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Chapter 7: Problem 23

In an experiment, the local heat transfer over a flat plate were correlated in the form of local Nusselt number as expressed by the following correlation $$ \mathrm{Nu}_{x}=0.035 \mathrm{Re}_{x}^{0.8} \operatorname{Pr}^{1 / 3} $$ Determine the ratio of the average convection heat transfer coefficient \((h)\) over the entire plate length to the local convection heat transfer coefficient \(\left(h_{x}\right)\) at \(x=L\).

Short Answer

Expert verified
Answer: The ratio of the average convection heat transfer coefficient to the local convection heat transfer coefficient at x=L is given by: $$ \frac{h_{avg}}{h_x} = \frac{Nu_{avg}}{Nu_L} $$ Find the values of \(Nu_{avg}\) and \(Nu_L\) by integrating the given Nusselt number correlation and using the resulting expressions for local and average convection heat transfer coefficients.

Step by step solution

01

Understanding Nusselt number

The Nusselt number (Nu) is a dimensionless number that represents the ratio of convective heat transfer to conductive heat transfer. The higher the Nusselt number, the better the heat transfer performance. In this exercise, we have the local Nusselt number \(Nu_x\) given as a function of the local Reynolds number, \(Re_x\), and the Prandtl number, Pr: $$ Nu_x = 0.035Re_x^{0.8}Pr^{1/3} $$
02

Express local convection heat transfer coefficient in terms of Nusselt number

To find the local convection heat transfer coefficient (\(h_x\)) in terms of the Nusselt number, we can use the following formula: $$ h_x = \frac{Nu_x \cdot k}{x} $$ Where \(k\) is the thermal conductivity and \(x\) is the distance from the leading edge of the flat plate.
03

Calculate the average Nusselt number using the given correlation

To find the average Nusselt number (\(Nu_{avg}\)) for the entire flat plate length, we need to integrate the given Nusselt number correlation over the plate length and divide by the plate length: $$ Nu_{avg} = \frac{1}{L} \int_0^L Nu_x \, dx $$ Substitute the given correlation for \(Nu_x\) into the formula above: $$ Nu_{avg} = \frac{1}{L} \int_0^L \big( 0.035Re_x^{0.8}Pr^{1/3} \big) \, dx $$
04

Express average convection heat transfer coefficient in terms of Nusselt number

Similar to the local convection heat transfer coefficient, we can express the average convection heat transfer coefficient (\(h_{avg}\)) in terms of the average Nusselt number using the following formula: $$ h_{avg} = \frac{Nu_{avg} \cdot k}{L} $$
05

Find the ratio of \(h_{avg}\) to \(h_x\)

We now have the expressions for both the local and average convection heat transfer coefficients, so we can calculate the desired ratio: $$ \frac{h_{avg}}{h_x} = \frac{\frac{Nu_{avg} \cdot k}{L}}{\frac{Nu_x \cdot k}{x}} $$ Simplify the equation by canceling out \(k\) and then substitute \(x=L\): $$ \frac{h_{avg}}{h_x} = \frac{Nu_{avg}}{Nu_L} $$ Determine the ratio after integrating in step 3 and substituting the results into this equation.

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Most popular questions from this chapter

Liquid mercury at \(250^{\circ} \mathrm{C}\) is flowing in parallel over a flat plate at a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). Surface temperature of the \(0.1-\mathrm{m}\)-long flat plate is constant at \(50^{\circ} \mathrm{C}\). Determine \((a)\) the local convection heat transfer coefficient at \(5 \mathrm{~cm}\) from the leading edge and \((b)\) the average convection heat transfer coefficient over the entire plate.

Consider a hot automotive engine, which can be approximated as a \(0.5-\mathrm{m}\)-high, \(0.40\)-m-wide, and \(0.8-\mathrm{m}\)-long rectangular block. The bottom surface of the block is at a temperature of \(100^{\circ} \mathrm{C}\) and has an emissivity of \(0.95\). The ambient air is at \(20^{\circ} \mathrm{C}\), and the road surface is at \(25^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of \(80 \mathrm{~km} / \mathrm{h}\). Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block.

For laminar flow of a fluid along a flat plate, one would expect the largest local convection heat transfer coefficient for the same Reynolds and Prandl numbers when (a) The same temperature is maintained on the surface (b) The same heat flux is maintained on the surface (c) The plate has an unheated section (d) The plate surface is polished (e) None of the above

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

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