Calculation of chilled water consumption for heat exchange in tank jackets
Thread Content
Suppose there is a tank with a capacity of 1000 L, containing 1000 L of water at 60 degrees Celsius. A full jacket is used. Chilled water is circulated inside the jacket, entering at 7 degrees and returning at 12 degrees. Requirement: Water at 60 degrees must be reduced to 20 degrees within 2 hours. Generally, it is calculated as follows: Frozen water: 5 (temperature difference) * 4.1 (specific heat) * M (mass) = Material: 1 kg (mass) * 40 (temperature difference) * 4.1 (specific heat). If M = 8 kg, then the flow rate of frozen water is 8/2 (hours) = 4 m3/h. However, this calculation method is often incorrect. This is because the heat exchange area of the jacket and the heat transfer coefficient are very low; in other words, even with a large amount of chilled water, it is not possible to reduce the temperature of 1000 liters of water by 20 degrees within 2 hours. At this point, the heat that can be provided by the jacket is equal to K (the heat transfer coefficient of stainless steel) * A (the heat transfer area) * (T1 – T2) / (Ln(T1/T2)), which equals Q_jacket. Here, T1 = 60 degrees and T2 = 20 degrees. The heat required to cool the material is given by Q_material = 1 kg (mass) * 40 (temperature difference) * 4.1 (specific heat). The ratio of Q_material to Q_jacket gives us T; for example, if T = 3 hours, it means that no matter how much chilled water is used, it will take at least 3 hours for the material to cool from 60 degrees to 20 degrees, and cooling it in 2 hours is not possible. Using the formula Q_chilled_water = Q_jacket, we can calculate the flow rate of chilled water per hour. However, the problem is that I don’t know the value of K for the jacket. What should this value be? Is my calculation description above correct?
Generally, stainless steel has a relatively high value; you can check the heat transfer coefficients of various materials in metal material tables. For carbon steel, 1500 is generally used.
Material is just one factor; it cannot determine the heat transfer coefficient. The geometric structure of the jacket flow channel, the properties of the medium within it, the flow velocity, and so on, all have a significant impact.
As long as there is enough cooling material, instant cooling can be achieved. Q = density * volumetric flow rate * time * heat capacity * temperature difference = 1 kg (mass of material) * 40 (temperature difference) * 4.1 (specific heat). It is possible to calculate this heat amount.
(T1-T2)/(LnT1/T2) – this is the logarithmic mean temperature difference. The temperature of the water inside the tank is between 60 and 20, while that of the chilled water is between 7 and 12. There seems to be an issue with this value
After reviewing what the others have said and performing calculations based on that, abnormal values were observed. Upon checking the force-temperature equation Q=KA△T/d, it was found that this formula lacks the term d (the heat transfer distance); once d was included in the calculation, the results became acceptable. So, could it be that the calculations presented by the others omitted d?
I looked at the heat transfer calculation problems and realized that I was wrong in what I wrote earlier; K should be the heat transfer coefficient, not the thermal conductivity of stainless steel. The heat transfer coefficient for the tube wall can be roughly calculated as K = thermal conductivity of the tube wall metal × d (thickness of the tube wall).