Objectives Of Conduction Analysis

Objectives of conduction analysis To determine the temperature field, T(x,y,z,t), in a body (i.e. how temperature varies with position within the body) ‰T(x,y,z,t) depends on: T(x,y,z) - boundary conditions - initial condition - material properties (k, cp, ρ …) - geometry of the body (shape, size) ‰Why we need T(x,y,z,t) ? - to compute heat flux at any location (using Fourier’s eqn.) - compute thermal stresses, expansion, deflection due to temp. etc. - design insulation thickness - chip temperature calculation - heat treatment of metals

Unidirectional heat conduction (1D)

Area = A

0

Solid bar, insulated on all long sides (1D heat conduction) qx

q&

x

x+Δx

x

A

= Internal heat generation per unit vol. (W/m3)

qx+Δx

Unidirectional heat conduction (1D) First Law (energy balance)

( E& in − E& out ) + E& gen = E& st q x − q x + Δx

E = ( ρ AΔx)u ∂E ∂t

= ρ AΔx

∂u ∂t

q

= ρAΔxc

∂T ∂t

q

x

∂E + A ( Δ x ) q& = ∂t ∂T

= − kA

x + Δ x

= q

∂x x

+

∂q ∂x

x

Δ x

Unidirectional heat conduction (1D)(contd…) ∂T ∂T ∂ ⎛ ∂T − kA + kA + A ⎜k ∂x ∂x ∂x ⎝ ∂x ∂ ⎛ ∂T ⎞ ∂T ⎜k ⎟ + q& = ρ c ∂x ⎝ ∂x ⎠ ∂t Longitudinal conduction

Internal heat generation

If k is a constant

∂T ⎞ ⎟ Δx + AΔxq& = ρ AcΔx ∂t ⎠

Thermal inertia

∂ 2T q& ρ c + = 2 ∂x k k

∂T 1 ∂T = ∂t α ∂t

Unidirectional heat conduction (1D)(contd…) ‰ For T to rise, LHS must be positive (heat input is positive) ‰ For a fixed heat input, T rises faster for higher α ‰ In this special case, heat flow is 1D. If sides were not insulated, heat flow could be 2D, 3D.

Boundary and Initial conditions: ‰ The objective of deriving the heat diffusion equation is to determine the temperature distribution within the conducting body. ‰ We have set up a differential equation, with T as the dependent variable. The solution will give us T(x,y,z). Solution depends on boundary conditions (BC) and initial conditions (IC).

Boundary and Initial conditions (contd…) How many BC’s and IC’s ? - Heat equation is second order in spatial coordinate. Hence, 2 BC’s needed for each coordinate. * 1D problem: 2 BC in x-direction * 2D problem: 2 BC in x-direction, 2 in y-direction * 3D problem: 2 in x-dir., 2 in y-dir., and 2 in z-dir. - Heat equation is first order in time. Hence one IC needed

1- Dimensional Heat Conduction The Plane Wall :

Hot fluid

…. . . ... ... .. .. .. .. . … k Ts,1 .. . .............. .. . ...... .. .. .. ....... ...... .. .. .. .. . . . ..... . . x=0 d dx

dT ⎛ ⎜ k dx ⎝

Ts,2

Cold fluid T∞,2

x=L ⎞ ⎟ = 0 ⎠

Const. K; solution is: dT kA T s ,1 − T s , 2 (T s ,1 − T s , 2 ) = q x = − kA = dx L L / kA

Thermal resistance (electrical analogy) OHM’s LAW :Flow of Electricity V=IR elect

Voltage Drop = Current flow×Resistance

Thermal Analogy to Ohm’s Law :

Δ T = qR therm Temp Drop=Heat Flow×Resistance

1 D Heat Conduction through a Plane Wall T∞,1

Hot fluid

…. . . ... ... .. .. ..... … k Ts,1 .. . .............. .. . ...... .. .. .. ....... ...... .. .. .. .. . . . ..... . . x=0

T∞,1

Ts,1

qx

R

1

t

Ts,2

L = + h1 A kA

T∞,2

x=L T∞,2

1 h2 A

L k A

1 h1 A

∑

Ts,2

Cold fluid

+

1 h2 A

(Thermal Resistance )

Resistance expressions THERMAL RESISTANCES • Conduction • Convection

Rcond = Δx/kA Rconv = (hA) • Fins

-1

Rfin = (hηΑ)−1 • Radiation(aprox) 1.5 -1 Rrad = [4AσF(T1T2) ]

Composite Walls : T∞,1 h1

A

B

C

KA

KB

KC

h2

T∞,2 T∞,1 qx

q

x

=

1 h1 A T∞

,1

∑

− T∞ R

,2

=

t

where, U =

LA

LB

LA kAA

LB kB A

h1 A 1 Rtot A

LC kC A

+

LA kA

1 h2 A

− T ∞ ,2 = UA Δ T LC LB 1 + + + kB kC h2 A

T∞ 1

T∞,2

LC

,1

= Overall heat transfer coefficient

Overall Heat transfer Coefficient U

=

1 R

total

A

=

1 h 1

1 L + Σ k

1 + h 2

Contact Resistance : TA TB

A

B R

t, c

=

Δ T q x

ΔT

=

U

L 1 + h1 k

A A

1 LC LB 1 + + + k B kC h2

Series-Parallel :

A T1

B KB

KA

C Kc

AB+AC=AA=AD

D KD

T2

LB=LC

Series-Parallel (contd…)

T1

LA kA A

LB kB A

LD kD A

LC kC A

T2

Assumptions : (1) Face between B and C is insulated. (2) Uniform temperature at any face normal to X.

Example: Consider a composite plane wall as shown: kI = 20 W/mk

qx

AI = 1 m2, L = 1m T1 = 0°C

Tf = 100°C kII = 10 W/mk

h = 1000 W/ m2 k

AII = 1 m2, L = 1m

Develop an approximate solution for th rate of heat transfer through the wall.

1 D Conduction(Radial conduction in a composite cylinder) h1

r1

T∞,1

r2

h2

T∞,2 r k 3 2

k1

qr = T∞,1

T∞,2

1 ( h 1 )( 2 π r1 L )

ln

1 ( h 2 )( 2π r2 L ) r1 r2

2 π Lk

ln 1

r2 r3

2 π Lk

2

T ∞ , 2 − T ∞ ,1

∑R

t

Critical Insulation Thickness : T∞ h

Insulation Thickness : r o-r i ri

Ti

Objective :

r0

R tot =

ln(

r0 ri

)

1 + 2 π kL ( 2 π r0 L ) h

decrease q , increases R

tot

Vary r0 ; as r0 increases ,first term increases, second term decreases.

Critical Insulation Thickness (contd…) Maximum – Minimum problem Set

dR tot = 0 dr 0 1 1 − 2 π kr 0 L 2 π hLr

r0 =

k h

Max or Min. ?

2

Take :

= 0 0

d 2 R tot = 0 2 dr 0

at

r0 =

k h

d 2 R tot −1 1 = + 2 π kr 2 0 L dr 2 0 π r 2 0 hL h2 = 2 π Lk

3

0

r0 =

k h

Critical Insulation Thickness (contd…) Minimum q at r0 =(k/h)=r c r (critical radius)

R tot

good for electrical cables

good for steam pipes etc. R c r=k/h

r0

1D 1D Conduction Conduction in in Sphere Sphere r2 r1 T∞,2

k Inside Solid:

Ts,2

Ts,1 T∞,1

1 d ⎛ ⎜ kr 2 r dr ⎝

2

dT ⎞ ⎟ = 0 dr ⎠

{T s ,1 − T s , 2 }⎡⎢⎣⎢ 11−−((rr // rr )) ⎤⎥⎦⎥ dT 4 π k (T s ,1 − T s , 2 ) = (1 / r1 − 1 / r 2 ) dr

→ T ( r ) = T s ,1

−

1

1

→ q r = − kA → R t , cond

1 / r1 − 1 / r 2 = 4π k

2

Conduction with Thermal Energy Generation E& q& = = Energy generation per unit volume V Applications: * current carrying conductors

* chemically reacting systems * nuclear reactors

Conduction with Thermal Energy Generation The Plane Wall : k Ts,1

q&

T∞,1

Ts,2

Assumptions: T∞,2

Hot fluid

Cold fluid x= -L

x=0

x=+L

1D, steady state, constant k, uniform q&

Conduction With Thermal Energy Generation (contd…) 2

d T dx

2

+

q&

=0

k

Boundary

x = −L,

cond . :

x = +L, Solution :

T = −

q& 2k

x

2

T = Ts , 1 T = Ts , 2

+C x +C 1 2

Conduction with Thermal Energy Generation (cont..) Use boundary conditions to find C1 and C2 2 ⎛ x2 ⎞ Ts , 2 −Ts ,1 x Ts , 2 +Ts ,1 q L & ⎜⎜1 − 2 ⎟⎟ + + Final solution : T = 2k ⎝ L ⎠ 2 L 2 No more linear

Heat flux :

dT ′ ′ = − qx k dx

Derive the expression and show that it is no more independent of x

Hence thermal resistance concept is not correct to use when there is internal heat generation

Cylinder with heat source T∞ h

Assumptions: 1D, steady state, constant k, uniform q&

ro r

Start with 1D heat equation in cylindrical co-ordinates:

Ts

q&

1 d ⎛ dT ⎜r r dr ⎝ dr

⎞ q& ⎟ + =0 ⎠ k

Cylinder With Heat Source Boundary cond. : r = r0 ,

T = Ts

dT =0 r = 0, dr q& 2 ⎛⎜ r2 ⎞⎟ Solution : T (r ) = r0 ⎜1 − 2 ⎟ +Ts 4k ⎝ r0 ⎠

Ts may not be known. Instead, T∝ and h may be specified. Exercise: Eliminate Ts, using T∝ and h.

Cylinder with heat source (contd…) Example:

A current of 200A is passed through a stainless steel wire having a thermal conductivity K=19W/mK, diameter 3mm, and electrical resistivity R = 0.99 Ω. The length of the wire is 1m. The wire is submerged in a liquid at 110°C, and the heat transfer coefficient is 4W/m2K. Calculate the centre temperature of the wire at steady state condition. Solution: to be worked out in class