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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

Consider the nanofluid of Example 2.2. (a) Calculate the Prandtl numbers of the base fluid and nanofluid, using information provided in the example problem. (b) For a geometry of fixed characteristic dimension \(L\), and a fixed characteristic velocity \(V\), determine the ratio of the Reynolds numbers associated with the two fluids, \(R e_{\text {wf }} / R e_{\mathrm{w}_{\mathrm{d}}-}\) Calculate the ratio of the average Nusselt numbers, \(\overline{N u}_{L, \text {, d }} / \overline{N u}_{\text {L, b }}\), that is associated with identical average heat transfer coefficients for the two fluids, \(\bar{h}_{\mathrm{mf}}=\bar{h}_{\mathrm{bd}}\). (c) The functional dependence of the average Nusselt number on the Reynolds and Prandtl numbers for a broad array of various geometries may be expressed in the general form $$ \overline{N u}_{L}=\bar{h} L / k=C R e^{w N} P r^{1 / 3} $$ where \(C\) and \(m\) are constants whose values depend on the geometry from or to which convection heat transfer occurs. Under most conditions the value of \(m\) is positive. For positive \(m\), is it possible for the base fluid to provide greater convection heat transfer rates than the nanofluid, for conditions involving a fixed geometry, the same characteristic velocities, and identical surface and ambient temperatures?

An industrial process involves evaporation of a thin water film from a contoured surface by heating it from below and forcing air across it. Laboratory measurements for this surface have provided the following heat transfer correlation: $$ \overline{N u_{L}}=0.43 R e_{L}^{0.58} P r^{0.4} $$ The air flowing over the surface has a temperature of \(290 \mathrm{~K}\), a velocity of \(10 \mathrm{~m} / \mathrm{s}\), and is completely dry \(\left(\phi_{\infty}=0\right)\). The surface has a length of \(1 \mathrm{~m}\) and a surface area of \(1 \mathrm{~m}^{2}\). Just enough energy is supplied to maintain its steady-state temperature at \(310 \mathrm{~K}\). (a) Determine the heat transfer coefficient and the rate at which the surface loses heat by convection. (b) Determine the mass transfer coefficient and the evaporation rate \((\mathrm{kg} / \mathrm{h})\) of the water on the surface. (c) Determine the rate at which heat must be supplied to the surface for these conditions.

If laminar flow is induced at the surface of a disk due to rotation about its axis, the local convection coefficient is known to be a constant, \(h=C\), independent of radius. Consider conditions for which a disk of radius \(r_{o}=100 \mathrm{~mm}\) is rotating in stagnant air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and a value of \(C=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. If an embedded electric heater maintains a surface temperature of \(T_{x}=50^{\circ} \mathrm{C}\), what is the local heat flux at the top surface of the disk? What is the total electric power requirement? What can you say about the nature of boundary layer development on the disk?

A wet-bulb thermometer consists of a mercury-in-glass thermometer covered with a wetted (water) fabric. When suspended in a stream of air, the steady-state thermometer reading indicates the wet-bulb temperature \(T_{\mathrm{ub}}\). Obtain an expression for determining the relative humidity of the air from knowledge of the air temperature \(\left(T_{\infty}\right)\), the wet-bulb temperature, and appropriate air and water vapor properties. If \(T_{\infty}=45^{\circ} \mathrm{C}\) and \(T_{w b}=25^{\circ} \mathrm{C}\), what is the relative humidity of the airstream?

A manufacturer of ski equipment wishes to develop headgear that will offer enhanced thermal protection for skiers on cold days at the slopes. Headgear can be made with good thermal insulating characteristics, but it tends to be bulky and cumbersome. Skiers prefer comfortable, lighter gear that offers good visibility, but such gear tends to have poor thermal insulating characteristics. The manufacturer decides to take a new approach to headgear design by concentrating the insulation in areas about the head that are prone to the highest heat losses from the skier and minimizing use of insulation in other locations. Hence, the manufacturer must determine the local heat transfer coefficients associated with the human head with a velocity of \(V=10 \mathrm{~m} / \mathrm{s}\) directed normal to the face and an air temperature of \(-13^{\circ} \mathrm{C}\). A young engineer decides to make use of the heat and mass transfer analogy and the naphthalene sublimation technique (see Problem 6.63) and casts head shapes of solid naphthalene with characteristic dimensions that are half-scale (that is, the models are half as large as the full-scale head). (a) What wind tunnel velocity \(\left(T_{\mathrm{x}}=300 \mathrm{~K}\right)\) is needed to apply the experimental results to the human head associated with \(V=10 \mathrm{~m} / \mathrm{s}\) ? (b) A wind tunnel experiment is performed for \(\Delta t=120 \mathrm{~min}, T_{x}=27^{\circ} \mathrm{C}\). The engineer finds that the naphthalene has receded by \(\delta_{1}=0.1 \mathrm{~mm}\) at the back of the head, \(\delta_{2}=0.32 \mathrm{~mm}\) in the middle of the forehead, and \(\delta_{3}=0.64 \mathrm{~mm}\) on the ear. Determine the heat transfer coefficients at these locations for the full-scale head at \(-13^{\circ} \mathrm{C}\). The density of solid naphthalene is \(\rho_{\mathrm{A}, \text { sol }}=1025 \mathrm{~kg} / \mathrm{m}^{3}\). (c) After the new headgear is designed, the models are fitted with the new gear (half-scale) and the experiments are repeated. Some areas of the model that were found to have small local heat transfer coefficients are left uncovered since insulating these areas would have little benefit in reducing overall heat losses during skiing. Would you expect the local heat transfer coefficients for these exposed areas to remain the same as prior to fitting the model with the headgear? Explain why.
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