According to the figure the melting section is divided up into two areas:

Melt film on the barrel wall with an underlying solid bed of the width: b*y

The principle is again the same as that in equation $$P = \int_0^z \int_{-\frac{b_{max}}{2}}^{+\frac{b_{max}}{2}} (\tau_{0x} r_{0x} + \tau_{0z} r_{0z}) dx \, dz$$. For the melt film, the shear stresses are to be replaced according to [3] as follows:

$$\tau_{0x} = K(T_{Fl}) \left(\frac{v_{rel}}{\delta}\right)^{n-1} \frac{v_{0x}}{\delta} \tag{1}$$

$$\tau_{0z} = K(T_{Fl}) \left(\frac{v_{rel}}{\delta}\right)^{n-1} \frac{v_{0z} - v_{Fz}}{\delta} \tag{2}$$

With this the relative speed of the melt film $v_{rel}$ will be formed using:

$$v_{rel} = \sqrt{(v_{0z} + v_{Fz})^2 + v_{0x}^2} \tag{3}$$

This results in the consumption:

$$P_3 = \frac{K(T_{Fl})by\Delta z v_{rel}^{n-1}}{n} (v_{0x}^2 + (v_{0z} - v_{Fz})v_{0z})k \tag{4}$$

A prerequisite of this way of modelling is the assumption of a pure drag flow in the melt film.