/*! 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 48 Consider wave-free laminar conde... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider wave-free laminar condensation on 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 new average heat transfer coefficient to the original one is equal to the number of smaller plates N: \(\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}} = N\).

Step by step solution

01

Find the heat transfer coefficient for each smaller plate

As we are dividing the plate into N equal parts with length \(L_N = \frac{L}{N}\) each, we can assume that the heat transfer coefficients for all smaller plates are equal. Thus, let's denote this heat transfer coefficient by \(h_N\). According to the information given, we can write the following relationship between the heat transfer coefficients of the original plate (\(\bar{h}_L\)) and the smaller plates (\(h_N\)): \[h_N \cdot L_N = \bar{h}_L \cdot L\] Now, let's solve for \(h_N\): \[h_N = \frac{\bar{h}_L \cdot L}{L_N}\]
02

Find the average heat transfer coefficient over all the smaller plates

Since all the N smaller plates have the same heat transfer coefficient, the new average heat transfer coefficient \(\bar{h}_{L,N}\) can be directly written as: \[\bar{h}_{L,N} = h_N\] Substitute the expression of \(h_N\) obtained in Step 1: \[\bar{h}_{L,N} = \frac{\bar{h}_L \cdot L}{L_N}\]
03

Find the ratio of the new average heat transfer coefficient to the original one

Now, we want to find the ratio \(\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}}\). Using our expression for \(\bar{h}_{L,N}\) from Step 2: \[\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}} = \frac{\frac{\bar{h}_L \cdot L}{L_N}}{\bar{h}_L}\] Simplify by canceling out the \(\bar{h}_L\) terms: \[\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}} = \frac{L}{L_N}\] As we know the relationship between \(L\) and \(L_N\), we can substitute the value of \(L_N = \frac{L}{N}\): \[\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}} = \frac{L}{\frac{L}{N}}\] Finally, simplify the expression: \[\frac{\bar{h}_{L,N}}{\bar{h}_{L,1}} = N\] So, the ratio of the new average heat transfer coefficient to the original one is equal to the number of smaller plates N.

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Key Concepts

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

Average Heat Transfer Coefficient
Understanding the average heat transfer coefficient is essential when analyzing heat exchange systems, such as those involving laminar condensation on a surface. The average heat transfer coefficient, often denoted as \( \bar{h} \), is a value that represents the overall ability of a heat exchanger surface to transfer heat between a solid boundary and a fluid. For an isothermal plate, this coefficient signifies the heat transferred per unit surface area per unit temperature difference between the plate and the fluid.

The exercise provided deals with the concept of dividing an isothermal plate into several smaller plates and assessing the change in the average heat transfer coefficient. The average coefficient over one plate \( \bar{h}_{L,1} \) is the basis for comparison upon subdivision. By dividing the plate into \( N \) smaller plates, the heat transfer coefficient for each of the smaller plates \( h_N \) changes due to the alteration in the surface area and heat conduction paths. Through careful manipulation of mathematical relationships, students can better understand how these coefficients relate to each other and how they are influenced by the physical dimensions of the plates in question.

Here's a simple analogy: if you imagine heat transfer as people passing through doors (the larger the door, the more people can pass), \( \bar{h} \) would be akin to measuring the flow of people per door opening. If the original door is divided into several smaller ones, each 'new door' has its own flow measurement which, when averaged, could give us insight into the efficiency of the smaller doors compared to the original one.
Isothermal Plate
An isothermal plate is fundamental to the field of heat transfer. This concept refers to a surface or plate that maintains a constant temperature across its entirety, despite heat being exchanged with surrounding fluids. The isothermal condition simplifies the analysis of heat transfer problems, as it allows the assumption that the temperature gradient driving the heat flow remains consistent. In the context of laminar condensation, which involves a fluid undergoing a phase change on the surface of the plate, the isothermal nature ensures that the temperature difference driving the condensation remains steady.

In the exercise at hand, the vertical isothermal plate not only simplifies calculations but also allows students to intuitively grasp how a constant temperature difference facilitates a predictable heat transfer scenario. As such, when the single isothermal plate is segmented into smaller plates of equal length, each maintains the same isothermal behavior. However, the effect on the condensation heat transfer rate and the average heat transfer coefficient for each of the smaller plates becomes a key aspect to explore.

Tip for Students

When dealing with problems associated with isothermal plates, always ensure that your calculations account for the constant temperature condition and how it specifically relates to the thermal properties and geometry of the problem at hand.
Heat Transfer Enhancement
Heat transfer enhancement techniques are crucial in engineering to improve the efficiency of thermal systems. They involve methods or modifications aimed at increasing the rate of heat transfer without necessarily altering the temperatures or the fluids involved. In the application discussed, by dividing the isothermal plate into multiple smaller sections, one can manipulate the heat flow characteristics and potentially enhance the condensation heat transfer.

The approach taken in the exercise effectively increases the average heat transfer coefficient by altering the surface geometry. Each smaller plate acts as an individual heat exchanger unit, with a higher heat transfer coefficient due to the reduction in plate length. This is no different from increasing the surface area in other heat transfer enhancement methods, like adding fins to a heat sink or roughening a surface to increase turbulence.

  • Impact of Design: Changing the physical dimensions of heat exchange surfaces can lead to significant improvements in heat transfer efficiency.
  • Practical Applications: These principles are applied in designing high-performance heat exchangers, such as intercoolers, radiators, or condensers.
Through understanding such enhancement techniques, students can tailor heat exchange systems for optimal performance, pushing the boundaries of what's possible in thermal design.

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Most popular questions from this chapter

Electrical current passes through a horizontal, 2-mmdiameter conductor of emissivity \(0.5\) when immersed in water under atmospheric pressure. (a) Estimate the power dissipation per unit length of the conductor required to maintain the surface temperature at \(555^{\circ} \mathrm{C}\). (b) For conductor diameters of \(1.5,2.0\), and \(2.5 \mathrm{~mm}\), compute and plot the power dissipation per unit length as a function of surface temperature for \(250 \leq T_{s} \leq 650^{\circ} \mathrm{C}\). On a separate figure, plot the percentage contribution of radiation as a function of \(T_{s}\).

The condenser of a steam power plant consists of AISI 302 stainless steel tubes \(\left(k_{s}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), each of outer and inner diameters \(D_{o}=30 \mathrm{~mm}\) and \(D_{i}=\) \(26 \mathrm{~mm}\), respectively. Saturated steam at \(0.135\) bar condenses on the outer surface of a tube, while water at a mean temperature of \(T_{m}=290 \mathrm{~K}\) is in fully developed flow through the tube. (a) For a water flow rate of \(\dot{m}=0.25 \mathrm{~kg} / \mathrm{s}\), what is the outer surface temperature \(T_{s, o}\) of the tube and the rates of heat transfer and steam condensation per unit tube length? As a first estimate, you may evaluate the properties of the liquid film at the saturation temperature. If one wishes to increase the transfer rates, what is the limiting factor that should be addressed? (b) Explore the effect of the water flow rate on \(T_{s, o}\) and the rate of heat transfer per unit length.

A silicon chip of thickness \(L=2.5 \mathrm{~mm}\) and thermal conductivity \(k_{s}=135 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is cooled by boiling a saturated fluorocarbon liquid \(\left(T_{\text {sat }}=57^{\circ} \mathrm{C}\right)\) on its surface. The electronic circuits on the bottom of the chip produce a uniform heat flux of \(q_{o}^{\prime \prime}=5 \times 10^{4} \mathrm{~W} / \mathrm{m}^{2}\), while the sides of the chip are perfectly insulated. Properties of the saturated fluorocarbon are \(c_{p, l}=\) \(1100 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, h_{f g}=84,400 \mathrm{~J} / \mathrm{kg}, \rho_{l}=1619.2 \mathrm{~kg} / \mathrm{m}^{3}\), \(\rho_{v}=13.4 \mathrm{~kg} / \mathrm{m}^{3}, \sigma=8.1 \times 10^{-3} \mathrm{~N} / \mathrm{m}, \mu_{1}=440 \times 10^{-6}\) \(\mathrm{kg} / \mathrm{m} \cdot \mathrm{s}\), and \(P r_{I}=9.01\). In addition, the nucleate boiling constants are \(C_{s, f}=0.005\) and \(n=1.7\). (a) What is the steady-state temperature \(T_{o}\) at the bottom of the chip? If, during testing of the chip, \(q_{o}^{\prime \prime}\) is increased to \(90 \%\) of the critical heat flux, what is the new steady-state value of \(T_{o}\) ? (b) Compute and plot the chip surface temperatures (top and bottom) as a function of heat flux for \(0.20 \leq q_{o}^{\prime \prime} / q_{\max }^{\prime \prime} \leq 0.90\). If the maximum allowable chip temperature is \(80^{\circ} \mathrm{C}\), what is the maximum allowable value of \(q_{e}^{\text {"? }}\) ?

10.71 Wetting of some metallic surfaces can be inhibited by means of ion implantation of the surface prior to its use, thereby promoting dropwise condensation. The degree of wetting inhibition and, in turn, the efficacy of the implantation process vary from metal to metal. Consider a vertical metal plate that is exposed to saturated steam at atmospheric pressure. The plate is \(t=1 \mathrm{~mm}\) thick, and its vertical and horizontal dimensions are \(L=250 \mathrm{~mm}\) and \(b=100 \mathrm{~mm}\), respectively. The temperature of the plate surface that is exposed to the steam is found to be \(T_{s}=90^{\circ} \mathrm{C}\) when the opposite surface of the metal plate is held at a cold temperature, \(T_{\alpha^{*}}\) (a) Determine \(T_{c}\) for 2024-T6 aluminum. Assume the ion-implantation process does not promote dropwise condensation for this metal. (b) Determine \(T_{c}\) for AISI 302 stainless steel, assuming the ion- implantation process is effective in promoting dropwise condensation.

The bottom of a copper pan, \(150 \mathrm{~mm}\) in diameter, is maintained at \(115^{\circ} \mathrm{C}\) by the heating element of an electric range. Estimate the power required to boil the water in this pan. Determine the evaporation rate. What is the ratio of the surface heat flux to the critical heat flux? What pan temperature is required to achieve the critical heat flux?

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