Partial derivaives
leaves the section through its right-hand cross-section at 𝑥 = 𝛼+ℎ is 𝑞(𝛼 + ℎ, 𝑡) . Therefore, the net rate at which thermal energy enters the section is given as follows:
𝑞(𝛼, 𝑡) − 𝑞(𝛼 + ℎ, 𝑡)
Assuming that no heat is lost to the surroundings, thermal energy must be conserved and so the rate of change of the thermal energy in the section is equal to the net rate at which thermal energy enters the section, this gives:
∫ 𝑐𝑇(𝑥, 𝑡)𝜌𝑑𝑥 𝛼+ℎ 𝛼
𝑑 𝑑𝑡
= 𝑞(𝛼, 𝑡) − 𝑞(𝛼 + ℎ, 𝑡)
Assuming that 𝑇 𝑡 is continuous, we can bring the 𝑑 𝑑𝑡
inside the integral and rearrange the equation
into a form that will allow us to take the limit ℎ→0 , this gives:
𝜌𝑐∫ 𝑇 𝑡 (𝑥,𝑡) 𝑑𝑥 𝛼+ℎ 𝛼
+ 𝑞(𝛼 + ℎ, 𝑡) − 𝑞(𝛼, 𝑡) = 0
∫ 𝑇 𝑡 (𝑥,𝑡) 𝑑𝑥 𝛼+ℎ 𝛼
𝜌𝑐 ℎ
𝑞(𝛼 + ℎ, 𝑡) − 𝑞(𝛼, 𝑡) ℎ
lim ℎ→0
(
+
)= lim ℎ→0
(0)
To take the limit ℎ→0 , we apply the Fundamental theory of calculus to the first term and use the definition of a partial derivative to classify the second term to get:
𝜌𝑐𝑇 𝑡 (𝛼, 𝑡) + 𝑞 𝑥 (𝛼, 𝑡)=0
We now adopt Fourier’s Law 9 , which is given by:
𝑞 =−𝑘𝑇 𝑥
Where k is the thermal conductivity of the rod, which we also assume to be constant throughout the rod. The negative sign in Fourier’s Law means that thermal energy flows from high to low temperature areas. Substituting Fourier’s Law into our pr evious equation we get:
𝜕 2 𝑇 𝜕𝑥 2
𝜕𝑇 𝜕𝑡
𝜌𝑐
−𝑘
=0
Rearranging this gives the Heat Equation 10 :
𝜕 2 𝑇 𝜕𝑥 2
𝜕𝑇 𝜕𝑡
= 𝐾
Where the thermal diffusivity 𝐾 = 𝑘 𝜌𝑐
. The Heat equation is a second order linear partial differential
equation.
Solving the Heat Equation
𝜕 2 𝑇 𝜕𝑥 2
𝜕𝑇 𝜕𝑡
= 𝐾
9 Wikipedia n.d. 10 Oliver et al (2021), 32-34.
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