05 Test Chamber Insulation (Cylindrical Wall)

Test Chamber Insulation (Cylindrical) ME433 COMSOL INSTRUCTIONS

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TEST CHAMBER INSULATION (Cylindrical) Problem Statement The wall of a small cylindrical test chamber is made of stainless steel and glass, as shown in the diagram below. It is desired to reduce the heat loss from a chamber by adding an insulation layer of 85% magnesia. Of interest is determining where to insert the insulation layer so that the rate of heat loss is lowest. One student recommended adding the 85% magnesia layer on the cold side (glass) while a second student suggested the hot side (steel). A third student recommended inserting the insulation layer between the steel and glass. A fourth student who did not seem to pay much attention to the discussion claimed that it does not matter where to add the insulation. You are asked to carry out a study to determine the option that will result in the lowest rate of heat loss from the chamber. Known quantities: Chambers inside radius, cm 1 8.5r Stainless steel thickness, cm 1.5sL Glass thickness, cm85% magnesia thickness,

1gL cm

tbook for .

4mL T = 100 C , steel

C

2C

se appendices in your tex

bservations

Three principal cases will be considered to determine where the rate of heat loss is

The amount of material in a layer depends on where it is placed. An inner layer of

Upon a close inspection of the thermal circuit equation 3.57, it is clear that by

h steel = 8 W/m2T, glass = 10 C h glass = 16 W/m Uthe properties of materials considered O

lowest (a) 85% magnesia layer on the outside of the chamber, (b) 85% magnesia layer between steel and glass layers, and (c) 85% magnesia layer on the inside of the chamber.

thickness L will have less material than an outer layer of the same thickness.

changing the position of the insulation layer in the wall, the denominator valuechanges [Hint: Examine the arguments of logarithmic terms and other terms involving radii]. We therefore expect the total q to change as we place the insulation layer at different locations.

Chamber Wall Steel inside, glass outside


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Assignment

1. Using COMSOL, show the temperature distribution for the 3 cases. 2. Using COMSOL, compute the heat per unit length, for the 3 cases.

3. Compute theoretical . Compare the results with those in the above question. Are

COMSOL results valid? [Hint: COMSOL results for heat per unit length q were computed for half the geometry]

4. Using COMSOL, plot graphs for each of the cases showing T(x) for the three

layers. Each graph should show clearly the boundaries of the layers and the corresponding temperature distribution.

5. Pick one of the 3 cases and compute the temperatures of the inner and outer

boundaries of the chamber analytically. Compare these results with COMSOL solution and claim whether or not they are valid.

6. [Extra Credit]: Would the rate of heat transfer vary in an analogous multi layer

rectangular wall as one of the insulation layers is moved to a different position? Explain.

7. [Extra Credit]: If the design requirement is not to exceed 20 C at the outer

surface, which of the three cases would you recommend? Can you use less than 3 layers? Take financial considerations into account.

8. [Extra Credit]: Verify the results of example 3.2 using COMSOL. Include

COMSOL solution in your report for this module.

'q ,

'q


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Modeling with COMSOL Multiphysics MODEL NAVIGATOR

are not tending to look at the field flow outside the chamber walls, we will only need to work

Heat

We will model this problem with 2 dimensional semicircular disks. Since we inwith the Heat Transfer Module. We are also assuming the process to be in steady state. For this setup:

1. Start COMSOL Multiphysics.

2. In the Space dimension list select 2D (under the New Tab).

3. From the list of application modes select Heat Transfer Module General Transfer Steady State Analysis

4. Click OK.


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DEFINING CONSTANTS Continue by creating a database for constants the model uses:

2. lick OK.

EOMETRY MODELING

1. From the Options menu select Constants, and in the resulting dialog box

define the following names and expressions.

C G In this section, we will create a 2 dimensional geometry that will be used as a model in our problem. According to problem statement, we will need to create 3 multi layered circular disks (1 multi layered disk per case considered in problem statement). To reduce computational time, we will make 3 semicircular disks instead of full disks.

1. Create a circular disk by going to the Draw menu, selecting Specify Objects Circle.

2. Start by entering information for first circle, or C1, from the following table.

C1 C2 C3 C4 C5 C6

RADIUS 0.085 0.1 0.11 0.15 0.085 0.1 BASE CENTER CENTER CENTER CENTER CENTER CENTER

X 0 0 0 0 0.4 0.4 C7 C8 C9 C10 C11 C12

RADIUS 0.14 0.15 0.085 0.125 0.14 0.15 BASE CENTER CENTER CENTER CENTER CENTER CENTER

X 0.4 0.4 0.8 0.8 0.8 0.8

3. When done with step 2, click OK and repeat steps 1 to 3 for the other 11 circles.

4. Click on Zoom Extents button in the main toolbar to zoom into the

geometry. You should now have created 12 circular disks for the 3 cases considered. The geometry at this point should look like the one shown below.

NAME EXPRESSION VALUE DESCRIPTION

k_s 43[W/(m*degC)] 43[W/(mK)] 1% Carbon Steel @ 20C

k_g 0.7[W/(m*degC)] 0.7[W/(mK)] Window Glass @ 20C

k_m 0.065[W/(m*degC)] 0.065[W/(mK)] 85% Magnesia @ 20C

T_inf1 100[degC] 373.15[K] T Inside Chamber

T_inf4 10[degC] 283.15[K] T Outside Chamber

h_1 8[W/(m^2*degC)] 8[W/(m2K)] h Inside Chamber

h_4 16[W/(m^2*degC)] 16[W/(m2K)] h Outside Chamber


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Proceed by creati porary) rectang

5. Go to the Draw m e.

6. Enter info o CTAN from

E 1

ng the following (tem les:

enu and select Specify Objects Rectangl

rmati n for RE GLE 1 the following table.

RECTANGL (R1) WIDTH 0.15 HEIGHT 0.3 BASE CORNER

X -0.15 Y -0.15

7. With rectangle R1 selected (light red color), select Edit Copy.

8. Select Edit Paste.

9. Enter 0.4 as the displacement value in the x

direction in the Paste displacement window.

11. Select Edit Paste for the second time.

12. Enter tdisplacem

13. Click OK.

You should no r g s red. They will be used to construct semicircu disks e geo ry at point uld look like

e one shown below.

10. Click OK.

0.8 as he displacement value in the x direction in the Paste ent window.

w have created 3 tempora y rectan les for the 3 case considelar . Th met this sho

th


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To construct multilayered semicircular disks,

14. Select Draw Create Composite Object option.

15. In the Set formula field, type (C1+C2+C3+C4) R1 (without quotation marks).

16. Click Apply.

17. In the Set formula field, type (C5+C6+C7+C8) R2 and click Apply.

) R3 and click OK.

19. Choose Select All from the Edit menu.

20. Select objects CO2 and CO4 by left clicking them in the model space while

22. Enter -0.2 as the displacement value in the x e Move displacement window.

where in the white model space in COMSOL to deselect the geometry objects CO2 and CO4.

25. Select object CO4 by left clicking it in the model space.

26. Select Draw Modify Move.

18. In the Set formula field, type (C9+C10+C11+C12

holding the Shift key on your keyboard.

21. Select Draw Modify Move.

direction in th

23. Click OK.

24. Left click any


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27. Enter -0.2 as the displacement value in the x direction in the Move displacement window.

28. Click OK.

29. Choose Select All from the Edit menu.

30. Choose

Split Object from the Draw menu.

31. Left click anywhere in the white model space in COMSOL to deselect the entire

geometry.

32. Select objects C7, C11, and C15 by left clicking them in the model space while holding the Shift key on your keyboard.

33. Press the Delete key on your keyboard to delete objects C7, C11, and C15.

34. Click on Zoom Extents button in the main toolbar to zoom into the

geometry. You should now see your completed 2D chamber wall. The leftmost semicircular disk

presents case A (as described in observations), the middle semicircular disk represents

y its aterial name (as opposed to generic names, such as CO6). To do so,

35. Double click object CO6 and rename it as steelA.

recase B, and the rightmost semicircular disk represents case C. Let us further visually distinguish these cases in COMSOL by naming each layer bm

36. Click OK

37. Use the table given below to rename the rest of the objects i steps 35 an

n the same way as instructed in d 36.

OBJECT CO5 CO3 CO10 CO9 CO8 CO14 CO13 CO12

NAME glassA magnesiaA steelB magnesiaB glassB steelC magnesiaC glassC


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When completed, your final geometry should look like the

one shown below:

Try to investigate the geometry you just made. In particular, try to explore what the Geometric Properties button does: as you select the parts of the geometry. See how

the function of this button changes as you switch the program mode with . PHYSICS SETTINGS Physics settings in COMSOL consist of b

two parts: (1) Subdomain settings and (2)

2. Select subdomains 1, 5, and 9 in the Subdomain selection window while holding the Control (ctrl) key on your keyboard. (These subdomains correspond to the magnesia layer).

3. Enter k_m in the field for thermal conductivity k.

oundary conditions. The subdomain settings let us specify material types, initial conditions and modes of heat transfer (i.e. conduction and/or convection). The boundary conditions settings are used to specify what is happening at the boundaries of the geometry. Subdomain Settings:

1. From the Physics menu select Subdomain Settings (equivalently, press F8).


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4. Select subdomain 2 (do not hold the ctrl key in this step)

5. Select subdomains 4 and 7 in the Subdomain selection window while holding the Control (ctrl) key on your keyboard. (These subdomains correspond to the glass layer).

6. Enter k_g in the field for thermal conductivity k.

7. Select subdomain 3 (do not hold the ctrl key in this step)

8. Select subdomains 6 and 8 in the Subdomain selection window while holding

the Control (ctrl) key on your keyboard. (These subdomains correspond to the steel layer).

window.

RY BOUNDARY CONDITION COMMENTS

9. Enter k_s in the field for thermal conductivity k. 10. Click OK to apply these properties and close the Subdomain Settings

Boundary Conditions:

1. From the Physics menu open the Boundary Settings (F7) dialog box.

2. Apply the following boundary conditions:

BOUNDA

1 through 18 Insulation/Symmetry Not physical boundaries

22,23,30,31,38,39 Heat Flux Enter h = h_1; TINF = T_inf1

19,26,27,34,35,42 Heat Flux Enter h = h_4; TINF = T_inf4

Boundaries 20,21,24,25,28,29,32,33,36,37,40,41 are internal boundaries. COMSOL utomatically applies the contina

auity boundary condition by default. Internal boundaries

ible to the user. Leave them as

Boundaries 1 18 are not real boundaries in the sense that we use a semicircular model. The correct boundary condition of insulation/symmetry is applied by default to these boundaries.

3. Click OK to close the Boundary Settings window.

ppear grayed out in the selection window and are inaccessey are! th


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MESH GENERATION To minimize the computational time without compromising much accuracy of the solution, we must change the default meshing parameters. To do this,

1. Go to the Physics menu and select Selection Mode Subdomain Mode.

2. enu, choose Select All.

3. Left click the Mesh Selected (Mapped)

From the Edit m

button on the left hand side

hould get the following radial mesh:

toolbar. As a result of these steps, you s

We are now ready to compute our solution.


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COMPUTING AND SAVING THE SOLUTION

ionary herefore, no

odifications need to be made. To enable the solver, proceed with the following steps:

1. From the Solve menu select Solve Problem. (Allow few seconds for solution)

2. Save your work on desktop by choosing File Save. Name the file according ics

r map:

In this step we define the type of analysis to be performed. We are interested in statanalysis here, which we previously selected in the Model Navigator. Tm

to the naming convention given in the Introduction to COMSOL Multiphysdocument document.

The result that you obtain should resemble the following surface colo

By default, your immediate result will be given in Kelvin instead of degrees Celsius as shown above. The next section (Postprocessing and Visualization) will help you in obtaining the above and other diagrams, plotting graphs for each of the cases showing radial T(x) for the three layers, and computing heat per unit length, through COMSOL).

'q


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POSTPROCESSING AND VISUALIZATION After solving the problem, we would like to be able to look at the solution. COMSOL offers us a number of different ways to look at our temperature (and other) fields. Inproblem we will deal with 2D color maps, heat per unit length, 'q computation, and 1D

this

mperature distribution plots. You will then address the questions of COMSOL solution

fficiently in COMSOL alone.

1. From the Postprocessing menu, open Plot Parameters dialog box (F12).

2. Under the Surface tab, change the unit of temperature to degrees Celsius from the drop down menu in the Unit field.

tevalidity and compare the results to theoretical predictions. All of these steps can be done e Changing Temperature Units:

3. Click OK. The 2D temperature distribution will be displayed with degrees Celsius as the units of temperature.


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Saving Color Maps: After you have selected a view that shows the results clearly, you may want to save it aan image for future discussion. This may be done as follows:

1. Go to the File menu and select Export Image. This will bring up an Export Image window.

s

or a 4 by 6 image, acceptable image quality settings are given in the figure below. If ncrease the DPI value.

Fyou need higher image quality, i

2. Change your Export Image value settings to the ones in the above figure. 3. Click the Export button.

4. Name and save the image.

Generating 1D Plot of Radial T(x): In this section, we will make 3 separate plots of radial temperature distribution for each of the 3 cases. Lets choose the most convenient line along which to plot the temperature distribution the central line running at y = 0 with the slope of 0.

1. From Postprocessing menu select Cross Section Plot Parameters option.

degrees Celsius.

2. Switch to the Line/Extrusion tab.

3. Change the Unit of temperature to


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4. Change the x axis data from arc length to x.

line data: x0 = 0.085 and x1 = 0.15; y0 = 0 and y1 = 0.

ber wall) for case (a). To save is plot,

7. Click the save

5. Enter the following coordinates in the Cross section

6. Click Apply.

These steps produce a plot of T(x) at y = 0, from x = 8.5 cm (inside surface of the chamber wall) to x = 15 cm (outside surface of the chamth

button in your figure with results. This will bring up an Export Image window.

8. Follow steps 2 4 as instructed on page 13 to finish with exporting the image.

Repeat steps 5 8 on this page using the table of coordinates for Cross section line data field for cases (b) and (c) given below:

COORDINATE CASE B CASE C X0 0.285 0.485 X1 0.35 0.55

Y0, Y1 0 0

Computing Heat per Unit Length q:

menu select Boundary Integration option.

field, select boundaries 19 and 26 while holding the oard.

dialog window, change Predefined Quantities lux.

To compute the heat per unit length, 'q for case (a),

1. From the Postprocessing 2. In the Subdomain selection

Control key on your keyb

3. In the Subdomain Integrationsetting to Normal total heat f


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4. Click the OK button. The value of the integral (solution) is displaye t d a

programs prompt on the bottom. For case (a), 'q = 41.53 W/m.

Note the value of the heat per unit length, . This is the value for case (a).

epeat steps 1 4 for cases (b) and (c). For step 2, the corresponding boundaries for ases (b) and (c) are tabulated below,

BC SELECTION

'q

Rc

CAS B 27 and 34 E CASE C 35 and 42

Note the values of the heat per unit length, for these cases as well. You will use these

er wall.

'qvalues to answer assignment question 3. Keep in mind that these values (and the one for case (a) are only valid for half the chamb


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ANALYTIC VALIDATION In this section, we will discuss the validation of COMSOL solution. This includes comparing both COMSOL heat per unit length, and COMSOL temperature distribution. The analytic solution for is given by equation 3.57:

'q to this problem'q

1 42 1 3 2 4 31 1 1 2 3 4 4

ln ln ln1 12 2 2 2 2

rT Tq

r r r r r rr h L k L k L k L r h

L

Note that this equation contains terms involving length L. The value of L is not given in this problem (L corresponds to the depth of the object in 3 rd dimension). To compare COMSOL to the according to equation 3.57, this fact must be taken into account. 'q rq

Simply rearrange equation 3.57 to get r rq qL

. In this form, rq has the same dimensions s COMS s and OL 'q . Thus 'q rqa are readily comparable. Compute analytic and

comp . Do both results agree well? MSOL solution valid? With the knowledge of q, apply equation 3.27 to find the wall temperatures (both inner nd outer) of the chamber for any of the configurations. Compare these results with th

rqare with COMSOLs 'q Is CO

a e 1D radial graphs of T(x) you made earlier. Do both results agree well? This completes COMSOL modeling procedures for this problem.

Documents

05 Test Chamber Insulation (Cylindrical Wall)