Happy Thesis Tuesday to you all! Today we'll be continuing our discussion of the First Law of Thermodynamics by examining its final form- the differential form. For an incompressible, Newtonian fluid in Cartesian coordinates, this is:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -(d/dx(qx) + d/dy(qy) + d/dz(qz)) + mu*Phiv
where
qx = d/dx(k*T), etc.
Phiv = 2*((d/dx(u))*(d/dx(u)) + (d/dy(v))*(d/dy(v)) + (d/dz(w))*(d/dz(w))) + (d/dx(v) + d/dy(u))*(d/dx(v) + d/dy(u)) + (d/dy(w) + d/dz(v))*(d/dy(w) + d/dz(v)) + (d/dx(w) + d/dz(u))*(d/dx(w) + d/dz(u)) - (2/3)*(d/dx(u) + d/dy(v) + d/dz(w))*(d/dx(u) + d/dy(v) + d/dz(w))
Typically, Phiv is neglected. It is only important for flows with large velocity gradients or very large viscosities. Furthermore, it should be noted that if the fluid has a constant thermal conductivity, the first term on the right-hand side of the equation becomes:
(d/dx(qx) + d/dy(qy) + d/dz(qz)) = k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
So the final equation that we will be dealing with is:
rho*cp*(d/dt(T) + u*d/dx(T) + v*d/dy(T) + w*d/dz(T)) = -k*(d/dx(d/dx(T)) + d/dy(d/dy(T)) + d/dz(d/dz(T)))
In these equations, T is the temperature of the fluid and q is the heat flux. Next week we'll conclude our look at the first law with an example employing this differential equation.
