If the pipe is reasonably long in relation to diameter, I wouldn't worry too much about unequal heating.
Pipe material conductivity = k
Inner radius = r
Outer radius = R
Water temp at x from start = T(x)
Oven temp = H
Specific heat of water/unit vol = s
velocity of water = V
A = π.r^2
A full analysis gets messy because there will be heat flow along the pipe (back towards point of entry into oven) as well as radially, but let's ignore that.
Taking a slice through the pipe thickness dx, the radial heat flow depends on the temperature difference and the inner and outer radii:
F(x) = dx.2π.k.(H-T(x))/ln(R/r)
In time dt, heat entering dx of the pipe = F.dt
In that time, a volume of water V.A.dt passes any given point.
So in distance dx the water warms by dT = F/V.A.s
dT/dx = 2π.k.(H-T(x))/(ln(R/r).V.A.s)
ln((H-T)/c) = -2π.k.x/(ln(R/r).V.A.s), where c = H - T(0).
If the pipe is length L, water emerges at temp T(L) :
ln((H-T(L))/(H-T(0))) = -2π.k.L/(ln(R/r).V.A.s)
Check the math, plug in the constants and solve for L.