/*! 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 36 A \(5 \mathrm{~cm}\)-high stainl... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 36

A \(5 \mathrm{~cm}\)-high stainless steel truncated cone has a base diameter of \(10 \mathrm{~cm}\) and a top diameter of \(5 \mathrm{~cm}\). The sides are insulated, and the base and top temperatures are \(100^{\circ} \mathrm{C}\) and \(50^{\circ} \mathrm{C}\), respectively Assuming one-dimensional heat flow, estimate the heat flow. Take \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel.

Short Answer

Expert verified
The heat flow through the truncated cone is calculated using the formula with the given parameters.

Step by step solution

01

Understand the Problem

We have a truncated cone made of stainless steel with given top and base temperatures. We need to calculate the heat flow through this cone using its geometry and the thermal conductivity of the material.
02

Define the Formula for One-Dimensional Heat Flow

The formula for heat flow in a one-dimensional system is given by \[ q = -k imes A_m imes \frac{螖T}{螖x} \]where:- \( q \) is the heat flow,- \( k \) is the thermal conductivity,- \( A_m \) is the mean cross-sectional area,- \( 螖T \) is the temperature difference,- \( 螖x \) is the thickness of the material (height in this case).
03

Calculate the Mean Cross-sectional Area

The mean cross-sectional area \( A_m \) for a truncated cone is the average of the top and base areas. The base and top radii are 5 cm and 2.5 cm, respectively:- Base area, \( A_b = \pi (0.05)^2 \)- Top area, \( A_t = \pi (0.025)^2 \)\[ A_m = \frac{A_b + A_t}{2} = \frac{\pi (0.05)^2 + \pi (0.025)^2}{2} \].
04

Substitute Values and Calculate Heat Flow

Substitute the values into the heat flow formula to calculate \( q \).Given:- \( k = 15 \, \text{W/m K} \),- \( 螖T = 100 - 50 = 50 \, \text{K} \),- \( 螖x = 0.05 \, \text{m} \),- Use the calculated mean area from Step 3.\[ q = -15 \times \left( \frac{\pi (0.05)^2 + \pi (0.025)^2}{2} \right) \times \left( \frac{50}{0.05} \right) \].

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Key Concepts

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

Understanding Thermal Conductivity
Thermal conductivity plays a crucial role in understanding how materials transfer heat. It is a material-specific property that measures the ability of a material to conduct heat. In mathematical terms, thermal conductivity, denoted by \( k \), is expressed in units of watts per meter per Kelvin (W/m K). This unit describes how much heat passes through a material of a certain thickness with a given temperature difference across it.
For example, in the problem, the stainless steel has a thermal conductivity of \( k = 15 \, \text{W/m K} \). This value tells us how efficiently heat is transferred through the steel material when exposed to a temperature gradient.
  • High thermal conductivity implies quicker heat transfer, like in metals.
  • Low thermal conductivity indicates slower heat transfer, as in insulators.
Understanding this concept is essential when dealing with problems involving heat flow and temperature differences across materials.
The Role of a Truncated Cone
A truncated cone is a geometric shape defined by slicing a cone parallel to its base, resulting in a flat top. In the context of heat transfer, a truncated cone potentially affects how heat flows through the shape due to its varying cross-sectional area from base to top.
The given problem involves a stainless steel truncated cone with different diameters at its base and top. These dimensions affect the heat flow calculation, as the area through which the heat moves changes slightly.
  • Base diameter: \(10 \, \text{cm}\)
  • Top diameter: \(5 \, \text{cm}\)
The mean cross-sectional area is crucial for simplifying the calculation by averaging the areas of the base and the top. This approach allows us to apply the one-dimensional heat flow formula despite the geometric complexity.
One-Dimensional Heat Flow Simplification
The concept of one-dimensional heat flow is applied to simplify complex heat transfer problems by assuming heat moves in a single direction. This assumption is valid when the heat transfer primarily occurs along the thickness of the material.
In the problem, the sides of the truncated cone are insulated, allowing us to assume a one-dimensional heat flow vertically through the height of the cone. This simplification helps in applying the formula:
\[ q = -k \times A_m \times \frac{螖T}{螖x} \]
Here, \( q \) represents the heat flow, \( 螖T \) is the temperature difference, \( A_m \) is the mean area, and \( 螖x \) is the height. By applying this assumption, we can focus on the heat transfer perpendicular to the base and top, ignoring lateral heat exchanges along the sides.
Properties of Stainless Steel
Stainless steel is a popular choice in many engineering applications due to its excellent properties. Several characteristic features make it suitable for the given heat transfer problem.
First, stainless steel is known for its good thermal conductivity. With a value of \(15 \, \text{W/m K}\), it facilitates significant heat transfer relative to other materials. Moreover, it maintains strength and integrity even at elevated temperatures.

Additional benefits of stainless steel include:
  • Corrosion resistance, making it durable in various environments.
  • Mechanical strength, ensuring structural robustness under thermal stress.
Understanding these properties helps us appreciate why stainless steel is used in applications requiring reliable heat transfer and durability.

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Most popular questions from this chapter

Inconel-X-750 straight rectangular fins are to be used in an application where the fins are \(2 \mathrm{~mm}\) thick and the convective heat transfer coefficient is \(300 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Investigate the effect of tip boundary condition on estimated heat loss for fin length \(L\) varying from \(6 \mathrm{~mm}\) to \(20 \mathrm{~mm}\). Take \(T_{B}=800 \mathrm{~K}, T_{e}=300 \mathrm{~K}, k=18.8\) \(\mathrm{W} / \mathrm{m} \mathrm{K}\).

(i) Derive an expression for the relation between heat loss and temperature difference across the inner and outer surfaces of a hollow sphere, the conductivity of which varies with temperature in manner given by \(k=k_{0}\left[1+a\left(T-T_{0}\right)\right]\), where \(T_{0}\) is a reference temperature. (ii) Find the corresponding result for a hollow cylinder. (iii) Compare the expressions derived for parts (i) and (ii) for the special case of the outside radius becoming infinite. Explain the different values obtained.

The thermal conductivity of a solid may often be assumed to vary linearly with temperature, \(k=k_{0}\left[1+a\left(T-T_{0}\right)\right]\), where \(k=k_{0}\) at a reference temperature \(T_{0}\) and \(a\) is a constant coefficient. Consider a solid slab, \(0

A space radiator is made of \(0.3 \mathrm{~mm}\)-thick aluminum plate with heatpipes at a pitch of \(8 \mathrm{~cm}\). The heatpipes reject heat at \(330 \mathrm{~K}\). The back of the radiator is insulated, and the front sees outer space at \(0 \mathrm{~K}\). If the aluminum surface is hard-anodized to give an emittance of \(0.8\), determine the fin effectiveness of the radiator and the rate of heat rejection per unit area. Take \(k=200 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the aluminum. (Hint: We do not expect the plate temperature to vary more than a few kelvins: assume \(q_{\mathrm{rad}}\) is constant at an average value to obtain an approximate analytical solution.)

A \(2 \mathrm{~mm}\)-diameter resistor, for an electronic component on a space station, is to have a sheath of thermal conductivity \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). It is cooled by forced convection with \(\bar{h}_{c} \simeq 1.1 D^{-1 / 2} \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), for diameter \(D\) in meters, and by radiation with \(q_{\mathrm{rad}}=\sigma \varepsilon\left(T_{s}^{4}-T_{e}^{4}\right)\). Determine the radius of the sheath that maximizes the heat loss when the resistor is at \(400 \mathrm{~K}\) and the surroundings are at \(300 \mathrm{~K}\). Take the value of the surface emittance \(\varepsilon\) as (i) \(0.9\). (ii) \(0.5\).
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