| High precision water temperature control |
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| choosing and calculating methods |
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| Method 1: calculate according to laser photoelectricity conversion rate |
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Calculating formula：Q＝P（1-n） |
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Q：heating capacity of lase（W) |
e.g.: |
1200w laser’s input power is 20000w and photoelectricity conversion rate is 25%, so
Q= 20000（1-25％）= 20000x0.75= 15000W= 12900Kcal/h （1W＝0.86Kcal/h） |
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P：output power of laser（W） |
Q |
＝20000（1-25％） |
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n：photoelectricity exchanging rate（％） |
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＝20000x0.75 |
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＝15000W |
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＝12900Kcal/h （1W＝0.86Kcal/h） |
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| Method 2: calculate according to temperature difference of entering water and returning water |
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Calculating formulaQ=CpXPXqvXAT |
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| Q：heating productivity（W） |
| Cp：the cooling medium specific heat under rated pressure（J/Kg℃） |
| ......water：4186.8J/Kg℃ |
| p：the proportion of cooling medium（Kg/m3） |
| ......水1000Kg/m3 |
| qv：the circulating flow of cooling medium（m3/h） |
AT：the temperature difference of cooling medium（℃）..T2-T1 |
| e.g.the water flow rate is 8 and the temperature difference is 3℃, so Q=4186.8x1000x8x3/3600=27912W) =24000Kcal/h (1W=0.86Kcal/h) |
| Q＝4186.8x1000x8x3/3600=27912W) |
| ＝24000Kcal/h (1W=0.86Kcal/h) |
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| Method 3: calculate according to water temperature rising rate and seek the rate max. |
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Calculating formulaQ=CpXPXqvXAT
| Q：heating capacity（W） |
| Cp：the cooling medium specific heat under rated pressure（J/Kg℃） |
| ......water：4186.8J/Kg℃ |
| p：the proportion of cooling medium（Kg/m3） |
| ......水1000Kg/m3 |
| qv：the circulating flow of cooling medium（m3/h） |
AT： the temperature difference of cooling medium（℃）..T2-T1 |
e.g. the water flow rate is 8 and the temperature difference is 3℃, so Q=4186.8x1000x8x3/3600=27912W) =24000Kcal/h (1W=0.86Kcal/h) |
| Q＝4186.8x1000x8x3/3600=27912W) |
| ＝24000Kcal/h (1W=0.86Kcal/h) |
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