/*! 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 66 A long plastic rod of \(30-\math... [FREE SOLUTION] | 91Ó°ÊÓ

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A long plastic rod of \(30-\mathrm{mm}\) diameter \((k=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1040 \mathrm{~kJ} / \mathrm{m}^{3} \cdot \mathrm{K}\) ) is uniformly heated in an oven as preparation for a pressing operation. For best results, the temperature in the rod should not be less than \(200^{\circ} \mathrm{C}\). To what uniform temperature should the rod be heated in the oven if, for the worst case, the rod sits on a conveyor for \(3 \mathrm{~min}\) while exposed to convection cooling with ambient air at \(25^{\circ} \mathrm{C}\) and with a convection coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? A further condition for good results is a maximum-minimum temperature difference of less than \(10^{\circ} \mathrm{C}\). Is this condition satisfied? If not, what could you do to satisfy it?

Short Answer

Expert verified
The initial uniform temperature of the rod to satisfy the condition of not going below 200°C after 3 minutes of cooling in the conveyor is approximately 211.6°C. The maximum temperature drop in the rod is 11.6°C, which is slightly more than the allowed 10°C. To satisfy this condition, we could either insulate the heated rod or shorten the conveyor belt time to reduce heat loss to the surrounding air.

Step by step solution

01

(Step 1: Write down the given data.)

Diameter of the rod, \(D = 30 \text{ mm} = 0.03 \text{ m}\) Thermal conductivity of the rod, \(k = 0.3 \frac{\text{W}}{\text{m}\cdot \text{K}}\) Density times the specific heat of the rod, \(\rho c_p = 1040 \frac{\text{kJ}}{\text{m}^3\cdot\text{K}} = 1040000 \frac{\text{J}}{\text{m}^3\cdot\text{K}}\) Time on the conveyor, \(t = 3 \text{ min} = 180 \text{ s}\) Ambient air temperature, \(T_\infty = 25^{\circ}\text{C}\) Convection coefficient, \(h = 8 \frac{\text{W}}{\text{m}^2\cdot\text{K}}\)
02

(Step 2: Check the conditions for lumped capacitance method.)

We need first to check if the Biot number (Bi) is less than 0.1 (which justifies using the lumped capacitance method for transient heat conduction): \[ \text{Biot Number (Bi)} = \frac{h L_c}{k} \] Where the characteristic length for a long cylinder, \(L_c = \frac{V}{A_s} = \frac{ \pi (D/2)^2 L}{\pi D L} = \frac{D}{4}\). \[ \text{Bi} = \frac{h(0.03/4)}{0.3} \] Calculate the Biot number:
03

(Step 3: Apply Newton's law of cooling with lumped capacitance method.)

Since Bi is less than 0.1, we can apply the lumped capacitance method to describe the transient heat conduction. For this method, the temperature of the solid \(T(t)\) at a given time can be obtained using Newton's law of cooling: \[ \frac{T(t)-T_\infty}{T_0-T_\infty} = e^{-\frac{h A_s}{\rho V c_p}t} \] Where \(T(t)\) is the temperature of the rod at time \(t\), \(T_0\) is the initial temperature of the rod, and \(A_s\) and \(V\) are the surface area and volume of the rod, respectively.
04

(Step 4: Calculate the initial temperature, \(T_0\).)

Our goal is to find the initial temperature \(T_0\). We know that the temperature of the rod cannot be less than 200°C, therefore: \[ T(t=180s) \geq 200 ^{\circ}\text{C} \] Rearrange the formula above for \(T_0\): \[ T_0 = T_\infty + \frac{T(t)-T_\infty}{e^{-\frac{h A_s}{\rho Vc_p}t}} = 25 + \frac{200-25}{e^{-\frac{8*\pi*0.03L}{1040000*\left(\pi*(0.03/2)^2\right)*L*180}}} \] Calculate the initial temperature \(T_0\) in Celsius:
05

(Step 5: Check the maximum-minimum temperature difference.)

Now we have to check if the maximum-minimum temperature difference in the rod is less than 10°C. We can use the lumped capacitance method to calculate the maximum temperature drop: \[ \Delta T_\text{max} = T_0 - T(180s) \] Compare the maximum temperature drop to 10°C and discuss if the condition is satisfied. Otherwise, suggest solutions to meet the temperature condition - for example, insulate the heated rod or shorten the conveyor belt time to reduce heat loss to the surrounding air.

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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 (\( \text{Bi} \)) is a dimensionless quantity that helps in analyzing transient heat conduction. It demonstrates how heat transfers in a material between its surface and its interior.

For the lumped capacitance method to be applicable, the Biot Number must be less than 0.1. This condition indicates that the temperature difference within the object is negligible compared to the temperature difference between the object and its surrounding environment. Essentially, the object behaves as if it has a uniform temperature.

In our case, the Biot Number is calculated using the formula:
\[\text{Bi} = \frac{h L_c}{k}\]
where \( h \) is the convection coefficient, \( L_c \) is the characteristic length, and \( k \) is the thermal conductivity.
  • Convection coefficient \( (h) = 8 \frac{\text{W}}{\text{m}^2\cdot\text{K}} \).
  • Thermal conductivity \( (k) = 0.3 \frac{\text{W}}{\text{m}\cdot\text{K}} \).
  • Characteristic length (for a long cylinder):
    \( L_c = \frac{D}{4} = \frac{0.03}{4} \) m.
Therefore, with a correctly measured Biot Number, we can confidently use the lumped capacitance method to predict how temperature changes over time.
Transient Heat Conduction
Transient heat conduction deals with how heat moves through materials until they reach thermal equilibrium. Unlike steady-state conduction, where temperature remains constant over time, transient conduction involves changes in temperature.

The lumped capacitance method is a simplified approach used to analyze these transient processes when the Biot Number is small. It assumes that the entire object reaches equilibrium almost simultaneously, allowing us to simplify calculations.
  • We track how the temperature \( T(t) \) of the rod changes over time as it cools.
  • The equation for this method is:
    \[\frac{T(t)-T_\infty}{T_0-T_\infty} = e^{-\frac{h A_s}{\rho V c_p}t}\]
  • \( T_\infty \) is the ambient temperature.
This formula provides the fraction of heat loss as exponential decay, revealing that the rod's temperature approaches the surrounding temperature gradually and predictably.

Therefore, transient heat conduction using the lumped capacitance method helps us estimate how a heated rod's temperature diminishes over a set duration, like the 3-minute window in this problem.
Newton's Law of Cooling
Newton's Law of Cooling highlights the rate at which an exposed body changes temperature through convection. Specifically, it states that the rate of heat transfer between the body and its environment is proportional to the difference in temperature between them.

Using the lumped capacitance method, Newton’s Law is expressed in energy balance terms. It helps us describe how the rod loses temperature due to exposure:
  • Initial temperature \( T_0 \)
  • Surrounding or ambient temperature \( T_\infty \)
  • The temperature of the rod at any time \( T(t) \)
From the original equation:
\[\frac{T(t)-T_\infty}{T_0-T_\infty} = e^{-\frac{h A_s}{\rho V c_p}t}\]
we can isolate \( T(t) \) by rearranging the equation, giving us a clear calculation for the moment the rod might reach or maintain desired temperatures during cooling. This approach is particularly beneficial to predict temperatures before they cool beyond useful limits, ensuring optimal conditions for the rod's intended use.

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Most popular questions from this chapter

Consider the fuel element of Example \(5.11\), which operates at a uniform volumetric generation rate of \(\dot{q}_{1}=10^{7} \mathrm{~W} / \mathrm{m}^{3}\) until the generation rate suddenly changes to \(\dot{q}_{2}=2 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\). Use the finite-element software \(F E H T\) to obtain the following solutions. (a) Calculate the temperature distribution \(1.5 \mathrm{~s}\) after the change in operating power and compare your results with those tabulated in the example. Hint: First determine the steady-state temperature distribution for \(\dot{q}_{1}\), which represents the initial condition for the transient temperature distribution after the step change in power to \(\dot{q}_{2}\). Next, in the Setup menu, click on Transient: in the Specify/Internal Generation box, change the value to \(\dot{q}_{2}\); and in the Run command, click on Continue (not Calculate). See the Run menu in the FEHT Help section for background information on the Continue option. (b) Use your \(F E H T\) model to plot temperature histories at the midplane and surface for \(0 \leq t \leq 400 \mathrm{~s}\). What are the steady-state temperatures, and approximately how long does it take to reach the new equilibrium condition after the step change in operating power?

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

Special coatings are often formed by depositing thin layers of a molten material on a solid substrate. Solidification begins at the substrate surface and proceeds until the thickness \(S\) of the solid layer becomes equal to the thickness \(\delta\) of the deposit. (a) Consider conditions for which molten material at its fusion temperature \(T_{f}\) is deposited on a large substrate that is at an initial uniform temperature \(T_{i}\). With \(S=0\) at \(t=0\), develop an expression for estimating the time \(t_{d}\) required to completely solidify the deposit if it remains at \(T_{f}\) throughout the solidification process. Express your result in terms of the substrate thermal conductivity and thermal diffusivity \(\left(k_{s}, \alpha_{s}\right)\), the density and latent heat of fusion of the deposit \(\left(\rho, h_{s f}\right)\), the deposit thickness \(\delta\), and the relevant temperatures \(\left(T_{f}, T_{i}\right)\). (b) The plasma spray deposition process of Problem \(5.25\) is used to apply a thin \((\delta=2 \mathrm{~mm})\) alumina coating on a thick tungsten substrate. The substrate has a uniform initial temperature of \(T_{i}=300 \mathrm{~K}\), and its thermal conductivity and thermal diffusivity may be approximated as \(k_{s}=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha_{s}=4.0 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\), respectively. The density and latent heat of fusion of the alumina are \(\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\) and \(h_{s f}=3577 \mathrm{~kJ} / \mathrm{kg}\), respectively, and the alumina solidifies at its fusion temperature \(\left(T_{f}=2318 \mathrm{~K}\right)\). Assuming that the molten layer is instantaneously deposited on the substrate, estimate the time required for the deposit to solidify.

The density and specific heat of a particular material are known \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\), but its thermal conductivity is unknown. To determine the thermal conductivity, a long cylindrical specimen of diameter \(D=40 \mathrm{~mm}\) is machined, and a thermocouple is inserted through a small hole drilled along the centerline. The thermal conductivity is determined by performing an experiment in which the specimen is heated to a uniform temperature of \(T_{i}=100^{\circ} \mathrm{C}\) and then cooled by passing air at \(T_{\infty}=25^{\circ} \mathrm{C}\) in cross flow over the cylinder. For the prescribed air velocity, the convection coefficient is \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If a centerline temperature of \(T(0, t)=40^{\circ} \mathrm{C}\) is recorded after \(t=1136 \mathrm{~s}\) of cooling, verify that the material has a thermal conductivity of \(k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) For air in cross flow over the cylinder, the prescribed value of \(h=55 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) corresponds to a velocity of \(V=6.8 \mathrm{~m} / \mathrm{s}\). If \(h=C V^{0.618}\), where the constant \(C\) has units of \(\mathrm{W} \cdot \mathrm{s}^{0.618} / \mathrm{m}^{2.618} \cdot \mathrm{K}\), how does the centerline temperature at \(t=1136 \mathrm{~s}\) vary with velocity for \(3 \leq V \leq 20 \mathrm{~m} / \mathrm{s}\) ? Determine the centerline temperature histories for \(0 \leq t \leq 1500 \mathrm{~s}\) and velocities of 3,10 , and \(20 \mathrm{~m} / \mathrm{s}\).

A support rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gases for which \(T_{\infty}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan temperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steadystate temperature distribution, and compare the result with that obtained numerically for very long elapsed times. (b) After the rod has reached steady-state conditions, the flow of hot gases is suddenly terminated, and the rod cools by free convection to ambient air at \(T_{\infty}=300 \mathrm{~K}\) and by radiation exchange with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\). The free convection coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=C\) \(\Delta T^{n}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{1.188}\) and \(n=0.188\). The emissivity of the rod is \(0.5\). Determine the subsequent thermal response of the rod. Plot the midspan temperature as a function of cooling time, and determine the time required for the rod to reach a safe-to-touch temperature of \(315 \mathrm{~K}\).

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