Calculation of heat absorption and cooling in die manufacturers

There are two types of heat exchange processes: heat absorption and cooling in the mold. Therefore, the mold can actually be regarded as a heat exchanger. In many cases, we do not calculate how much heat or cooling water is required in the design. Sometimes it needs to be calculated. How to calculate it?

1. Calculate the heat absorbed by the rubber mixture

Sometimes we need to know the amount of heat needed to heat the compound or the target temperature of the mold, and how to calculate it. First, we need to know some terms. 1. Specific heat capacity (Cp) refers to the heat required to raise the temperature of a high-quality material by 1 ° C, and the water required to raise the temperature by 1 ° C (the specific heat of water is 1). Z The specific heat of common plastics ranges from 0.25 to 0.55. Generally, the unit is cal/g. ° C or BTU/lb ° F.

Since the specific heat capacity of each temperature is not a fixed value, it is usually determined by the specific heat capacity obtained by raising the temperature from 14.5 ° C to 15.5 ° C. One gallon is the energy required per liter of water from 14.5 ° C to 15.5 ° C 1cal=4.1868 J

For example, the mass of PS material product (Cp=0.34) is 35g, the mold is 16 cavities, and the injection molding is carried out in 16S cycle. Calculate the total amount of plastic required for all plastics from 280 ° C to 22 ° C per hour at room temperature.

First calculate the total amount of rubber: 35gX16=560g

Injection time per hour: 3600S/6S=600 times/h

Total glue injection per hour: 560gX600 times/h=336000g/h=336kg/h.

Energy required to increase these weights by 1 ° C:

336000g / hX0.34cal /（g。℃）= 114240cal / h，

Converted into joules: 114240cal/hX4.1868J=478.3KJ/h.

Difference between target temperature and room temperature: 280 ℃ - 22 ℃=258 ℃.

The total heat required is calculated as follows:

478300j /（h。℃）X258℃＝ 123401400j / h ＝ 123MJ / h。

Because 1MJ=1/3.6KW h. Then 123MJ/h/3.6MJ/KW h = 34.17KW。

The total heat required for all plastics to rise from 280 ° C at room temperature to 22 ° C per hour is 34.17KW.

2. Calculation of rubber cooling demand

We already know that the molten compound in the mold cavity must be cooled to the temperature that can be sprayed. At this time, the heat in the plastic must be taken away. Then, let's continue with the above example. 336KG molten rubber needs to be cooled from 280 ° C to safe spraying temperature of 50 ° C, and the temperature difference is 230 ° C. As with the previous calculation method, the calculation formula of heat to be removed from the mold is:

48300j /（h。℃）X230℃= 110000KJ / h = 110MJ/ H。

This heat is transferred to the cooling tower equipment through water:

110MJ / h / 3.6MJ / KW. h = 30.55kw。

In mold and freezing technology, heat is usually expressed as "cold ton", which is defined as the heat required to melt 1 ton (T) of 0 ° C ice into 0 ° C water, and its conversion unit is:

1 cold ton=3.516KW 1KW=3413Btu/h

In this example, we need 30.55KW/3.516KW/cold ton=8.7 cold tons.

Cooling water required for heat calculation:

Continue with the previous example: it is known that 110000000J/h (or 110MJ/h) of heat needs to be removed from the mold. If the inlet water temperature is 5 ℃ and the outlet water temperature is 10 ℃, the temperature difference between them is 5 ° C. How many liters of water do you need to take away this heat?

Assume that the energy required to heat X liters of water is:

X ＝ 110000000J /（h。℃）/ 5℃＝ 22000000J / h。

Convert to heat:

X = 22000000J / h / 41868J / cal = 5254609cal / h。

Because the specific heat of water is 1:

5254609cal/h/1cal/g=5254609g/h=5254L/h water.

Because the flow of cooling water is usually measured in minutes: 5254L/h/60min/h=88L/min=23.3U Sgal/min (gallons).

In other words, 88 liters of water are needed to consume 110 million J/h.