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Research on Defrosting Using Air Source Heat Pump Refrigerant Subcooling and Exothermic Heat
The air source heat pump is in winter heating conditions. When the surface temperature of the heat source side heat exchanger is lower than the air dew point temperature and lower than 0℃, the heat exchanger surface may be frosted. Many scholars have conducted research on defrosting, such as air defrosting, electric heating defrosting and hot gas defrosting. The author proposes a new defrosting method for selection.
When is heating, the refrigerant coming out of the condenser still has a higher temperature. For the cold and hot air unit, it is generally not lower than 32℃ (20℃ inlet air, 40℃ condensing temperature); for the cold and hot water unit, it is generally not lower than 40℃ (hot water inlet 40℃). The evaporation temperature of frost is lower than 0°C. The section where the refrigerant enters the evaporator through the condenser has larger sensible heat which can be used for defrosting. The normal refrigerant cycle is 12341 (see Figure 1).
If the refrigerant at state point 3 directly enters the frosting part of the evaporator, and is subcooled to point 5, then throttling to point 6, enters the evaporator, evaporates to point 1, and enters the compressor Compress to point 2, enter the condenser and condense to point 3. This constitutes a heating cycle with additional subcooling and defrosting.
2·Calculation and analysis
First calculate whether the heat released by the refrigerant supercooling can meet the heat required for defrosting.
R22 refrigerant mass flow rate is calculated at 1kg/s.
The possible range of air frosting is -12.8℃
According to Figure 1, the heat used for defrosting is
where: h3 and h5 are the enthalpy values u200bu200b(kJ/kg) corresponding to points 3 and 5, respectively. According to the assumed conditions, the temperatures corresponding to points 3 and 5 are 32°C and 20°C, respectively. Using the software Solkane5.0 to calculate h3u003d239.35kJ/kg, h5u003d224.29kJ/kg, so
in the evaporator The heat of evaporation is qou003dh1-h6(3) where: h1 and h6 are the corresponding enthalpy values u200bu200b(kJ/kg) of point 1 and point 6, respectively. The average logarithmic temperature difference between the refrigerant and the air in the evaporator is calculated at 10°C, and the calculated refrigerant evaporation temperature is -7.4°C. Assuming that the evaporative superheat of the refrigerant is 11℃, h1u003d415.13kJ/kg calculated by Solkane5.0. The enthalpy value remains unchanged before and after throttling: h5u003dh6, substituting into equation (3), you get
Use Tianzheng HVAC 7.5 to calculate the standard atmospheric pressure The air condition parameter is:
hru003d18.76kJ/kg dry air
dru003d5.14g/kg dry air
hcu003d9.44kJ/kg dry Air
dcu003d3.77g/kg dry air
where: hr and dr are the enthalpy and moisture content at the entrance of the evaporator respectively; hc and dc are the evaporator respectively The enthalpy and moisture content at the outlet. When the flow rate of R22 is 1kg/s, the corresponding evaporator air volume is:
When the evaporator is above 0℃, the moisture in it begins to precipitate, but it is continuously cooled during the downward flow. Suppose it has been frosted before it flows to the water tray, and the amount of frost is
The freezing heat of water is qsu003d335kJ/ kg, the defrosting heat required when the corresponding R22 flow rate is 1kg/s is
Substitute the relevant value into the formula (5 )~Equation (7) gives
Compare Eq. (2) with Eq. (8) to get qf>qm, that is, the refrigerant's excess The heat is enough to melt the frost on the surface of the evaporator.
Figure 2 shows a finned heat exchanger defrosting system with 4 processes.
When is heating, only one of the 4 solenoid valves (L, M, N and P) is opened. The liquid refrigerant from the heat exchanger (condenser at this time) is from the opened solenoid valve Enter the fin heat exchanger for subcooling and heat release defrosting, and then enter the gas-liquid separator (E, F, G or H) corresponding to the opening of the solenoid valve. The refrigerant from the outlet of the gas-liquid separator enters the liquid collection pipe J, then enters the distributor K through the throttle valve R, and enters the remaining three pipelines of a, b, c and d through the one-way valve. The evaporator evaporates, and the gaseous refrigerant enters the corresponding gas-liquid separator, and then collects from the gas outlet to the gas collector I.
If the solenoid valve P is opened and L, M and N are closed, the liquid refrigerant from the heat exchanger enters the lower part of the fin heat exchanger from the pipe d to defrost, and from the pipe p The path enters the gas-liquid separator H, exits the liquid outlet through the check valve, enters the liquid collection pipe J from the pipeline r, and then enters the distributor K through the throttle valve R. Because the one-way valve in the pipeline v is closed, The throttled refrigerant flows through the one-way valve through pipelines s, t, and u, and enters the fin heat exchanger from pipelines a, b, and c, respectively, and the evaporated gas enters the gas from pipelines e, h, and m, respectively. The liquid separators E, F and G are collected from their gas outlets to the gas collector I via pipelines g, k and o respectively.
When is cooling, all 4 solenoid valves are opened, and the high-temperature gaseous refrigerant enters the fin heat exchanger through the gas-liquid separators E, F, G and H from the pipelines e, h, m and p . The condensed liquid refrigerant enters the throttle valve S through the pipelines a, b, c and d, solenoid valves L, M, N, and P. The decompressed refrigerant enters the heat exchanger, and then returns through the four-way reversing valve. To the compressor.
The price of the refrigeration solenoid valve is more expensive, and it can be replaced by a combination of a cheaper four-way reversing valve and a one-way valve. The gas-liquid separator in Figure 2 must work reliably and not be too expensive. It can be designed and manufactured according to the structure shown in Figure 3.
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