The thickness of the wall is only 0.218" , the average diameter is 5.454", and the average radius is 2.727". So the wall thickness is less than 10% of the average radius, and we can neglect the curvature of the wall. We can do this by slitting the pipe along its length, and then laying the pipe wall flat. This leaves us with a flat slab ##h=0.218(2.54)=0.554\ cm## thick, with cross section ##\pi (5.454)(2.54)=W=43.5## cm wide and ##L=50## cm long. If Q is the magnitude of the power source, the heat flux at the outer surface of the pipe (x = h) is ##\frac{Q}{WL}##, and the heat flux at the inner surface of the pipe (x = 0) is zero (the inner surface is assumed to be insulated).
The solution for the temperature distribution within the slab is a function of time t and distance above the inner surface x. In the present problem, this temperature distribution will feature a short term transient that dies out after a short amount of time, and a long time asymptotic temperature distribution that grows linearly with time. We will focus here primarily of the long time asymptote.
The average temperature of the slab (averaged over the volume of the slab) and any time will be given by:
$$\bar{T}=T_i+\frac{Qt}{\rho C LWh}$$, where Q is the total heating rate, t is the time since heating began, ##T_i## is the initial slab temperature (60 C), ##\rho## is the density of the steel, and C is the heat capacity of the steel. At long times, in addition to the average temperature, there will also be a superimposed temperature variation with location x through the slab, ##\hat{T}(x)##. So the overall long term temperature distribution will be given by $$T=\bar{T}+\hat{T}(x)=T_i+\frac{Qt}{\rho C LWh}+\hat{T}(x)\tag{1}$$
More generally, the temperature variations within the slab are described by the transient heat conduction equation:$$\rho C \frac{\partial T}{\partial t}=k\frac{\partial^2T}{\partial x^2}\tag{2}$$
If we substitute Eqn. 1 into Eqn. 2, we obtain:$$\frac{Q}{LWh}=k\frac{d^2\hat{T}}{d x^2}\tag{3}$$If we integrate Eqn. 3 once with respect to x, we obtain:$$k\frac{d\hat{T}}{dx}=\frac{Q}{LWh}x\tag{4}$$where use has been made here of the zero heat flow boundary condition at x = 0. If we next integrate Eqn. 4 with respect to x, we obtain: $$\hat{T}=\hat{T}(0)+\frac{Q}{kLWh}\frac{x^2}{2}\tag{5}$$where ##\hat{T}(0)## is the constant of integration. The value of this constant of integration can be established by noting that, since ##\bar{T}## is the average temperature of the slab, the average value of ##\hat{T}## must be equal to zero. This leads to the result that:
$$\hat{T}(0)=-\frac{Qh}{6kLW}\tag{6}$$
So, our solution for the long-term temperature distribution is given by: $$T=T_i+\frac{Qt}{\rho C LWh}-\frac{Qh}{6kLW}+\frac{Q}{kLWh}\frac{x^2}{2}$$
So, the temperature at the inner surface of the pipe is: $$T(0)=T_i+\frac{Q}{\rho C LWh}\left(t-\frac{1}{6}\frac{h^2}{\alpha}\right)$$where ##\alpha## is the thermal diffusivity ##\left(\frac{k}{\rho C}\right)##. And the temperature at the outer surface of the pipe is: $$T(h)=T_i+\frac{Q}{\rho C LWh}\left(t+\frac{1}{3}\frac{h^2}{\alpha}\right)$$The difference in temperature between the inner and outer surface will be:
$$T(h)-T(0)=\frac{Qh}{2kLW}$$
