1. UndersTand the Thermal Properties of Polymers and Learn The Method for Calculating Energy Values and Flow for Both AMorphous and SemiStalline Materials.
2. Calculating the (Cooling) Energy That Must be Removed from the Molded Parts to Change the Tempeature from the Processing (Melt) TEMPERATURE TO A SAFE EJECTION TEMPARATIURAARARARAARARAAAAOAOAAOAOAOOOOOOOOOOOOOOOOOOOOOsOOAAAAAAAAAAAAAAAsedEAEationationationationationationationationationationationationationationsEEEsEEEsssssssssssssssssssssssckesesateses self self self self self rep spirit self self self inanth, in inteary
E = w x cp x ?t in btu, where
W = Weight of Material in LB
CP = Heat Capacity in BTU/LB-° F, and
Gt = Change in Temperature, ° F.
° f and therefore:
E = 0.45 LB x 0.34 BTU/LB-° F X (464-212) ° F = 0.45 x 0.34 x 252 = 38.56 BTU
: SPH = 3600/15 = 240 SHOTS/HR. Thus, The Energy Flow Required to cool the abs part from our example above is:
Q = SPH x BTU/SHOT = 240 x 38.56 = 9254.4 BTU/HR
And this do not have a lax heat value.
During Cooling. Values for Latent Heat are expressed in btu/lb. Table 1 shows value for both Heat CaPacity and Latent Heat Common SemiStalline Materials.
The Energy Calculation for Semi-Crystalline Materials is Similar to the Abs Example Above, Except that We Must Also the Latent Heat. We will use.
FIRST, CALCULE ENERGY PER ShOT Assumping A Shot Weight of 0.30 Lb, CP of 0.61 BTU/LB-° F, ?T of (450-212), and a latent heat value (HL) of 89.1 btu/lb.
E/shot = w x ((CP x ?T) + HL) or:
E/shot = 0.30 x ((0.61 x 238) + 89.1) = 0.30 x (145.2 + 89.1) = 0.30 x 234.3 = 70.3 BTU/Shot.
Assigning A Cycle Time of 8 SEC, SPH = 3600/8 = 450. With this value we can call that the entrance to cool this process:
Q = 70.3 BTU/Shot x 450 SPH = 31,635 BTU/HR.
Notice that the cooling Energy Requirement in this example is quite large related to the amorphous abs part, design the smaller pp part weight, the latent-heled group.
Knowing the ?T of the water as it flows through the mold and the flow rate, we can determine the energy flow rate, for a particular cooling circuit:
Qw = ?t x gpm x 60 min/hr x 8.34 LB/GAL x 1 BTU/LB-° F.
Qw = ?t x gpm x 500.4.
Let & rsquo; s assnsume, for exmple, a ?T of 2.8 ° F and a Flow Rate of 0.85 GPM:
Qw = 2.8 x .85 x 500.4 = 1190.95 BTU/HR
We can also rearrange this expresion to class a gpm requirement, assuming we almedy know how much heat we need and assuming a value for ?t. This expression would:
Gpm = qw /(=t x 500.4).
T = 3.5 ° f.
GPM = 1800/(3.5 x 500.4) = 1.03 gpm.
Tools for Evaluating Existing Cooling Circuits and for Designing New Circuits. But An Important Question Remuans: What value of ?t Should we use?
And Shots Per Hour. This is a great way to improve your cooling design capability.
Good Method. The Sensors can be mounted in a pair of tees with quick-conftings on the other ports, creating a convenient way into cooling circuit)
COOLING CIRCUITS.
1) SPH = 3600/12 = 300
2) W
3) Cooling Energy Per PART, QCooling = W X ((0.61 BTU/LB-° F
x 320 ° F) + 89.1 BTU/LB) = 12.9 x (195.2 + 89.1) = 12.9 x 284.3
= 3667.5 btu/hr.
The-PANTS METHOD. For Our Example Let & RSQUO; S assign 55% to the b-side user that the part cools a little longer on the B -Side. Thore controlMent issues
Qcore = 0.55 x 3667.5 = 2017 BTU/hr.
Besides
Gpm = qw / (=t x 500.4).
& Delta; T = 0.20 ° F x Circuit LENGTH or 0.20 x 23 in. = 4.6 ° F.
Gpm required = 2000/(4.6 x 500.4) = 2000/2302 = 0.87 gpm.
Referring to table 2 and extrating a value for the effective circuit diameter of 0.335 in., Our 0.87 GPM Seems SUFFICIENT to Prive, Even Atmapement