / \mathrm{s})\). (c) Accounting now for the effect of evaporation, what is the lowest temperature the air can have without the water freezing? Assume the air to be dry.

An industrial process involves the evaporation of water from a liquid film that forms on a contoured surface. Dry air is passed over the surface, and from laboratory measurements the convection heat transfer correlation is of the form $$ \overline{N_{L}}=0.43 \operatorname{Re}_{L}^{0.58} P r r^{\Omega .4} $$ (a) For an air temperature and velocity of \(27^{\circ} \mathrm{C}\) and \(10 \mathrm{~m} / \mathrm{s}\), respectively, what is the rate of evaporation from a surface of \(1-\mathrm{m}^{2}\) area and characteristic length \(L=1 \mathrm{~m}\) ? Approximate the density of saturated vapor as \(\rho_{A, \text { sat }}=0.0077 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What is the steady-state temperature of the liquid film?

On a summer day the air temperature is \(27^{\circ} \mathrm{C}\) and the relative humidity is \(30 \%\). Water evaporates from the surface of a lake at a rate of \(0.10 \mathrm{~kg} / \mathrm{h}\) per square meter of water surface area. The temperature of the water is also \(27^{\circ} \mathrm{C}\). Determine the value of the convection mass transfer coefficient. 6.53 It is observed that a 230 -mm-diameter pan of water at \(23^{\circ} \mathrm{C}\) has a mass loss rate of \(1.5 \times 10^{-5} \mathrm{~kg} / \mathrm{s}\) when the ambient air is dry and at \(23^{\circ} \mathrm{C}\). (a) Determine the convection mass transfer coefficient for this situation. (b) Estimate the evaporation mass loss rate when the ambient air has a relative humidity of \(50 \%\). (c) Estimate the evaporation mass loss rate when the water and ambient air temperatures are \(47^{\circ} \mathrm{C}\), assuming that the convection mass transfer coefficient remains unchanged and the ambient air is dry.

The naphthalene sublimation technique involves the use of a mass transfer experiment coupled with an analysis based on the heat and mass transfer analogy to obtain local or average convection heat transfer coefficients for complex surface geometries. A coating of naphthalene, which is a volatile solid at room temperature, is applied to the surface and is then subjected to airflow in a wind tunnel. Alternatively, solid objects may be cast from liquid naphthalene. Over a designated time interval, \(\Delta t\), there is a discernible loss of naphthalene due to sublimation, and by measuring the surface recession at locations of interest or the mass loss of the sample, local or average mass transfer coefficients may be determined. Consider a rectangular rod of naphthalene exposed to air in cross flow at \(V=10 \mathrm{~m} / \mathrm{s}, T_{\mathrm{s}}=300 \mathrm{~K}\), as in Problem 6.10, except now \(c=10 \mathrm{~mm}\) and \(d=30 \mathrm{~mm}\). Determine the change in mass of the \(L=500\)-mm-long rod over a time period of \(\Delta t=30 \mathrm{~min}\). Naphthalene has a molecular weight of \(M_{\mathrm{A}}=128.16 \mathrm{~kg} / \mathrm{kmol}\), and its solid-vapor saturation pressure at \(27^{\circ} \mathrm{C}\) and \(1 \mathrm{ltm}\) is \(p_{\text {A, } a t}=1.33 \times 10^{-4}\) bar.
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