Yesterday, I simply received an opportunity to inspect a few of cooling tower equipment, comparable to motors, fans, fan stack, low oil stage swap, and so forth. It was nice. I got extra understanding about the system, which could be very beneficial for me as young chemical engineer.
Before I do the inspection, I tried to study something about cooling tower. I found from this linkabout how to evaluate a cooling tower. In this submit I want to give you an example of my own cooling tower.
There are a number of parameters to assess the performance of cooling towers. Note: CT = cooling tower; CW = cooling water
This is the distinction between cooling tower inlet and outlet temperature. A high cooling tower range implies that cooling tower has been in a position to reduce temperature effectively. The components is:
CT vary (oC) = CW inlet temperature (oC) CW outlet temperature (oC)
Inlet and outlet cooling water temperature of my cooling tower is 31 and 41oC, respectively. Then CT range is 41-31 = 10oC
That is the difference between the cooling tower outlet cold water temperature and ambient wet bulb temperature. The lower the approach the better the cooling tower performance. The ‘approachis a greater indicator of cooling tower performance.
CT method (oC) = CW outlet temperature (oC) CW wet bulb temperature (oC)
Wet bulb temperature = 27.4oC, subsequently CT method is 31-27.4 = 3.6oC
That is the ratio between the range and the best range (in proportion), i.e. distinction between cooling water inlet temperature and ambien wet bulb temperature. The upper this ratio, the upper the cooling tower effectiveness.
CT effectiveness (%) = one hundred x (CW in temp CW out temp) / (CW in temp WB temp)
CT effectiveness of my cooling tower = a hundred x (41-31) / (forty one-27.Four) = 73.53%
That is the heat rejected in kCal/hr, given as product of mass movement rate of water, specific heat and temperature difference.
Flow price of water = 1225 m3/hr
Density = a thousand kg/m3
Particular heat =4.2 kJ/kgoC
Temperature difference = 10 oC
Heat rejected = 1225 x one thousand x four.2 x 10 = 51,450,000 kJ/hr = 12,348,000 kCal/hr
That is the water quantity evaporated for cooling duty. Theoretically the evaporation quantity works out to 1.Eight m3for each 1,000,000 kCal heat rejected. The next method can be utilized:
Evaporation loss (m3/hr)= 0.00085 x 1.8 x circulation charge (m3/hr) x (CW inlet temperature CW outlet temperature)
Evaporation loss =0.00085 x 1.Eight x 1225 x (41-31) = 18.Seventy four m3/hr
Cycles of Focus
That is the ratio of dissolved solids in circulating water to the dissolved solids in make up water. Based on the vendor of my cooling tower, cycles of concentration is normally 5-7.