Questions about the wall thickness of a jacketed reactor vessel?
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New to the field, don’t understand the ins and outs! Please give some guidance: I saw an example in the \"Design Guide for Pressure Vessel Engineers\": For a certain reactor, the design parameters are as follows: the inner diameter of the reactor vessel is Di=1200mm, the maximum operating pressure of the material inside the vessel is p=5.4MPa, the pressure of saturated steam in the jacket is p=3.92MPa, and the design temperature of the reactor is 250℃ ； The material for the kettle body and head is 16MnR, with a corrosion margin of 1 mm. Dimensions are indicated as follows: the inner diameter of the kettle cylinder is 1200, the inner diameter of the jacket cylinder is 1300, the total height of the straight sides of the cylinder is 1024, and the length including the head portion is 1174 ； During the calculation for this instance, when determining the thickness of the reactor vessel: first, using an internal pressure of 5.4 MPa, an allowable stress of 138 MPa, and a weld coefficient of 1.0, the calculated thickness comes out to be 24 mm, with a negative deviation of 0.9; finally, the nominal thickness is rounded to 26 mm” ； Under external pressure conditions, assuming a nominal thickness of 30 mm for the cylinder, an outer diameter of 1200 + 2×30 = 1260, and a length of 1174. . . . . The allowable external pressure of 3.06 < 3.92 is obtained, which does not meet the stability requirements ； Thus, assuming a nominal thickness of 38 mm for the cylinder, the allowable external pressure is found to be 4.0, which is greater than 3.92; hence the requirement is met. Therefore, the nominal thickness of the cylinder is set at 38 mm. In the above process: 1. He does not clearly state what the final nominal thickness is – the results of the two calculations were 26 mm and 38 mm respectively – so the final value should be 38 mm, right? This difference is a bit too large, isn’t it? 2. The result from the previous step was 26 mm; how was it decided to try the parameter value of 30 mm first, and then to try 38 mm as well? Are there any guidelines that can be applied here? It’s truly amazing how, especially starting from the previous value of 26mm, suddenly choosing 30mm was done
You’re just looking at example problems. By actually using the vertical vessel module in SW6, it can be calculated in one go. It’s not really surprising; the wall thickness of this device is controlled by external pressure. However, to achieve better heat transfer, it is recommended to add reinforcing rings to the inner cylinder, or to use a semi-tube structure for the jacket, so that the wall thickness of the inner cylinder can be controlled by internal pressure. Also, the terms 16MnR and weld coefficient are no longer in use.
You’re just looking at example problems. By actually using the vertical vessel module in SW6, it can be calculated in one go. It’s not really surprising; the wall thickness of this device is controlled by external pressure. However, to achieve better heat transfer, it is recommended to add reinforcing rings to the inner cylinder, or to use a semi-tube structure for the jacket, so that the wall thickness of the inner cylinder can be controlled by internal pressure. Also, the terms 16MnR and weld coefficient are no longer in use.
You’re just looking at example problems. By actually using the vertical vessel module in SW6, it can be calculated in one go. It’s not really surprising; the wall thickness of this device is controlled by external pressure. However, to achieve better heat transfer, it is recommended to add reinforcing rings to the inner cylinder, or to use a semi-tube structure for the jacket, so that the wall thickness of the inner cylinder can be controlled by internal pressure. Also, the terms 16MnR and weld coefficient are no longer in use.
Thank you for the guidance from yj6946! ! !
The weld coefficient mentioned by the original poster should actually be referred to as the welding joint coefficient. It is specified in GB/T150-2011, Part 1, page 13. As mentioned above, an excessive wall thickness can affect the heat transfer efficiency of the container; using reinforcement rings is a good solution for this issue. Additionally, with a design temperature of 250 degrees, it is better to use heat transfer oil rather than steam, as heat transfer oil requires lower pressure and offers better safety prospects. When calculating SW6, there are two scenarios: one is where no thickness is specified, and the program calculates it itself; the other is where a thickness is input, and the program determines whether that thickness meets the requirements. In short, the purpose of example problems is to highlight a particular key point, while in practical work, design must be carried out based on actual operating conditions.
Thank you, Director Feng, for your guidance! The information seems a bit outdated; the correct term is indeed: welding joint factor!
Thank you, Director Feng, for your guidance! The information seems a bit outdated; the correct term is indeed: welding joint factor!
Thank you, Director Feng, for your guidance! The information seems a bit outdated; the correct term is indeed: welding joint factor!