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Dr. Primal Fernandoprimal@eng.fsu.eduPh: (850) 410-6323

Chapter 18 – Transient heat conduction

Thermal-Fluids I

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Transient heat conduction

• In general, The temperature of a body varies with time as well as position.

In rectangular co-ordinates this variation is expressed as T(x,y,z,t) x,y,z → variations in x,y,z directions t → variation with time

• The studies in this chapter is focused on Lumped system analysis Transient heat conduction in large plane walls, long cylinders and

spheres with spatial effects Transient heat conduction in semi-infinite solids Transient heat conduction in multi-dimensional systems

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BROAD OBJECTIVE: INVESTIGATE THE PROBLEM OF

HOW DO SPHERES COMING

OUT OF A OVEN COOL?

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Consider …

• An engineer, a psychologist, and a physicist were asked to make recommendations to improve the productivity of an under-producing dairy farm …

• Engineer: more technology

• Psychologist: improve environment

• Physicist …

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“Consider a spherical cow …”

T(t)

• Great engineers and physicists are able to appropriately simplify problems to extract the physics!

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Lumped system

• A lumped system is one in which the dependent variables of interest are a function of time alone. In general, this will mean solving a set of ordinary differential equations (ODEs)

• A distributed system is one in which all dependent variables are functions of time and one or more spatial variables. In this case, we will be solving partial differential equations (PDEs)

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Lumped system

• Consider a small hot copper ball coming out from an oven.– Temperature change with time.– Temperature does not change much with position at any given time.– Lumped system analysis are applicable to this system.

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Lumped system

• Consider a large roast in an oven.– Temperature distribution not even.– Temperature does change much with position at any given time.– Lumped system analysis are not applicable to this system.

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Consider a body of arbitrary shape of mass m, volume V, surface area As, density ρ, and specific heat Cp initially at a uniform temperature of Ti.

At time t=0, the body is placed into a medium at temperature T∞

Heat transfer take place between body and its environment

Temperature of the body change with the time and the temperature of the body at a given time T=T(t)

Heat transfer into the body at any given time T=T(t)

[ ])(tTThAQ s −= ∞

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)( TThAQ s −= ∞

Heat transfer into the body at temperature T

=

dtdt time during body the of energy the in increase The

period time a during body the intotransfer Heat

TmcdtTThA ps ∆=−∞ )(

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dTmcdtTThA ps =−∞ )(

Vm ρ=

dtVchA

TTdT

p

s

ρ−=

− ∞ )(

∫∫ −=− ∞

t

0p

stT

T

dtVchA

TTdT

iρ

)(

)(

t

0p

stT

Tt

VchATT

i ρ−=− ∞

)()ln(

tVchA

TTTtT

p

s

i ρ−=

−−

∞

∞)(lnt

VchA

i

p

s

eTTTtT ρ

−

∞

∞ =−−)(

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tVchA

i

p

s

eTTTtT ρ

−

∞

∞ =−−)( bt

i

eTTTtT −

∞

∞ =−−)(

wheres1units

VchAb

p

s →=ρ

Time constant

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Criteria for lumped system analysis

bt

i

eTTTtT −

∞

∞ =−−)(

s1units

VchAb

p

s →=ρ

Characteristic lengths

c AVL =

Biot number Bik

hLBi c=

body the of surface theat resistance Convectionbody the within resistance Conduction

==h1kLBi c

//

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body the of surface theat resistance Convectionbody the within resistance Conduction

==h1kLBi c

//

Small Bi number indicate low conduction resistance, and therefore small thermal gradient within the body

Lumped system is exact when Bi = 0

Generally accepted lumped system analysis when,

10Bi .≤If Bi < 0.1, there is a ± 5% error or less in estimating temperature throughout body as a single-valued function of time T(t)

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Temperature is uniform throughout sphere.

- Temperature gradients are small inside sphere.

- Resistance to conduction within solid much less than resistance to convection across fluid boundary layer.

Remember 1st Major Assumption

body the of surface theat resistance Convectionbody the within resistance Conduction

==h1kLBi c

//

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Rate of heat energy passing through sphereQ = - h As (Ts - T∞)

(W) = (W/[m2-Ko])(m2)(Ko)

Heat transfer coefficient isassumed not to be a function of ∆T.

Remember 2nd Major Assumption

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Problem: Steel balls 12 mm in diameter are annealed by heating to 1150 Kand then slowly cooling to 400 K in an air environment for which T∞=325 K and h=20 W/m2K. Assuming the properties of the steel to be k=40 W/mK, ρ=7800 kg/m3, and cp=600 J/kgK, estimate the time required for the cooling process.

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SolutionBiot number Bi k

hLBi c=

( )33

2

3

sc 10210

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3r

r4

r34

AVL −− ×=×==

==π

πCharacteristic length

( ) ( )100010

mKW

mKm

W

4010220

khLBi

23c .. <=

××==

−

Therefore, temperature of the steel balls remain approximately uniform: lumped system analysis applicable

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∞

∞

−−

=TtT

TThcL

t ipc

)(ln

ρ

bt

i

eTTTtT −

∞

∞ =−−)(

pcp

s

cLh

VchAb

ρρ==

∞

∞

−−

=TtT

TTb1t i

)(ln

( )

min.

ln)(

ln

70418s1122

KmW

kgKJm

mkg

3254003251150

206001027800

TtTTT

hcL

t2

33ipc

==

−−×××

=−−

=−

∞

∞ρ

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Bi number provide-Measure of temperature drop in solid relative to temperature difference between the surface and the fluid

body the of surface theat resistance Convectionbody the within resistance Conduction

==h1kLBi c

//

∞∞ −−

=

−

−

=TTTT

QTT

QTT

Bi2s

2s1s

2s

2s1s

,

,,

,

,,

Steady state system

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Transient heat conduction in large plane walls, long cylinders and spheres with spatial effects

In this section variation of temperature with time and position in one dimensional problems such as those associated with large plane wall, long cylinder and sphere.

A distributed system is one in which all dependent variables are functions of time and one or more spatial variables. In this case, we will be solving partial differential equations (PDEs)

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Transient heat conduction in large plane walls, long cylinders and spheres with spatial effects

0tatTT i =<∞

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large plane walls with spatial effects

Transient temperature distribution T(x,t) in a wall results in a partial differential equation, which can be solved using advanced mathematical techniques. The solution however, normally involves infinite series , which are inconvenient and time-consuming to evaluate. Therefore, there is a clear motivation to present the solution in tabular or graphical form. Solution involves so many parameters such as x, L, t, k, α, h, Ti and T∞. In order to reduce the number of parameters, it is defined dimensionless quantities

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Dimensionless parameters

Dimensionless temperature ∞

∞

−−

=TT

TtxTtxi

),(),(θ

Dimensionless distance from the center LxX =

Dimensionless heat transfer coefficient khLBi = (Biot number)

Dimensionless time 2Ltατ = (Fourier number)

Nondimensionalization enables us to present temperature data in terms of X, Bi and τ

The above defined dimensionless quantities can be used for cylinder or sphere by replacing the variable x by r and L by ro.

ro for cylinder and sphere (not V/A)

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One dimensional transient heat conduction problem: For above geometries, solutions involve in finite series, which are difficult to deal with.

Solutions using one dimensional approximation

20LxeATT

TtxTtx 11i

wall

21 .),/cos(),(),( >=

−−

= −

∞

∞ τλθ τλPlane wall:

20rrJeATT

TtrTtr 0101i

cyl

21 .),/(),(),( >=

−−

= −

∞

∞ τλθ τλCylinder:

20rr

rreATT

TtrTtr01

011

isph

21 .,

/)/sin(),(),( >=

−−

= −

∞

∞ τλλθ τλSphere:

A1 and λ1 are functions of Bi and their values are listed in Table 18-1

J0 is the Zeroth order Bessel function and values are listed in Table 18-2

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Temperature of the center of the plane wall, cylinder and sphere

τλθ21eA

TTTt0Tt0 1

iwall

−

∞

∞ =−−

=),(),(Plane wall:

τλθ21eA

TTTt0Tt0 1

icyl

−

∞

∞ =−−

=),(),(Cylinder:

τλθ21eA

TTTt0Tt0 1

isph

−

∞

∞ =−−

=),(),(

Sphere:

Once Bi number is known, these relations can be use to determine the temperature anywhere in the medium (interpolations may be required to determine intermediate values) .

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Heisler charts – M.P. Heisler 1947

• There are 3 charts associated with each geometry

– Chart 1: Determine the temperature of the of the center of the geometry (T0) at a given time t.

– Chart 2: Determine the temperature of another location (T) in terms of (T0) center temperature.

– Chart 3: Determine the total amount of heat transfer up to the time.

Note: these plots are valid to τ>0.2

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Heisler charts - Large plane walls (18-13)

(chart 1) Fourier number

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Heisler charts - Large plane walls (18-13)

(chart 2)

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Heisler charts - Large plane walls (18-13)

(chart 3)

Note: Qmax=mcp(T∞ - Ti)

Total mount of heat transfer during whole period

Q is amount of heat transfer at finite time period t

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Heisler charts – Long cylinder (18-14)

Heisler charts - Sphere (18-15)

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Transient heat conduction in semi-infinite solids

• Semi-infinite solid is an idealized body that has an single plane surface and extends to infinity in all other directions

– The earth– A thick wall

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Transient heat conduction in semi-infinite solids-Graphical representation

Nondimensionalized temperature

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Transient heat conduction in multidimensional systems• The presented charts can be used to determine the temperature

distribution and heat transfer in one dimensional heat conductionproblems associated with, large plane wall, a long cylinder, a sphere and a semi infinite medium.

• Using a superposition approach call product solution, these charts can also be used to construct solutions for two dimensional transient heat conduction problems encountered in geometries such as short cylinder, a long rectangular bar, or semi-infinite cylinder or plate and even three dimensional problems associated with geometries such as a rectangular prism or semi-infinite rectangular bar………… ⇒ provided that: all surfaces of the solid are subjected to convection to the same fluid at a T∞ with heat transfer coefficient h and no heat generation in the body.

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Product solution -short cylinder

−−

−−

−

−

∞

∞

∞

∞=

∞

∞

TTTtrT

TTTtxT

TTTtxrT

iiicylinder

initewallplane

cylindershort

),(),(),,(inf

Short cylinder

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Product solution –rectangular profile

−−

−−

−

−

∞

∞

∞

∞=

∞

∞

TTTtyT

TTTtxT

TTTtxrT

iiiwallplane

wallplane

bargularrec

),(),(),,(tan

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Look the examples in the book,

Friday.. work on problems, will give the home work, probably a quiz