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Calculate the oil cooler: fixed tube sheet, refrigerant oil on the shell side, Freon R22 on the tube side
The image isn’t very clear. It shows an input warning 1473: \"Interpret the results for this thermosiphon reboiler with care.\" The pressure loss in the inlet line is -0.42473 bar, while the pressure in the outlet line is 0.11701 bar at a flow rate of 2.2072 kg/s. These values are not compatible with the driving pressure of 0.19321 bar resulting from the liquid head. There is little to no pressure head available to drive the liquid through the exchanger.
Is this the tube side or the shell side? It’s that the inlet pressure is not sufficient to induce thermal siphonage
Is this the tube side or the shell side? It’s that the inlet pressure is not sufficient to induce thermal siphonage
The shell side contains refrigeration oil, while the tube side contains Freon R22. The numbers given in this explanation don’t match those I entered; I’m not sure where these numbers come from, so it’s a bit difficult to understand
The power is not sufficient to overcome the resistance; this may be due to 1) the inlet and outlet pipe diameters being too small (resulting in higher resistance), or 2) the liquid level in the container being too low (resulting in less driving force)
1. The pipe diameter is calculated based on the mass flow rate and flow velocity of the two fluids in the shell side and tube side; it basically cannot be adjusted. 2. Does the container liquid level refer to the liquid level height at the inlet of the tube side? Since this device is used on fishing boats, there is a limit on its total height – it can be at most 2200; it seems impossible to make any adjustments. :(
1. The suggested pipe diameter refers to the fact that the inlet and outlet pipes on the side of the medium being heated (usually the pipe side) are too small, not the diameter of the heat exchange tubes. Why can’t this be changed? It now indicates that the pipe flow velocity is 2.2 meters, resulting in too high a pressure drop in the pipe; why not use a larger diameter? 2. There must be a problem with your calculation model; it’s not possible to fix the flow rate – the flow rate should vary and adjust automatically to match the pressure drop. It’s a series of interconnected reactions: for example, if the pressure drop is too high, it causes the flow rate in the pipes to decrease, which in turn increases the vaporization rate of the heat exchanger (with the heat exchanger remaining unchanged). This again leads to an increase in the flow rate (as the flow resistance on the heat exchanger side decreases). The end result is a very high vaporization rate in the heat exchanger, which can cause vibrations and other negative effects. An important design parameter for siphon systems is the vaporization rate, which is generally around 20–30%
(1) The value of 2.2 is not the flow velocity in m/s; it is the mass flow rate in kg/s. The mass flow rate in the shell side is constant, as it comes from the compressor. When the compressor operates stably, this mass flow rate remains fixed. The mass flow rate in the tube side, which is calculated based on the required cooling load, is this same value. Therefore, it isn’t the high flow velocity that causes a large pressure drop. Moreover, the pressure drop at the inlet is negative – shouldn’t that indicate an increase in pressure? (2) Vaporization rate…… Well, what I have here is the circulation ratio; it’s set at 2 at the moment. This means that 2 parts of liquid refrigerant R22 go in, and 1 part of liquid plus gas comes out. Various textbooks and literature specify that the circulation ratio for R22 is 2. Let me briefly explain the calculation process for my heat exchanger. This device is used to cool the lubricating oil in compressors. The high-temperature, high-pressure gas discharged by the compressor contains refrigerant R22 as well as misted lubricating oil droplets. After passing through a separation unit, the lubricating oil turns into a liquid and enters the oil cooler. The separated refrigerant R22 gas then passes through a condenser, where it becomes high-temperature, high-pressure liquid. This liquid splits into two streams: one stream proceeds to the next stage, while the other stream enters the cooling coil, where it vaporizes through phase change, thereby cooling the lubricating oil. After that, it returns to the compressor to start the cycle again, and the lubricating oil also re-enters the circulation system through various oil injection points. That’s roughly how it goes.
And I tried to increase the diameter of the inlet pipe as you suggested, but it didn’t work:handshake