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FINITE ELEMENT ANALYSIS

FEA LAB REPORT

DUCT/DUCTLESS PIPEIndustrial & Manufacturing Engineering [I & ME], Semester-6

Raja Muhammad Zeeshan [IME 021]

Muhammad Faisal Amjad [IME 013]

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Finite Element Modeling & Analysis of Heat Conduction

In a Ducted & Ductless Cylinder

Temperature distribution in the cylinder with & without the duct.

A ducted cylinder isone having a duct for

fluid to pass in order

to absorb heat energyto keep the cylinder

under desired

temperature limits, the

fluid takes away the

heat, however in a

ductless cylinder ducts

are not present and so

the temperature passes

from the internalsurface to external viamaterial which mayraise the materials

temperature above

desired limitation and

therefore may harm

cylinder.

In lab experiment

where the inner

surface was kept at atemperature of 200

degree Celsius and

outer surface at 15degree Celsius for

both ducted ductless

cylinder with thermal

conductivity

10W/m/oC, and 1000

W/m2 heat flux

removal rate for

ducted cylinder, the

following resultsclearly show that the

minimum temperaturein ducted cylinder is

less than thetemperature kept at

outer surface which

indicates that the heat

removal kept the

temperature of the

cylinder below the

room temperature andthat the temperature distribution is such that the temperature drops rapidly as moved along from inner to outer

surface. Whereas in case of ductless cylinder the minimum temperature is room temperature which was applied in

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program which indicates that if minimum temperature was not restricted the minimum temperature in ductless

cylinder would have been higher. The temperature distribution also indicates a uniform decreasing trend as moved

along from inner surface to outer surface.

Therefore it can be concluded that in order to keep the temperature of cylinder within a desired limit, ductedcylinders with appropriate heat flux removal rate should be used.

Comparing the solution without the duct with the theoretical solution given by equation:

y To calculate temperature at center of line L3y Substitute to find T(r)

o T1 = 200 degree Celsiuso T2 = 15 degree Celsiuso r1 = 1 mo r2 = 2 mo r = 1.5 m

y T(r) at r = 1.5 m = 91.78 oCFollowing the temperature line along line L3, It can be checkedthat at distance = 0.5 m,The temperature T(r) is same as calculated

theoretically.

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The amount of heat generated

in the core of the vessel is such

that the maximum heat flux

through its inner surface is

5000 W/m2. If the maximum

allowable temperature in the

vessel is 250C, what will be

the required heat extraction

rate through each duct?

The solution to above questionwas obtained by hit and trial

method, different values of heatflux removal rates were used andtemperature at the inner surfacefor each value was checked via

nodal solution, the value thatkept the temperature at innersurface almost near to 250 degreeCelsius was then determined. Value ofq came out to be 3315 W/m2. The

inner surface temperature for this valueof q, was 249 oC.

In order to further reduce thetemperature at much higher value of q

can be used to keep the temperature incontrol or viceversa.

EXTRA WORK:

y First and Second Experiment bothwere repeated with fine meshing, for

temperature distribution, the

experiment was repeated using a

higher value of thermal

conductivity.

y ForSecond experiment the heat fluxremoval rate to keep the temperature

near or equal to 250 oC at inner

surface increased due to fine

meshing and required rate came out

to be 3352 W/m2

y For each work, thermal gradient fluxand temperature line plot along line

L3 were plotted to compare results.