How to calculate the small chiller for the ammonia tank area?
Thread Content
Ammonia tanks stored at normal pressure and low temperatures are equipped with refrigeration systems, but how is the size of the refrigeration unit determined? Since I have no experience with such units, I don’t know how to calculate it, and I also don’t know how to determine the amount of ammonia vapor that is released from the tank. Who can explain it?
Just calculate it based on the saturated vapor pressure of ammonia
Reply to 3# Yan Qiusheng: Could you be more specific? Isn’t the choice of ice machine based on cooling capacity? I still don’t quite understand
In this case, it is mainly about calculating the maximum evaporation rate of the ammonia tank; the required condition is the size of the ammonia tank； Insulation parameters ； Maximum ambient temperature ； The evaporation rate is determined by calculating the heat exchange between the ammonia tank and the surrounding environment; the capacity of the ice machine must be at least about twice the evaporation rate ； The flashing of ammonia feed is generally not considered
Reply to 5# a77sun: Thank you. I’m starting to understand a bit now, but how do I calculate the heat exchange between the ammonia tank and the surrounding environment? Where can I find information on this?
Reply to 6# lordbeatles: For the heat transfer calculation of larger ammonia tanks, a multi-layer flat-wall heat transfer approach can be used. The calculation should be carried out in two stages, taking into account different levels of liquid volume – the highest and lowest levels. It is also necessary to consider the thermal expansion and contraction of gaseous ammonia above the liquid surface, as this reflects changes in pressure. The temperature of the gas phase in the upper layer may exceed the equilibrium temperature; for precise calculations, further reference to relevant literature is required
When selecting a chiller, it is first necessary to determine the evaporation rate of the storage tank, that is, the amount of gaseous ammonia that evaporates per unit of time. If this cannot be determined, the cooling loss of the tank can be calculated using the formula in GBT15586; an appropriate adjustment can then be made, and from that the ammonia evaporation rate can be inferred. It is also important to determine the evaporation temperature and condensation temperature of the ammonia, and calculations can be carried out using the Carnot cycle. Essentially, this involves using a pressure-enthalpy diagram, and the enthalpy difference can be found by referring to the corresponding values on the diagram. The cooling capacity is given by Qm* (H1-H5). Basically, all gases obey the gas state equation. If necessary, we can communicate privately.
By the way, based on actual operational and computational experience, a superheat of about 5°C needs to be taken into account at the inlet of the ice machine. Additionally, when considering the power required by the ice machine, the volumetric efficiency must be factored in, and the compression process by the compressor should be carried out according to adiabatic isentropic principles
Seeking communication. . . . . . I work with pipes; I’m not very good at the technical aspects of it
Hello, could you provide a specific calculation example?
There is currently a 10,000 cubic meter storage tank for liquid ammonia at atmospheric pressure; to maintain the pressure, a pressure-preserving refrigeration compressor has been installed to condense the ammonia that evaporates inside the tank back into liquid form and send it back to the tank； The compressor rear condenser is an evaporative condenser. Compressor inlet ammonia parameters: -33°C, 0.103 MPa, 842 Kg/h; ammonia evaporation temperature: -33°C, condensation temperature: 40°C. Determine the cooling capacity of the compressor, the power required by the compressor, and the power required by the evaporative condenser. Compressor outlet parameters: =1825 kJ/kg (isentropic compression). Condenser outlet parameters: =40°C, =400 kJ/kg. Evaporator inlet parameters: =-33°C, =400 kJ/kg. Evaporator outlet parameters: =-33°C, =1425 kJ/kg. Qm=842 Kg/h. Cooling capacity of the compressor = (h5-h4)×Qm = 1025×0.2338888 = 239.736 KW (note unit conversion to international units). Theoretical compressor power = (h1-h5)×Qm = 400×0.2338888 = 93.556 KW; the actual compression power is obtained by dividing the theoretical value by the compressor efficiency. Condenser power (heat load) = (h1-h3)×Qm = 1425×0.233888 = 333.2917 KW. The values for enthalpy and other parameters mentioned above are for reference only; you should perform the calculations using your own values