Unsteady Heat Transfer in Semi-infinite Solids

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

S(t)

solid

liquidCoating with density , latent heat of fusion: hsf

Substrate, k,

From energy balance:

solidified mass during t) (energy input)

hdm

dt= q"A, where m = V = AS,

where S is solidified thickness

dS

dt= q"

sf

h m Q q A tsf ( "

ExampleAs 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 material is at the fusion temperature, Tf, and the substrate is maintained at a constant temperature Ti. Using the semi-infinite assumption,Derive an expression for the total time that is required to solidify the coating layer of thickness .

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

• Assume the molten layer stays at a constant temperature Tf throughout the process. The heat lost to the substrate is solely supplied by the release of the latent heat of fusion.

From an Energy Balance:

Example (cont.)

Recognizing that the previous situation corresponds to the case discussed in chapter 9-3 in the text as a semi-infinite transient heat transfer problem with a constant surface temperature boundary condition (NOTE: the case in the textbook corresponds to an external convection case. However, it can be modeled as constant surface temperature case by setting h=, therefore, Ts=T).

s

i

If the surface temperature is T and the initial temperature of

the bolck is T , the analytical solution of the problem can be found:

The temperature distribution and the heat transfer into the block a

2

s

i

0

ss

re:

T(x,t)-T, where erf( ) is the Gaussian error function.

T 2

2It is defined as erf(w)=

k(T )q "(t)=

s

w v

i

xerf

T t

e dv

T

t

Example (cont.)

f f

0 0

2

f2

From the previous equation

k(T ) k(T )=q"= , and

2k(T )( ) , therefore, t. Cooling time

4

ti i

sf

sf

sfi

f isf

dS T T dth dSdt t h t

hTt t t

k T Th

Use the following values to calculate: k=120 W/m.K, =410-5 m2/s, =3970 kg/m3, and hsf=3.577 106 J/kg, Tf=2318 K, Ti=300K, and =2 mm

Example (cont.)

( ))

.tT

ht ti

sf

2k(Tf 0 00304

0 0.2 0.4 0.6 0.8 10

0.001

0.002

0.003

0.004

( )t

t

• (t) t1/2

• Therefore, the layer solidifies very fast initially and then slows down as shown in the figure• Note: we neglected contact resistance between the coating and the substrate and assume temperature of the coating material stays the same even after it solidifies.

• It takes 0.43 seconds, to solidify a 2 mm thick coating !

Question: What assumptions, if any were unrealistic in our analysis ?

Example (cont.)

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

( , ),

2

( , ) 2318 (300 2318) 2318 2018 79.062

S

i S

T x t T xerf

T T t

x xT x t erf erf

t t

Question: What assumptions, if any were unrealistic in our analysis ?

Example (cont.)

For a fixed distance away from the surface, we can also examine the temperature variation as a function of time. E.g. 1 cm deep into the substrate the temperature should behave as:

T x t erfx

terf

t( . , ) .

. F

HGIKJ F

HGIKJ0 01 2318 2018 79 06 2318 2018

0 79

0 2 4 6 8 100

400

800

1200

1600

2000

x=1 cmx=2 cmx=3 cm

Time

Tem

pera

ture T1( )t

T2( )t

T3( )t

t

• At x=1 cm, the temperature rises almost instantaneously at a very fast rate. A short time later, the rate of temp. increase slows down significantly since the 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 erfx

terf x( , ) . . F

HGIKJ 1 2318 2018 79 06 2318 2018 79 06a f

0 0.01 0.02 0.03 0.04 0.050

1000

2000

3000

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

distance (m)

Tem

pera

ture

(K

)

T1( )x

T2( )x

T3( )x

x

• Heat penetrates into the substrate as shown for different time instants.• It takes more than 5 seconds 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.