How to determine the heat transfer coefficients inside and outside the heating coil

The heat transfer coefficient of the inner coil can be determined with reference to that of shell-and-tube heat exchangers. The outer coil is not very clear.

Within the pipe, it can be calculated using the formulas for straight pipes, and then corrections can be applied. Outside the pipe, it is calculated based on natural convection in a large space.

It’s not the external coil, but rather the heat transfer coefficient of the medium inside and outside the coil~;P

Is this calculated using the heat exchanger formula? Could you provide the source of that formula?~

This post was last edited by wiseboy on 2015-6-10 09:39, version 1. The inner wall coefficient of the pipe is easy to calculate: it is first determined using the formula for straight pipes, yielding aio; thereafter, an adjustment is made: ai = (1 + 3.5do/Dr) × aio, where do represents the outer diameter of the pipe, in meters ； Dr——Coil diameter, m ； 2. The difficulty lies in the calculation of the film coefficient outside the tube: it is calculated using the formula for the film coefficient in natural convection. This formula is not only complex, as it requires determining the flow pattern, but it also involves the property of the liquid’s expansion coefficient, which is difficult to find; one needs to know what type of liquid is being used. The hardest part is that the formula includes the temperature of the outer wall of the pipe, which is unknown at the beginning; you must use trial and error methods to determine this wall temperature multiple times. Theoretically, basic math knowledge is powerless. Therefore, due to space constraints, I cannot provide you with a more specific algorithm here. If it’s for engineering purposes, you can find someone to do the calculations (I believe professionals use software such as VHeater) ； If you want to learn *manual calculation, then look for literature. Lesson: Many people spend months struggling, working hard, and seeking advice, only to give up in the end because of the final issue – the unknown temperature of the outer wall of the pipe.

This post was last edited by wiseboy on 2015-6-10 09:49. Additionally, LZ did not clarify outside the coil: is it a flowing medium? It is also crucial that it be loaded once and then remain there, with no entry or exit. For example, a storage tank is used to transfer a fixed amount of fluid at once. So LZ’s question is so vague; it’s a waste of everyone’s time and effort. As for “K=CM/A”, it makes no sense.

Let me explain in more detail: The heating coil functions similarly to a heat exchanger. The inlet temperature of the fluid inside the container is 201.3, while the outlet temperature is 172.1. The length of the container’s barrel section is 4000 mm; it has a flat cover. The heat exchange tubes are inserted into the container; there is 1’ of tube per unit. The temperature during heat exchange is 145 at the inlet and 160 at the outlet. The container is insulated. To determine the length of the heat exchange tubes, I initially used the formula K=1/(1/Aw+δ/λ+1/An) in units of W/(m²·°C). This formula is used to calculate the heat transfer coefficient; the specific heat capacity of both the medium inside the heat exchange tubes and that in the container is 2.985 KJ/KG-°C. Then I use K=CM/A. Then, the heat transfer area is calculated using the overall heat transfer coefficient. But the problem is that I don’t know whether Aw and An are the heat transfer coefficients for the internal and external media; I’m not sure if it’s possible to calculate them. The material balance table provided by the client in HYSYS isn’t complete enough, as the values for thermal conductivity are left blank. It’s my first time doing this calculation, and I’m not sure if it’s correct. If it’s not necessary to use the heat transfer coefficients of the internal and external media, are there any other good methods? I don’t quite understand the formula you suggested, as I’m not majoring in fluid dynamics; I found this formula by looking at textbooks on heat transfer in higher education.

AW, An – that’s exactly the problem. Based on your level of knowledge, I won’t reply anymore.

My process engineer used HYSYS to do the calculations again for me; he determined the thermal conductivity values for both the inside and outside of the tube. I simply divided each of these values by the circumference of the tube to obtain the thermal conductivity coefficients. The final result I got differed by 2 meters from the value he obtained using the software. I calculated it to be 55, while he used software to get a result of 57. I think such a result is fairly conservative and acceptable~

The result is the same, but I still can’t agree to it: because the method is wrong. So similar results can only be a coincidence; such luck is not a pattern.