IHT: Interactive Heat Transfer

thoughtgreenpepperMechanics

Oct 27, 2013 (4 years and 16 days ago)

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ME 259
-

IHT: Interactive Heat Transfer

Computer Exercises


1. Basic Equation Solving


Suppose you wish to solve the following set of equations, where q
1
, q
2
, q
3
, and T are unknowns:


q
1

= 0.35(1000
-
T)

q
2

= 5.67

10
-
8
(T
4
-
300
4
)

q
3

= 1.4(T
-
300)
4/3

q
1

= q
2

+ q
3


Note that two equations are nonlinear and the system cannot be solved analytically.


STEP 1
-

Preprocessing


Simply type the equations as given onto the IHT workspace as shown below. You may insert comments or
headings by preceding text lines with a dou
ble slash // or by bracketing long sections of text with /* ……. */.


// Nonlinear Equation System Example


q1 = 0.35*(1000
-
T)

q2 = 5.67E
-
08*(T^4
-
300^4)

q3 = 1.4*(T
-
300)^(4/3)

q1 = q2 + q3


STEP 2
-

Solving


Click the
Solve
button to prepare the solver. If
your equations have syntax errors, an error warning will appear
-

click OK and usually the solver will highlight the word or equation that contains the error. If you do not supply
enough equations for the number of unknowns, another error warning will appe
ar
-

click OK and a summary of
equations and unknowns will appear to help you determine what is missing.


If there are no syntax or number of equations errors, an Initial Guesses table appears. Initial guesses are
required for each unknown and the solver
uses a value of 1 as a default guess. You may change these
guesses, however, the default values are often satisfactory. You may also insert Minimum and Maximum
solution values if you know the range of possible solution values. For example, we know that tem
perature in
Kelvin can never be less than zero; so inserting a minimum value of 0 for a temperature unknown will accelerate
convergence since the solver will then only search for positive values. In this example, set the minimum value
for T to 300 (why?).


After initial guesses have been set, click OK and wait for the Data Browser table to appear. (Note: if a previous
dataset exists, the solver will ask you whether you want to save it or discard it.) If the message "Equation set
successfully solved" appears

in the Data Browser, then the values for q1, q2, q3, and T represent a valid
solution. In general, there may exist several valid solution sets for a system of nonlinear equations. The
particular solution found by the solver depends upon the initial guesse
s and the search algorithm. In this
example, only one valid solution set exists since the equations are based upon a physical problem:


T

322.1

q1

237.3

q2

150.7

q3

86.59


STEP 3
-

Postprocessing


Once a valid solution has been found, you may want to copy
it to the clipboard for pasting. Highlight the
solution by clicking the upper left rectangle in the table and then click Copy. Options are available for
transposing rows/columns and including variable names. It is suggested that you paste your solution to

the
bottom of your equation worksheet before printing a hardcopy. This is done by locating the cursor at the bottom
and then clicking Paste from the Edit menu. Also, be sure to type in appropriate units next to each solution
value. You can now print your
entire problem by clicking Print from the File menu. If you wish to have greater
control of how your page looks in terms of font style/size, organization, bold/italic/underline, etc., copy and
paste your entire worksheet to Notepad or Microsoft Word.


Perf
orming parametric studies and plotting results are possible using the
Explore
and
Graph
functions. These
features will be used in the next example.


2. Solving a Heat Transfer Problem: Insulated Steam Pipe Undergoing Convection and Radiation


The following

worksheet example shows how to set up a typical heat transfer problem. Note that through the
use of commenting, the problem can be presented in a structured format where the given information and
knowns are segregated from the equations and unknowns. The
solver recognizes any variable that is set to a
numerical value as a known. You may use any alphanumeric name for a variable, however, please try to be
conventional. There are the usual intrinsic functions such as pi, sqrt, exp, ln, sin, cos, tan, etc. Ref
er to the
Help Index for a complete listing of the intrinsic functions. Also, IHT does not keep track of dimensional units


you must make sure that consistent units are used.


//
ME 259 Class Example

//
Given:

Insulated steam pipe undergoing convection an
d radiation


ri = 0.05

// m

ro = 0.075

// m

L = 25


// m

Ti = 423.15

// K

To = 298.15

// K

hi = 100


// W/m
2
-
K

ho = 10


// W/m
2
-
K

eps = 0.8

sigma = 5.67E
-
08 // W/m
2
-
K
4


//
Find:


a) Total heat loss from pipe (q)

//


b) Outside surface temperature of insulation (Ts)

//

c) Effect of varying ro from 0.06 to 0.25 m

//

d) Effect of varying eps from 0.05 to 1.0



//
Assumptions:


i) steady
-
state conditions

//


i) one
-
dim
ensional radial conduction through insulation

//







iii) constant properties

//






iv) large surroundings, Tsur = To

//





v) pipe wall has negligible thermal resistance



//
Properties:

// 85% magnesia insulation at

360 K, from Table A.3,

p.835

kins = 0.055 // W/m
-
K


//
ANALYSIS:


//
Thermal resistances

Rtotal = Rconvi + Rins + Rconvo*Rrado/(Rconvo+Rrado)

Rconvi = 1/(hi*2*pi*ri*L)

Rins = ln(ro/ri)/(2*pi*kins*L)

Rconvo = 1/(ho*2*pi*ro*L)

Rrado = 1/(hr*2*pi*ro*L)

hr = eps*sigma*(Ts^2+To^2)
*(Ts+To)


//
Heat rate & temperature equations

q = (Ti
-
To)/Rtotal

q = (Ti
-
Ts)/(Rconvi+Rins)

Ts_C = Ts
-

273.15


The solution to this problem is given below
:



Rconvi


0.001273


K/W


Rconvo

0.008488


K/W


Rins


0.04693


K/W


Rrado


0.01654


K/W


Rtotal


0.0
5381


K/W


Ts


311.2


K


Ts_C


38.03


C


hr


5.133


W/m
2
-
K


q



2323



W


Now, try performing a parametric study where you vary the insulation thickness (i.e., the value of ro) over the
range of 0.06 to 0.25 m in steps of 0.01 m To do this, click on
Explor
e

and choose the Variable to Sweep as
ro and input the range parameters. Then click OK and wait for the Data Browser to appear. Here, you will see
the results corresponding to each value of ro. Suppose you are interested in the effect of ro on the heat los
s, q.
This is best viewed by plotting q vs. ro. Click on
Graph,
then

select ro as coordinate X and q as coordinate Y1.
Click OK and a plot should appear. You can add titles, change scales, change lines, change data markers, and

add a legend if you wish. Th
e graph can be printed as a full page or copied, sized, and pasted to your
worksheet. Next, try plotting Ts versus ro and then repeat for a parametric study with 0.05 < eps < 1.


There are many other features in IHT, including 1
st
-
order differential equati
ons, integrals, material property
functions, user
-
defined functions, and pre
-
written heat transfer correlations, special equations, and models (see
Help

menu). Not bad for a $15 program !!