5 Cooling and Trimming the Part

5.1 Introduction5.2 Overall Cooling Heat Balance5.3 Cooling the Formed Shape5.4 Steady State Heat Balance

Interfacial ResistanceShape FactorConvection Heat Transfer Coefficient

5.5 Cyclic Heat BalanceCooling the Free Surface of the SheetCooling Thin Sheet in Ambient AirTransient Heat Removal From the SheetQuiescent Ambient AirMoving Ambient AirCooling on Nonmetallic Molds

5.6 Transient Heat Transfer During Sheet Cooling on the Mold SurfaceComputer ModelsInterfacial Air

5.7 ShrinkageUnconstrained ShrinkageConstrained Shrinkage

5.8 TrimmingTrimming Heavy-Gage PartsTrimming Thin-Gage Parts

5.9 Mechanics of CuttingThe Trim RegionRegistering the Trim SiteThe Nature of the CutFracture MechanicsMechanical ChippingMultiple-Edged Tool or Toothed Saw PerformanceAbrasive Cut-Off WheelToothless or Shear and Compression CuttingFracture Mechanics in TrimmingNibblingBrittleness, Orientation and Trim Temperature

5.10 Steel Rule DieResharpeningTabbing and Notching

5.11 Punch and Die TrimmingForged and Machined Dies

5.12 Drilling5.13 Other Cutting Techniques

Thermal CuttingWater Jet Cutting

5.14 TrimmingA Summary5.15 References

5.1 Introduction

Once the part has been heated and formed to shape by contact with the cool moldsurface, it must be cooled and rigidified. Then the part and its web must be separatedby trimming. When thermoforming a reactive polymer such as thermosettingpolyurethane or a crystallizing one such as nucleated CPET1, the formed shape isrigidified by holding it against a heated mold to continue the crosslinking reaction orcrystallization. In most cases, rigidifying implies cooling while in contact with acolder mold. For automatic thin-gage formers, the molds are usually actively cooledwith water flowing through channels. Free surfaces of medium- and heavy-gage sheetare frequently cooled with forced air, water mist or water spray. For amorphouspolymers, cooling of the formed part against a near-isothermal mold rarely controlsthe overall thermoforming cycle. For crystalline and crystallizing polymers such asPET, PP and HDPE, the cooling cycle can be long and can govern overall cycle time.

When the sheet has cooled sufficiently to retain its shape and, to a large extent,its dimension, it is stripped from the mold and transferred to a trimming station,where the web is separated from the product. Requisite holes, slots and cut-outs aredrilled, milled or burned into the part at this time. Thin-gage roll-fed sheet can beeither trimmed on the mold surface immediately after forming or on an in-linemechanical trimming press. Heavier-gage formed sheet is usually removed from themold and manually or mechanically trimmed on a remote station. Trimming is reallya solid phase mechanical process of crack propagation by brittle or ductile fracture.Care is taken when trimming brittle polymers such as PS or PMMA to minimizemicrocracks. With brittle polymers, the very fine sander, saw or microcrack dustgenerated by mechanical fracture can be a serious problem. Other polymers such asPP and PET are quite tough and require special cutting dies. Dull steel-rule diescause fibers or hairs at the cutting edge of thin-gage fiber-forming polymers such asPET and PP. Slowly crystallizing polymers pose registry problems when in-linetrimming presses are used. The speed of trim cutting and the nature of the cuttingsurface control the rate of crack propagation through the plastic. There are noexhaustive studies of the unique trimming and cutting characteristics of thermo-formed polymers and so much information must be inferred from other sourcesincluding extensive studies on the machining of plastics.

5.2 Overall Cooling Heat Balance

As a first approximation, as the sheet touches the mold, it is assumed to be at anaverage, equilibrated temperature, Tequil, as described in Section 3.13. When the

1 The unique processing conditions for crystallizing polyethylene terephthalate (CPET) are de-

scribed in detail in Chapter 9, Advanced Thermoforming Processes.

average sheet temperature reaches the set temperature, Tset, as given in Table 2.5,the sheet is assumed to be sufficiently rigid to be removed from the moldsurface. Typically, the amorphous polymer set temperature is about 200C or400F below its glass transition temperature, Tg. The crystalline polymer set temper-ature is about 200C or 400F below its melting temperature, Tmelt. During thecooling time, the mold temperature is assumed to be essentially constant atTmold. The amount of heat to be removed by the coolant flowing through the moldis given as:

(5.1)where V* is the volume of plastic, p is its density and cp is its heat capacity1. Heatis removed from the free sheet surface by convection to the environmental air. Heatis removed from the sheet surface against the mold surface by conduction throughthe mold to the coolant. The coolant fluid removes the heat by convection. The heatload at any point on the mold surface depends on the sheet thickness, tlocal, at thatpoint. Sheet thickness, as noted in Chapter 4, is not uniform across the mold surface.The heat load at any point is given as:

(5.2)This is the heat to be removed from a given region during the cooling portion of thetotal cycle, 9cool. The local heat flux then is:

(5.3)

The units on q['ocal are kW/m2 or Btu/ft2 h 0F. The total heat load during this timeis given as:

(5.4)

The total heat load per unit time on a steady-state process is given as:

(5.5)where N is the number of parts produced per unit time. The units on Qsteady state arekW or Btu/h.

At steady-state conditions, this energy is removed by the coolant system and byconvection to the environmental air. Mold, coolant and air temperatures increaseuntil the steady state is reached.

1 The last equality relates the amount of heat removed to the differential enthalpy, AH, of the

polymer. This expression should be used if the polymer is crystalline and molten during theforming step and is solidifying during the cooling step.

5.3 Cooling the Formed Shape

Consider a typical cooling step. The sheet of variable thickness but known tempera-ture is pressed against a slightly irregular surface of a mold. The properties of themold material are known and uniform throughout its volume, Chapter 6. Coolant ofknown properties flows through uniformly spaced channels in the mold. The freesurface of the part is also cooled. At any instant, the temperature profile through thevarious layers of material is as shown in schematic in Fig. 5.1. The rate at which theenergy is removed from the plastic to ambient air and coolant depends on the sumof resistances to heat transfer through each of these layers. Heat removal by thecoolant is the primary way of cooling the formed part. Transfer to the environmentalair is a secondary method but it can be quite important when trying to optimize cycletime. There are two aspects to heat removal from the formed polymer sheet to thecoolant. The first deals with the overall heat transfer at steady state conditions. Thesecond focuses on certain aspects of cyclical transient heat transfer.

Series Thermal Resistances

Figure 5.1 Schematic of various thermal resistances for sheet cooling against thermoforming mold

Ambient EnvironmentMold

T o

Free Surface Film Resistance

Air Gap

Plastic SheetMold

Surface Irregularities

Coolant Film Resistance

Coolant *

T0

Tp

L

Figure 5.2 Typical serpentine coolant flow channel through thermoforming mold

5.4 Steady State Heat Balance

Consider the limiting case where all the energy is transferred directly to the mold andthence to the coolant. Typically, coolant lines are drilled or cast on a discrete, regularbasis parallel to the mold surface (Fig. 5.2). Heat removal by coolant depends onconvection heat transfer, or fluid motion1. The total amount of energy removed bythe coolant embedded in the mold is given as:

Q = UA AT (5.6)where U is the overall heat transfer coefficient, A is the coolant surface area and ATis the increase in coolant temperature between inlet and outlet portions of the flowchannel. The overall heat transfer coefficient includes all flow resistances between thesheet and the coolant. It is usually written as:

where R1 represents the ith resistance to heat transfer. As seen in Fig. 5.1, for flowingfluids, there is a convective film resistance at the conduit surface, l/7iDhc, where TTDis the circumference of the conduit. If the coolant fluid is not kept clean, the coolantchannel can become coated with residue, thus increasing thermal resistance. This

1 This section deals only with the convection heat transfer coefficient of coolant flowing through

lines in the mold. Section 6.4 considers coolant pressure drop-flow rate relationships for thespecification of coolant line size.

Baffle Mold Insert Flow Channel

Mold Base

Coolant

1 Information extracted from [1] by permission of copyright holder

resistance is called a fouling factor, ff. Fouling factors are given in Table 5.1 [I].The resistance through the mold depends on the relative shapes of the mold surfaceand the coolant channels, and is usually described as 1/Skm, where S is a shape factorand km is the thermal conductivity of the mold material. Since polymer sheet doesnot press tightly against the mold surface, there is a conductive resistance owing totrapped air, l/ha. Although there may be other thermal resistances, these are theprimary ones. So the overall heat transfer coefficient, U, is written as:

Interfacial Resistance

In most heat transfer processes, intimate or perfect contact between the hot and coldsolids is assumed. Imperfect contact causes resistance to heat flow. Mold surfacewaviness and microscopic roughness or asperities reduce physical contact (Fig. 5.3).Increasing pressure against the sheet increases physical contact and reduces theresistance to heat transfer. In general, energy is transmitted across the interstices bya combination of:

Table 5.1 Fouling Factors for Coolant Lines1

Coolant

Treated make-upcooling tower water

Treated make-upcooling tower water

City waterCity waterRiver waterRiver waterTreated boiler

feedwaterTreated boiler

feedwaterIndustrial heat

transfer oilEthylene glycolGlycerine-waterBrineBrineSteam

Condition

Water < 125F

Water > 125F

Water < 125FWater > 125FWater < 125FWater > 125FWater < 125F

Water > 125F

Temperature < 125FTemperature > 125F

Fouling factor(Btu/f t -h^F)- 1Velocity < 3 ft/s

0.001

0.002

0.0010.0020.0020.0030.001

0.001

0.001

0.0010.0020.0020.0030.001

Velocity >3 ft/s

0.001

0.002

0.0010.0020.0010.0020.0005

0.001

0.001

0.0010.0010.0010.0020

Figure 5.3 Interfacial resistance between sheet and thermoforming mold surface. Left shows sub-stantial thermal resistance owing to large air gap. Right shows reduced thermal resistance

Conduction at the asperities, Conduction through the interstitial fluid, and Radiation.The resistance is thus a function of: The contacting material properties such as

Relative hardness,Thermal conductivity,Surface roughness andFlatness,

The conductivity and pressure of the interstitial fluid, and The pressure applied against the free surface of the sheet.The interface coefficient, ha, is a measure of thermal resistance across the gap. It issimilar in concept to the confection heat transfer in that resistance to heat flowdecreases with increasing value of ha. For perfect contact, ha- oo.

In thermoforming, the interstitial fluid is air, perhaps at a substantially reducedpressure. If the interface is a uniform air gap of 8 = 0.025 cm or 0.010 in and airthermal conductivity is kair = 0.029 W/m 0C or 0.0167 Btu/ft h 0F, the value forha is about 114 W/m2 0C or 20 Btu/ft2 h 0F. Contact heat transfer coefficientvalues between flowing polymer melts and mold surfaces of about ha = 568W/m2 0C or 100 Btu/ft2 h 0F have been reported in injection molding [2-4].Similar values are expected here. For two surfaces in contact, ha = haopn, where p isthe applied pressure and hao depends on the relative waviness and roughness of thetwo surfaces, Table 5.2 [5]. As seen, n has a value of about 2/3 for both rigid-rigidand rigid-flexible material contact in vacuum. Values for hao are typically 50 timesgreater in air than in hard vacuum.

There are no available data for interfacial resistance during thermoforming ofsoftened plastic sheet against various types of mold surfaces. For hot sheet pressedagainst a relatively smooth, heated mold surface at a relatively high differentialpressure, it is expected that an appropriate value for ha would be about 568W/m2 0C or 100 Btu/ft2 h 0F. For rapidly cooling and rigidizing plastic sheetpressing against a highly textured, cold mold surface with a modest differentialpressure, an appropriate value range for ha of 114 to 284 W/m2 0C or 20 to

Cold Mold Moid Asperities Hot Mold

Cold, Rigid or Heavy-Gage Sheet Trapped Air Hot, Flexible or Thin-Gage Sheet

50 Btu/ft2 h 0F should be considered for first cooling time estimates. Section 5.6 oncomputer simulation of the cooling process explores the relative effect of interfacialresistance on time-dependent sheet cooling.

Shape Factor

For thin metal molds, heat is conducted very rapidly from the plastic to the coolant.For relatively thick molds of: Plaster, Wood, Epoxy, Glass fiber-reinforced unsaturated polyester resin (FRP), Pressed fiberboard, or Any other nonmetallic material,heat transfer is slowed by low mold material thermal conductivity. If the coolantsystem is considered to be coplanar with the mold surface (Fig. 5.4), the resistance toheat transfer per unit length (L = 1), across a mold D units thick is given as:

Rn, = ^- (5.9)where km is the mold material thermal conductivity. Thermal conductivity values formany mold materials are given in Table 2.7. For round discrete conduits (Fig. 5.5)[6], a shape factor, S, is used:

(5.10)

(5.11)where S is given by:

Table 5.2 Contact Resistance and Conductance [5] Ji0 = Ii00P11

(Units on hc = W/m2 0C or Btu/ft2 h 0F)(Units on p are MPa or lbf/in2)

Materialcontact

Elastic deformationtheory

Hard-to-hardHard-to-hardHard-to-soft

Innerlayer

None

VacuumAirVacuum

Contact coefficient, Ji00

(W/m2 C)

35.8

113.65680

170

(Btu/ft2 h 0F)

6.3

201000

30

n

2/3

2/31/62/3

Figure 5.5 Geometric factors for mold shape factor analysis

Figure 5.6 gives this equation in graphic form. It is apparent that the thermalresistance of the mold decreases with increasing value of S, which is achieved withmany large-diameter coolant lines placed relatively close to the mold surface. Thetypical value range for S is 2 < S < 3. Example 5.1 illustrates the relative effects ofthese parameters on mold thermal resistance.

Example 5.1 Shape Factors and Mold Thermal ResistanceDetermine the relative thermal resistances for the following two molds:Mold 1: Thin-walled aluminum mold with km = 131 Btu/ft h 0F, having d= 1/2-inwater lines on P = 2 in centers, with the center line being D=I in from the moldsurface.

Hot Sheet

MoldCoolant Channel P D

d

Figure 5.4 Schematic of coplanar coolant flow channel and mold surface, concept frequently calledflooded cooling

Coolant

Mold

Mold

Mold 2: Thick-walled plaster mold with km= LO Btu/ft h 0F, having d= 1/2-inwater lines on P = 4 in centers, with the centerline being D = 2 in from the moldsurface.

Mold 1P/d = 4, D/d = 2. From Fig. 5.6, S = 2.

Mold 2P/d = 8, D/d = 4. From Fig. 5.6, S = 1.6.

^ i = U T T o = 0-625The plaster mold has more than 160 times the thermal resistance to heattransfer than the aluminum mold.

Coolant Channel Spacing to Diameter Ratio, P/d

Figure 5.6 Effect of coolant line location on mold shape factor

Convection Heat Transfer Coefficient

The metal molds on automatic, roll-fed thin-gage thermoformers and on manyheavy-gage forming operations are actively cooled, with water being the primary

Shap

e Fa

ctor, S

D/d = 1

coolant. There is a thermal resistance between the cool bulk flowing fluid and thewarmer tube wall (Fig. 5.1). The primary dimensionless group used in fluid mechan-ics is the Reynolds number, Re:

Re = ^ P (5.12)where D is the tube diameter, v is the fluid velocity, p is the density of the fluid and|i is its Newtonian viscosity [7]. The Reynolds number is the ratio of inertial toviscous forces for the fluid. Slowly moving fluids are laminar when Re < 2000.Convection heat transfer to slowly moving fluids is poor. Rapidly flowing fluids arefully turbulent when Re > 10,000 and heat transfer is very rapid. Example 5.2illustrates the interaction between flow rate and Reynolds number.

Example 5.2 Water as CoolantFlow RatesConsider 21 C or 700F water flowing through 0.5-in or 1.27-cm diameter coolantchannels. Determine the Reynolds number and the flow characteristic if the velocityis a) 0.52 ft I s or 0.16 m/s and b) 2.6 ft/s or 0.79 m/s. What are the volumetric flowrates at these velocities?

The water density is 62.4 lb/ft3. The viscosity is 0.658 x 10"3 lbm/ft s. TheReynolds number is:

,^_ft.4^._l__3*,.vFor v = 0.52 ft/s: Re = 2050 and the water is laminar.For v = 2.6 ft/s: Re = 10,300 and the water is turbulent.

The volumetric flow rate is given as:

V = ^ ! . v = 0 .00136-v^ = 0.612-v-^-4 s mmFor v = 0.52 ft/s, the flow rate is 0.32 GPM.For v = 2.6 ft/s, the flow rate is 1.6 GPM.

As discussed in Section 3.6, energy interchange between solid surfaces andflowing fluids is by convection. The proportionality between heat flux and thermaldriving force is the convection heat transfer coefficient. The convection heat transfercoefficient is obtained from standard heat transfer theory and experiments. There aremany methods for calculating values of hc. In general, however, the Chilton-Colburnanalogy between resistance to fluid flow and resistance to thermal energy flow yieldsadequate results [8]. The analogy states:

S t . P r 2 / 3 = f / 8 ( 5 1 3 )

where St is the Stanton number, Pr is the Prandtl number and f is the coefficient offriction or friction factor. The Stanton and Prandtl numbers are:

where h is the convective heat transfer coefficient, p is the fluid density at theappropriate temperature, cp is the fluid heat capacity, v is the average fluid velocity,v = u/p, is the kinematic viscosity, and a = k/pcp, is the fluid thermal diffusivity. Inessence, Pr is the ratio of inertial to thermal properties and St is the ratio of fluid tothermal resistances. Table 5.3 gives appropriate Prandtl number values for severalcoolants [9,10]. As examples, Pr ^ 7 for room temperature water and Pr 300 for

(5.14)

(5.15)

Table 5.3 Prandtl Number Values For Several Coolants[9,10]

Coolant

AirAirSteamSteamWaterWaterWaterWaterWater

SAE 30 OilSAE 30 OilSAE 30 OilSAE 30 OilSAE 30 OilSAE 30 OilGlycerineGlycerineGlycerineGlycerineGlycerineAirAirAirLight OilLight OilLight OilLight Oil

Temperature(0F) (0C)

32100212400

3270

10015020060

100150200250300507085

10012032

300600105150250300

038

100204

02138669316386693

12114910212938490

1503154065

120150

Prandtl no.

0.720.720.960.94

13.76.824.522.741.88

1170340122623522

31,00012,5005,4002,5001,600

0.720.710.685

340623522

10O0C oil coolant. Examples 5.3 and 5.4 show the relative effectiveness of convectionheat transfer to water and oil, respectively, as coolants1.

Example 5.3 Water as CoolantTemperature IncreaseA 3 ft x 3 ft x 0.125-in thick plastic sheet is initially at 31 5F and is to be cooledto 2000F. The sheet specific heat is 0.5 Btu/lb 0F and the density is 70 Ib I ft3.Twelve sheets are thermoformed per hour. Determine the increase in coolanttemperature.

Heat load from sheet:Q (per cycle) = pcpV*AT = 574 Btu/cycle

Q = 574 12 = 6888 Btu/h to be removed by coolant.Consider 700F water flowing through a 3/4-in diameter coolant channel atv = 4 ft/s. From Example 6.1, the Re = 50,000. The flow is turbulent andf = 0.0248. Pr = 7.02.

St Pr2/3 = f/8

h^ R ^ = 770Btu/f t2-h-F

Consider a 3 ft x 3 ft aluminum mold containing four waterlines spacedevenly. Consider the water lines to be I^ in from the mold surface. ThusP/d = 12 and D/d = 2. From Fig. 5.6, the shape factor, S 3. The thermalconductivity of aluminum, from Table 2.12, km = 72.5 Btu/ft h 0F. Con-sider ha = oo. As a result, the total thermal resistance, U is given as:

U = 0.75Tr.770 + 3^7l5 = 0 - 0 1 1 2

or: U = 89.3 Btu/ft2 h 0F.Surface area of coolant channels, A = TtDLN, where D is the coolant

diameter ( = 3/4-in), L is the channel length (=10 ft), and N is the numberof channels ( = 4). A = 7.85 ft2.

The increase in coolant temperature is given as:_ ^

= _ 6 8 8 8 _UA 89.3-7.85

This is considered to be a typical temperature increase for turbulent coolantflow.

1 These examples use fluid flow data of Examples 6.1 and 6.2 discussed in detail in Chapter 6 on

mold design.

Example 5.4 Oil as CoolantTemperature IncreaseA 3 ft x 3 ft x 0.125-in thick plastic sheet is initially at 375F and is to be cooledto 2000F. The sheet specific heat is 0.5 Btu/lb 0F and the density is 70 Ib I ft3.Twelve sheets are thermoformed per hour. Determine the increase in coolanttemperature. Comment on the results.

Heat load from sheet:Q (per cycle) = pcpVAT = 574 Btu/cycle

Q = 574 12 = 6888 Btu/h to be removed by coolant.Consider 1500F SAE 10-like oil flowing through a 3/4-in diameter coolantchannel at v = 4 ft/s. From Example 5.2, the Reynolds number, Re = Dvp =1925. The flow is laminar and f = 0.0333. From Table 5.3, the Prandtlnumber, Pr = 290 and Pr2/3 = 44.

St Pr2/3 = f/8

h - (pCpV) - 38 Btu/ft2 h 0Fc ~ 8 (29O)2/3"" '

Consider a 3 ft x 3 ft aluminum mold containing four waterlines spacedevenly. Consider the water lines to be \\ in from the mold surface. ThusP/d = 12 and D/d = 2. From Fig. 5.6, the shape factor, S 3. The thermalconductivity of aluminum, from Table 2.12, km = 72.5 Btu/ft h 0F. Con-sider h a = oo. As a result, the total thermal resistance, U is given as:

U = 0.757T 38 + 3 72.5 = ' ! 3 9

or: U = 7.22 Btu/ft2 h 0F.Note that this value is less than one tenth that of water flowing at the

same velocity. Surface area of coolant channels, A = 71DLN, where D is thecoolant diameter ( = 3/4-in), L is the channel length (=10 ft), and N is thenumber of channels ( = 4). A = 7.85 ft2.

The increase in coolant temperature is given as:_ 6888UA 7.22-7.85

Since the sheet must be cooled to 2000F, this is an unacceptable increase inoil temperature. One solution is to increase the oil flow rate. Doubling theflow rate will increase the pressure drop by a factor of just less than four. Asecond alternative is to increase the number of coolant lines. Doubling thenumber of coolant lines will increase the coolant surface area but the oilvelocity will drop to half its current value, thus driving the flow even deeperinto the laminar region. Since the friction factor is inversely proportional tothe fluid velocity, the friction factor will double. However, since the heattransfer coefficient is essentially independent of velocity, the heat transfercoefficient will not be affected. In addition, the decreased spacing will

increase the shape factor value, S. As a result, doubling the number ofcoolant lines results in a more than double increase in the overall heattransfer coefficient and a resulting more than halving of the increase in oiltemperature.

More exact calculations require knowledge of the roughness and geometriccharacteristics of the flow channel. In general, the Nusselt number, Nu, a dimension-less convective heat transfer coefficient is a product of the Reynolds number and thePrandtl number, as:

Nu = ^ = C Rem Prn (5.16)

For fully developed turbulent flow in very smooth tubes, C = 0.023, m = 0.8 andn = 0.4, and the equation is called the Dittus-Boelter equation. For developingturbulent flow in smooth tubes, m = 0.8, n = 0.33 and C is given as:

/D\0.055

C = 0.036 - (5.17)where D is the channel diameter, L is the channel length and 10 < L/D < 400. Thisequation is usually called the Nusselt equation.

For laminar flow in smooth tubes, the Sieder-Tate version is frequently used. Itis much more dependent on entrance length effects and so has m = 1/3, n = 1/3 andC as:

/DY/3C =1.86 - (5.18)

where Re Pr (d/L) > 10. Example 5.5 illustrates heat transfer coefficient valuesobtained from these equations for a relatively simple flow channel design. Fornoncircular channels, the diameter is replaced with the hydraulic diameter, Dh, givenas:

D h = ^ (5.19)where A is the cross-sectional area of the flow channel and P is the wetted perimeter.

Example 5.5 Convective Heat Transfer Coefficients for Serpentine MoldChannelFigure 5.7 shows an example of a serpentine mold channel that has a roughnessvalue, e = 0.001 D. Two coolants are to be evaluatedwater at 700F and oil at1500F. Determine the relative heat transfer effectiveness of these coolants in this

flow channel.

Water as a coolant

v, kinematic viscosity= 1.06 x 10~5 ft2/sp, density = 62.4 lb/ft3

Figure 5.7 Coolant flow channel around mold insert

k, thermal conductivity = 0.35 Btu/ft h 0Fv, average velocity = 4 ft/s

Prandtl number = 7Reynolds number, R e = 15,700Flow is turbulentThe friction factor-Reynolds number equation is:

f = 0.0204 + 4.212 Re" 0 6 4 2 = 0.0289

Stanton/Colburn analogy:

h = (P^)_ fp r2/38

h, convective heat transfer coefficient = 886 Btu/ft2 h 0FNu, Nusselt number = hD/k = 106

Dittus-Boelter fully developed flow:Nu = 0.023 Re0-8 Pr0-4 = 114.h = 958 Btu/ft2 h 0F

e Nusselt developing flow (L = 20):/ I \0.055

Nu = 0.036 Re0 8 Pr0-33 = 127.\4oy

h = 1061 Btu/ft2 h - 0 F Nusselt developing flow (L = 6):

N u = 136, h = 1137 Btu/ f t 2 -h- 0 F Weighted average:

Nu = 131, h = 1095 Btu/ft2 h 0F

Coolant

Mold Base

Mold Cavity

Flow Channel

Plug

The effect of developing flow is higher convective heat transfer.

Oil as a coolant

v, kinematic viscosity= 1.35 x 10~4 ft2/sp, density = 54.5 lb/ft3k, thermal conductivity = 0.082 Btu/ft h 0Fv, average velocity = 4 ft/s

Prandtl number, Pr = 290Reynolds number, Re = 1235Flow is laminarThe friction factor-Reynolds number equation is:

f = 16/Re = 0.0518

Stanton/Colburn analogy:

8

h, convective heat transfer coefficient = 58 Btu/ft2 h 0FNu, Nusselt number = hD/k = 29.5

Sieder-Tate for developing flow (L = 20):Nu = 1.86 [Re Pr (D/L)]2/3 = 20.8h = 41 Btu/ft2 h - 0F

Sieder-Tate for developing flow (L = 6):Nu = 46.4, h = 91 Btu/ft2 h 0F

Averaged developing flow:Nu = 32.1, h = 63 Btu/ f t 2 -h- 0 F

Again, convective heat transfer in developing flow is higher than with fullydeveloped flow.

The arithmetic is considerably simplified for water at 210C or 700F. For laminarflow, the heat transfer coefficient is obtained from [H]:

hlam = 3.52 ^ -j Re1 / 2( j-J (5.20)where k is the thermal conductivity of the fluid, and D and L are the diameter andlength of the flow channel, respectively. For turbulent flow in water at 210C or 700F,the heat transfer coefficient is:

/k\ / r> \ 0.055

hturb = 0.068(^-jReO8^ rJ (5.21)The relative values for laminar and turbulent water flow in a typical mold are givenin Example 5.6.

Example 5.6 Laminar and Turbulent Heat TransferDetermine the relative effect on heat transfer coefficient when the flow rate isincreased so that flow moves from laminar to turbulent flow. Consider aserpentine coil imbedded in an epoxy mold, where Lid 100. From Equations5.20 and 5.21, determine the relative effect on the Biot number, the relativeheat transfer coefficient. Then determine the actual heat transfer coefficientsfor water flowing in a 0.500-in or 1.27 cm diameter coil. The thermalconductivity of water is 0.35 Btu/ft - h -0F. Then, determine the heat transfercoefficient in the transition region, 2000 < Re < 10,000.

For laminar flow, Re < 2000. Consider Re = 2000. From Equation 5.20, theBiot number, Bi = hD/k is given as:

hD /D\ 1 / 2Bi = = 3.52 Re1/2 - = 3.52 20001/2 (0.01)1/2 = 15.7k Vw

or fully turbulent flow, Re > 10,000. Consider Re = 10,000. From Equation5.21, the Biot number is:

hD /)\0.055Bi = - = 0.068 Re0-8 - = 0.068 100008 (0.0I)0-055 = 83.7

k VWNote that increasing the flow rate by a factor of five increases the relativeheat transfer coefficient by a factor of 83.7/15.7 = 5.3. The dimensionalconvection heat transfer coefficient for the coolant is given as:

l W i a m = 15.7 - 0.35 f t . h . o F Qjjt = 132 ft2 fa o p = 750 ^ - ^

hwater,turb = 83.7 0.35 12 = 700 ft2 ^ o p = 4000 ^ ^

Approximate values for heat transfer coefficients in the transition region,2000 < Re < 10,000 are obtained by averaging. For example, for Re = 5000,the approximate heat transfer coefficient is:

Since the heat is conducted perpendicular to the mold plane, the fluid resistancemust be corrected for the flow channel diameter and the number of flow channels ina given mold area (Fig. 5.1). If there are N flow channels of diameter d, then the arearatio is TTDNL/L2 , and the effective resistance is:

R

= s ^

5.5 Cyclic Heat Balance

The total amount of energy to be removed from the plastic sheet by the coolant iseasily determined for any given cycle from Equation 5.1. At steady state, this energyis usually transferred to: The metal mold, and thence to the coolant, Cooler metal surfaces such as stripper bars and sheet clamps, and thence to

surrounding air, Surrounding air, and/or Sprayed liquids such as water mist that then evaporate.Consider the limiting case where a plastic sheet of uniform temperature Tp anduniform thickness tp is brought in contact with a mold having an initially uniformtemperature Tm and thickness tm. Consider the mold to be made of a high thermalconductivity metal that is in contact with a coolant having a uniform temperatureTm. The free surface of the plastic sheet is insulated. The time-dependent temperatureprofile through the mold and the plastic sheet are shown in schematic in Fig. 5.8.Typically, the initial interfacial temperature between the sheet and the mold is muchcloser to that for the mold than that for the plastic. As cooling proceeds, theinterfacial temperature falls.

Tem

pera

ture

Time

Figure 5.8 Characteristic time-dependent temperature profiles of thermoformed part cooling againsta metal mold

Thin Section on Free Surface

Mold Surface Temperature

Sheet Contacting Mold Surface

Thickest Section on Free Surface

Initial Sheet Temperature

Cooling the Free Surface of the Sheet

There are four general methods for cooling the free surface of the sheet:

No cooling. In this case, the surface is essentially insulated from the environment, Natural air cooling. That is, no fans or forced air cooling methods are used. Forced air cooling. This can be as simple as clock-timer actuated shop floor fans

or as complex as special-purpose high-velocity blowers, and Water fog or mist cooling. The mist is usually clock-timer controlled to shut off

some time prior to sheet removal from the mold surface. This allows the finewater drops to evaporate before the part is removed.

Free surface cooling effectiveness increases as one progresses down this list. Themeasure of effectiveness of heat removal is the convective heat transfer coefficient, h.Table 5.4 summarizes the relative values of convective heat transfer coefficients forthe last three free surface cooling methods. The convective heat transfer coefficientfor an insulated surface is h = 0. Matched metal molds are used if the sheet is foam,reinforced or highly filled, or if close tolerance is needed in certain regions. In thiscase, the effective heat transfer coefficient, heff= oo. These methods are shown inschematic in Fig. 5.9.

Cooling Thin Sheet in Ambient Air

When a thin sheet of plastic is heated to the forming condition, it must be quicklytransferred to the mold surface to minimize heat loss to the surrounding air. For verythin sheet, less than 0.005 in or 0.01 cm in thickness, the sheet is usually heated by

Table 5.4 Cooling a Part on the MoldRelative Values of Thermal Resistance

Physicalresistance

Free surface coolingAmbient airForced airWater spray

Interface gap**MoldAluminumPlaster

CoolantLaminar water (L/7idNhCOOI)Turbulent water (L/7tdNhcool)

Form forresistance

1/hair1/hairl/hmistl/hc

D*/kmdto

Typical reciprocal value

(W/m2 0C)

2.84 to 5.6828.4 to 56.8284 to 568114 to 284

34,33021

4544000

(Btu/ft2 h 0F)

0.5 to 15 to 10

50 to 10020 to 50

60003.7

80700

D* is the effective mold thickness and includes shape factor** See Table 5.2 for these values

Figure 5.9 The effect of various free-surface cooling techniques on time-dependent temperatureprofiles of thermoformed part cooling against a metal mold

contact heat transfer, then blown from the heater directly onto the mold surface.There is an extensive discussion of transient heat transfer to thin sheet in Chapter 3.This lumped-parameter analysis is directly applicable to the cooling of thin sheet inambient air. The dimensionless temperature, Y, is given as:

(5.23)where T1 is the initial sheet temperature, Tair is the air temperature, 6 is time, a isthermal diffusivity, k is thermal conductivity, h is convection heat transfer coefficientbetween the sheet and the ambient air, and L is the sheet thickness. Note that thecooling time, 0 is proportional to the sheet thickness to the first power. In contrast,for conduction-controlled heat transfer, the cooling time is proportional to thesquare of sheet thickness. The lumped parameter cooling time curves for various filmthicknesses are compared with experimental data in Fig. 5.10. Since nothing isknown about the experimental processing conditions or sheet material parameters,the curves are fit at L = 0.15 cm or 0.060 in by selecting values for the convection

Tem

pera

ture

Tem

pera

ture

h=O,lnsulated Cool AirInitial Sheet

TemperatureIncreasingTime

Mold SurfaceTemperature

Mold Free Surface Mold Free Surfaceh= oo, Matched MoldsWater Spray

Mold Free Surface Mold Free Surface

heat transfer coefficient, h. The shapes of these convection-controlled curves agreebetter with the data than do the conduction-controlled curves.

Transient Heat Removal From the Sheet

The energy in the sheet is removed by conduction to the mold and free surfaces. Thetransient one-dimensional conduction heat transfer equation, Equation 3.4, applies:

(5.24)

(5.25)(5.26)(5.27)

Equation 5.25 is the heat flux from the free sheet surface to the ambient air, withhConv a s the convection heat transfer coefficient, Table 3.2. T1 is the initial sheettemperature at the time of contact with the mold surface. If draw-down onto thesheet surface is very rapid, T1(X) is represented by the equilibration temperatureprofile discussed in Section 3.13. As a first approximation, T1(X) = T1, an averagesheet temperature. Two limiting conditions bound the solution of this equation andits attendant boundary conditions.

Quiescent Ambient Air

When heat transfer to air is very small relative to the rate of energy conducted tothe mold, h->0, the free surface is approximated by an insulated surface. This isusually the case for natural convection of heat to quiescent air. Figure 5.11 is used

Figure 5.10 Effect of sheet thickness on cooling timefor con vection-controlled thin sheet. Solid lines arecalculated. Dashed lines are experimental

Time,

s

Sheet Thickness, ft

Calculated h = 10 Rtu/ft2hFAmbient Air

Forced Air

Water Spray

subject to the following boundary conditions:

Dimensionless Time, Fourier Number

Figure 5.11 Time-dependent average sheet temperature as a function of the rate of heat loss fromthe free surface. Dimensionless temperature, Y = (T T0V(Tj T0) where Tj is the initial sheettemperature and T0 is the mold surface temperature and ambient air temperature. Biot number,Bi = hL/k, where h is the convection heat transfer coefficient and k is polymer thermal conductivity.Fourier number, Fo = a0/L2 where a is the thermal diffusivity, 9 is time and L is sheet thickness

to obtain the dimensionless time, Fo = a9/L2, with the dimensionless temperature, Y,given as:

( T - Tmold)CTi-Tmold)

where T1 is the average sheet temperature. Since hconv = O, Bi = hL/k, the Biot numberis zero. L is the actual thickness of the sheet. This is illustrated in Example 5.7.

Example 5.7 Cooling Time for Heavy-Gage PS SheetI

A polystyrene sheet 0.438 in or 1.11 cm thick initially at 375F or 191C is cooledby pressing it against a mold at 75 F or 24 C. Determine the time required to coolthe sheet to an average temperature of 175F or 79 C. PS thermal conductivity is0.073 Btu I ft -h-F or 0.0003 cal/g s 0C. PS density is 65.5 Ib /ft3 or 1.05g/cm3. PS specific heat is 0.5 Btu jib 0F= 0.5 cal/g 0C. The free surface isexposed only to ambient air. Assume Zz00 = O.

Dim

ensi

onle

ss Te

mpe

ratu

re

Free Surface,ConvectionMold, Conduction

T

ToT o

Biot Number, hL/k

The dimensionless average temperature, Y is given as:

(175-75)Y =

(375-75) = -3 3 3

From Fig. 5.11, the Fourier number at this value of Y for Bi = O isFo = 0.345. The polymer thermal diffusivity is given as:

a = = 5.7x 1 0 - 4 = 2.2 x 10~3 ^p-c p s h

The cycle time is given as:r/fl 1

F o = j - 2 = 5.7 XlO"4 - ^ 5 - 6[S] = 0.345

or 8 = 746 s. This is shown as the top line of Table 5.5.

Moving Ambient Air

Figure 5.11 is also used to approximate the relative effect of air movement on thecooling time, so long as T a i r Tmold. Now the dimensionless temperature, Y, is givenas:

v ~ mold) (T Tair)(T,-Tmoldr(T,-Tair) p - 2 y )

Note that as Bi increases, through increase in the value for the convection heattransfer coefficient, the cooling time drops rapidly. When Bi = oo, the dimensionlesstime to reach the same dimensionless temperature drops to one-fourth that for thequiescent case (Bi = 0). This directly supports the discussion in Chapter 3 that thevalue of L should be half the actual sheet thickness when energy interchange issymmetric across the sheet. Example 5.8 illustrates the effect of heat transfercoefficient value on cooling time.

Example 5.8 Cooling Time for Heavy-Gage Polystyrene SheetIIUsing the data from Example 5.7, determine the reduction in cooling time if the

following free surface coolants are used:a) Ambient air with a heat transfer coefficient of

h = 1 Btu/ft2 -h'F or 5.68 W/m2 0C,b) Forced air with a heat transfer coefficient of

h=10 Btu/ft2 -h 0F or 57 W/m2 0C, andc) Water mist with a heat transfer coefficient of

h = 100 Btu/ft2 -h-F or 570 W/m2 0C.

For each case, the Biot number Bi = hL/k.. 1 Btu 0.438 1 ft h 0F

A ca ) B l =Fl^-I^ f t '0^-Btu- = 0-5

The value of Fo for Y = 0.333 and Bi = 0.5 from Fig. 5.11 is Fo = 0.256 or0 = 554 s. This is about 26% reduction in cycle time from the insulatedexample, Bi = 0 of Example 5.7.

b) Bi = 5. The value of Fo from Fig. 5.11 is 0.123 and the cycle time9 = 266 s. This is a 64% reduction in cycle time.

c) Bi = 50. The value of Fo from Fig. 5.11 for Bi = oo is 0.086. The bestcycle time possible is 0 = 187 s. This is a 75% reduction in cycle time fromthe insulated example.

These values are tabulated in Table 5.5.

The effect of altering free surface cooling conditions is seen in Fig. 5.12 [12]. Thenature of the polymer and the details about the processing conditions are unknown.For L = 0.15 cm or 0.060 in sheet, water spray reduces the cooling time by about60%. For thicker sheet, L = 0.45 cm or 0.180 in, water spray reduces the cooling timeby about 65%. As expected, the cooling time increases with increasing sheet thick-ness. As expected, the effect is most obvious with ambient air cooling.

On the other hand, if conduction heat transfer through the plastic sheet controlsthe cooling time, the effect of sheet thickness is easily determined. At any givenaverage sheet temperature, the dimensionless time, Fo, is a simple function of theBiot number:

Fo = f(Bi) (Y fixed) (5.30)

Figure 5.12 The effect of various cooling tech-niques on cooling time as function of sheet thick-ness, theory from Fig. 5.11. Experimental datafrom [12]Sheet Thickness, ft

Time,

s

Ambient Air

Forced Air

Water Spray

Calculated

Initial sheet temperature, T1 = 190.60C or 375FMold ambient temperature, T0 = 23.90C or 75FFinal average temperature, Ta = 79.4C or 175FThermal conductivity = 0.0003 cal/gm s 0C or 0.073 Btu/ft h 0FThermal diffusivity = 0.00057 cm2/s or 0.0022 ft2/hFrom these values, the following are found:Y = 0.333Bi = hL/kFo = oc9/L2 = 0.0863[l + 3 exp( - 0.667 Bi0-667)]

As seen in Fig. 5.11, the approximate form for Equation 5.30 is:Fo = Fo00[I + p exp(-aBia)] (5.31)

Fo00 is the value of Fo when Bi = oo, the limiting case where both sides of the sheetcontact solid surfaces. When Bi = 0, Fo = 4Fo00. Since Fo = oc9/L2, the cooling time9 is proportional to the square of the sheet thickness for both limiting cases. It isapparent then, that when conduction from the plastic sheet controls heat transfer,doubling the sheet thickness always increases the cooling time by a factor of four.See also Table 5.5 where Equation 5.31 is used to compare cycle times for variousfree surface cooling modes.

As seen in Fig. 5.13, actual cooling times show a relationship that is more linearwith thickness. The observed values are substantially higher than the values obtainedfrom Fig. 5.12. This indicates that, at least in this case, the experimental sheet is notbeing cooled in a conduction-controlled environment. Again, the exact conditionsused to obtain the experimental data of Fig. 5.13 are unknown.

Cooling on Nonmetallic Molds

For metal molds, the time-dependent conduction resistance through the mold duringthe heat removal process from the sheet is usually negligible. Heavy-walled, non-

Table 5.5 Calculated Cooling Times for Polystyrene Sheet (Combined Convection/ConductionHeat Transfer)Type offree surfacecooling

InsulatedAmbient airForced airWater sprayDirectcontact ormatched die

Convection heattransfer coefficient

(W/m2 0C)

05.7

57570OO

(Btu/ft2 h 0F)

01

10100OO

Thin-gage sheet

L-0 .111 cm-0.0438 in

Biotno.

00.050.55.0OO

Fourierno.

0.3450.3230.2560.1230.086

Cycletime(S)

7.466.985.542.661.87

Heavy-gage sheet

L = 1.11 cm = 0.438 in

Biotno.

00.55.0

50OO

Fourierno.

0.3450.2560.1230.0860.086

Cycletime(S)

746554266187187

metallic molds alternately store and liberate heat during each forming cycle. Anestimate of the rate at which heat penetrates the mold is obtained by integrating thetime-dependent heat conduction equation, as discussed below. As with energy inputto plastic sheet, Chapter 3, there are two limiting conditions that are important. Ifthe energy transfer is constant heat flux and the mold is considered as a semi-infiniteslab of thermal diffusivity, oc, the depth of penetration of energy, 5, is approximately:

(Constant Heat Flux) 6 = v/6oc6 (5.32)

The penetration dimensionless time, Fo = aO/L2= 1/6 [13]. If the surface is raisedinstantaneously to a constant temperature, the depth of penetration of energy, 5, isapproximately:

(Constant Surface Temperature) 5 = ^/24ae (5.33)and the penetration dimensionless time, Fo = a0/L2 = 1/24. In thermoforming, theactual condition is closer to the constant flux approximation. If D is the effectivethickness of the mold, then the time for the heat to be felt at the coolant interface is:

D2 A D2

2 ^ < 0 < 6 ^ ^ 3 4>Example 5.9 illustrates how the type of mold material dictates the penetration timeof heat into the mold. For many thermoforming operations where actively coolednonmetallic molds are used, the coolant probably would not see all the energy removedfrom the sheet until after the sheet had been removed from the mold. The mold, ineffect, is storing the sheet thermal energy during the cooling cycle, then transferringit to the coolant after the cycle. The heat is also being convected away from the moldsurface during the times when no sheet contacts the mold surface, as shown in Fig.5.14. The approximate increase in mold temperature during the cooling cycle and theapproximate time required to return the mold to within 2% of its initial temperatureare illustrated in Example 5.10. For all practical cases where low-conductivity moldsmade of wood, plaster, fiberglass and even certain metal-filled epoxies are used, theaverage mold temperature continues to increase in value during the forming process

Tjme, s

Figure 5.13 Measured sheet thickness-dependentcooling times for various cooling techniques. Figureadapted from [23]Sheet Thickness, in

Ambient AirForced Air

Water Spray

Figure 5.14 Schematic thermoforming mold temperatureprofiles when mold is in contact with hot polymer [top] andwith ambient air [bottom]

until the energy added during the cooling portion of the cycle just equals thatextracted during the entire cycle time. From Equation 5.2, the amount of heat addedper unit area of the mold for a given cycle is:

(5.35)Example 5.9 Transient Heat Penetration into MoldsConsider two molds, a prototype plaster mold with a relative thickness D = 2 cm or5.1 cm, and a production aluminum mold with a relative thickness D = 0.2 in or0.51 cm. Determine the elapsed time for a heat pulse to travel to the reverse sideof each of these mold materials. The thermal diffusivity for aluminum is 0.49Cm2Is= 1.9ft2/h and for plaster is 3Ox 10~4 cm2/s or 11.6 x 10~3 ft2/h.From Equation 5.34, the penetration time is bracketed by D2/c a, wherec = 24 for constant surface temperature and c = 6 for constant heat flux. Foraluminum, D2/oc = 0.0816 s. For plaster, D2/oc = 1333 s. Thus the penetrationtime for aluminum is:

0.0034 s

Example 5.10 Approximate Increases in Cycle Time for Prototype MoldsConsider the case where a prototype plaster mold heats during contact with theformed sheet. If the initial mold temperature is 75F or 24C and the final moldtemperature is 125F or 52C, estimate the average sheet temperature at 746 s (seeExample 5.7) for an insulated free surface. Then estimate the time required to coolthe sheet to an average temperature of 175F or 79C. Then determine theapproximate time required to cool the mold back to its initial temperature.

The correct approach to this problem is to solve the transient heat conduc-tion equations for the cooling sheet and the heating mold, as done in Section5.6. The bounds on the answer are obtained by assuming an initial moldtemperature of 125F or 52C. At Fo = 0.345 and Bi = 0, Y = 0.333.

(T-T0) (T-125)Y =

O W J = (375- 125) = 0 '3 3 3

Tave = 2080F = 98C

In other words, the average sheet temperature is somewhere between 175For 79C as determined with an isothermal mold and 2080F or 98C asdetermined by an artificially high mold temperature of 125F or 52C. Thetime needed to attain an average sheet temperature of 175F or 79C isobtained as follows:

For Bi = 0, Fo = 0.53 from Fig. 5.11. Therefore 0 = 1145 s. The actual timeis somewhere between 746 s for 75F or 52C mold temperature and 1145 sfor 125F or 52C mold temperature. The increase of more than 50% incycle time is a strong indication of the importance of maintaining constantmold surface temperature.

Figure 5.11 is used to determine the cooling rate of the prototype mold.Consider cooling the mold to within 2% of its initial temperature. That isTmold,final = 76For24C. Now:

For Bi = 0, Fo = 1.4. The cooling time is given as:

I - *

For plaster, oc = 30 x 10~4 cm2/s and D = 2 cm. Therefore the cooling timeis given as 0 = 1870 s. It is apparent that temperatures in prototype moldsincrease with the number of parts molded.

where tavg is the average sheet thickness, for this illustration. T1 is the initial sheettemperature, Tf is the final sheet temperature and T* is the time-average sheettemperature, such as TJ = (T1 + Tavg)/2. This amount of heat is to be removed in thecooling time, 0C. There are N forming cycles in the total time 6T. The total amountof heat to be removed from the sheet in the time 0T is thus given as:

QtOtHi = N-PCp-^(T 1 -Tf) (5.36)

The heat convected away by the cooled air in contact with the sheet in the time 9Tis:

AO = h ( T * - T ) COQl (5 37)Vconv l xairVAa A air/

A \~>.J>i)

The time when no sheet contacts the mold surface is given as (1 6COOI/6T)- Duringthis time, energy is being transferred from the mold surface to the cooling air:

/1 A \Qconv = h a i r (Tm o l d a v g Ta i r) - (5.38)

UxThat absorbed by the mold is given as:

Qmold = Qavg -Qconv (5-39)

The amount of heat transferred to the coolant during the entire cooling portion ofthe cycle is:

Qcooi ~ h c o o l(Tm o l d avg Tcool) (5.40)The overall heat balance is then given as:

^cool ^coolvA mold,avg A cool/

= N pcp U8(T1 - Tf) - hair(T* - Tair) % ^ - hair(Tmold,avg - Tair) ~ 9 coo l ) (5.41)Ux Ux

As is apparent, Tmoldavg is the only unknown temperature. Example 5.11 illustratesthe relative temperature increase for cyclic cooling on a plaster mold. As expected,average equilibrium mold temperature values increase with decreased time betweenforming steps and decrease with higher coolant flow rates, lower coolant tempera-tures and more efficient convection cooling at the free surface. A more exact analysisis presented below when all aspects of transient heat removal from plastic sheet areincorporated in a general computer program in the next section.

Example 5.11 Equilibrium Mold Temperature

Make a steady-state heat balance on a plaster mold to determine its equilibriummold temperature. The mold is cooling a 0.060 in or 1.5 mm PSpart in 20 s. Thereis a 20 s delay for part removal, insertion and draw-down of the next part. Theinitial average sheet temperature is 375F or 1900C. The final average sheettemperature is 175F or 79C. The initial mold temperature is 75F or 24C. ThePS specific heat is 0.45cal/g 0C or 0.45Btujib 0F. The PS density is 1.05g/cm3

or 65.5 Ib I ft3. The convection heat transfer coefficient to forced air is hair =28.4 W/m2 0C or 5 Btu/ft2 h 0F. The air temperature is 75F or 24C. The heattransfer coefficient to the coolant is hcool = 284 W/m2 0C or 50 Btu/ft2 h 0F. Thecoolant temperature is 75F or 24C.

All the values for Equation 5.41 are known except the equilibrium moldtemperature, Tmold avg. The first term on the right is:

N rcp tavg(Tx - Tf) = 90 65.5 0.45 ~ (375 - 175) = 2650 J ^

This is the heat that must be removed from the sheet. The second term onthe right is the amount of heat convected from the sheet to the air. Here0cool/0T = 0.5. The average sheet temperature is Tavg = 375 175 = 275F.

ha i r(Tavg-Ta i r)-0.5 = 5 0 0 ^

The average heat loss from the uncovered mold is given as the third part ofthe right side of Equation 5.41:

hair(Tmold,avg - Tair)(l - 0.5) = 5 0.5 (TmokUvg - 75)And the average heat loss to the coolant is given as the term on the left inEquation 5.41:

hcool ' ( l m o l d , a v g ^ cool) = ^ ' V ^ mold,avg ' ^ )

The mold temperature is then obtained from:

(50 + 2.5) (TmokUvg - 75) = 2650 - 500 = 2150 J ^OrTm o l d , a v g=116For46.6C.

5.6 Transient Heat Transfer During Sheet Coolingon the Mold SurfaceComputer Models

As shown in Fig. 5.1, the convective and conductive elements discussed aboverepresent resistances to heat removal from the sheet. Time-dependent sheet tempera-ture is determined by simultaneously solving the transient heat conduction equationsfor both the formed plastic sheet and the mold1:

(5.42)

(5.43)

1 The general transient one-dimensional heat conduction equation was described in detail in

Chapter 3.

Next Page

Front MatterTable of Contents5. Cooling and Trimming the Part5.1 Introduction5.2 Overall Cooling Heat Balance5.3 Cooling the Formed Shape5.4 Steady State Heat BalanceInterfacial ResistanceShape FactorConvection Heat Transfer Coefficient

5.5 Cyclic Heat BalanceCooling the Free Surface of the SheetCooling Thin Sheet in Ambient AirTransient Heat Removal from the SheetQuiescent Ambient AirMoving Ambient AirCooling on Nonmetallic Molds

5.6 Transient Heat Transfer during Sheet Cooling on the Mold Surface - Computer ModelsInterfacial Air

5.7 ShrinkageUnconstrained ShrinkageConstrained Shrinkage

5.8 TrimmingTrimming Heavy-Gage PartsTrimming Thin-Gage Parts

5.9 Mechanics of CuttingThe Trim RegionRegistering the Trim SiteThe Nature of the CutFracture MechanicsMechanical ChippingMultiple-Edged Tool or Toothed Saw PerformanceAbrasive Cut-Off WheelToothless or Shear and Compression CuttingFracture Mechanics in TrimmingNibblingBrittleness, Orientation and Trim Temperature

5.10 Steel Rule DieResharpeningTabbing and Notching

5.11 Punch and Die TrimmingForged and Machined Dies

5.12 Drilling5.13 Other Cutting TechniquesThermal CuttingWater Jet Cutting

5.14 Trimming - a Summary5.15 References

Index