Cooling Requirements at Die (Pellets)

Reference ID: MET-8E56 | Process Engineering Reference Sheets Calculation Guide

Introduction & Context

This calculation determines the cooling requirements for a die (or pellet) extrusion process by sizing the internal cooling jacket that removes the sensible heat carried by the molten thermoplastic. Accurate jacket sizing ensures the melt reaches the desired outlet temperature, maintains product quality, prevents overheating of the equipment, and optimizes energy consumption. The methodology is standard in process engineering for heat-exchanger design where a hot process fluid (melt) is cooled by a secondary coolant (water) flowing in a concentric jacket.

Methodology & Formulas

The procedure follows an energy-balance approach combined with convective heat-transfer correlations. All temperatures are expressed in kelvin or degree Celsius (differences are identical), specific heats are converted to consistent units, and hydraulic-diameter-based correlations are used for turbulent internal flow.

1. Sensible Heat Load

The rate of heat that must be removed from the melt is given by \[ \dot{Q}= \dot{m}_{\text{melt}}\,c_{p,\text{melt}}\,(T_{\text{melt,in}}-T_{\text{melt,out}}) \] where \(\dot{m}_{\text{melt}}\) is the melt mass flow rate and \(c_{p,\text{melt}}\) its specific heat.

2. Coolant Mass Flow Rate

Assuming the coolant absorbs all the heat, its required mass flow rate follows from an energy balance on the water side: \[ \dot{m}_{\text{water}}= \frac{\dot{Q}}{c_{p,\text{water}}\,(T_{\text{water,out}}-T_{\text{water,in}})} \]

3. Internal Flow Characteristics

The hydraulic diameter \(D_h\) defines the flow cross-section \[ A_c = \frac{\pi D_h^{2}}{4} \] The mean velocity of the water is \[ v_{\text{water}} = \frac{\dot{m}_{\text{water}}}{\rho_{\text{water}}\,A_c} \] Reynolds and Prandtl numbers are then \[ \text{Re}= \frac{\rho_{\text{water}}\,v_{\text{water}}\,D_h}{\mu_{\text{water}}},\qquad \text{Pr}= \frac{c_{p,\text{water}}\,\mu_{\text{water}}}{k_{\text{water}}} \]

4. Internal Convective Coefficient (Dittus-Boelter)

For turbulent flow (\(\text{Re}>10^{4}\) and \(0.6<\text{Pr}<160\)) the Nusselt number is \[ \text{Nu}=0.023\,\text{Re}^{0.8}\,\text{Pr}^{0.4} \] and the internal heat-transfer coefficient follows as \[ h_i = \frac{\text{Nu}\,k_{\text{water}}}{D_h} \]

5. Overall Heat-Transfer Coefficient

The overall coefficient accounts for internal convection, conduction through the jacket wall, and external convection: \[ \frac{1}{U}= \frac{1}{h_i}+ \frac{t_{\text{wall}}}{k_{\text{wall}}}+ \frac{1}{h_o} \] where \(t_{\text{wall}}\) and \(k_{\text{wall}}\) are the wall thickness and thermal conductivity, and \(h_o\) is the external convective coefficient.

6. Log-Mean Temperature Difference (Counter-Flow)

The temperature driving forces at the two ends of the exchanger are \[ \Delta T_1 = T_{\text{melt,in}}-T_{\text{water,out}},\qquad \Delta T_2 = T_{\text{melt,out}}-T_{\text{water,in}} \] The log-mean temperature difference is \[ \text{LMTD}= \frac{\Delta T_1-\Delta T_2}{\ln\!\left(\dfrac{\Delta T_1}{\Delta T_2}\right)} \] (with safeguards against zero or negative arguments as implemented in the code).

7. Required Heat-Transfer Area

Finally, the area needed to transfer the heat load is \[ A = \frac{\dot{Q}}{U\,\text{LMTD}} \]

Validity Checks

The following conditions are examined to ensure the applicability of the correlations used:

Check	Criterion	Action if Not Met
Reynolds number regime	\(\text{Re} \ge 10^{4}\)	Warning: Dittus-Boelter may be invalid (laminar or transitional flow)
Prandtl number range	\(0.6 < \text{Pr} < 160\)	Warning: Dittus-Boelter outside validated Pr range
Temperature differences for LMTD	\(\Delta T_1 > 0\) and \(\Delta T_2 > 0\)	Warning: Non-positive temperature differences; check inlet/outlet temperatures
Log-mean temperature difference	\(\text{LMTD} > 0\)	Warning: LMTD non-positive; review temperature specifications

The minimum flow is governed by the pellet heat load you must remove before the cut reaches the rotary dryer. Key variables:

Pellet throughput (kg h⁻¹) and specific heat (J kg⁻¹ °C⁻¹)
Target ΔT between melt temperature and pellet center at die exit (typically 30–50 °C for semi-crystalline resins)
Latent heat of crystallization for semi-crystalline polymers
Heat-transfer coefficient between pellet and water film (depends on die-hole velocity and water temperature)

Use the steady-state energy balance Q = ṁ c ΔT + ṁ ΔH_cryst to size the flow; then add 15 % safety margin for fouling.

Sticking is rarely a bulk-flow issue; it is a local surface-cooling problem. Typical causes:

Non-uniform water distribution—check manifold pressure drop < 0.3 bar across die width
Steam pocket at the die face due to inadequate venting; install 0.5–1 mm vent grooves at pellet landing zone
Water temperature > 45 °C for amorphous resins—lower to 30–35 °C
Die-plate temperature too high; reduce by 5–10 °C or add internal cooling channels

Verify with IR camera; surface should be < 90 °C within 50 mm of die face.

Freeze-off occurs when cold water contacts the die before polymer flow is established. Counter-measures:

Pre-heat die to 10 °C above melt temperature before opening water valve
Use timed sequence: start extruder, wait for steady strand exit, then ramp water flow from 30 % to 100 % over 15 s
Install temperature-controlled water bypass loop; keep die face dry until polymer emerges
For PET or PA, add 5 bar nitrogen purge through die holes for 10 s to displace moist air

Target values for closed-loop die-cooling water:

Total hardness < 60 ppm as CaCO₃; use softener or RO polish
Chlorides < 50 ppm to protect 17-4 PH die plates
pH 7.0–8.5; add 200 ppm sodium nitrite corrosion inhibitor
Total suspended solids < 10 ppm; 50 µm side-stream filter
Biocide dose to maintain aerobic count < 10³ CFU ml⁻¹

Monitor conductivity on-line; flush loop if > 1200 µS cm⁻¹.

Worked Example: Cooling of Polymer Melt in a Die

A polymer extrusion line uses a water-cooled die to solidify the melt exiting the barrel. The melt enters the die at 180 °C and must be cooled to 130 °C. A water jacket circulates cooling water from 20 °C to 40 °C. The design engineer must determine the required heat-transfer area of the die wall to meet the cooling duty.

Knowns

π = 3.142
Inlet melt temperature, \(T_{m,in}\) = 180 °C
Outlet melt temperature, \(T_{m,out}\) = 130 °C
Melt mass flow rate, \(\dot m_{m}\) = 0.5 kg s\(^{-1}\)
Specific heat of melt, \(c_{p,m}\) = 2000 J kg\(^{-1}\) K\(^{-1}\)
Inlet water temperature, \(T_{w,in}\) = 20 °C
Outlet water temperature, \(T_{w,out}\) = 40 °C
Specific heat of water, \(c_{p,w}\) = 4180 J kg\(^{-1}\) K\(^{-1}\)
Density of water, \(\rho_{w}\) = 998 kg m\(^{-3}\)
Dynamic viscosity of water, \(\mu_{w}\) = 0.001 Pa s
Thermal conductivity of water, \(k_{w}\) = 0.6 W m\(^{-1}\) K\(^{-1}\)
Wall thickness, \(t_{wall}\) = 0.005 m
Thermal conductivity of wall material, \(k_{wall}\) = 16 W m\(^{-1}\) K\(^{-1}\)
Hydraulic diameter of the water channel, \(D_{h}\) = 0.01 m
External heat-transfer coefficient (water side), \(h_{o}\) = 500 W m\(^{-2}\) K\(^{-1}\)
Overall heat-transfer coefficient, \(U\) (to be calculated)
Heat-transfer duty, \(\dot Q\) = 50 000 W

Step-by-Step Calculation

Calculate the temperature differences for the melt and water streams:
\(\Delta T_{melt}=T_{m,in}-T_{m,out}=180-130=50\;{\rm K}\)
\(\Delta T_{water}=T_{w,out}-T_{w,in}=40-20=20\;{\rm K}\)
Determine the required water mass flow rate from the energy balance:
\(\dot m_{w}= \dfrac{\dot Q}{c_{p,w}\,\Delta T_{water}}= \dfrac{50\,000}{4180\times20}=0.598\;{\rm kg\,s^{-1}}\)
Cross-sectional area of the water channel (circular):
\(A_{c}= \dfrac{\pi D_{h}^{2}}{4}= \dfrac{3.142\times(0.01)^{2}}{4}=7.85\times10^{-5}\;{\rm m^{2}}\)
Mean water velocity:
\(v = \dfrac{\dot m_{w}}{\rho_{w}A_{c}}= \dfrac{0.598}{998\times7.85\times10^{-5}}=7.63\;{\rm m\,s^{-1}}\)
Reynolds number for the water flow:
\(\displaystyle Re=\frac{\rho_{w}vD_{h}}{\mu_{w}}= \frac{998\times7.63\times0.01}{0.001}=7.62\times10^{4}\approx 76151\)
Prandtl number for water:
\(\displaystyle Pr=\frac{c_{p,w}\mu_{w}}{k_{w}}= \frac{4180\times0.001}{0.6}=6.967\)
Nusselt number (turbulent flow, Dittus-Boelter correlation):
\(\displaystyle Nu=0.023\,Re^{0.8}Pr^{0.4}=402.047\)
Internal heat-transfer coefficient (water side):
\(h_{i}= \dfrac{Nu\,k_{w}}{D_{h}}= \dfrac{402.047\times0.6}{0.01}=24\,123\;{\rm W\,m^{-2}\,K^{-1}}\)
Overall heat-transfer coefficient, accounting for wall conduction:
\[ \frac{1}{U}= \frac{1}{h_{i}}+\frac{t_{wall}}{k_{wall}}+\frac{1}{h_{o}} \] \(\displaystyle \frac{1}{U}= \frac{1}{24\,123}+ \frac{0.005}{16}+ \frac{1}{500}=0.002354\;{\rm m^{2}K\,W^{-1}}\)
Hence, \(U = \dfrac{1}{0.002354}=424.817\;{\rm W\,m^{-2}\,K^{-1}}\)
Log-mean temperature difference (LMTD) for counter-flow arrangement:
\(\Delta T_{1}= (T_{m,in}-T_{w,out})=180-40=140\;{\rm K}\)
\(\Delta T_{2}= (T_{m,out}-T_{w,in})=130-20=110\;{\rm K}\)
\[ LMTD = \frac{\Delta T_{1}-\Delta T_{2}}{\ln(\Delta T_{1}/\Delta T_{2})} =124.398\;{\rm K} \]
Required heat-transfer area:
\[ A = \frac{\dot Q}{U\,LMTD}= \frac{50\,000}{424.817\times124.398}=0.946\;{\rm m^{2}} \]

Final Answer
The die wall must provide a heat-transfer area of 0.946 m² to cool the polymer melt from 180 °C to 130 °C using the specified water jacket.

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