Question

Assume you are injection molding plastics parts using a polylactic acid. In the process, the resin...

Assume you are injection molding plastics parts using a polylactic acid. In the process, the resin is heated from room temperature (25 oC) to 200 oC. The cycle time is 5 second, and shot weight is 0.05 kg. Please calculate the heat input per hour of the process. The heat capacity data of PLA can be found from the excel file. The PLA used has 10, 000g/mol molecular weight (weight-average), Tg around 59.35 oC and melting point Tm at 160 oC with heat of fusion of 34 J/g. (8 pt)

T (K) Cp (J/k/mol)
5 0.31
6 0.6
7 0.95
8 1.34
9 1.78
10 2.25
15 4.85
20 7.74
25 10.585
30 13.15
40 18.06
50 22.585
60 26.575
70 30.455
80 34.195
90 37.77
100 41.145
110 44.4
120 47.52
130 50.52
140 53.41
150 56.2
160 58.98
170 61.71
180 61.4
190 67.08
200 69.75
210 72.35
220 74.96
230 77.51
240 80.13
250 82.8
260 84.25
270 87.11
280 89.82
290 92.48
298.15 94.69
300 95.3
310 98.13
320 101.59
330 112.16
332.5 145.44 (Tg)
340 146.01
350 146.77
360 147.53
370 148.29
380 149.05
390 149.81
400 150.57
410 151.33
420 152.09
430 152.85
440 153.61
450 154.37
460 155.13
470 155.89
480 156.65
490 157.41
500 158.17
510 158.93
520 159.69
530 160.45
540 161.21
550 161.97
560 162.73
570 163.49
580 164.25
590 165.01
600 165.77

Homework Answers

Answer #1

above values are given in the question, cycle time = 5s. shot weight = 0.05kg

=> in an hour total mass of material heated = 3600/5 * 0.05 kg = 36kg

Now, molar weight of PLA is 10kg, => 36kg = 3.6mol

now, total heat required = heat to attain melting point + heat of fusion + heat to attain injection point

Heat to attain melting point (433K):

mC​t now c value changes with temp. for PLA

therefore, total heat required H(​298-433) = 3.6*(1.85*94.69 + 10*(95.3+98.13+101.59) + 2.5*112.16 +7.5*145.44+ 10*(146.01+146.77+147.53+148.29+149.05+149.81+150.57+151.33+152.09+152.85) +3*153.61) = 71641.5J

Heat of fusion = 34*36000J = 1224000J

Heat to attain injection temperature (473) , H(433-473) =

3.6*(7*152.85+10*(153.61+154.37+155.13) +3*(155.89)) = 22207.4J

therefore total heat =71641.5J +​ ​1224000J + 22207.4J = 1317849J

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