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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 \(25.82 \mathrm{~W/(m^2 \cdot K)}\).

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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.

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Most popular questions from this chapter

Standards for firewalls may be based on their thermal response to a prescribed radiant heat flux. Consider a \(0.25\)-m-thick concrete wall \(\left(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=880 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is at an initial temperature of \(T_{i}=25^{\circ} \mathrm{C}\) and irradiated at one surface by lamps that provide a uniform heat flux of \(q_{s}^{\prime \prime}=10^{4} \mathrm{~W} / \mathrm{m}^{2}\). The absorptivity of the surface to the irradiation is \(\alpha_{s}=1.0\). If building code requirements dictate that the temperatures of the irradiated and back surfaces must not exceed \(325^{\circ} \mathrm{C}\) and \(25^{\circ} \mathrm{C}\), respectively, after \(30 \mathrm{~min}\) of heating, will the requirements be met?

A solid steel sphere (AISI 1010 ), \(300 \mathrm{~mm}\) in diameter, is coated with a dielectric material layer of thickness \(2 \mathrm{~mm}\) and thermal conductivity \(0.04 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The coated sphere is initially at a uniform temperature of \(500^{\circ} \mathrm{C}\) and is suddenly quenched in a large oil bath for which \(T_{\infty}=100^{\circ} \mathrm{C}\) and \(h=3300 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Estimate the time required for the coated sphere temperature to reach \(140^{\circ} \mathrm{C}\). Hint: Neglect the effect of energy storage in the dielectric material, since its thermal capacitance \((\rho c V)\) is small compared to that of the steel sphere

As part of a heat treatment process, cylindrical, 304 stainless steel rods of \(100-\mathrm{mm}\) diameter are cooled from an initial temperature of \(500^{\circ} \mathrm{C}\) by suspending them in an oil bath at \(30^{\circ} \mathrm{C}\). If a convection coefficient of \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained by circulation of the oil, how long does it take for the centerline of a rod to reach a temperature of \(50^{\circ} \mathrm{C}\), at which point it is withdrawn from the bath? If 10 rods of length \(L=1 \mathrm{~m}\) are processed per hour, what is the nominal rate at which energy must be extracted from the bath (the cooling load)?

When a molten metal is cast in a mold that is a poor conductor, the dominant resistance to heat flow is within the mold wall. Consider conditions for which a liquid metal is solidifying in a thick-walled mold of thermal conductivity \(k_{v}\) and thermal diffusivity \(\alpha_{w}\). The density and latent heat of fusion of the metal are designated as \(\rho\) and \(h_{s f}\), respectively, and in both its molten and solid states, the thermal conductivity of the metal is very much larger than that of the mold. Just before the start of solidification \((S=0)\), the mold wall is everywhere at an initial uniform temperature \(T_{i}\) and the molten metal is everywhere at its fusion (melting point) temperature of \(T_{f}\). Following the start of solidification, there is conduction heat transfer into the mold wall and the thickness of the solidified metal \(S\) increases with time \(t\). (a) Sketch the one-dimensional temperature distribution, \(T(x)\), in the mold wall and the metal at \(t=0\) and at two subsequent times during the solidification. Clearly indicate any underlying assumptions. (b) Obtain a relation for the variation of the solid layer thickness \(S\) with time \(t\), expressing your result in terms of appropriate parameters of the system.

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?

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