Optimum channel depth

It is interesting to point out that for a die with a pressure flow through a slit, or sets of slits, the optimum channel depth is directly proportional to the die gap. Decreasing the die gap by a certain percentage will result in an optimum channel depth that is reduced by the same percentage. To determine the optimum helix angle we can re-write eqn. (6.23) for this specific application,... 

Next we turn to Eq. 9.2-5 and derive the optimum channel depth for the maximum pressure rise at fixed screw speed and barrel diameter. We rewrite Eq 9.2-5, neglecting the effect of the flight clearance and the shape factors, as follows... 

Next, we take the derivative of this equation with respect to channel depth to obtain the optimum channel depth, H0p... 

From the preceding equations, we can now select a barrel diameter, calculate the optimum channel depth, then the screw speed, the mean shear rate, and the power per unit volume. In the following table, the results are shown with barrel diameters from 8 in to 16 in. 

This results in the following optimum channel depth H H =... 

At the optimum channel depth H, the optimum helix angle becomes ... 

The melt conveying theory discussed in Section 7.4 can be used to determine the optimum screw geometry for melt conveying. This optimum geometry will not normally be used in the metering section of the extruder screw. The optimum channel depth for output can be determined from ... 

The optimum channel depth for output rate H can now be determined from Eqs. 8.28 through 8.30. 

The corresponding optimum channel depth can be found by inserting Eq. 8.39 into Eq. 8.31. The optimum channel depth resulting from simultaneous optimization is simply ... 

Unfortunately, the optimum channel depth is dependent on many more variables than the optimum helix angle. The latter depends only on the power law index and reduced flight width. In addition to these variables, the optimum channel depth also depends on the screw speed, screw diameter, consistency index, and pressure gradient. This means that it is not possible to design a universally optimum screw geometry. Thus, one has to determine the most likely operating parameters that the screw is likely to encounter and design for those parameters. 

An expression for pressure P can be found by rewriting Eq. 8.29. The optimum channel depth is ... 

The optimum channel depth and helix angle can also be determined from the melt conveying theory of power law fluids when the flow is considered to be one-dimensional see also Section 7.4.2. By combining Eqs. 7.256 and 7.257, the output can be written as ... 

The optimum channel depth can again be determined by taking the partial derivative of output V and setting the result equal to zero. The resulting expression is ... 

Summarizing, it can be concluded that for simultaneous optimization of the channel depth and helix angle, the modified Newtonian analysis yields reasonably accurate results when compared to the two-dimensional power law analysis. The important equations are Eq. 8.39 for the optimum helix angle and Eq. 8.40 for the optimum channel depth. The results of simultaneous optimization from a one-dimensional power law analysis are less accurate than the modified Newtonian analysis. 

It was discussed in Section 7.2.2 that there appears to be an optimum channel depth for which the solids conveying rate reaches a maximum. At low values of the pressure increase over the solids conveying section, this optimum channel depth is indeed apparent because this optimum channel depth does not occur when the channel curvature is taken into account. At higher values of the pressure increase, however, there is an actual optimum channel depth even when the channel curvature is taken into account. This is shown in Fig. 8.20 for a 114-mm (4.5-in) extruder running at 60 rpm the coefficient of friction is 0.5 on the barrel and 0.3 on the screw. When the pressure gradient increases, the optimum channel depth decreases. 

Equation 8.66 does not have a simple closed form solution. The optimum channel depth can be evaluated by using a numerical or graphical method. The optimum... 

This equation was derived by assuming a zero pressure gradient in the metering section and by using the Newtonian throughput-pressure relationship, Eq. 7.198. The optimum channel depth in the pump section Hp can be obtained by setting ... 

This results in the following expression for the optimum channel depth in the pump section ... 

At each pressure gradient, there is an optimum channel depth for which the output reaches a maximum value. This is similar to the situation in single screw extruders. The optimum channel depth in these examples ranges from about 0.05 to 0.010 D. As the pressure gradient increases, the optimum channel depth decreases. 

In more recent years, combining optimum channel depth with reduced screw pitch matched the feed zone s specific output rate to the melting and mixing capacity of the remainder of the screw. 

Channel depth, H at a given flow rate there is an optimum channel depth for a maximum pressure i.e., it depends on the extruder output rate and the die restriction a drinking straw (high) or a water pipe (low). 

A.3 Scaleup of Solids Conveying Section. The optimum channel depth for the solids conveying zone of a single-screw extruder having a barrel diameter of 5.0 cm was found to be 0.6 cm. At 100 rpm the mass flow rate of nylon was found to be 10.45 g/s and P2/P = 100./s = 0.3 and = 0.5. Determine the screw rpm (N), the channel depth (//), and the mass flow rate (G) in scaling up to a single-screw extruder with Db = 11.4 cm. 

B.2 Optimum Channel Depth in the Solids Conveying Zone. Starting with the expression for the mass flow rate given in Eq. 8.23, And an expression for the optimum channel depth (i.e., maximize G with respect to H). Although the value of H must be determined numerically from the expression, specify what parameters H depends on. 

C.1 Calculation of the Optimum Channel Depth for Solids Conveying. In Problem 8B.2 an expression for the optimum channel depth was obtained. For a 11.4 cm diameter extruder running at 60 rpm with values of fi, = 0.5 and fs = 0.3 and P2/P1 = 200, determine the optimum value of H (i.e.. And the value of H that makes G/p a maximum). It should be noted that in practice it is difficult to obtain accurate values of the friction coefficients, and hence, one must use results of the nature asked for here only as a guideline. 

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