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 |
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):
mCt 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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