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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

Parallel flow of atmospheric air over a flat plate of length \(L=3 \mathrm{~m}\) is disrupted by an array of stationary rods placed in the flow path over the plate. Laboratory measurements of the local convection coefficient at the surface of the plate are made for a prescribed value of \(V\) and \(T_{x}>T_{x}\). The results are correlated by an expression of the form \(h_{x}=0.7+13.6 x-3.4 x^{2}\), where \(h_{x}\) has units of \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(x\) is in meters. Evaluate the average convection coefficient \(\bar{h}_{L}\) for the entire plate and the ratio \(\bar{h}_{L} / h_{L}\) at the trailing edge.

Experiments to determine the local convection heat transfer coefficient for uniform flow normal to a heated circular disk have yielded a radial Nusselt number distribution of the form $$ N u_{D}=\frac{h(r) D}{k}=N u_{o}\left[1+a\left(\frac{r}{r_{o}}\right)^{n}\right] $$ where both \(n\) and \(a\) are positive. The Nusselt number at the stagnation point is correlated in terms of the Reynolds \(\left(R e_{D}=V D / v\right)\) and Prandtl numbers $$ N u_{o}=\frac{h(r=0) D}{k}=0.814 \operatorname{Re}_{D}^{1 / 2} \mathrm{Pr}^{0.36} $$ Obtain an expression for the average Nusselt number, \(\overline{N u}_{D}=\bar{h} D / k\), corresponding to heat transfer from an isothermal disk. Typically, boundary layer development from a stagnation point yields a decaying convection coefficient with increasing distance from the stagnation point. Provide a plausible explanation for why the opposite trend is observed for the disk.

A 2-mm-thick layer of water on an electrically heated plate is maintained at a temperature of \(T_{w}=340 \mathrm{~K}\), as dry air at \(T_{\infty}=300 \mathrm{~K}\) flows over the surface of the water (case A). The arrangement is in large surroundings that are also at \(300 \mathrm{~K}\). (a) If the evaporative flux from the surface of the water to the air is \(n_{\mathrm{A}}^{\prime \prime}=0.030 \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}^{2}\), what is the corresponding value of the convection mass transfer coefficient? How long will it take for the water to completely evaporate? (b) What is the corresponding value of the convection heat transfer coefficient and the rate at which electrical power must be supplied per unit area of the plate to maintain the prescribed temperature of the water? The emissivity of water is \(\varepsilon_{w}=0.95\). (c) If the electrical power determined in part (b) is maintained after complete evaporation of the water (case B), what is the resulting temperature of the plate, whose emissivity is \(\varepsilon_{p}=0.60\) ?

An object of irregular shape has a characteristic length of \(L=1 \mathrm{~m}\) and is maintained at a uniform surface temperature of \(T_{s}=400 \mathrm{~K}\). When placed in atmospheric air at a temperature of \(T_{x}=300 \mathrm{~K}\) and moving with a velocity of \(V=100 \mathrm{~m} / \mathrm{s}\), the average heat flux from the surface to the air is \(20,000 \mathrm{~W} / \mathrm{m}^{2}\). If a second object of the same shape, but with a characteristic length of \(L=5 \mathrm{~m}\), is maintained at a surface temperature of \(T_{s}=400 \mathrm{~K}\) and is placed in atmospheric air at \(T_{\infty}=300 \mathrm{~K}\), what will the value of the average convection coefficient be if the air velocity is \(V=20 \mathrm{~m} / \mathrm{s}\) ?

For laminar flow over a flat plate, the local heat transfer coefficient \(h_{x}\) is known to vary as \(x^{-1 / 2}\), where \(x\) is the distance from the leading edge \((x=0)\) of the plate. What is the ratio of the average coefficient between the leading edge and some location \(x\) on the plate to the local coefficient at \(x\) ?
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