/*! 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 54 An electric heater consists of a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 54

An electric heater consists of a thin ribbon of metal and is used to boil a dielectric liquid. The liquid temperature \(T_{e}\) is uniform at its boiling point, and the heat transfer coefficient on the ribbon can be assumed to be uniform as well. A resistance measurement allows the average temperature of the ribbon \(\bar{T}\) to be determined. Obtain an expression for \(\bar{T}-T_{e}\) in terms of the ribbon dimensions (width \(W\), thickness \(2 t\) ), the ribbon thermal conductivity \(k\), the heat transfer coefficient \(h_{c}\), and the ribbon electrical conductivity \(\sigma\left[\Omega^{-1} \mathrm{~m}^{-1}\right]\).

Short Answer

Expert verified
\(\bar{T} - T_e = \frac{2 J^2 \sigma t}{h_c}\)

Step by step solution

01

Identify Key Parameters

The heater consists of a ribbon with dimensions width (\(W\)) and thickness (\(2t\)). The relevant properties are the thermal conductivity (\(k\)), electrical conductivity (\(\sigma\)), and heat transfer coefficient (\(h_c\)). The liquid temperature is \(T_e\), and we need to find the difference in temperature (\(\bar{T} - T_e\)).
02

Consider Heat Transfer Balance

The temperature difference \(\bar{T} - T_e\) across the ribbon is driven by the heat generated due to electrical resistance, which is dissipated by heat transfer to the dielectric liquid. Using the heat transfer balance: \(P = h_c \, A \, (\bar{T} - T_e)\), where \(P\) is the power per unit length, \(A\) is the surface area, and \(h_c\) is the heat transfer coefficient.
03

Determine Power Per Unit Length

The power per unit length \(P\) generated by the ribbon can be written based on Ohm's Law and resistive heating as \(P = I^2 R\), where \(R = \frac{L}{\sigma \cdot (Wt)}\) is the resistance. Let \(J\) be the current density, \(P = I^2 R = J^2 \cdot \sigma \cdot (W \cdot 2t)\).
04

Express Resistance in Terms of Dimensions

The resistance of a very thin metal ribbon with length \(L\), width \(W\), and thickness \(2t\) is \(R = \frac{L}{\sigma \cdot (W \cdot 2t)}\). The electrical current density \(J = I/(W \cdot 2t)\) allows the resistance-based power expression: \(P = J^2 \sigma \cdot (W \cdot 2t)\).
05

Setup Heat Transfer Equation

Equate the power generation to the heat loss due to convection \(J^2 \sigma \cdot (W \cdot 2t) = h_c \cdot (W \cdot L) \cdot (\bar{T} - T_e)\). This equation balances the electrical power generated in the ribbon with the convective heat transfer away from it.
06

Simplify Expression for Temperature Difference

Rearrange the above equation to get the expression for \(\bar{T}-T_e\): \(\bar{T} - T_e = \frac{J^2 \sigma \cdot (W \cdot 2t)}{h_c \cdot W} = \frac{2 J^2 \sigma t}{h_c}\). This gives the temperature difference based on the ribbon's properties and conditions.

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 91影视!

Key Concepts

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

Thermal Conductivity
Thermal conductivity is an important property when studying heat transfer. It represents how well a material conducts heat. Higher thermal conductivity means heat flows easily through the material. For metals, which typically have high thermal conductivity, they can effectively transfer heat across their body. This is crucial in heating applications, like the electric heater ribbon used for boiling a dielectric liquid.

Understanding thermal conductivity helps in determining how quickly a material reaches thermal equilibrium with its surroundings. In mathematical terms, if a material has a thermal conductivity \(k\), it plays a role in heat transfer equations, balancing the generated heat and the heat transferred to the liquid.
  • Strong thermal conductivity ensures efficient heat transfer.
  • Metal ribbons used in heaters depend on this property for effective operation.
  • It's factored into the equation for the temperature difference \((\bar{T} - T_e)\).
Heat Transfer Coefficient
The heat transfer coefficient \(h_c\) is a crucial factor that quantifies the convective heat transfer between surfaces and surrounding fluids. In our exercise, the coefficient determines how effectively the ribbon's thermal energy transfers to the boiling dielectric liquid.

The heat transfer coefficient reflects how vigorously heat is moving away from the ribbon surface. The greater the value of \(h_c\), the more efficient the heat dissipation to the liquid. This can affect the ribbon's temperature, as shown by the equation for \(\bar{T} - T_e\).
  • Higher \(h_c\) leads to faster thermal equilibrium.
  • It impacts how quickly the material under thermal stress can cool down.
  • The coating or surface texture of materials influences \(h_c\).
Electrical Conductivity
Electrical conductivity \(\sigma\) refers to a material's ability to conduct electric current. It's an essential property for the heating ribbon, as it determines how much electrical power can be converted into heat. This property affects the resistance of the ribbon, leading to changes in the power generated per unit length.

The relationship between electrical conductivity and generated power is critical: by knowing \(\sigma\), we can calculate the power dissipated due to resistance. This is central to the energy balance equation that helps determine the temperature difference of the ribbon: \(\bar{T} - T_e\). A high electrical conductivity reduces resistance, increasing the current and power.
  • Key for determining resistance in materials.
  • Influences how much energy converts from electrical to thermal.
  • Higher conductivity results in more efficient conductive heating.

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

Liquid oxygen at \(90 \mathrm{~K}\) flows inside a \(3 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 2 \mathrm{~mm}\)-wall-thickness, AISI 303 stainless steel tube. The tube is insulated with \(3 \mathrm{~cm}\)-thick fiberglass insulation of thermal conductivity \(0.021 \mathrm{~W} / \mathrm{m} \mathrm{K}\). Will atmospheric water vapor condense on the outside of the insulation when the ambient air temperature is \(300 \mathrm{~K}\) and the dewpoint temperature is \(285 \mathrm{~K}\) ? Take the outside convective and radiative heat transfer coefficients as \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

A spherical metal tank has a \(2.5 \mathrm{~m}\) outside diameter and is insulated with a \(0.5\) \(\mathrm{m}\)-thick cork layer. The tank contains liquefied gas at \(-60^{\circ} \mathrm{C}\) and the ambient air is at \(20^{\circ} \mathrm{C}\). The inside heat transfer coefficient can be assumed to be large, and the combined convection and radiation outside heat transfer coefficient is estimated to be \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Atmospheric water vapor diffuses into the cork, and a layer of ice forms adjacent to the tank wall. Determine the thickness of the layer. Assume that the cork thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \mathrm{K}\) is unaffected by the ice and water, but comment on the validity of this assumption.

A \(2 \mathrm{~mm}\)-diameter resistor, for an electronic component on a space station, is to have a sheath of thermal conductivity \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). It is cooled by forced convection with \(\bar{h}_{c} \simeq 1.1 D^{-1 / 2} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), for diameter \(D\) in meters, and by radiation with \(q_{\mathrm{rad}}=\sigma \varepsilon\left(T_{s}^{4}-T_{e}^{4}\right)\). Determine the radius of the sheath that maximizes the heat loss when the resistor is at \(400 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). Take the value of the surface emittance \(\varepsilon\) as (i) \(0.9\). (ii) \(0.5\).

Electronic components are attached to a \(10 \mathrm{~cm}\)-square, \(2 \mathrm{~mm}\)-thick aluminum plate, and the backface is cooled by a flow of air. The backface has rectangular aluminum fins \(25 \mathrm{~mm}\) long, \(0.3 \mathrm{~mm}\) thick, at a pitch of \(3 \mathrm{~mm}\). If the cooling air is at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), what is the allowable heat dissipation rate if the plate temperature should not exceed \(70^{\circ} \mathrm{C}\). Take \(k=180 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum.

A \(1 \mathrm{~mm}\)-diameter resistor has a sheath of thermal conductivity \(k=0.12 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and is located in an evacuated enclosure. Determine the radius of the sheath that maximizes the heat loss from the resistor when it is maintained at \(450 \mathrm{~K}\) and the enclosure is at \(300 \mathrm{~K}\). The surface emittance of the sheath is \(0.85\).
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Fluids

Read Explanation

Work Energy and Power

Read Explanation

Electricity

Read Explanation

Kinematics Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
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.