Film die gap for a two-metering-zone screw extruder: 23-Chem-B8 Polymer Engineering, December 2015, Q5

Question 5: Film die gap for a two-metering-zone screw extruder (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Newtonian melt $\eta=250$ Pa·s. Screw: $D=75$ mm, $N=40$ rpm $=0.667$ s$^{-1}$, helix angle $\phi=17.7^\circ$, flight width $e=7.5$ mm. Metering zone 1: $L_1=250$ mm, $h_1=4$ mm; zone 2: $L_2=500$ mm, $h_2=6$ mm. Slit die: width $T=1$ m, land length $L_d=10$ mm, gap $H$ (unknown).

Symbol	Value	Symbol	Value
$D$	0.075 m	$h_1,h_2$	4 mm, 6 mm
$N$	0.667 s$^{-1}$	$L_1,L_2$	0.25 m, 0.50 m
$\phi$	17.7°	$T,L_d$	1 m, 0.010 m
$e$	0.0075 m	$\eta$	250 Pa·s

Find. The die gap $H$ for satisfactory extrusion with the two metering zones just matched.

Unwrapped screw channel: a shallow first metering zone feeds a deeper second zone, then the slit film die. “Just matched” is taken as zero pressure at the junction, so zone 1 delivers its pure drag flow and zone 2 develops all of the die pressure.

Approach. Each metering zone carries the same output $Q=$ drag flow $-$ pressure flow. “Just matched” means the pressure at the junction between the two zones is zero, so the shallow zone runs at its pure drag flow and the deep zone alone builds the pressure the die needs; sizing the die to pass that flow at that pressure gives $H$.

Channel width and drag flow. The channel width perpendicular to the flights is $W=\pi D\sin\phi-e=\pi(0.075)(0.304)-0.0075=0.0641$ m. Drag flow of a metering zone is $Q_d=\tfrac12\pi D N\cos\phi\,W h$: $$Q_{d1}=1.92\times10^{-5},\qquad Q_{d2}=2.88\times10^{-5}\ \text{m}^3/\text{s}$$ (the deeper zone can drag more melt forward).
Pressure-flow coefficients. For each zone $Q_p=\dfrac{W h^3\sin\phi}{12\eta L}\,\Delta P$; the coefficients are $$B_1=\frac{W h_1^3\sin\phi}{12\eta L_1}=1.66\times10^{-12},\qquad B_2=\frac{W h_2^3\sin\phi}{12\eta L_2}=2.81\times10^{-12}\ \frac{\text{m}^3}{\text{s\,Pa}}$$
Apply the matched condition. Zero junction pressure means zone 1 develops no net pressure, so it delivers pure drag flow and this sets the throughput: $$Q=Q_{d1}=1.92\times10^{-5}\ \text{m}^3/\text{s}\;(=1.15\ \text{L/min})$$ Zone 2 then converts its surplus drag capacity into the die pressure: $$P_{die}=\frac{Q_{d2}-Q}{B_2}=\frac{(2.88-1.92)\times10^{-5}}{2.81\times10^{-12}}=\boxed{3.42\ \text{MPa}}$$
Size the die gap. The slit-die output is $Q=\dfrac{T H^3}{12\eta L_d}\,P_{die}$. Solving for $H$, $$H=\left(\frac{12\eta L_d\,Q}{T\,P_{die}}\right)^{1/3} =\left(\frac{12(250)(0.010)(1.92\times10^{-5})}{(1)(3.42\times10^{6})}\right)^{1/3} =\boxed{0.55\ \text{mm}}$$

Quantity	Value
Matched output $Q=Q_{d1}$	$1.92\times10^{-5}$ m$^3$/s (1.15 L/min)
Die-entry pressure $P_{die}$	3.42 MPa
Film die gap $H$	$\approx0.55$ mm

Check

“Just matched” is interpreted as zero pressure at the junction between the two metering zones — the balanced design point at which the shallow zone neither builds nor loses head and simply meters its drag flow into the deeper pressure-generating zone (Crawford, screw-characteristic analysis). This yields $Q=Q_{d1}$ and a physically sensible film speed $v=Q/(TH)\approx35$ mm/s. An alternative reading (equal pressure development in both zones, $\Delta P_1=\Delta P_2$) gives $Q\approx5.2\times10^{-6}$ m$^3$/s and $H\approx0.21$ mm; the exam's Note 1 invites the candidate to state such an assumption.

Topic: Single-screw extruder metering-zone analysis and die matching.

Key relations: drag flow $Q_d=\tfrac12\pi D N\cos\phi\,W h$; pressure flow $Q_p=W h^3\sin\phi\,\Delta P/(12\eta L)$; slit-die conductance $Q=T H^3 P/(12\eta L_d)$; channel width $W=\pi D\sin\phi-e$.

Why this works: In a metering zone the barrel drags melt forward (drag flow) while the back-pressure from the die pushes some of it back (pressure flow); the net is the output, identical in every section by continuity. The die imposes its own linear $Q$–$P$ line. With two zones of different depth in series the operating point is fixed once a design condition is named; “matched” (zero junction pressure) makes the shallow zone the metering element and the deep zone the pump, and the die gap follows from the cubic conductance law.

Common pitfall: Forgetting to subtract the flight width from the channel width, or mixing the down-channel length with the axial length (the factors of $\sin\phi$ and $\cos\phi$ track this). The die gap enters as $H^3$, so a small gap change swings the output strongly.

Source: Crawford, Plastics Engineering 3rd ed., §4.2–4.3 (analysis of flow in extruders and dies).

Quantity	Value
Matched output \(Q=Q_{d1}\)	\(1.92\times10^{-5}\) m\(^3\)/s (1.15 L/min)
Die-entry pressure \(P_{die}\)	3.42 MPa
Film die gap \(H\)	\(\approx0.55\) mm

Symbol	Value	Symbol	Value
\(D\)	0.075 m	\(h_1,h_2\)	4 mm, 6 mm
\(N\)	0.667 s\(^{-1}\)	\(L_1,L_2\)	0.25 m, 0.50 m
\(\phi\)	17.7°	\(T,L_d\)	1 m, 0.010 m
\(e\)	0.0075 m	\(\eta\)	250 Pa·s

Question 5 of 6: Film die gap for a two-metering-zone screw extruder

Question 5: Film die gap for a two-metering-zone screw extruder (25 marks)