/*! 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 8 The heat transfer coefficient fo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper. The sphere, which is \(12.7 \mathrm{~mm}\) in diameter, is at \(66^{\circ} \mathrm{C}\) before it is inserted into an airstream having a temperature of \(27^{\circ} \mathrm{C}\). A thermocouple on the outer surface of the sphere indicates \(55^{\circ} \mathrm{C} 69 \mathrm{~s}\) after the sphere is inserted into the airstream. Assume and then justify that the sphere behaves as a spacewise isothermal object and calculate the heat transfer coefficient.

Short Answer

Expert verified
The heat transfer coefficient (h) for air flowing over the copper sphere can be calculated using Newton's law of cooling, assuming that the sphere behaves as a spacewise isothermal object. By analyzing the temperature-time history and using the thermal properties of copper, the heat transfer coefficient is found to be approximately \(25.82 \mathrm{~W/(m^2 \cdot K)}\).

Step by step solution

01

Compute the initial temperature difference

The initial temperature of the sphere (Ts_initial) is \(66^{\circ} \mathrm{C}\), and the airstream temperature (T_infinity) is \(27^{\circ} \mathrm{C}\). The initial temperature difference between the sphere and the airstream is: ΔT_initial = Ts_initial - T_infinity = \(66^{\circ} \mathrm{C} - 27^{\circ} \mathrm{C} = 39^{\circ} \mathrm{C}\)
02

Calculate the temperature difference at the given time

At 69 seconds after the sphere is inserted into the airstream, the temperature of the sphere (Ts_69s) is \(55^{\circ} \mathrm{C}\). Calculate the temperature difference between the sphere and the airstream at this time: ΔT_69s = Ts_69s - T_infinity = \(55^{\circ} \mathrm{C}- 27^{\circ} \mathrm{C} = 28^{\circ} \mathrm{C}\)
03

Justify the spacewise isothermal assumption

Since the temperature difference between the sphere and the airstream decreased by a relatively small amount (ΔT_initial - ΔT_69s = \(11^{\circ} \mathrm{C}\)) over 69 seconds, we can assume that the sphere behaves as a spacewise isothermal object. This means that the temperature of the sphere is uniform at any given time.
04

Use Newton's law of cooling to find the heat transfer coefficient

Newton's law of cooling states that the rate of heat transfer (Q) between an object and its surroundings is proportional to the temperature difference between the two. Using this law and the given data, we can write an equation to solve for the heat transfer coefficient (h): Q = h * A * ΔT A - Surface area of the copper sphere with diameter (D) of 12.7mm A = \(4 \pi (D/2)^2 = 4 \pi (6.35 \mathrm{~mm})^2 = 506.97 \mathrm{~mm^2}\) Now, let's rewrite the equation with the fact that the sphere's temperature decreases over time: \(\frac{dT_s}{dt} = - \frac{h * A}{V * \rho * C_p} (T_s - T_\infty)\) We need to find the heat transfer coefficient (h) given the thermal properties and data: \(\rho\) - copper density = \(8960 \mathrm{~kg/m^3}\) \(C_p\) - copper specific heat = \(385 \mathrm{~J/(kg \cdot K)}\) \(V\) - Volume of copper sphere = \(\frac{4}{3} \pi (\frac{D}{2})^3 = \frac{4}{3} \pi (6.35 \mathrm{~mm})^3 = 1.07 \times 10^{-6} \mathrm{~m^3}\)
05

Rearrange the equation for h

Rearrange the equation to isolate the heat transfer coefficient (h): h = -\(\frac{V * \rho * C_p \cdot \frac{dT_s}{dt}}{A * (T_s - T_\infty)}\)
06

Calculate the derivative of the temperature

The rate of temperature change with respect to time (\(\frac{dT_s}{dt}\)) can be approximated by the difference between the initial temperature and the temperature after 69 seconds, divided by the elapsed time: \(\frac{dT_s}{dt} = \frac{T_s(0) - T_s(69)}{69} = \frac{66^{\circ} \mathrm{C} - 55^{\circ} \mathrm{C}}{69 \mathrm{~s}} = 0.15942 \mathrm{~K/s}\) (We will assume temperature change is linear for this problem.)
07

Calculate the heat transfer coefficient h

Substitute the values into the rearranged equation for h: h = -\(\frac{1.07 \times 10^{-6} \mathrm{~m^3} \cdot 8960 \mathrm{~kg/m^3} \cdot 385 \mathrm{~J/(kg \cdot K)} \cdot 0.15942 \mathrm{~K/s}}{506.97 \mathrm{~mm^2} \cdot 28^{\circ} \mathrm{C}}\) = \(25.8157 \mathrm{~W/(m^2 \cdot K)}\) The heat transfer coefficient for air flowing over the copper sphere is approximately \(25.82 \mathrm{~W/(m^2 \cdot K)}\).

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.

Newton's Law of Cooling
Understanding heat transfer mechanisms is crucial in various fields, from engineering to environmental science. One fundamental principle governing these mechanisms is Newton's law of cooling. It describes how the temperature of an object changes due to heat transfer. Newton's law suggests that the rate of heat loss of a body is proportional to the difference between its own temperature and the ambient temperature, i.e., the surroundings. In mathematical terms, it is often expressed as:
\( \frac{dT}{dt} = -k(T - T_{\text{env}}) \).
Here, \( T \) represents the temperature of the object, \( T_{\text{env}} \) the temperature of the environment, \( k \) is a proportionality constant related to the heat transfer coefficient, and \( \frac{dT}{dt} \) indicates the rate of change of temperature over time. This relationship helps in solving practical problems like calculating the heat transfer coefficient for a copper sphere cooled by an airstream, which is essential for optimizing thermal systems in industrial applications.
Spacewise Isothermal Assumption
In heat transfer analysis, simplifying assumptions can make complex problems more tractable. The spacewise isothermal assumption is one such simplification. It posits that at any given moment, the temperature within an object is uniform, which means there are no temperature gradients across the material. This assumption is particularly valid when dealing with small objects or those with high thermal conductivity, like copper. It allows using simpler mathematical models to describe the thermal behavior of the object.
Justifying this assumption involves showing that temperature changes are relatively uniform over time, and that the material in question can quickly distribute heat throughout its volume. In the example of the copper sphere, the small change in temperature over a short timespan lends credibility to the isothermal assumption, simplifying the task of calculating the heat transfer coefficient.
Thermal Properties of Materials
The thermal properties of materials play a pivotal role in determining how heat is transferred. Key properties include thermal conductivity, specific heat capacity, and density.
  • Thermal conductivity is a measure of a material's ability to conduct heat.
  • Specific heat capacity indicates how much energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius.
  • Density is the mass per unit volume of a material.
In the context of heat transfer coefficient calculation, knowing the thermal properties of copper—a material with high thermal conductivity and specific heat capacity—allows for precise assessments of heat transfer rates. These properties are used in conjunction with Newton's law of cooling to quantify the rate at which the sphere will equilibrate to the airstream temperature, ultimately affecting the heat transfer coefficient.
Rate of Heat Transfer
The rate of heat transfer is a quantifiable measure of the amount of thermal energy exchanged between a system and its surroundings per unit time. In the context of the specific exercise, the rate of heat transfer can be understood through the equation derived from Newton's law of cooling:
\( Q = h \times A \times (T - T_{\text{env}}) \).
In this equation, \( Q \) is the rate of heat transfer, \( h \) represents the heat transfer coefficient, \( A \) is the surface area through which heat is being transferred, and \( (T - T_{\text{env}}) \) is the temperature difference between the system and the environment. Calculating the rate informs us on how quickly a system, such as a copper sphere, loses heat to its surroundings. The heat transfer coefficient, critical in this calculation, is deduced from the changes in temperature over time and helps predict how efficiently a material can cool or heat under specific conditions.

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

Consider the series solution, Equation \(5.42\), for the plane wall with convection. Calculate midplane \(\left(x^{*}=0\right)\) and surface \(\left(x^{*}=1\right)\) temperatures \(\theta^{*}\) for \(F o=0.1\) and 1 , using \(B i=0.1,1\), and 10 . Consider only the first four eigenvalues. Based on these results, discuss the validity of the approximate solutions, Equations \(5.43\) and \(5.44\).

In a manufacturing process, long rods of different diameters are at a uniform temperature of \(400^{\circ} \mathrm{C}\) in a curing oven, from which they are removed and cooled by forced convection in air at \(25^{\circ} \mathrm{C}\). One of the line operators has observed that it takes \(280 \mathrm{~s}\) for a \(40-\mathrm{mm}\) diameter rod to cool to a safe-to-handle temperature of \(60^{\circ} \mathrm{C}\). For an equivalent convection coefficient, how long will it take for an 80 -mm-diameter rod to cool to the same temperature? The thermophysical properties of the rod are \(\rho=2500 \mathrm{~kg} / \mathrm{m}^{3}, c=900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Comment on your result. Did you anticipate this outcome?

Common transmission failures result from the glazing of clutch surfaces by deposition of oil oxidation and decomposition products. Both the oxidation and decomposition processes depend on temperature histories of the surfaces. Because it is difficult to measure these surface temperatures during operation, it is useful to develop models to predict clutch-interface thermal behavior. The relative velocity between mating clutch plates, from the initial engagement to the zero-sliding (lock-up) condition, generates heat that is transferred to the plates. The relative velocity decreases at a constant rate during this period, producing a heat flux that is initially very large and decreases linearly with time, until lock-up occurs. Accordingly, \(q_{f}^{\prime \prime}=q_{o}^{\prime \prime}=\left[1-\left(t / t_{\mathrm{lu}}\right)\right]\), where \(q_{o}^{\prime \prime}=1.6 \times 10^{7} \mathrm{~W} / \mathrm{m}^{2}\) and \(t_{1 \mathrm{u}}=100 \mathrm{~ms}\) is the lock-up time. The plates have an initial uniform temperature of \(T_{i}=40^{\circ} \mathrm{C}\), when the prescribed frictional heat flux is suddenly applied to the surfaces. The reaction plate is fabricated from steel, while the composite plate has a thinner steel center section bonded to low- conductivity friction material layers. The thermophysical properties are \(\rho_{s}=\) \(7800 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{s}}=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{s}=40 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the steel and \(\rho_{\mathrm{im}}=1150 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{fm}}=1650 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k_{\mathrm{fm}}=4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) for the friction material. (a) On \(T-t\) coordinates, sketch the temperature history at the midplane of the reaction plate, at the interface between the clutch pair, and at the midplane of the composite plate. Identify key features. (b) Perform an energy balance on the clutch pair over the time interval \(\Delta t=t_{\mathrm{lu}}\) to determine the steadystate temperature resulting from clutch engagement. Assume negligible heat transfer from the plates to the surroundings. (c) Compute and plot the three temperature histories of interest using the finite-element method of FEHT or the finite-difference method of IHT (with \(\Delta x=0.1 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~ms}\) ). Calculate and plot the frictional heat fluxes to the reaction and composite plates, \(q_{\mathrm{rp}}^{\prime \prime}\) and \(q_{\mathrm{cp}}^{\prime \prime}\), respectively, as a function of time. Comment on features of the temperature and heat flux histories. Validate your model by comparing predictions with the results from part (b). Note: Use of both \(F E H T\) and \(I H T\) requires creation of a look-up data table for prescribing the heat flux as a function of time.

Thermal energy storage systems commonly involve a packed bed of solid spheres, through which a hot gas flows if the system is being charged, or a cold gas if it is being discharged. In a charging process, heat transfer from the hot gas increases thermal energy stored within the colder spheres; during discharge, the stored energy decreases as heat is transferred from the warmer spheres to the cooler gas. Consider a packed bed of \(75-\mathrm{mm}\)-diameter aluminum spheres \(\left(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}, c=950 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=\right.\) \(240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) and a charging process for which gas enters the storage unit at a temperature of \(T_{g, i}=300^{\circ} \mathrm{C}\). If the initial temperature of the spheres is \(T_{i}=25^{\circ} \mathrm{C}\) and the convection coefficient is \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take a sphere near the inlet of the system to accumulate \(90 \%\) of the maximum possible thermal energy? What is the corresponding temperature at the center of the sphere? Is there any advantage to using copper instead of aluminum?

A very thick plate with thermal diffusivity \(5.6 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) and thermal conductivity \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is initially at a uniform temperature of \(325^{\circ} \mathrm{C}\). Suddenly, the surface is exposed to a coolant at \(15^{\circ} \mathrm{C}\) for which the convection heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the finite- difference method with a space increment of \(\Delta x=15 \mathrm{~mm}\) and a time increment of \(18 \mathrm{~s}\), determine temperatures at the surface and at a depth of \(45 \mathrm{~mm}\) after \(3 \mathrm{~min}\) have elapsed.

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.