My model is YPC-ST-400 meaning its TonsOfRefrigeration (TR) is 400
As can be seen we have total refrigerant (water) amount of 550kg



Using steam tables at (~5celsius) saturated properties of water:

Code:
SATURATED
Sat. pressure            0.8718 kPa ~6.5mmHg
Sat. temperature         5.0000 C
Quality                  1.0000
Moisture                 0.0000
Enthalpy
  hf                    21.0067 kJ/kg
  hfg                 2489.7275 kJ/kg
  hg                  2510.7342 kJ/kg
Internal energy
  uf                    21.0058 kJ/kg
  ufg                 2361.4265 kJ/kg
  ug                  2382.4323 kJ/kg
Entropy
  sf                     0.0762 kJ/kg/C
  sfg                    8.9507 kJ/kg/C
  sg                     9.0269 kJ/kg/C
Specific volume
  vf                     0.0010 m3/kg
  vfg                  147.1621 m3/kg
  vg                   147.1631 m3/kg
Thus at evaporator per kg water being evaporated @6.5mmHg will remove ~2489.7275kJ
of enthalpy from the chilled water. Assuming that ~60% of 550kg water is continously
circulated to evaporator we can calculate max. amount of enthalpy that is being
removed & then translate it to TR ...1TR=~3.5kW ...1kJ/sec=1kW

TR = 2489.7275kJ/kg x (550x0.6)kg/hr x 1hr/3600sec x 1TR/3.5kW = 65.2TR

Even if I had 100% water circulation to evaporator which is nonsense I cannot get the value 400TR?