Stirring power issue, heat calculation
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I encountered an energy calculation problem in the design: it involves a high-speed mixer with a power of 75 KW and a rotation speed of 1500 rpm. There is no chemical reaction taking place inside the reactor, and solid materials account for about 40% of the contents. The goal is to use chilled water to cool the heat generated during mixing. My idea is that, according to the law of conservation of energy, the heat inside the reactor is generated when electrical energy is converted into kinetic energy, which is then converted into thermal energy. In other words, there is a certain ratio between thermal energy and electrical energy. I want to find a conversion factor that allows me to directly relate electrical energy to heat, thereby simplifying the calculations. However, I’m not sure how to choose this factor or what factors should be taken into consideration. Also, the amount of chilled water needed should be calculated based on the maximum heat generation. Could the effective work done by the mixer be directly converted into heat? From a thermodynamic perspective, work can indeed be completely converted into heat. Moreover, after the mixer operates for a while, the increase in the kinetic energy of the materials becomes very small. Is it possible to consider things this way? Additionally, the effective power of the mixer is determined based on the motor’s power. So, what value should this conversion factor be?
I don’t understand what device you have and what you want to do. The answer is: thermal equivalent, 1KW=860KCAL. If you want to perform the conversion, it’s not the motor power that should be used, but rather the actual power consumption. The rotation speed is 1500 rpm; I’ve never seen such a high mixing speed before, and I’m not sure what kind of equipment this is.
Regarding the calculation of the heat dissipation from process equipment in clean rooms, the formula for calculating the heat dissipation of such equipment is: Q = 1000n1n2n3n4SN/η (W). Here, Q represents the total amount of heat dissipated by the process equipment; n1 is the efficiency coefficient of the motor, that is, the ratio of the maximum actual power consumption to the installed power. This coefficient reflects the degree of utilization of the rated power N, and it generally ranges from 0.7 to 0.9 ； n2---------Coefficient of simultaneous use, that is, the ratio of the installed power of motors in use simultaneously in a room to the total installed power; it depends on the usage pattern of the equipment in the manufacturing process, and is generally between 0.5 and 0.8 ； n3---------Load factor, which is the ratio of the average actual power consumption per hour to the designed maximum power consumption. It indicates to what extent the average load reaches a new level corresponding to the maximum load; it can generally be taken as around 0.5 ； n4---------Coefficient considering the heat carried away by exhaust air; generally, 0.5 can be used ； S--------- is the heat storage coefficient, which is the ratio of the maximum instantaneous load for motor cooling to the actual power consumption per hour; it is set at 0.95 for a three-shift operation, 0.9 for a two-shift operation, and 0.80 for a one-shift operation ； N---------Rated power of the electric motor (installation power) ； η---------Motor efficiency (usually 85) ； So, now I would like to ask: If there is a clean room without exhaust ventilation, operating on a two-shift basis, and there are two process devices inside that operate simultaneously, with each device having an installed power of 6 KW, we take n1 (0.7~0.9) as 0.8; for n2 (0.5~0.8), we take it as 1 since the devices operate at the same time); we take n3 as 0.5, and n4 as 1 because there is no exhaust ventilation. We also take S as 0.9. Therefore, the total heat dissipation of the process equipment in this clean room should be: Q=1000n1n2n3n4SN/η. Q=1000×0.8×1×0.5×1×0.9×(6×2)/85. Q=360×12/85. Q=50.82 (W)
Heat generation of the generator set: The heat dissipation of a generator set comes from two sources. One is heat transfer through the cover plates of the generator set and its enclosure structure, and the other is the heat resulting from air leakage in the cooling circulation air of the generator set. Large and medium-sized generator sets typically use a closed air self-circulation cooling system; the losses in the generator windings are transferred to the cooling air, and the heat from this air is then removed by the cooling water through the unit’s water coolers. According to the measured data, the temperature of the air discharged from the stator generally does not exceed 65°C, while the temperature of the air entering the rotor is generally not lower than 5°C. The heat dissipation from the generator housing can be calculated using the following formula:
w
Where:
—— The heat transfer coefficient of the generator housing, in w/㎡•℃
—— The area of the generator housing, in ㎡
—— The average temperature of the air used for cooling the generator, in ℃
—— The indoor air temperature, in ℃
The heat dissipation due to air leakage from the generator can be calculated using the following formula:
w
Where:
—— The air leakage coefficient; 0.3% for steel covers
—— The volume of air used for cooling the generator, in m3/h
—— The specific heat capacity of air, in w/kg•℃
—— The density of air, taken as 1.2 kg/m3
—— The temperature of the air leaking from the generator, in ℃
—— The indoor air temperature, in ℃
Based on the temperature of the cooling air inside the generator set and the surface area of the generator, it is not difficult to calculate the heat transfer rate of the generator housing. However, there are significant differences in the calculation of the heat lost through leakage. As mechanical manufacturing technology continues to improve, particularly the efficiency of air coolers, there are substantial variations among different manufacturers regarding the volume of air used in the cooling cycle of generator sets. For example, according to the mechanical and electrical design manuals, the cooling air volume for a 300,000 KW unit is approximately 200 m³/h, but the cooling air volume provided by most international manufacturers is around 120 m³/h, which results in significant differences in the calculated values. Generally, the lower the temperature of the cooling air, the lower the temperature of the generator’s coils, and thus the higher the efficiency of the generator. However, the temperature of the cooling air is influenced by the size of the cooler; larger coolers increase the complexity of manufacturing the unit and reduce its economic viability. It is not possible to reduce the cooling air temperature indefinitely, and manufacturers take into account an economic range when designing the unit in order to achieve the best cost-performance ratio. Therefore, in actual design calculations, the generator manufacturer should provide the parameters for the cooling air flow rate in order to calculate the heat loss due to air leakage. Transformer heat generation: Transformer heat dissipation refers primarily to the energy losses within the transformer. These losses consist of copper loss (resistive loss) and iron loss (ferromagnetic loss). Copper loss varies depending on the load level, whereas iron loss is independent of the load and can be considered a constant value. Typically, the copper loss at rated load is defined as short-circuit loss, while the iron loss at rated voltage is defined as no-load loss. The losses of self-cooled, air-cooled, and dry-type transformers are all dissipated into the surrounding air. The heat dissipation of an air-cooled transformer can be simply calculated using the following formula: Kw. Where: —— The no-load loss of the transformer, in Kw; —— The short-circuit loss of the transformer, in Kw. As for the heating generated by busbars and cables, in power plants, self-cooling enclosed busbars are commonly used for connecting generators and transformers. The heat generation of the busbar includes two components: the heat generated by power loss in the busbar and the heat dissipated due to induction in the enclosure. Since both ends of the main bus are connected to a generator and a transformer device respectively, the air between the bus and the enclosure is essentially sealed; the enclosure serves to protect against and shield electromagnetic waves, thereby reducing the impact of the bus’s electromagnetic field on surrounding electrical equipment and the environment, without reducing the bus’s heat dissipation. The power loss heat of the busbar is dissipated into the air between the busbar and the enclosure, and then transferred to the environment through the enclosure shell. While the heat dissipation via the shell’s sensing is transferred directly to the environment. The heat dissipation caused by busbar power loss can be calculated using the following formula: Kw. The heat dissipation due to induction in the busbar enclosure can be calculated using the following formula: Kw. Where: —— Phase current of the busbar (A) ; —— DC resistance of the busbar at operating temperature (Ω/m); —— DC resistance of the busbar enclosure at operating temperature (Ω/m); —— Skin effect coefficient of the busbar; —— Skin effect coefficient of the busbar enclosure; —— Length of the busbar (m). Heat generation in reactors: Reactors are used in distribution systems with high capacity to limit short-circuit currents; they can also be used as filtering reactors in rectifier systems. The heat dissipation of a reactor can be calculated using the following formula: Kw. Where: —— is the utilization factor of the reactor, usually taken as 0.95; —— is the load factor of the reactor, usually taken as 0.75; —— is the power loss of the reactor at its rated power (Kw), which is determined based on the rated current, rated reactance, and model of the reactor. A reactor is composed of windings, and its heating characteristics involve a large heat capacity and high heat generation; it takes some time to reach a steady level of heat generation. For reactors that operate continuously, their heat generation is stable; whereas for reactors that operate intermittently, the heat generation should be determined based on the operating time and the reactor’s heat generation characteristic curve
w
Where:
—— The heat transfer coefficient of the generator housing, in w/㎡•℃
—— The area of the generator housing, in ㎡
—— The average temperature of the air used for cooling the generator, in ℃
—— The indoor air temperature, in ℃
The heat dissipation due to air leakage from the generator can be calculated using the following formula:
w
Where:
—— The air leakage coefficient; 0.3% for steel covers
—— The volume of air used for cooling the generator, in m3/h
—— The specific heat capacity of air, in w/kg•℃
—— The density of air, taken as 1.2 kg/m3
—— The temperature of the air leaking from the generator, in ℃
—— The indoor air temperature, in ℃
Based on the temperature of the cooling air inside the generator set and the surface area of the generator, it is not difficult to calculate the heat transfer rate of the generator housing. However, there are significant differences in the calculation of the heat lost through leakage. As mechanical manufacturing technology continues to improve, particularly the efficiency of air coolers, there are substantial variations among different manufacturers regarding the volume of air used in the cooling cycle of generator sets. For example, according to the mechanical and electrical design manuals, the cooling air volume for a 300,000 KW unit is approximately 200 m³/h, but the cooling air volume provided by most international manufacturers is around 120 m³/h, which results in significant differences in the calculated values. Generally, the lower the temperature of the cooling air, the lower the temperature of the generator’s coils, and thus the higher the efficiency of the generator. However, the temperature of the cooling air is influenced by the size of the cooler; larger coolers increase the complexity of manufacturing the unit and reduce its economic viability. It is not possible to reduce the cooling air temperature indefinitely, and manufacturers take into account an economic range when designing the unit in order to achieve the best cost-performance ratio. Therefore, in actual design calculations, the generator manufacturer should provide the parameters for the cooling air flow rate in order to calculate the heat loss due to air leakage. Transformer heat generation: Transformer heat dissipation refers primarily to the energy losses within the transformer. These losses consist of copper loss (resistive loss) and iron loss (ferromagnetic loss). Copper loss varies depending on the load level, whereas iron loss is independent of the load and can be considered a constant value. Typically, the copper loss at rated load is defined as short-circuit loss, while the iron loss at rated voltage is defined as no-load loss. The losses of self-cooled, air-cooled, and dry-type transformers are all dissipated into the surrounding air. The heat dissipation of an air-cooled transformer can be simply calculated using the following formula: Kw. Where: —— The no-load loss of the transformer, in Kw; —— The short-circuit loss of the transformer, in Kw. As for the heating generated by busbars and cables, in power plants, self-cooling enclosed busbars are commonly used for connecting generators and transformers. The heat generation of the busbar includes two components: the heat generated by power loss in the busbar and the heat dissipated due to induction in the enclosure. Since both ends of the main bus are connected to a generator and a transformer device respectively, the air between the bus and the enclosure is essentially sealed; the enclosure serves to protect against and shield electromagnetic waves, thereby reducing the impact of the bus’s electromagnetic field on surrounding electrical equipment and the environment, without reducing the bus’s heat dissipation. The power loss heat of the busbar is dissipated into the air between the busbar and the enclosure, and then transferred to the environment through the enclosure shell. While the heat dissipation via the shell’s sensing is transferred directly to the environment. The heat dissipation caused by busbar power loss can be calculated using the following formula: Kw. The heat dissipation due to induction in the busbar enclosure can be calculated using the following formula: Kw. Where: —— Phase current of the busbar (A) ; —— DC resistance of the busbar at operating temperature (Ω/m); —— DC resistance of the busbar enclosure at operating temperature (Ω/m); —— Skin effect coefficient of the busbar; —— Skin effect coefficient of the busbar enclosure; —— Length of the busbar (m). Heat generation in reactors: Reactors are used in distribution systems with high capacity to limit short-circuit currents; they can also be used as filtering reactors in rectifier systems. The heat dissipation of a reactor can be calculated using the following formula: Kw. Where: —— is the utilization factor of the reactor, usually taken as 0.95; —— is the load factor of the reactor, usually taken as 0.75; —— is the power loss of the reactor at its rated power (Kw), which is determined based on the rated current, rated reactance, and model of the reactor. A reactor is composed of windings, and its heating characteristics involve a large heat capacity and high heat generation; it takes some time to reach a steady level of heat generation. For reactors that operate continuously, their heat generation is stable; whereas for reactors that operate intermittently, the heat generation should be determined based on the operating time and the reactor’s heat generation characteristic curve
Reply to 2# DXJ122: It is a high-speed disperser used for producing coatings, which is why its rotation speed is very high.
Reply to 3# Hai Zhi Lan 12: Thank you very much. I’m learning slowly; I only have one mixer. Power 75KW
In China, this is indeed a new challenge that requires numerous experiments; each type of impeller consumes different amounts of energy, and the mixing intensity also varies. I think you can verify it later using the operating current and the temperature reached per unit time. There is a small amount of energy loss in this process, which should be negligible. There is also the issue of the material’s specific heat capacity. Looking forward to further discussion!
What the people upstairs said makes sense, but I think there isn’t a direct one-to-one relationship between mixing and heat generation; mixing is efficient, and mechanical equipment is efficient as well – it’s not simply a matter of energy calculation.
I think the basic shaft power is essentially the heat generation power, as most of the energy used for stirring ends up as heat, with only a small portion being dispersed outside the tank due to vibrations and other factors
Theoretically: Heat output Q = Actual power consumption of the motor * Motor efficiency * Gearbox efficiency * Correction factor k. The actual power consumption of the motor can be measured using electricity meters; motor efficiency and gearbox efficiency can be found in relevant documentation. The correction factor k takes into account factors such as the transmission efficiency of the frame and seals, as well as the thermal radiation from the reaction tank, and it needs to be determined through multiple experiments