Pipe heat dissipation calculation

There’s no information on the ambient temperature, nor is there any data on the medium temperature. Many parameters are still missing, so it’s impossible to carry out the calculation

1. As mentioned by the friend on the second floor, some conditions are lacking, so it is not possible to provide exact figures. 2. Some assumptions are made here to give an approximate value: 2.1. The pipeline is in a wind-free environment at 20°C, with a steam pressure of 1 MPa. 2.2. At an external surface temperature of 150°C, the heat dissipation rate is approximately 2245 W/m2; therefore, the heat dissipated over a length of 20 meters is about 60,060 W. 2.3. This amount of heat dissipation is equivalent to the heat generated by 0.32 tons per hour of steam at 160°C and 1 MPa pressure. 3. Note: In reality, a heat dissipation rate of 60 KW only reduces the enthalpy of the steam inside the pipeline, resulting in a decrease in temperature, rather than representing a loss of steam. The greater the steam flow rate in the pipe, the smaller the temperature drop, and so on. Unless the very low steam flow rate is further reduced in temperature, it cannot meet the user’s requirements regarding steam parameters.

Actually, I want to calculate how much steam will be lost under these conditions. My goal is to heat the material to around 145°C, which is its boiling point; moreover, some of the pipes lack insulation, and I want to determine, in terms of steam, how much steam is lost per hour due to this heat loss.

1. Thank you for the flowers. 2. More conditions are missing in your new idea; initially I calculated based on maintaining a temperature of 145°C, considering only heat dissipation and not the heating aspect； The new idea builds on the existing foundation; it is necessary to specify that the material needs to be raised from a certain temperature to 145°C. What’s missing is information regarding the low-temperature value as well as the specific heat capacity of the medium. 3. The heat lost is converted into steam, and if this conversion is based on the enthalpy difference between water at 20°C and steam at 160°C, then the amount of steam lost is relatively small. If the low-temperature value of the steam is higher, the enthalpy difference decreases, resulting in an increased amount of steam lost. So there needs to be a preset premise. 4. Instead of that, it is simpler and clearer to express it in terms of power loss. 5. Since we are not aware of your specific circumstances, the above description is for reference only

This post was last edited by zhengfaqi on 2019-10-19 at 21:13. Is it possible to calculate only the heat loss due to heat dissipation, that is, the steam loss that may occur? For example, if the temperature outside the pipe is 145°C and the ambient temperature is 20°C, with a pipe diameter of 426 mm, we should consider only the heat loss resulting from heat dissipation and ignore everything else. In the formula Q = a(tw-to)F, Q represents the heat dissipation amount, with the unit being watts. Tw is the surface temperature of the heat dissipation surface, in °C. to represents the ambient temperature, in °C. F is the area of the heat dissipation surface, in m^2. a is the overall heat transfer coefficient, in W/(°C×m^2). Can it be calculated using this formula?

Estimating heat transfer for gaseous media in this way may lead to significant errors. If heat dissipation through the air is taken into account, the overall heat transfer coefficient may not exceed 30. If heated with saturated steam, the overall heat transfer coefficient could be over 2000. Therefore, the process should also be determined first. I guess it’s because you’re considering the heat loss of the pipe in air, and then want to heat it again with steam?

By considering only the losses due to heat dissipation, how much steam will be wasted?

In my previous calculations, I also took into account only the heat emitted by the metal at 145°C when it is directly exposed to air

Large-diameter pipes generally have less heat loss in air. Due to the large deviations in the empirical formulas for the air flow velocity in natural convection outside the tube, conservative empirical estimates are generally used. If indoors, it can be 0.5~0.8 m/s. When it’s outdoors, the impact of wind has to be taken into account. Is it to reduce costs by not using insulation? Heat dissipation also incurs costs. By providing insulation to reduce the heat transfer coefficient, the heat loss can be brought within acceptable limits.

Heat loss from the tank wall: Heat loss from the tank wall = heat transfer coefficient for heat loss * external surface area of the tank wall * (outer wall temperature of the tank wall – ambient temperature). Heat loss from the tank top: Heat loss from the tank top = Heat transfer coefficient of the tank top * Surface area of the tank top * (Outer wall temperature of the tank wall – Ambient temperature). Heat loss from the bottom of the tank: Heat loss from the bottom of the tank = Heat transfer coefficient of the bottom of the tank * Surface area of the top of the tank * (Outer wall temperature of the tank walls – Surface temperature during the coldest month in the area where the tank is located). Total heat loss of the storage tank: Total heat loss of the storage tank = Heat loss from the tank walls + Heat loss from the tank top + Heat loss from the tank bottom. This is what I found; I’m not sure if it will be helpful to you