/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Consider laminar flow about a ve... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider laminar flow about a vertical isothermal plate of length \(L\), providing an average heat transfer coefficient of \(\bar{h}_{L}\). If the plate is divided into \(N\) smaller plates, each of length, \(L_{N}=L / N\), determine an expression for the ratio of the heat transfer coefficient averaged over the \(N\) plates to the heat transfer coefficient averaged over the single plate, \(\bar{h}_{L, N} / \bar{h}_{L, 1}\).

Short Answer

Expert verified
The ratio of the average heat transfer coefficient for the N smaller plates to the single plate, \(\frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}}\), can be expressed as: \[ \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} = \frac{N \int_0^{\frac{L}{N}} h(x)\,dx}{\int_0^L h(x) \,dx} \]

Step by step solution

01

Write the general expression for the average heat transfer coefficient over length L

The average heat transfer coefficient over a length L, \(\bar{h}_L\), can be expressed as: \[ \bar{h}_L = \frac{1}{L} \int_0^L h(x) \,dx \] where h(x) is the heat transfer coefficient as a function of the position x along the plate, and L is the length of the plate.
02

Divide the plate into N smaller plates

Divide the plate of length L into N smaller plates, each of length \(L_N = L/N\). Note that the total length is still L: \[ L = N \cdot L_N \]
03

Find the expression for the average heat transfer coefficient for N smaller plates

Now, we will find the average heat transfer coefficient over the N smaller plates, \(\bar{h}_{L, N}\): \[ \bar{h}_{L, N} = \frac{1}{L_N} \int_0^{L_N} h(x) \,dx \]
04

Calculate the ratio \(\bar{h}_{L, N} / \bar{h}_{L, 1}\)

Finally, we can calculate the ratio of the average heat transfer coefficients for the N smaller plates and the single plate as: \[ \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} = \frac{\frac{1}{L_N} \int_0^{L_N} h(x)\,dx}{\frac{1}{L} \int_0^L h(x) \,dx} \]
05

Simplify the expression for the ratio

Using the relationship \(L_N = L/N\) and factoring out the constant term in the integrals, we can simplify the ratio as: \[ \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} = \frac{\frac{1}{\frac{L}{N}} \int_0^{\frac{L}{N}} h(x)\,dx}{\frac{1}{L} \int_0^L h(x) \,dx} \] \[ \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} = \frac{N \int_0^{\frac{L}{N}} h(x)\,dx}{\int_0^L h(x) \,dx} \] So, the desired ratio of \(\bar{h}_{L, N}\) to \(\bar{h}_{L, 1}\) can be expressed as: \[ \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} = \frac{N \int_0^{\frac{L}{N}} h(x)\,dx}{\int_0^L h(x) \,dx} \]

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Key Concepts

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

Average Heat Transfer Coefficient
A heat transfer coefficient represents the heat transfer rate per unit area per degree of temperature difference between the surface and the fluid. It's an essential concept in understanding how efficiently heat is conducted in different fluid flow scenarios.
When considering an isothermal plate, the average heat transfer coefficient, denoted as \( \bar{h}_L \), provides insight across the length of this plate. It's calculated by integrating the variable local heat transfer coefficient \( h(x) \) over the entire length \( L \):
  • The formula is \( \bar{h}_L = \frac{1}{L} \int_0^L h(x) \,dx \).
  • In essence, it averages out the variations in heat transfer along the plate.

This concept is crucial when you want to predict the overall heat transfer performance of the surface more practically.
Isothermal Plate
An isothermal plate has a uniform temperature across its entire surface.
It's an ideal surface condition where all its points are at the same temperature, making analysis more straightforward. For example,
  • It simplifies calculations as the thermal boundary conditions remain constant across the plate.
  • In laminar flow conditions, a uniform temperature helps in examining the change and distribution of the heat transfer coefficient.

The uniformity provides reliable data for experimentation and modelling, especially in controlled environments where precise temperature management is crucial.
Heat Transfer Coefficient Ratio
The ratio of heat transfer coefficients plays a significant role when dividing a larger system, such as a plate, into smaller units for analysis. This ratio helps to understand the difference in heat transfer efficiency when a plate is segmented compared to when it is considered as a whole.
The expression for the ratio \( \frac{\bar{h}_{L, N}}{\bar{h}_{L, 1}} \) is calculated through:
  • Breaking the plate down into smaller sections, each of length \( L_N = \frac{L}{N} \).
  • Averaging the heat transfer coefficients over these smaller sections and comparing them to the whole plate.
This concept is extremely helpful for optimizing surface area use and assessing efficiency in various segments. It ensures that different sections of a system perform up to the expected heat transfer standards, especially in processes like heating, cooling, and ventilation.

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Most popular questions from this chapter

Consider a large vertical plate with a uniform surface temperature of \(130^{\circ} \mathrm{C}\) suspended in quiescent air at \(25^{\circ} \mathrm{C}\) and atmospheric pressure. (a) Estimate the boundary layer thickness at a location \(0.25 \mathrm{~m}\) measured from the lower edge. (b) What is the maximum velocity in the boundary layer at this location and at what position in the boundary layer does the maximum occur? (c) Using the similarity solution result, Equation \(9.19\), determine the heat transfer coefficient \(0.25 \mathrm{~m}\) from the lower edge. (d) At what location on the plate measured from the lower edge will the boundary layer become turbulent?

A computer code is being developed to analyze a 12.5-mm-diameter, cylindrical sensor used to determine ambient air temperature. The sensor experiences free convection while positioned horizontally in quiescent air at \(T_{\infty}=27^{\circ} \mathrm{C}\). For the temperature range from 30 to \(80^{\circ} \mathrm{C}\), derive an expression for the convection coefficient as a function of only \(\Delta T=\) \(T_{s}-T_{\infty}\), where \(T_{s}\) is the sensor temperature. Evaluate properties at an appropriate film temperature and show what effect this approximation has on the convection coefficient estimate.

Hot air flows from a furnace through a \(0.15\)-m-diameter, thin-walled steel duct with a velocity of \(3 \mathrm{~m} / \mathrm{s}\). The duct passes through the crawlspace of a house, and its uninsulated exterior surface is exposed to quiescent air and surroundings at \(0^{\circ} \mathrm{C}\). (a) At a location in the duct for which the mean air temperature is \(70^{\circ} \mathrm{C}\), determine the heat loss per unit duct length and the duct wall temperature. The duct outer surface has an emissivity of \(0.5\). (b) If the duct is wrapped with a \(25-\mathrm{mm}\)-thick layer of \(85 \%\) magnesia insulation \((k=0.050 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) having a surface emissivity of \(\varepsilon=0.60\), what are the duct wall temperature, the outer surface temperature, and the heat loss per unit length?

The maximum surface temperature of the \(20-\mathrm{mm}-\) diameter shaft of a motor operating in ambient air at \(27^{\circ} \mathrm{C}\) should not exceed \(87^{\circ} \mathrm{C}\). Because of power dissipation within the motor housing, it is desirable to reject as much heat as possible through the shaft to the ambient air. In this problem, we will investigate several methods for heat removal. (a) For rotating cylinders, a suitable correlation for estimating the convection coefficient is of the form $$ \begin{gathered} \overline{N u}_{D}=0.133 \operatorname{Re}_{D}^{2 / 3} \operatorname{Pr}^{1 / 3} \\ \left(\operatorname{Re}_{D}<4.3 \times 10^{5}, \quad 0.7<\operatorname{Pr}<670\right) \end{gathered} $$ where \(R e_{D} \equiv \Omega D^{2} / \nu\) and \(\Omega\) is the rotational velocity (rad/s). Determine the convection coefficient and the maximum heat rate per unit length as a function of rotational speed in the range from 5000 to \(15,000 \mathrm{rpm}\). (b) Estimate the free convection coefficient and the maximum heat rate per unit length for the stationary shaft. Mixed free and forced convection effects may become significant for \(R e_{D}<4.7\left(G r_{D}^{3} / P r\right)^{0.137}\). Are free convection effects important for the range of rotational speeds designated in part (a)? (c) Assuming the emissivity of the shaft is \(0.8\) and the surroundings are at the ambient air temperature, is radiation exchange important? (d) If ambient air is in cross flow over the shaft, what air velocities are required to remove the heat rates determined in part (a)?

As is evident from the property data of Tables A.3 and A.4, the thermal conductivity of glass at room temperature is more than 50 times larger than that of air. It is therefore desirable to use windows of double-pane construction, for which the two panes of glass enclose an air space. If heat transfer across the air space is by conduction, the corresponding thermal resistance may be increased by increasing the thickness \(L\) of the space. However, there are limits to the efficacy of such a measure, since convection currents are induced if \(L\) exceeds a critical value, beyond which the thermal resistance decreases. Consider atmospheric air enclosed by vertical panes at temperatures of \(T_{1}=22^{\circ} \mathrm{C}\) and \(T_{2}=-20^{\circ} \mathrm{C}\). If the critical Rayleigh number for the onset of convection is \(R a_{L} \approx 2000\), what is the maximum allowable spacing for conduction across the air? How is this spacing affected by the temperatures of the panes? How is it affected by the pressure of the air, as, for example, by partial evacuation of the space?

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