Steady-state conduction through a plane wall without heat generation

Question 5 Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was T0 higher than expected for a linear temperature distribution.

Assuming that the thermal conductivity has a linear dependence on the temperature k = k0(1 + αT) where α is a constant, develop a relationship to evaluate α in terms of T0, T1, and T2. Solution: Write Temperature Balance Equation : Since the system is in steady state and there is no accumulation or generation the heat balance equation reduces to:

∂ ∂T ×k ∂x ∂x

= 0 (Eq. 5.1)

Heat conductivity is changing linearly and is given as:

k ( T )=k 0 ( 1+αT ) ( Eq . 5.2) Integrate Eq. 5.1

k (T )

∂T =C 1 (Eq. 5.2) ∂x

Separate variables.

k ( T ) ∂ T =C 1 ∂ x

(Eq. 5.3)

Substitute k(T) and integrate both sides

∫ k 0 ( 1+αT ) ∂ T =C1 x+ C2 αT 2 k0 ¿

(Eq.5.4) 

2

T+

)= C x 1+C 2

(Eq.5.5)

α T 12 T 1+ 2 )= C 1 x (0)+ C2 T 0 =T 1 =k 0 ¿

(Eq.5.6)

2

αT2 T2+ 2 )= C 1 x ( L)+C 2 (Eq. 5.7) T L =T 2 =k 0 ¿ Subtract T1 from T2: 2

α T2 α T1 T 2+ T1+ 2 2 )T 2 −T 1=k 0 ¿ k 0¿

2

) = C1 x ( L )−C 1 x (0)

(Eq.5.8)

Then C1 can be calculated as follows:

C1 =

T 2−T 1 L

2

2

αT2 α T1 k 0 (T 2+ )−k 0 (T 1 + ) 2 2 = L

(Eq.5.9)

Then C2 can be calculated by substituting C1 in Eq. 5.7 with Eq. 5.9: 2

α T2 T2+ 2 C 2=k 0 ¿

)-

(

(k 0 T 2 +

α T 22 αT 2 −k 0 T 1 + 1 )L 2 2 α T 12 =k 0 T 1+ L 2

) (

)

(

)

(Eq.5.10)

Then from Eq.5.4 Temperature Profile T(x) can be derived as:

k 2

T ( x )=¿

(¿ ¿ 0(T 2 +

α T2 α T 12 )−k 0 (T 1+ ) )x 2 2 L C1 x+C 2=¿

+

(

k0 T 1 +

α T1 2

2

)

(Eq.5.11)

Re-arrange to find α:

k (¿¿ 0 x)(T −T 21−T 12 xL) (Eq.5.12) 2 ( T ( x ) × L ) −(k 0 x )(2T 2 +2T 1−2 T 1 xL) α= ¿ 2 2

We know that:

T L= 2

T 1+ T 2 +∆T0 2

(Eq.5.13)

Therefore either T1 or T2 can be expressed though equation to compute α

∆T0

and substituted into the