Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Consider steady one-dimensional heat transfer through a multilayer medium. If the rate of heat transfer \(\dot{Q}\) is known, explain how you would determine the temperature drop across each layer.

Short Answer

Expert verified
Question: Given the rate of heat transfer, \(\dot{Q}\), the thickness and thermal conductivity of each layer in a multilayer medium, determine the temperature drop across each layer using the concept of thermal resistance and Fourier's Law of heat conduction. Answer: To find the temperature drop across each layer in a multilayer medium, follow these steps: 1. Identify the layers, their thicknesses, and their thermal conductivity. 2. Calculate the thermal resistance of each layer using the formula \(R = \frac{d}{kA}\) where \(d\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the area through which heat is being transferred. 3. Find the total thermal resistance of the system by adding the individual thermal resistances: \(R_{total}=R_1+R_2+\cdots+R_n\). 4. Apply Fourier's Law to the entire system and calculate the total temperature drop using ΔT = \(\frac{\dot{Q}R_{total}}{A}\). 5. Determine the temperature drop across each layer by multiplying the individual thermal resistance of each layer by the known rate of heat transfer: \(\Delta T_i = \frac{\dot{Q}R_{i}}{A}\) where \(\Delta T_i\) is the temperature drop across layer \(i\).

Step by step solution

01

To solve this problem, we first need to identify the layers in the multilayer medium, their thicknesses and their thermal conductivity. The thermal resistance of each layer, \(R\), can be calculated using the formula \(R = \frac{d}{kA}\) where \(d\) is the thickness of the layer, \(k\) is the thermal conductivity, and \(A\) is the area through which heat is being transferred. #Step 2: Find the total thermal resistance of the system#

Once we know the thermal resistance of each layer, we can calculate the total thermal resistance of the system by adding the individual thermal resistances: \(R_{total}=R_1+R_2+\cdots+R_n\). #Step 3: Apply Fourier's Law to the entire system#
02

Fourier’s law of heat conduction states that \(\delta{Q} = -kA \frac{\delta{T}}{\delta{x}}\), which means \(\dot{Q} = -kA \frac{dT}{dx}\). In this scenario, we have a known rate of heat transfer (\(\dot{Q}\)) and we want to determine the temperature drop across each layer. Rewriting the formula in terms of the temperature difference, we get \(\Delta T = \frac{\dot{Q}R_{total}}{A}\). #Step 4: Determine the temperature drop across each layer#

Now that we have calculated the total temperature drop across the medium, we can determine the temperature drop across each layer by multiplying the individual thermal resistance of each layer by the known rate of heat transfer: \(\Delta T_i = \frac{\dot{Q}R_{i}}{A}\) where \(\Delta T_i\) is the temperature drop across layer \(i\). Remember to follow these steps for each layer in the multilayer medium to determine the temperature drop across each layer.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Circular cooling fins of diameter \(D=1 \mathrm{~mm}\) and length $L=30 \mathrm{~mm}\(, made of copper \)(k=380 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$, are used to enhance heat transfer from a surface that is maintained at temperature \(T_{s 1}=132^{\circ} \mathrm{C}\). Each rod has one end attached to this surface \((x=0)\), while the opposite end \((x=L)\) is joined to a second surface, which is maintained at \(T_{s 2}=0^{\circ} \mathrm{C}\). The air flowing between the surfaces and the rods is also at \(T_{\infty}=0^{\circ} \mathrm{C}\). and the convection coefficient is $h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. (a) Express the function \(\theta(x)=T(x)-T_{\infty}\) along a fin, and calculate the temperature at \(x=L / 2\). (b) Determine the rate of heat transferred from the hot surface through each fin and the fin effectiveness. Is the use of fins justified? Why? (c) What is the total rate of heat transfer from a \(10-\mathrm{cm}\) by \(10-\mathrm{cm}\) section of the wall, which has 625 uniformly distributed fins? Assume the same convection coefficient for the fin and for the unfinned wall surface.

Hot air is to be cooled as it is forced to flow through the tubes exposed to atmospheric air. Fins are to be added in order to enhance heat transfer. Would you recommend attaching the fins inside or outside the tubes? Why? When would you recommend attaching fins both inside and outside the tubes?

One wall of a refrigerated warehouse is \(10.0 \mathrm{~m}\) high and $5.0 \mathrm{~m}\( wide. The wall is made of three layers: \)1.0$-cm-thick aluminum \((k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), 8.0\)-cm-thick fiberglass \((k=0.038\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\), and \(3.0\)-cm-thick gypsum board \((k=0.48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The warehouse inside and outside temperatures are \(-10^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, and the average value of both inside and outside heat transfer coefficients is \(40 \mathrm{~W} / \mathrm{m}^{2}\). \(\mathrm{K}\). (a) Calculate the rate of heat transfer across the warehouse wall in steady operation. (b) Suppose that 400 metal bolts ( $k=43 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(, each \)2.0 \mathrm{~cm}\( in diameter and \)12.0 \mathrm{~cm}$ long, are used to fasten (i.e., hold together) the three wall layers. Calculate the rate of heat transfer for the "bolted" wall. (c) What is the percent change in the rate of heat transfer across the wall due to metal bolts?

A room at \(20^{\circ} \mathrm{C}\) air temperature is losing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4-m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\)-thick insulation with a conductivity of $0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surfaces of the wall, the room air temperature, and the outdoor air temperature remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)

A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken off. Will the rate of heat transfer from the pipe increase or decrease for the same pipe surface temperature?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free