Transient Heat Conduction

Lecture 01

Governing Heat Conduction Equation •

Assuming k = constant,

∇ 2 T + q& = 1 ∂ T k α ∂ t •

For For 1-D 1-D cond conduc ucti tion on (x) (x) and and no int inter erna nall heat heat g gen ener erat atio ion, n,

∂ 2T 1 ∂T = 2 α ∂t ∂ x •

Solution, T(x,t) , requires requires two two BCs and an initia initiall condition condition

Transient Conduction •

Many Many heat heat tr tran ansfe sferr pro probl blem ems s are are time time depe depend nden entt

•

Chang Cha nges es with in in oper operati ating ng con tions ns iin n a system sys tem a cause cau se tempe teuntil mperat ure steady variation time, as condi wellditio as location within solid, arature new state (thermal equilibrium) is obtained.

•

In this this chapte chapterr we will will devel develop op pro proced cedure ures s fo forr de deter termin mining ing the tim time e dependence of the temperature distribution Real Real prob problem lems s may may inclu include de finite finite and semi-i semi-infi nfinit nite e solid solids, s, or or compl complex ex geometries, as well as two and three dimensional conductio conduction n Solut Solution ion techn techniqu iques es involv involve e the the lum lumped ped capaci capacitan tance ce method method,, exac exactt and approximate solutions, and finite difference methods. We will focus on the Lumped Capacitance Capacitance Method, which can be used

• •

for solids within which temperature gradients are negligible (Sections 5.1-5.3)

Transient Conduction •

Tran Transi sien entt = Unst Unste eady ady (ti (time me-d -dep epen ende dent nt )

•

E –xaVery mpleshort s time s scale: cale: hot wire anemome anemometry try (< 1 ms) – Short time scale: quen quenching ching of metallic parts (se (seconds) conds) – Intermediate time scale: ba baking king cookies (minutes) – Long time scale – daily heating/cooling of atmos atmosphere phere (hours) – Very long time sca scale: le: seasonal heating/cooling of the earth’s earth’s surface (months)

•

Lumped Capacitance Method

Cons Consid ider er a soli solid d th that at is is init initia ialllly y at a uni unifo form rm tem tempe pera ratu ture re,, Ti , and at t=0 is quenched by immersion in a cool liquid, of lower temperature T ∞ The temperature of the solid willsolid-liquid decrease interface, for time t>0, due to convection heat transfer at the until it reaches T ∞

T t=0

T ( x, 0) = T i

x

•

•

Lumped Capacitance Method

If the the therm thermal al con conduc ductiv tivity ity of the the solid solid is is very very high high,, resi resista stance nce to conduction within within the solid solid will be small compared compared to resistance resistance to heat transfer between solid and surroundings. Temper Temperatu ature re gradie gradients nts within within the solid solid will will be be n negl egligi igible ble,, ii.e. .e. the temperature of the solid is spatially uniform at any instant. T

T ( x, 0) = T i

x

Lumped Capacitance Method & & Starting from an overall energy balance on the solid: − E out = E st

− hA (T − T ) = ρVc s

θ θi

=

∞

T − T ∞

hAs t = exp −

T i − T ∞

ρVc

dT dt

(1)

where

θ = T − T ∞ i − T ∞ θi = T

Let’s define a thermal time constant

1 ( ρVc ) = Rt C t τt = hAs

(2)

Rt is the resistance to convection heat transfer Ct is the lumped thermal capacitance of the solid

Transient Temperature Response

Based on eq. (1) the temperature difference between solid and fluid decays exponentially. exponentially.

Transient Temperature Response From eq. (1) the time required for the solid to reach a temperature temperatu re T is:

t = ρVc ln θi θ hAs

(3)

The total energy transfer, Q, occurring up to some time t is:

Q =

∫

t 0

q dt = hAS t θ dt =( ρ Vc )θ i [1 − exp(− t / τ t )]

∫

0

θ = T − T ∞ θi = T i − T ∞

(4)

Validity of Lumped Capacitance Method Surface energy balance:

Ts,1

qcond

qconv

q cond = q conv kA (T s .1 − T s ,2 ) = hA (T s ,2 − T ∞ ) L

Ts,2 T∞

Validity of Lumped Capacitance Method (Rearranging (Rearrangin g the equation in the previous slide)

T s ,1 − T s ,2 = ∆T solid ( due to conduction) = (L / kA) = R cond = hL ≡ Bi T s ,2 − T ∞ ∆T solid / liquid ( due to convection) (1 / hA) R conv k

? What What is the the re rela lati tive ve ma magn gniitu tude de of ∆T solid versus ∆T solid/liquid for the lumped capacitance method to be valid?

Biot Number •

The The Biot Biot numb number er is is a dime dimens nsio ionl nles ess s para parame mete terr that that ind indic icat ates es the the rel relat ativ ive e importance of conduction and convection convection heat transfer processes:

Bi

=

hL k

= L kA 1 hA •

= R

R

t , cond t , conv

Practi Practical cal impl implica icatio tions ns of of Bi << 1: 1: objec objects ts may may hea heat/c t/cool ool in an an is isoth other ermal mal manner if – they are small and meta metallic llic – they are cooled/heated by natural convection in a gas (e.g., air)

Biot Bi ot an and dF Fou ouri rier er N Num umbe bers rs

The lumped capacitance method is valid when

Bi =

hLc

k

where the characteristic length: Lc=V/As=Volume of solid/surface area

< 0.1

We can also define a “dimensionless time”, the Fourier number:

αt Fo = 2

where

Lc Eq. (1) becomes:

θ = T − T ∞ = exp[− Bi ⋅ Fo] θi T i − T ∞

k α = ρ c

True or False?

•

A hot hot s soli olid dw will ill c cool ool dow down n fa faste sterr w when hen it iis sc cool ooled ed by by forc forced ed c conv onvect ection ion in water rather than in air.

•

For For th the e sa same me s sol olid id,, th the e lu lump mped ed c cap apac acit itan ance ce m met etho hod d is lilike kely ly mor more e ap appl plic icab able le when it is being cooled by forced convection in air than in water.

•

The lum lumped ped cap capaci acitan tance ce m meth ethod od is llike ikely ly mor more ea appl pplica icable ble for co cooli oling ng of of a hot solid made of aluminum (k=237 W/m.K) than copper (k=400 W/m.K)

•

The tra trans nsien ientt re resp spons onse e is ac accel celera erated ted b by y a dec decrea rease se in in th the e sp spec ecific ific heat heat of tthe he solid.

•

The The p phy hysi sica call m mea eanin ning g of th the eB Bio iott nu numb mber er iis s th that at iitt rrep epre rese sents nts th the e rrel elat ativ ive e magnitude of resistance due to conduction and resistance due to convection.

General LCM •

LCM can be applie applied d to to syst system ems s with with convec convectio tion, n, ra radia diatio tion, n, and heat flux boundary conditions and internal heat generation – – – – –

Forced conv convection, ection, h = constant Natural convection convection,, h h = = C(T-T ∞ )1/4 Radiation (eqn. 5.18) Convection and rad radiation iation (requires numerical integr integration) ation) Forced conv convection ection with constant sur surface face heat flux or internal hea heatt generation (eqn. 5.25)

(Note that the Biot number must must be redefined redefined when when effects other than convection are included)

Spatial Effects (Bi > 0.1) •

LCM LCM not not vali valid d sinc since e temp temper erat atur ure e grad gradie ient nt wit withi hin n soli solid d is significant

•

Need Need to solv solve e hea heatt con condu duct ctio ion n equ equat atio ion n wit with h app applilied ed boundary conditions

•

1-D 1-D tran transi sien entt cond conduc ucti tion on “fam “familily” y” of solu soluti tion ons: s: – Uniform, symmetric convection applied to plane w wall, all, long cylinder, sphere (sections 5.4-5.6) – Superposition Semi-infinite solid with various for B BCs Cs (section 5.7) conduction – of 1-D solutions multidimensional conduction (section 5.8)

Example (Problem 5.7 Textbook)

The heat transfer coefficient for air flowing over a sphere is to be determined by observing the temperature-time history of a sphere fabricated from pure copper.it The sphere, which is 12.7 mmhaving in diameter, is at 66°C 66° C before is inserted inser ted into an air stream a temperature of 27°C. 27°C. A thermocouple on the outer o uter surf surface ace of the sphere indicates indi cates 55°C, 55°C, 69 s after a fter the sphere is inserted inserte d in the air a ir

stream. Calculate the heat transfer coefficient, assuming that the sphere behaves as as a spacewise spacewise isothermal object. Is Is your assumption reasonable?

17

What if? •

What What happ happen ens s to to the the rate rate of cool coolin ing g ifif h in incr crea ease ses? s?

•

What What hap happen pens s to the ra rate te of coolin cooling g ifif th the ed diam iamete eterr of of the the spher sphere e increases?

•

What What happ happen ens s ifif we we have have a huge huge sphe sphere re? ?

General Lumped Capacitance Analysis In the general case we may have convection, radiation, internal energy Tsur

generation and an applied heat flux. The energy balance becomes: & , E & E gen st

q

”

s

q

”

" " q s " As ,h + E & g − (q conv + q rad )As ( c ,r ) = ρ Vc dT dt

rad

T∞, h ”

Numerical solutions are generally required Simplified solutions exist for no imposed heat flux or generation.

As,h

q conv As(c,r)

Example 5.2 • •

Calcu Calculat lation ion of tthe he ste steady ady sta state te ttemp empera eratur ture eo off th the e therm thermoco ocoupl uple e junct junction ion.. How muc much h tim time e is n need eeded ed ffor or th the e te tempe mperat rature ure tto o in incre crease ase fro from m 25 25° °C to w withi ithin n 1°C from its st steady eady st state ate val value? ue?

Example 5.2

Example (5.33)

Microwave ovens operate by rapidly aligning and reversing water molecules within the food, resulting in volumetric energy generation. (a) Cons Consider ider a froze frozen n 1-kg sphe spherical rical pie piece ce of ground be beef ef at an initial temp temperatu erature re of Ti=-20°C. Ti=-20°C. Determi Determine ne how long it will take the be beef ef to reach a uniform unif orm temperature temperatu re of T=0° T=0°C, C, with all the water in th the e form of ice. Assume tthat hat 3% of the oven power (P=1kW total) is absorbed by the food. (b) After a allll the ice is co conver nverted ted to liqu liquid, id, deter determine mine how lo long ng it will tak take e to heat the beef to Tf=80°C, if 95% of the oven power iis s absorbe absorbed. d.

Other transient problems • • • •

When When the the lum lumpe ped d cap capac acit itan ance ce anal analys ysis is is n not ot va valilid, d, we must must so solv lve e the the partial differential equations analytically or numerically Exac Exactt and and appr approx oxim imat ate e sol solut utio ions ns may may be be use used d Tabula Tabulated ted values values of coeff coeffici icient ents s used used in the the s solu olutio tions ns of the these se equations are available Transi Transient ent temper temperatu ature re distri distribut bution ions s for for com commo monly nly encou encounte ntered red problems involving semi-infinite solids can be found in the literature

Summary •

The lumpe lumped d capa capacit citanc ance e anal analysi ysis s can can be be used used when when the the tem tempe perat rature ure of the solid is spatially uniform at any instant during a transient process

• •

Temperature gradients within the solid are negligible Resistance to conduction within the solid is small compared to the resistance to heat transfer between the solid and the surroundings

The The Bio Biott numb number er must must be be les less s 0.1 0.1 for for the the lump lumped ed capa capacit citan ance ce ana analy lysi sis s to be valid. Tran Transi sien entt cond conduc ucti tion on pro probl blem ems s are are cha chara ract cter eriz ized ed by by tthe he Bio Biott and and the the Fourier numbers.

Transient Heat Conduction

Lecture 02

TRANSIENT (UNSTEADY) HEAT TRANSFER Review Many heat transfer problems require the understanding of the complete time history of the temperature variation. For example, in metallurgy, the heat treatment process can be controlled to produce specified characteristics of the processed materials. Annealing (slow cooling) can soften metals and improve ductility. On the other hand, quenching (rapid pid cooling) can harden the boundary and increase strength.(ra In order to characterize this transient behavior, the full unsteady equation should be modeled:

1 ∂T ∂T 2 ρ c = k ∇ T , or = ∇ 2T α ∂t ∂t k where α = is the thermal diffusivity ρ c

TRANSIENT (UNSTEADY) HEAT TRANSFER “A heated/cooled body at T i is suddenly exposed to fluid at T ∞ ∞ with a known heat transfer coefficient coefficient . Either Either evaluate the temperature temperature at a given time, or or find time for a given temperature.” temperature.”

Q: How good an approximation approximation would it be to say the bar is isothermal? A: Depends on the relative importance of the thermal conductivity in the thermal circuit compared to the convective heat transfer coefficient.

TRANSIENT (UNSTEADY) HEAT TRANSFER Biot iot Numb Number er ( Bi Bi) Defined to describe the relative resistance in a thermal circuit

hL c

Lc / kA

Internal conduction resistance within solid

Bi = k = 1 / hA = External convection resistance at body surface L is a characteristic length of the body c Bi→0: No co cond nduc ucti tion on re resi sist stan ance ce at al all. l. The The bod body y iiss iiso soth theermal rmal.. Small Bi Bi:: Conduction resistance is less important. The body may still be approximated as isothermal (purple (purple temp. plot in figure) capac capacitance itance analysis can be performed. Bi:: Lumped Conduction re resistance sistance is ssignificant. ignificant. The body cannot be treated treated as Large Bi isothermal (blue (blue temp. plot in figure).

TRANSIENT (UNSTEADY) HEAT TRANSFER Transient heat transfer with with no internal resistance: Lumped Parameter Analysis Valid for Bi < 0.1

Solid

Total Resistance= Rexternal + Rinternal

dT GE:

hA

BC:

∞ dt = − mc p (T − T )

Solution:

let

Θ

d Θ dt

T (t = 0 ) = T i

= T − T , therefore

= −

∞

hA

mc

Θ p

Lumped Parameter Analysis Θ i = T i − T ∞ ln Θ = − hA t Θi mc p hA

Θ Θi

=e

− mc p t

mc p

T − T ∞ = e − t T i − T ∞

hA

- To determine determine the temperature temperature at a given time, or - To determin determinee the time time requi required red for the the temperature to reach a specified value.

Note: Temperature Temperature is a fun function ction of tim timee only and not of sp space! ace!

Lumped Parameter Analysis

T = T − T ∞ = exp( − hA t ) T 0 − T ∞ ρ cV hA

α hLc k 1 1 t = t = Bi 2 t

ρ cV

c c k c L L ρ Thermal diffusivity: k

α ≡ ρ c

Lc -1

(m² s )

Lumped Parameter Analysis Define Fourier number [Fo] (a dimensionless time)

Fo ≡

Lc

2

and d Bio Biott numb number er t an

Bi ≡ hLC k

The temperature variation can be expressed as

T ~ exp(-Bi*Fo) where L c is characteri stic length scale : realte to the size of the solid invloved in the problem r L c = o (half - radius) when the solid is a cylinder. 2 r L c = o (one - third radius) when the solid is sphere 3 L c = L (half thickness) when the solid is aplane wal l with a 2L thickness

Spatial Effects and Analytical Solutions The Plane Wall: Wall: Solution to to the He Heat at Equation for a Plan Planee Wall with Symmetrical Symmetric al Convection Conditions

1 ∂T ∂2T

⋅

=

2

a ∂τ ∂ x T ( x, 0) = T i

∂T ∂ x x=0 = 0 ∂T − k ∂ x x= L = h[T ( L , t ) − T ∞ ]

Spatial Effects and Analytical Solutions The Plane Wall:

Note: Once spatial variability of temperature is included, there are seven different independent variables variables that need to resolved.

How does one simplify the functional dependence on 7 variables? The answer is to use Non-dimensional Parameters. We fir first st nee need d tto o understand theprocess, physicsand behind thethem phenomenon, identifynonparameters governing the group into meaningful dimensional numbers.

Spatial Effects and Analytical Solutions Dimensionless temperature difference: difference: *

x

θ * =

θ T − T ∞ = θ i T i − T ∞

Dimensionless coordinate: x = L * α t t = 2 = Fo Dimensionless time: L c

Bi = hL k solid The solution of the temperature distribution will now be a function of the other non-dimensionall quantities non-dimensiona * * The Bio Biott Number Number::

Exact Solution:

θ = f ( x , Fo, Bi ) ∞ * θ = ∑ C n exp (− ζ n 2 Fo )cos(ζ n x * ) 4 sin ζ

C n = 2ζ + sin(n2ζ ) n n

n =1

ζ n tan ζ n = Bi

The roots (eigenvalues) of the equation can be obtained from tables given in standard textbooks.

Spatial Effects and Analytical Solutions The One-Term Approximation Fo > 0.2 Variation of mid-plane temperature with time Fo

( x * = 0)

θ 0* = T − T ∞ ≈ C 1 exp(− ζ 12 Fo ) T i − T ∞ From tables given in standard textbooks, one can obtain

C 1

and

ζ 1

as a function of Bi. Variation of temperature with location ( x * ) *

* 0

and time ( Fo ):

*

θ = θ = cos ζ 1 x Change in thermal energy storage with time:

∆ E st = −Q

1 − sin ζ 1 θ 0 Q = Q0

*

ζ 1 Q0 = ρ cV (T i − T ∞ )

Spatial Effects and Analytical Solutions Graphical Representation of the One-Term Approximation:

Thee Heisle Th Heislerr Chart Charts: s: Midpl Mi dplane ane Temper Temperatu ature re

Spatial Effects and Analytical Solutions Temperature Distribution Change in Thermal Energy Storage

Assumptions in u Assumptions using sing Heisler Heisler charts: charts: • Co Cons nsta tan nt T i and thermal properties over the body • Consta Constant nt boundary boundary fluid fluid temperat temperature, ure, T ∞ • Simp Simple le geometry: geometry: slab, cylinder cylinder or sphere sphere

Radial Systems Long Rods or Spheres Heated or Cooled by Convection

Similar Heisler Heisler charts are available for radial systems in standard textbooks.

Important tips: Pay attention to the length scale used in those thos e charts charts,, and calcul calculate ate your Bio Biott numbe numberr accordingly. accordingly.

Unsteady Heat Transfer in Semiinfinite Solids

Solidification process of the coating layer during a thermal spray Solidification operation is an unsteady heat transfer problem. As we discuss earlier, thermal spray process deposits thin layer of coating materials on surface for protection and thermal resistant purposes, as shown. The heated, heated, molten materials will attach attach to the substrate substrate and cool down rapidly. The cooling process is important to prevent the accumulation of residual thermal stresses in the coating layer.

Unsteady Heat Transfer in Semi-infinite Solids (cont.)

liquid Coating with density ρ, latent heat of fusion: hsf

S(t) δ solid

Substrate, k, α

Example As described in the previous slide, the cooling process can now be modeled as heat loss through a semi-infinite solid. (Since the substrate is significantly thicker than the coating layer) The molten molten material material is at the fusion temperature temperature T f and the substrate is maintained at a constant temperature T i. Derive an expression for the total time that is required to solidify the coating layer of thickness d.

Example (cont.)

Assume the molten layer stays at a constant temperature Tf

throughout the process. The heat loss to the substrate is solely supplied by the release of the latent heat of fusion. From energy balance: hsf ∆m (solidified mass during ∆t) = ∆Q = q" A∆t (ene rgy input) dm h sf = q" A, where m = ρV = ρ AS, dt

where S is solidified thickness dS ρ = q" dt

Heat transfer from the molten material to the substrate (q = q ”A) ”A)

Example (cont.) Identify that the previous situation corresponds to the case of a semi-infinite transient heat transfer problem with a constant surface temperature boundary condition. This boundary condition can be modeled as a special case of convection boundary condition case by setting h = ∞, therefore, T s=T ∞).

Example (cont.) If the surface temperature temperature is Ts and the in initial itial tem temperature perature o off the bolck bolck is Ti , the analytical analytical solution o off the problem can be found: found: re: The temperature distribution and the heat transfer into the block are: erf f x T(x,t)-Ts = er , wh where ere erf( ) is the Gaussian error function. Ti − T s 2 α t

It is defined as erf(w)= qs"(t)=

k(Ts − T i )

πα t

2

∫

w

0

π

2

e−v dv

Example (cont.) From the previous equation

ρ hsf

dS dt

δ (t ) =

=q"=

k(Tf − Ti )

πα t

2k(Tf − T i )

ρ hsf πα

δ

∫

, and dS = 0

k(Tf − Ti )

ρ hsf πα

t

dt

∫ 0

t

πα δρ hsf 2 t , therefore, δ ∝ t . Cooling time t = 2 4k T f − T i

Use the following values to calculate: k=120 W/m.K, α=4×10-5 m2 /s, ρ=3970 kg/m3, and hsf =3.577 ×106 J/kg, Tf =2318 K, Ti=300K, and δ=2 mm

Example (cont.)

δ (t ) =

f − T i ) 2k(T

ρhsf πα

t = 0.00304 t δ(t) ∝ t1/2 Therefore, the layer solidifies very fast initially and then slows

0.004

0.003

δ( t )0.002 0.001 0

0

0.2

0.4

0.6

0.8

1

down as shown in the figure Note: we neglect contact resistance between the coating and the substrate and assume temperature of the coating material stays the same even after it solidifies.

t

To solidify 2 mm thickness, it takes 0.43 seconds.

Example (cont.)

What will be the substrate temperature as it varies in time? The temperature distribution is:

T ( x, t ) − TS Ti − T S

x , 2 α t

= erf

T ( x, t ) = 2318 + (300 − 2318)erf x = 2318 − 2018erf 2 α t

7 9.06 x t

Example (cont.) For a fixed distance away from the surface, we can examine the variation of the temperature as a function of time. Example, 1 cm deep into the substrate the temperature should behave as:

T ( x = 0.01, t ) = 2318 − 2018erf 7 9 .06 x = 2318 − 2018erf 0.79 t t

Example (cont.) 2000

At x=1 cm, the temperature rises almost instantaneously at a

1600

e r u t a r e p m e

T1 ( t )

T

T3 ( t )

1200 T2 ( t )

very fast fast rate. rate. A short time time later,

800

the rate of temp. increase slows

400 0

down significantly since the 0

2

4

6 t Time

x=1 cm x=2 cm x=3 cm

8

10

energy has to distribute to a very large mass.

At deeper depth (x=2 & 3

cm), the temperature will not respond to the surface condition until much later.

Example (cont.)

We can also examine the spatial temperature distribution at any given time, say at t=1 second. T ( x , t = 1) = 2318 − 2018erf 79.06 xt = 2318 − 2018erf 79.06 x 3000

) K T1 ( x ) 2000 ( e r u t a r T2 ( x ) e p m e T3 ( x ) 1000 T

0

Heat

penetrates

into

the

substrate as shown for different time instants. It takes more than 5 seconds

0

0.01

0.02

0.03

x distance (m)

t=1 s. t=5 s. t=10 s.

0.04 0.05

for the energy to transfer to a depth of 5 cm into the substrate The slopes of the temperature profiles indicate the amount of conduction heat transfer at that instant.

Unsteady Heat Transfer in Semi-infinite Solids (cont.) The governing equation (GE):

∂ 2T 1 ∂T = 2 ∂t ∂ x Three cases of boundary conditions (BC):

Unsteady Heat Transfer in Semi-infinite Solids (cont.) Case 1: Constant surface temperature

Case 2: Constant surface heat flux

Case 3: Surface convection

Numerical Methods for Unsteady Heat

Transfer

Unsteady heat transfer equation, no generation, constant k, two-

dimensional in Cartesian coordinate:

∂ 2T ∂ 2T = 2 + 2 α ∂t ∂x ∂y 1 ∂T

We have learned how to discretize the Laplacian operator into system of finite difference equations using nodal network. For the unsteady problem, the temperature variation with time needs to be discretized too. To be consistent with the notation from the book, we choose to analyze the time variation in small time increment ∆t, such that the real time t=p∆t. The time differentiation can be approximated as:

∂T ≈ TmP,n+1 − T mP,n , while m & n correspond to nodal location location ∂t m,n ∆t such su ch that that x=m x=m∆x, an and y= y=n∆y as int ntro rod duced uced earl earlie ier. r.

Finite Difference Equations m,n+1 m-1,n

m,n

m+1, n

m,n-1

From difference the nodal network in finite form: to the left, the heat equation can be written

Finite Difference Equations (cont.) T P +1 − T P m ,n

1

α

∆t

m ,n

TP

=

m +1,n

+TP

− 2T P

m −1,n 2

( ∆x )

m ,n

TP m ,n +1

+

+T P

− 2T P

m ,n −1 2

( ∆y )

m ,n

α ∆t

Assum Ass umee ∆x=∆y and the discret discretized ized Fouri Fourier er number number Fo= Fo= +

(

)

2

( ∆x )

TmP,n1 = Fo TmP+1,n + TmP−1,n + TmP,n +1 + TmP,n −1 + (1 − 4 Fo )TmP,n

This is the explicit, finite difference equation for a 2-D, unsteady heat transfer equation. e quation. The temperature te mperature at time p+1 is explicitly expressed as a function of neighboring neighboring temperatures at an earlier time p

Nodal Equations

Some common nodal configurations are listed in a table for your reference. On the third column of the table, there there is a stability stability criterion for each nodal configuration. configuration. This criterion criterion has to be satisfied for for the finite difference difference solution to be stable. stable. Otherwise, the solution may diverge and never reach the final solution.

Nodal Equations (cont.) For example, Fo≤1/4. That is, α∆t/(∆x)2 ≤1/4 and time increment increment has to be small ∆t≤(1/4α)(∆x)2. Therefore, the time enough in order to maintain stability of the solution.

This criterion can also be interpreted as that we should require the coefficient for TPm,n in the finite difference equation to be greater than or equal to zero.

Question: Why this can be a problem? Can we just make time increment as small as possible to avoid it?

Finite Difference Solution Question: How do we solve the finite difference equation?

First, by specifying initial conditions conditions for all points in the nodal

network. That is to specify values values for aall ll temperature temperature at time level level p=0.

Important: check stability criterion for each point.

From

the explicit equation, we can determine the nodal

temperatures at the next time time level level p+1=0+1=1. The following following transient response can then be determined at time steps; p+2, p+3, and so on.

Example

Example: A flat plate at an initial temperature of 100 C. is suddenly immersed into a cold temperature bath of 0˚C. Use the unsteady finite difference equation to determine the transient response of the temperature of the plate. L(thickness)=0.02 m, k=10 W/m.K, α=10×10-6 m2 /s, ˚

2

x

1 2

3

h=1000 W/m .K, Ti=100°C, T2∞=0°C, ∆x=0.01 m Bi=(h∆x)/k=1, Fo=(α∆t)/(∆x) =0.1 There are three nodal points: 1 interior and two exterior points:inFor node 2, it satisfies the case 1 configuration table.

P +1

T2

= Fo(T1P + T3P + T2P + T2P ) + (1 − 4 Fo)T2P = Fo(T1P + T3P ) + (1 − 2 Fo)T2P

= 0.1(T1P + T3P ) + 0.8T 2P Stability criterion: 1-2Fo ≥ 0 or Fo=0.1 ≤

1 2

, it is satisfied

Example For nodes 1 & 3, they are consistent with the case 3 in table. Node 1: T1P +1 = Fo( 2T2P + T1P + T1P + 2 BiT∞ ) + (1 − 4 Fo − 2 BiFo)T1P P 2

P 1

P 2

P 1

) + (1 − 2 Fo − 2 BiFo)T = 0.2T + 0.6T = Fo( 2T + 2 P +1 Node 3: T3 = 0.2T2P + 0.6T 3P BiT∞

Stability criterion: (1-2Fo-2BiFo) ≥ 0,

1

≥ Fo(1 + Bi ) = 0.2 and it is satisfied

System of equations

2 Use initial condition, T10 = T20 = T30 = 100,

T1P +1 = 0.2T2P + 0.6T 1P

T1 = 0.2T2 + 0.6T 1 = 80

P +1 2

P 1

P 3

P 2

T = 0.1(T + T ) + 0.8T P +1 P P T3 = 0.2T2 + 0.6T 3

1

0

1 2

0 1

0

0 3

0 2

T = 0.1(T + T ) + 0.8T = 100 1 0 0 T3 = 0.2T2 + 0.6T 3 = 80

Marching in time, T11 = T31 = 80, T21 = 100 T12 = 0.2T21 + 0.6T 11 = 0.2(100) + 0.6(80) = 68 2

1

1

1

T2 = 0.1(T1 + T3 ) + 0.8T2 = 0.1(80 + 80) + 0.8(100) = 96