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TABLE OF CONTENTS Page No

COOLING TIME CALCULATIONS ...........................................................................................................2

WALL THICKNESS.......................................................................................................................................3

COOLING CHANNEL...................................................................................................................................5

GATE SIZE......................................................................................................................................................7

INJECTION TIME .........................................................................................................................................8

PILLAR SPACING.........................................................................................................................................9

RUNNER SIZE..............................................................................................................................................10

CORE DESIGN .............................................................................................................................................12

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Cooling Time Calculations The following formulae are used for the Cooling time calculations:

Molding Shape Cooling Time at Centre of Mold Average Cooling Time Across the Mold Plate S0 - plate thickness

S0

2 .ln 4 (TM - Tc) aeff. π2 π (TD - Tc)

S0

2 .ln 8 (TM - Tc) aeff. π2 π2 (TD - Tc)

Cylinder d0 - diameter l0 - ∞

d0

2 .ln 1.599 (TM - Tc) aeff. 23.132 (TD - Tc)

d0

2 .ln 0.975 (TM - Tc) aeff. 23.132 (TD - Tc)

Cylinder d0 - diameter l0 - length

1 .ln 6.396 (TM - Tc) 23.14 + π2 aeff π (TD - Tc) d0

2 l0

2

1 .ln 7.802(TM - Tc) 23.14 + π2 aeff π2 (TD - Tc) d0

2 l0

2

Sphere d0 - diameter

d0

2 .ln 2 (TM - Tc) 4.aeff. π2 (TD - Tc)

d0

2 .ln 1.178(TM - Tc) 4.aeff. π2 (TD - Tc)

Parallelepiped b0 - width h0 - height l0 - length

1 .ln 64 (TM - Tc) 1 + 1 + 1 π2 .aeff π3 (TD - Tc) b0

2 h0

2 l0

2

1 .ln 512 (TM - Tc) 1 + 1 + 1 π2 .aeff π6 (TD - Tc) b0

2 h0

2 l0

2

Parallelepiped b0 - width h0 - height l0 - ∞

1 .ln 16 (TM - Tc) 1 + 1 π2 .aeff π2 (TD - Tc) b0

2 h0

2

1 .ln 64 (TM - Tc) 1 + 1 π2 .aeff π4 (TD - Tc) b0

2 h0

2

Cube b0 - width

b0

2 .ln 64 (TM - Tc) 3.aeff. π2 π3 (TD - Tc)

b0

2 .ln 512(TM - Tc) 3.aeff. π2 π6 (TD - Tc)

aeff = Thermal diffusivity, TD = Deflection temperature, TM = Material temperature Tc = Mold temperature

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Wall Thickness

This calculation uses the narrowest wall as the one from which the "narrow beam" is created. A vertical beam is treated as a beam fixed at one end & supported at the other. A "horizontal beam" is treated as fixed at both ends. For h1: To calculate the wall thickness, imagine a strip 1cm wide cut out of the open end of the cavity. If Y < Z, assume fixed ends, along a length of Ycm, and assuming a cavity pressure of 4000N, A force of P = Y . 1cm . 4000 N then acts on the 1cm wide strip. The wall is subject to a bending load. For the load calculation. The following calculations then apply: W = P .Y 12σb

where W = moment of resistance

P = total load (N) Y = length subject to load σb = bending load of alloy steel (= 200 N/mm2)

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If Y > Z, assume fixed one end, along a length of Z cm, and assuming a cavity pressure of 4000N, A force of P = Z . 1cm . 4000 N then acts on the 1cm wide strip. The wall is subject to a bending load. For the load calculation. The following calculations then apply: W = P .Z 8σb The following equations are next used: W = b .h12 ∴ h1 = √ 6W/b 6 where W = moment of resistance b = width of "strip" = 1cm for h2: If X < Z, assume fixed ends, along a length of Xcm, and assuming a cavity pressure of 4000N, A force of P = X . 1cm . 4000 N then acts on the 1cm wide strip. The wall is subject to a bending load. For the load calculation. The following calculations then apply: W = P .X 12σb

where W = moment of resistance P = total load (N) X = length subject to load σb = bending load of alloy steel (= 200 N/mm2) If X > Z, assume fixed one end, along a length of Zcm, and assuming a cavity pressure of 4000N, A force of P = Z . 1cm x 4000 N ( = 200,000 N) then acts on the 1cm wide strip. The wall is subject to a bending load. For the load calculation. The following calculations then apply: W = P .Z 8σb The following equations are next used: W = b .h22 ∴ h2 = √ 6W/b 6 where W = moment of resistance b = width of "strip" = 1cm

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Cooling Channel For the calculation of Cooling Channel dimensions, the following calculations are used: The heat content of the material being molded is:

q = A . th . dens . sph . (Tc - Tm) where q is the heat content of the material being molded (J) A is the mold part area th is the part thickness dens is the density of the material molded sp is the specific heat of the material molded Tc is the melt temperature Tm is the mold temperature The power of removing the heat in the cycle time is given by:

P = q/t where P = power (W)

q is the heat content of the material being molded t is the cycle time

The water flow rate required to remove this heat within the cycle time while only increasing the water temperature a specific amount: Flow = P litres/min densw . spw . (Twout - Twin) where densw is the density of water (=1000kg/m3)

spw is the specific heat of water (=4183J/kgK) Twout is the water temperature out Twin is the water temperature in

The diameter to achieve the minimum Reynolds number for full turbulance is: Re = Flow . densw .4 Viscw . π.d where Re = Reynolds number (= 4000 minimum for turbulant flow)

densw is the density of water (=1000kg/m3) d is the diameter of the cooling channel Viscw is the viscosity of water

(note: PlasWin's calculations use the viscosity @ 30°C = 80x105 kg/ms) The total length of cooling channel within the mold, required to extract the heat, given the temperature rise within the cooling channel:

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l = P ThCw . Twr . π. 0.023 .Re where l is the total length of the cooling channel

Re = Reynolds number ThCw is the wall temperature rise of the cooling channel (=0.58 J/msk)

(note: PlasWin's calculations assume a wall temp rise of 4°C) P is the diameter of the cooling channel Twr is the viscosity of water

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Gate Size For a circular die, the Newtonian shear rate is given by:

ϒ

= 4Q πr3 where ϒ is the shear rate

Q is volumetric flow rate in cm3s-1 r is the radius of the die in cm

For a slit die, the Newtonian shear rate is given by:

ϒ

= 6Q lt2 where ϒ is the shear rate

l is the length of the slit in cm (ie gate width) t is the width, or thickness of the slit in cm (ie gate depth)

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Injection Time The filling time of the mold is determined by the following calculations:

t = (dwpf/pfr)3

8. [(Tx-Tc)/(Tm-Tc)]3 where t is the filling time (secs)

dw is the material thickness (mm) pf is the material flow length (mm) pfr is the material flow path ratio

Tx is the deflection temperature of the material (°C) Tc is the mold temperatue (°C) Tc is the material temperatue (°C)

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Pillar Spacing The following calculations are used as part of determining pillar spacing: First load case-

dmax = w.L4

384EI E = Modulus of Elasticity, I = Moment of Inertia, w = Pressue N/mm If the deflection is excessive, place a support midspan:

dmax

= w.(L/2)4 384EI

If this deflection is excessive, continue placing supports mid way until acceptable deflection is achieved.

d

w

L

d

w

L/2 L/2

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Runner Size Consider the following runner design - To calculate pressure drop from across a runner layout, an assumption is made that all runners are full-round type. For each runner, the volume is given by: V = π r2 L where r is the runner radius

L is the runner length The total shot volume = (the sum of all runner volumes + part volumes) The shear rate is: Sr = 4Q where Q is the volumetric flow rate π r3

For the Primary runner, Q= (total shot volume)/(injection time) Next, calculate the shear stress Ss = µ Sr where µ is the melt viscosity of the material The pressure drop through the Primary runner segment is: P = Ss (2L)/r Now the next runner segment is considered - for the Secondary runner, the total volumetric flow rate

Q= (total shot volume - the volume in the primary runner)/2.(injection time) Note: this is divided by 2 because the flow splits in half again at the secondary runner. As before, the shear rate, shear stress and hence pressure drop through the Secondary runner are calculated.

Tertiary runner

Secondary runner

Primary runner

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For the tertiary runner segment - the total volumetric flow rate through each tertiary runner: Q= (total shot volume - volume in the primary runner - the volume in the secondary runner)/[2.(injection time)] The shear rate, shear stress and hence pressure drop through the Tertiary runner are then calculated. The total pressure loss from the sprue to each gate is then the sum of the pressure losses through each runner segment.

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Core Design

The maximum defletion experienced by the core is given by: fact = fB + fS + fES + fEM where fact is the total deflection at reference pressure

fB is the result of the bending moment fS is the result of the shear moment fes is the shift resulting from the shear force (through angle a) fem is the lowering of the core tip

fB = 32. P. D . L

15π E. D4 where P is the reference pressure of 10Mpa E is the modulus of elasticity of the core material

L is the core length D is the core diameter

fs = 4.62 x P. D . L2

3π E. D2

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the point xs where the resulting force over the core acts is: xs = L (D - L.tan a) / (3.D - 2.L.tan a) where a is the taper of the core

L is the core length D is the core diameter

The maximum contact pressure from shear is: ps = P. L (3.D - 2.L.tan a)

6.D.W then fes = ps. M/E (3.D - 2.L.tan a)

6.D.W The maximum contact pressure from bending moment is: Pm = P. L . (2xs + W) (3.D - 2.L.tan a)

6.D.W and fem = pm. M (2.L + W)

EW