/*! 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

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.

Carbon steel (AISI 1010) shafts of 0.1-m diameter are heat treated in a gas- fired furnace whose gases are at \(1200 \mathrm{~K}\) and provide a convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the shafts enter the furnace at \(300 \mathrm{~K}\), how long must they remain in the furnace to achieve a centerline temperature of \(800 \mathrm{~K}\) ?

A tile-iron consists of a massive plate maintained at \(150^{\circ} \mathrm{C}\) by an embedded electrical heater. The iron is placed in contact with a tile to soften the adhesive, allowing the tile to be easily lifted from the subflooring. The adhesive will soften sufficiently if heated above \(50^{\circ} \mathrm{C}\) for at least \(2 \mathrm{~min}\), but its temperature should not exceed \(120^{\circ} \mathrm{C}\) to avoid deterioration of the adhesive. Assume the tile and subfloor to have an initial temperature of \(25^{\circ} \mathrm{C}\) and to have equivalent thermophysical properties of \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1.5 \times 10^{6}\) \(\mathrm{J} / \mathrm{m}^{3} \cdot \mathrm{K}\) Tile, 4-mm thickness Subflooring (a) How long will it take a worker using the tile-iron to lift a tile? Will the adhesive temperature exceed \(120^{\circ} \mathrm{C} ?\) (b) If the tile-iron has a square surface area \(254 \mathrm{~mm}\) to the side, how much energy has been removed from it during the time it has taken to lift the tile?

A long cylinder of \(30-\mathrm{mm}\) diameter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is suddenly quenched in a large, constant- temperature oil bath at \(350 \mathrm{~K}\). The cylinder properties are \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection coefficient of \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how would the temperature history change?

A microwave oven operates on the principle that application of a high- frequency field causes electrically polarized molecules in food to oscillate. The net effect is a nearly uniform generation of thermal energy within the food. Consider the process of cooking a slab of beef of thickness \(2 L\) in a microwave oven and compare it with cooking in a conventional oven, where each side of the slab is heated by radiation. In each case the meat is to be heated from \(0^{\circ} \mathrm{C}\) to a minimum temperature of \(90^{\circ} \mathrm{C}\). Base your comparison on a sketch of the temperature distribution at selected times for each of the cooking processes. In particular, consider the time \(t_{0}\) at which heating is initiated, a time \(t_{1}\) during the heating process, the time \(t_{2}\) corresponding to the conclusion of heating, and a time \(t_{3}\) well into the subsequent cooling process.

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