Constant conductivity and No Heat Generation Assignment 2022

ASSIGNMENT
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1.
a) The general heat conduction equation for a large plane wall under variable
conductivity is given below. All symbols have their usual meaning in this context.
𝜕𝜕𝑥(𝑘𝜕𝑇𝜕𝑥)+ 𝑒̇𝑔𝑒𝑛=𝜌𝑐𝜕𝑇𝜕𝑡
Describe and show how this general equation will change under the following different
conditions while elaborating the reasons behind such changes.
(i) Constant conductivity, k
(ii) Steady state, no heat generation
Solution
i. For constant conductivity
From the general equation ∂
∂ x (k ∂ T
∂ x )+ e.
gen= ρc ∂T
∂ t
k =It is thethermal conductivity constant
Taking k out of the bracket
k ∂2 T
∂ x2 + e.
gen =ρc ∂T
∂ t
Dividing the constant through the equation
k
k
∂2 T
∂ x2 + e.
gen
k = ρc
k
∂ T
∂t
We get the final equation of constant conductivity as
∂2 T
∂ x2 + e.
gen
k = ρc
k
∂ T
∂ t
ii. Steady state, no heat generation
For a steady state, the amount of heat generated is zero e.
gen=0 and thus ∂T
∂ t =0
Substituting the values of the steady state to the constant conductivity equation
∂2 T
∂ x2 + ( 0 )
k = ρc
k ( 0 )
k Is a constant and thus the equation reduces to
∂2 T
∂ x2 =0

b) If right side of the plane wall (x = L, where L is the thickness of the wall) is
exposed to outside air of temperature T∞ with convective heat transfer coefficient
h while the left side (x = 0) is kept at a constant temperature TO, obtain an
expression for the temperature profile across the wall as a function of distance (x)
given that the values h, L, k, T∞, TO are constant. You can assume the heat
transfer is at steady state, with no heat generation.
Solution
Assumptions
 No generation of heat .
 Steady state
 Constant material properties
 1 Dheat transfer
 Equal surface area
Figure 1
The convection heat transfer
̇Q
A = ̇q=h(T ∞ ¿−T s ) ¿
heat transfer across the wall
̇Q
A = k
L (T s−T o )
heat transfer on the¿ hand side
̇Q
A = ̇h(T ∞ ¿−T s )¿
Rearranging the expression since ̇Q
A is the same
The overal temperature drop ̇(T ∞ ¿−T o) ¿
̇
(T ∞ ¿−T 0)=
̇
(T ∞ ¿−T s )+ ̇
(T s ¿−T o )= ̇Q
A [ 1
h + L
k ] ¿ ¿ ¿

Thermal resistance
Figure 1.1
Q= T ∞−T o
R
R= 1
hA + L
kA
Evaluating the surface temperature T s
T s=T ∞ − Q
hA
¿ T ∞−
T ∞ −To
R ∗1
hA
expression is
T s=T ∞ − T ∞−T o
1
hA + L
kA
2. Consider the following composite wall of a building. It is made out of plaster board
(𝐿𝑃=30 𝑚𝑚,=0.85 𝑊/𝑚𝐾), glass fibre insulation (𝐿𝑏=180 𝑚𝑚,𝑘=0.050 𝑊/𝑚𝐾) and
plywood (𝐿𝑆=50 𝑚𝑚,𝑘=0.70 𝑊/𝑚𝐾) as shown below. On a certain day, the convective
heat transfer coefficients inside and outside of the building are 35 𝑊/𝑚2𝐾 and 65 𝑊/𝑚2𝐾
respectively. The inside and outside temperatures are 240C and 360C respectively. The
wall has a surface area of 240 𝑚2. Consider that radiation heat transfer is negligible and
implement the following tasks.
Solution