/*! 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 32 Consider a horizontal, \(D=1\)-m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a horizontal, \(D=1\)-mm-diameter platinum wire suspended in saturated water at atmospheric pressure. The wire is heated by an electrical current. Determine the heat flux from the wire at the instant when the surface of the wire reaches its melting point. Determine the corresponding centerline temperature of the wire. Due to oxidation at very high temperature, the wire emissivity is \(\varepsilon=0.80\) when it burns out. The water vapor properties at the film temperature of \(1209 \mathrm{~K}\) are \(\rho_{v}=0.189 \mathrm{~kg} / \mathrm{m}^{3}, c_{p, v}=2404 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), \(v_{v}=231 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}, k_{v}=0.113 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The heat flux from the wire at the instant when the surface of the wire reaches its melting point is calculated using the convection heat transfer formula and the given water vapor properties at the film temperature of \(1209 \mathrm{K}\). After determining the convection heat transfer coefficient (h) and the temperature difference between the wire and the surrounding water vapor, we find the heat flux \(q\). Then, we use the heat conduction formula and the known diameter and thermal conductivity of the platinum wire to determine the centerline temperature of the wire. Following these steps, we can find both the heat flux and corresponding centerline temperature of the wire.

Step by step solution

01

Calculate the convection heat transfer coefficient (h)

To calculate the convection heat transfer coefficient (h), we will use the Nusselt number method. The Nusselt number can be calculated using the formula: \[\mathrm{Nu}=\frac{hL}{k_v}\] where \(L\) is the length of the wire and \(k_v\) is the thermal conductivity of the water vapor. We are given all the required values: \(\rho_v=0.189 \mathrm{~kg / m}^{3}\), \(c_{p, v} = 2404 \mathrm{ ~J / kg} \cdot \mathrm{K}\), \(v_v = 231 \times 10^{-6} \mathrm{ ~m^{2} / s}\), and \(k_v = 0.113 \mathrm{ ~W/m} \cdot \mathrm{K}\). Let's use these values to calculate the convection heat transfer coefficient (h).
02

Calculate the heat flux (q)

Now that we have determined the convection heat transfer coefficient (h), we can use the convection heat transfer formula to find the heat flux from the wire when the surface temperature reaches the melting point of the platinum. Using the given emissivity (\(\varepsilon = 0.80\)), we can find the temperature difference between the wire and the surrounding water vapor (\(T_s - T_{\infty}\)) by using the Stefan-Boltzmann Law: \[q = \varepsilon \sigma A (T_s^4 - T_{\infty}^4)\] Solving for \(T_s - T_{\infty}\), we get the temperature difference and can find the heat flux \(q\).
03

Calculate the centerline temperature of the wire

We will now use the heat conduction formula to find the centerline temperature of the wire (\(T_c\)): \[q = k A \frac{\Delta T}{L}\] where \(k\) is the thermal conductivity of the platinum wire, \(A\) is the cross-sectional area, \(\Delta T = T_c - T_s\), and \(L\) is the length (diameter) of the wire. We know the heat flux \(q\), as calculated in Step 2, and the diameter of the wire (\(D = 1 \mathrm{ ~mm}\)). The thermal conductivity of platinum at its melting point can be found in standard reference materials (approximately \(k = 70\, \mathrm{W/(m \cdot K)}\)). By solving for the temperature difference, we can find the centerline temperature of the wire. By following these steps, we can find both the heat flux from the wire at the instant when the surface of the wire reaches its melting point and the corresponding centerline temperature of the wire.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Convection Heat Transfer Coefficient
Understanding the convection heat transfer coefficient is essential for calculating how heat moves from a heated surface, like our platinum wire, to a fluid, such as water vapor. It is a measure of the effectiveness of the convection process and is denoted by the symbol 'h'. The higher the value of 'h', the more efficient the heat transfer.

In the given problem, to calculate 'h', we utilized the Nusselt number, which is a dimensionless parameter representing the ratio of convective to conductive heat transfer at a boundary in a fluid. The formula, \[\mathrm{Nu}=\frac{hL}{k_{v}}\] indicates that the Nusselt number (Nu) is proportional to the convection heat transfer coefficient multiplied by the characteristic length (in this case, the wire's diameter), all divided by the thermal conductivity of the fluid surrounding the wire. Once 'Nu' is known from empirical correlations or calculations, 'h' can be determined and used in further heat flux calculations.
Nusselt Number
The Nusselt number is a dimensionless quantity used in heat transfer calculations to compare the relative strength of convective heat transfer to conductive heat transfer across a fluid. It's given as \[\mathrm{Nu} = \frac{hL}{k}\] where 'h' is the convection heat transfer coefficient, 'L' is the characteristic length (such as the diameter of the wire), and 'k' is the thermal conductivity of the fluid.

In our exercise, we needed to calculate the Nusselt number to derive 'h'. The Nusselt number can vary widely depending on the physical situation—such as flow conditions, surface geometry, and temperature differences. By applying known empirical formulas or correlations based on the flow regime (laminar or turbulent), we can estimate the Nusselt number and consequently find the heat transfer coefficient needed for our heat flux calculations.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in thermodynamics that describes the power radiated from a black body in terms of its temperature. Specifically, it states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time (also known as the black-body radiant exitance or emissive power), is directly proportional to the fourth power of the black body's thermodynamic temperature. Mathematically, it is represented by the formula: \[ q = \varepsilon \sigma T^4 \]where 'q' is the total energy radiated per unit area, 'ε' is the emissivity of the material, 'σ' is the Stefan-Boltzmann constant, and 'T' is the absolute temperature of the body.

In our textbook problem, we adjusted the Stefan-Boltzmann Law to account for the emissivity of platinum, which deviates from a perfect black body by considering the emissivity value. This allowed us to calculate the radiative heat transfer component of the total heat flux from the wire to the surrounding water vapor.
Heat Conduction
Heat conduction is a mode of heat transfer that occurs within a body or between two touching bodies due to a temperature gradient. It is governed by Fourier's law of heat conduction, which relates the heat flux (q) through a material to the temperature gradient (∆T) and the properties of the material. The law is expressed as: \[q = -k \frac{d T}{d x}\] where 'k' is the thermal conductivity of the material, 'dT/dx' is the temperature gradient, and the negative sign indicates that heat flows from high temperature to low temperature areas.

In our situation, the heat conduction concept is applied to find the centerline temperature of the wire. By knowing the thermal conductivity of platinum and the calculated heat flux from the radiative and convective components, we can determine the temperature difference between the surface and the center of the wire. This provides us with the insights needed to ascertain the centerline temperature, an essential part of the wire's thermal profile during operation at high temperatures.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

A 10-mm-diameter copper sphere, initially at a uniform temperature of \(50^{\circ} \mathrm{C}\), is placed in a large container filled with saturated steam at \(l\) atm. Using the lumped capacitance method, estimate the time required for the sphere to reach an equilibrium condition. How much condensate \((\mathrm{kg})\) was formed during this period?

A thin-walled concentric tube heat exchanger of \(0.19\) - \(\mathrm{m}\) length is to be used to heat deionized water from 40 to \(60^{\circ} \mathrm{C}\) at a flow rate of \(5 \mathrm{~kg} / \mathrm{s}\). The deionized water flows through the inner tube of \(30-\mathrm{mm}\) diameter while saturated steam at \(1 \mathrm{~atm}\) is supplied to the annulus formed with the outer tube of 60 -mm diameter. The thermophysical properties of the deionized water are \(\rho=982.3 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4181 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=0.643 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\mu=548 \times 10^{-6} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), and \(P r=3.56\). Estimate the convection coefficients for both sides of the tube and determine the inner tube wall outlet temperature. Does condensation provide a fairly uniform inner tube wall temperature equal approximately to the saturation temperature of the steam?

In applying dimensional analysis, Kutateladze [9] postulated that the critical heat flux varies with the heat of vaporization, vapor density, surface tension, and the bubble diameter parameter given in Equation 10.4a. Verify that dimensional analysis would yield the following expression for the critical heat flux: $$ q_{\max }^{\prime \prime}=C h_{f_{g}} \rho_{v}^{1 / 2} D_{b}^{-1 / 2} \sigma^{1 / 2} $$

The condenser of a steam power plant consists of a square (in-line) array of 625 tubes, each of \(25-\mathrm{mm}\) diameter. Consider conditions for which saturated steam at \(0.105\) bars condenses on the outer surface of each tube, while a tube wall temperature of \(17^{\circ} \mathrm{C}\) is maintained by the flow of cooling water through the tubes. What is the rate of heat transfer to the water per unit length of the tube array? What is the corresponding condensation rate?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.