|
|
|
Now, the best way to design the cooling tower is based on the actual sizes of tower and is to find a proper L/G satisfying such sizes of cooling tower. The L/G is the most important factor in designing the cooling tower and related to the construction & operating cost of cooling tower. The fooling example will explain about the procedure of determining the L/G ratio. Example 14-1. Determine the L/G ratio under the assumption that the water flow rate was increased to 13,750 GPM and the wet bulb temperature remains unchanged from design conditions given in the example 13-1. (Solution) First, find a dry bulb temperature for 80% of relative humidity corresponding 80oF of wet bulb temperature at the tower inlet. First Step: Find a dry bulb temperature at the tower inlet. Second Step: Find an exit air temperature and air volume of fan. The procedure is exactly same as the contents described in the example 13-1. Net Fan Power = Motor HP
x (1 - Motor Margin) x Power Transmission Efficiency The iteration is continued until the value of Net Fan Power equals to Fan bhp varying air volume at the fan and the pressure drops corresponding to the air volume at each location of cooling tower. Net Fan Power = Fan bhp The air volume is finally obtained from this relationship. Then, the L/G ratio is obtained from below relations. Water Flow Rate in gpm through
Tower = Design Water Flow Rate x (1 - % By-Pass Water
). Water Flow Rate in Lb/Min
= 13,300.4 x (500/60) = 110,836.7 L/G = 110,836.7 / 69,283.3 = 1.59976
Download
the example file, Version ID-THERMAL/TOWER (idthermal.zip)
Example 14-2. The value of slope in the tower characteristic was just estimated like the above examples. Determine the actual slope using the design conditions of the example no.13-1 and 14-1. (Solution) NTU = C x (L/G)-m
------------------------------ Eq.14-1 The fact, which that the value of C for a designed cooling condition is same regardless the change of water flow rate, was already mentioned previously. From this rule, the value of slope can be derived as follows; Log C @ 100% Water = Log(NTU
@ 100% Water) + m x Log(L/G @ 100% Water) -- Eq. 14-4 Eq. 14-4 and Eq. 14-5 can
be written as below using the relation of Log C @ 100%
Water = Log C @ 110% Water. This form can be changed to: Log(NTU @ 100% Water / NTU @ 110% Water) = m x Log (L/G @ 110% Water / L/G @ 100% Water) Finally, this equation can be solved for m as follows; m = Log (NTU @ 100% Water / NTU @ 110% Water) / Log (L/G @ 110% Water / L/G @ 100% Water)
Slope = Log (1.5149 / 1.3863)
/ Log (1.5998 / 1.4413) = 0.8506 |