How to solve temperature evolution PDE for a water heater?

Question

I am trying to understand a PDE model for a heater.

Please kindly guide in the correct direction so that I can understand and possibly solve this equation?

The below is a heat equation for a water heater with two heating element and sensors.

$\eqalignno{&{\partial T\over \partial t}+V{\partial T\over \partial x} =({D_{c}+D_{a}}){\partial^{2}T\over \partial x^{2}} +{Q_{\rm ele}\over \rho c_{v}}+{U\cdot A_{s}\over AH\rho c_{v}}({T_{\rm amb} -T})\cr &&\hbox{(1)} }$

Can you please advise how can I solve such an equation?

(2) V = m ˙ / A

Qele=Q/(Az) AS=2π×d×H is the total surface area of the water tank, where d is the diameter of the water tank. The effect of heat transfer through thermal conduction is considered using the thermal diffusivity Dc=k/(ρ×cv), where k is the thermal conductivity of water, k=1.8 W/(m⋅K), and Dc=0.143×10−6m2/s.

The effect of heat transfer by natural convection is modeled using the buoyancy factor given by Da [9]

$D_{a} = ({\theta d})^{2}\sqrt {-g\beta {\partial T\over\partial x}} {\hbox{(3)}}$ where θ=0.46 is the Von Kármán constant. g=9.8 m/s2 is the acceleration due to gravity and β=207×10−6/K is the thermal expansion coefficient of water. The minus sign in (3) is due to the fact that natural convection only occurs when the instantaneous temperature gradient is negative, i.e., (∂T)/(∂x)<0. If the water tank has a diameter of d=0.5 m, (3) can be further simplified to $D_{a} =\gamma \sqrt {-{\partial T\over\partial x}} \eqno{\hbox{(4)}}$

Answer 1

Since your diffusion coefficient depends on the unknown temperature you should not hope to find an analytical solution. You will need to resort to numerical methods suitable for this kind of problems such as finite differences methods especially Crank Nicolson.