Questions regarding thermodynamics

According to heat transfer theory, the rate of heat transfer is proportional to the heat exchange area, the temperature difference, and the heat transfer coefficient. Therefore, we need to first calculate the current heat transfer rate, and then determine the additional area of heat exchange tubes required based on the condition of reducing the outlet temperature below 100°C. The specific calculation process is as follows: 1. Calculate the current heat transfer rate. Heat transfer rate = Heat exchange tube area × Heat transfer coefficient × Temperature difference. Here, the heat exchange tube area is 3.5 square meters, and the temperature difference is 260°C minus the ambient temperature (assumed to be 25°C), which is 235°C. The heat transfer coefficient is complex, as it requires taking into account various factors such as the properties of the fluid and the material of the heat exchange tubes; it generally needs to be determined through experiments or numerical simulations. Here we simplify the problem by assuming the heat transfer coefficient to be 100 W/(square meter·°C). Then the current heat transfer rate = 3.5 × 100 × 235 = 82125 W. 2. Determine the additional area of heat exchange tubes required. According to the requirements, the outlet temperature needs to be reduced below 100°C. Assuming the target temperature is 90°C, the new temperature difference is 90°C - 25°C = 65°C. Heat transfer rate = Heat exchange tube area × Heat transfer coefficient × Temperature difference. If we set the heat transfer rate in this equation equal to the target value, we get: 82125 = Heat exchange tube area × 100 × 65. Solving this equation yields a heat exchange tube area of 12.65 square meters. Therefore, the additional area of heat exchange tubes required is approximately 12.65 - 3.5 = 9.15 square meters. .

This needs to be calculated; if it is necessary to maintain a certain level of resistance, the heat transfer coefficient can be taken as 25.

It’s impossible to have such a high heat transfer coefficient! Your algorithm probably won’t meet the resistance requirements!

Thank you for your reply. There is one more question: since the capacity of the waste heat recovery system isn’t being changed, could the high heat exchange area cause the water inside to boil? The known conditions are that the total amount of water used for waste heat recovery is 1.6T + 1T (with the water consumption by the water-using equipment being 1T/h and the natural gas consumption being 75 m3/h); moreover, the air volume for the burner mentioned above seems to be incorrect. I wonder how much waste gas is produced per cubic meter of natural gas burned? How much heat can be generated when it drops from 260° to 90°

Thank you for your reply. Is the heat transfer coefficient the thermal conductivity? The material is Q235B, and its thermal conductivity is 43 W/(m·K). Can the formula from above be used for a rough calculation?

The heat transfer coefficient refers to the rate at which heat is transferred across the heat exchange interface, and it includes the thermal conductivity of the heat exchange tube material as well as the convective heat transfer coefficient of the fluid. In calculations, empirical formulas or experimental data are usually used to obtain the values. Since the thermal conductivity of Q235B is approximately 43 W/(m·K), empirical formulas can be used to estimate the heat transfer coefficient. Based on the information you provided, the volume of exhaust gas generated by burning 1 cubic meter of natural gas can be calculated using its calorific value. Assuming the calorific value of natural gas is 40 MJ/m3 and the temperature of the exhaust gas after combustion is 200°C, the exhaust gas flow rate is 76.5 m3/h. In waste heat recovery, the amount of heat that can be recovered during the process of reducing the temperature from 260°C to 90°C can be calculated using the following formula: Heat = Heat capacity × Mass of water × Temperature difference. Here, the heat capacity is 4.18 J/(g·°C), the mass of water is the total amount of water used for waste heat recovery, which is 17 T/h, and the temperature difference is 260°C – 90°C = 170°C. Calculations show that the heat that can be recovered during the cooling from 260°C to 90°C is approximately 11.9 MW. It should be noted that during waste heat recovery, if the area of the heat exchange tubes is too large, it may cause the temperature of the recovered water to rise, leading to boiling. Therefore, proper matching is required when selecting heat exchangers to ensure the safe and stable operation of the system. .