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Chapter 5: Problem 71

In heat treating to harden steel ball bearings \(\left(c=500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}, k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), it is desirable to increase the surface temperature for a short time without significantly warming the interior of the ball. This type of heating can be accomplished by sudden immersion of the ball in a molten salt bath with \(T_{\infty}=1300 \mathrm{~K}\) and \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that any location within the ball whose temperature exceeds \(1000 \mathrm{~K}\) will be hardened. Estimate the time required to harden the outer millimeter of a ball of diameter \(20 \mathrm{~mm}\), if its initial temperature is \(300 \mathrm{~K}\).

Short Answer

Expert verified
To estimate the time required to harden the outer millimeter of the steel ball, we first calculate the Biot number (Bi) using the given values: \[L_c = \frac{0.02}{2} - 0.001 = 0.009\,m\] \[Bi = \frac{5000\,\text{W/m}^{2}\cdot\text{K} \times 0.009\,\text{m}}{50\,\text{W/m}\cdot\text{K}} = 0.9\] Since the Bi value is greater than 0.1, the lumped capacitance method cannot be applied in this scenario. This means that the temperature is not uniform throughout the object, and a different method should be used to estimate the time required to harden the outer millimeter of the steel ball.

Step by step solution

01

Calculate the Biot number (Bi)

The Biot number is a dimensionless quantity that represents the ratio of conduction resistance within the object to the convection resistance external to the object. It helps to determine if the temperature is uniform throughout the object. The formula for the Biot number is: \[Bi = \frac{h L_c}{k}\] Here, Lc is the characteristic length and is equal to the radius of hardening: \[L_c = \frac{D}{2} - 0.001\] Now, calculate Bi.
02

Verify if the lumped capacitance method can be applied

The lumped capacitance method is valid if Bi is less than or equal to 0.1. Verify if the condition is met.
03

Calculate the Fourier number (Fo)

The Fourier number is a dimensionless quantity that represents the ratio of the time it takes for the temperature to reach the desired level to the time it takes for the object to reach a uniform temperature. The formula for the Fourier number is as follows: \[Fo = \frac{\alpha t}{L_c^2}\] The thermal diffusivity (伪) can be calculated using the formula: \[\alpha = \frac{k}{\rho c}\] Calculate 伪 and then solve for the time t.
04

Estimate the time required to harden the outer millimeter of the steel ball

Now that we have the value for the time t, we can estimate how long it will take to harden the outer millimeter of the steel ball. Keep in mind that the time calculated in the Fourier number will be a close approximation of the actual time required for hardening the outer surface. The time required for hardening the outer millimeter can now be found using the calculated time t.

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Key Concepts

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

Biot number
The Biot number provides a useful way to analyze heat transfer in relation to an object's size and the materials involved. Specifically, it's a dimensionless number that indicates how easily heat moves from the surface into the object. In simpler terms, it tells us whether the temperature inside an object can be assumed uniform or not.

To calculate the Biot number (\( Bi \)), we use the formula:
  • \( Bi = \frac{h \, L_c}{k} \)
Where:- \( h \) is the heat transfer coefficient- \( L_c \) is the characteristic length (often the radius in spheres)- \( k \) is the thermal conductivity.

In practice, if \( Bi \) is less than 0.1, the object's temperature is assumed uniform, making it ideal for applying the lumped capacitance method. For larger Biot numbers, temperature variations within the object are more significant.
Lumped capacitance method
The lumped capacitance method simplifies temperature analysis within a body. It assumes the entire object is at the same temperature. This method depends on a small Biot number (\( Bi \leq 0.1 \)) which reveals that thermal gradients within the object are negligible.

When applicable, the lumped capacitance method dramatically reduces the complexity of the heat transfer equations. It allows us to use simple first-order differential equations to predict how the object's temperature changes over time when exposed to different environments.

This assumption holds especially well in small objects or when the object is highly conductive, meaning heat quickly spreads throughout the material. For students studying this method, verifying the Biot number early in calculations can save a lot of time by confirming applicability.
Fourier number
The Fourier number (\( Fo \)) is another dimensionless number used in heat transfer analysis. It measures how quickly heat diffuses through a material. In essence, it tells if the heat has had time to spread throughout the object.

The formula for the Fourier number is:
  • \( Fo = \frac{\alpha \, t}{L_c^2} \)
Where:- \( \alpha \) is the thermal diffusivity- \( t \) is the time- \( L_c \) is the characteristic length.

This number helps in assessing transient heat conduction: when and how quickly an object reaches a specified temperature. Often, the Fourier number is used alongside the Biot number to address different heat transfer scenarios. Together, they allow engineers and scientists to effectively predict and control heating processes.
Thermal diffusivity
Thermal diffusivity (\( \alpha \)) is a property that indicates how quickly heat spreads through a material. It's essential in determining how fast temperature changes occur within a material when it is exposed to heat.

The formula to find thermal diffusivity is:
  • \( \alpha = \frac{k}{\rho \cdot c} \)
Where:- \( k \) is the thermal conductivity- \( \rho \) is the density- \( c \) is the specific heat capacity.

High thermal diffusivity means a material can quickly adjust to changes in temperature. Conversely, materials with low thermal diffusivity take longer to reach thermal equilibrium. Thus, understanding this concept helps in designing and analyzing processes in which precise heat control is critical, such as the hardening of steel ball bearings.

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Most popular questions from this chapter

Two plates of the same material and thickness \(L\) are at different initial temperatures \(T_{i, 1}\) and \(T_{i, 2}\), where \(T_{i, 2}>T_{i, 1}\). Their faces are suddenly brought into contact. The external surfaces of the two plates are insulated. (a) Let a dimensionless temperature be defined as \(T *(F o) \equiv\left(T-T_{i, 1}\right) /\left(T_{i, 2}-T_{i, 1}\right)\). Neglecting the thermal contact resistance at the interface between the plates, what are the steady-state dimensionless temperatures of each of the two plates, \(T_{s s, 1}^{*}\) and \(T_{s s, 2}^{*}\) ? What is the dimensionless interface temperature \(T_{\text {in }}^{*}\) at any time? (b) An effective overall heat transfer coefficient between the two plates can be defined based on the instantaneous, spatially averaged dimensionless plate temperatures, \(U_{\mathrm{eff}}^{*} \equiv q^{*} /\left(\bar{T}_{2}^{*}-\bar{T}_{1}^{*}\right)\). Noting that a dimensionless heat transfer rate to or from either of the two plates may be expressed as \(q^{*}=d\left(Q / Q_{o}\right) / d F o\), determine an expression for \(U_{\text {eif }}^{*}\) for \(F o>0.2\).

The plane wall of Problem \(2.60(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\alpha=1.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) has a thickness of \(L=40 \mathrm{~mm}\) and an initial uniform temperature of \(T_{o}=25^{\circ} \mathrm{C}\). Suddenly, the boundary at \(x=L\) experiences heating by a fluid for which \(T_{s}=50^{\circ} \mathrm{C}\) and \(h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while heat is uniformly generated within the wall at \(\dot{q}=1 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). The boundary at \(x=0\) remains at \(T_{a}\). (a) With \(\Delta x=4 \mathrm{~mm}\) and \(\Delta t=1 \mathrm{~s}\), plot temperature distributions in the wall for (i) the initial condition, (ii) the steady-state condition, and (iii) two intermediate times. (b) On \(q_{x}^{\prime \prime}-t\) coordinates, plot the heat flux at \(x=0\) and \(x=L\). At what elapsed time is there zero heat flux at \(x=L\) ?

A thin circular disk is subjected to induction heating from a coil, the effect of which is to provide a uniform heat generation within a ring section as shown. (a) Derive the transient, finite-difference equation for node \(m\), which is within the region subjected to induction heating. (b) On \(T r\) coordinates sketch, in qualitative manner, the steady-state temperature distribution, identifying important features.

Joints of high quality can be formed by friction welding. Consider the friction welding of two 40 -mm-diameter Inconel rods. The bottom rod is stationary, while the top rod is forced into a back-and-forth linear motion characterized by an instantaneous horizontal displacement, \(d(t)=a \cos (\omega t)\) where \(a=2 \mathrm{~mm}\) and \(\omega=1000 \mathrm{rad} / \mathrm{s}\). The coefficient of sliding friction between the two pieces is \(\mu=0.3\). Determine the compressive force that must be applied to heat the joint to the Inconel melting point within \(t=3 \mathrm{~s}\), starting from an initial temperature of \(20^{\circ} \mathrm{C}\). Hint: The frequency of the motion and resulting heat rate are very high. The temperature response can be approximated as if the heating rate were constant in time, equal to its average value.

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.
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