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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

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=x_{\mathrm{cr}}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (f) Average convection heat transfer coefficient, W/m虏 \(\cdot \mathrm{K}\) (g) Rate of convective heat transfer, W

Exhaust gases at \(1 \mathrm{~atm}\) and \(300^{\circ} \mathrm{C}\) are used to preheat water in an industrial facility by passing them over a bank of tubes through which water is flowing at a rate of \(6 \mathrm{~kg} / \mathrm{s}\). The mean tube wall temperature is \(80^{\circ} \mathrm{C}\). Exhaust gases approach the tube bank in normal direction at \(4.5 \mathrm{~m} / \mathrm{s}\). The outer diameter of the tubes is \(2.1 \mathrm{~cm}\), and the tubes are arranged in- line with longitudinal and transverse pitches of \(S_{L}=S_{T}=8 \mathrm{~cm}\). There are 16 rows in the flow direction with eight tubes in each row. Using the properties of air for exhaust gases, determine \((a)\) the rate of heat transfer per unit length of tubes, \((b)\) and pressure drop across the tube bank, and \((c)\) the temperature rise of water flowing through the tubes per unit length of tubes. Evaluate the air properties at an assumed mean temperature of \(250^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\). Is this a good assumption?

Consider a house that is maintained at a constant temperature of \(22^{\circ} \mathrm{C}\). One of the walls of the house has three singlepane glass windows that are \(1.5 \mathrm{~m}\) high and \(1.8 \mathrm{~m}\) long. The glass \((k=0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\) thick, and the heat transfer coefficient on the inner surface of the glass is \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Now winds at \(35 \mathrm{~km} / \mathrm{h}\) start to blow parallel to the surface of this wall. If the air temperature outside is \(-2^{\circ} \mathrm{C}\), determine the rate of heat loss through the windows of this wall. Assume radiation heat transfer to be negligible. Evaluate the air properties at a film temperature of \(5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

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.

Wind at \(30^{\circ} \mathrm{C}\) flows over a \(0.5\)-m-diameter spherical tank containing iced water at \(0^{\circ} \mathrm{C}\) with a velocity of \(25 \mathrm{~km} / \mathrm{h}\). If the tank is thin-shelled with a high thermal conductivity material, the rate at which ice melts is (a) \(4.78 \mathrm{~kg} / \mathrm{h} \quad\) (b) \(6.15 \mathrm{~kg} / \mathrm{h}\) (c) \(7.45 \mathrm{~kg} / \mathrm{h}\) (d) \(11.8 \mathrm{~kg} / \mathrm{h}\) (e) \(16.0 \mathrm{~kg} / \mathrm{h}\) (Take \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and use the following for air: \(k=\) \(0.02588 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7282, v=1.608 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \mu_{\infty}=\) \(\left.1.872 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, \mu_{\mathrm{s}}=1.729 \times 10^{-5} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right)\)
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