Example 13-2. Check the result if the cold water temperature obtained from example 13-1 is correct.

(Solution)
The actual cold water temperature obtained from example 13-1 is exactly same as the combine temperature of cold water temperature through the tower and hot water temperature of by-pass wall water.

Cold Water Temp. through Tower (Ctemp) = Wet Bulb Temp. + Approach + Design Range - New Range

Hot Water Temp. of By-Pass Wall Water (Btemp) = Ctemp + New Range

Final Cold Water = (Ctemp x Water Flow through Tower + Btemp x By-Pass Wall Water Flow) / Water Flow
= (Ctemp x Water Flow x (1 - %BP) + Btemp x Water Flow x %BP)/ Water Flow
= Ctemp x (1 - %BP) + Btemp x %BP
= Ctemp x (1 - %BP) + (Ctemp + New Range) x %BP
= Ctemp - Ctemp x %BP + Ctemp x %BP + New Range x %BP
= Ctemp + New Range x %BP
= WBT + Approach + Design Range - New Range + New Range x %BP
= WBT + Approach + Design Range - New Range x ( 1 - %BP)
= WBT + Approach + Design Range - Design Range / (1 - %BP) x (1 - %BP)
= WBT + Approach

• Cold Water Temperature Through Tower (Ctemp) = Actual Cold Water Temp. + Range - Actual Range = 88.633 + 15.0 - 15.507 = 88.126^oF
• Hot Water Temperature into Tower (Btemp) = Ctemp + Actual Range or = Actual Cold Water Temp + Range = Ctemp + 15.507 = 103.633^oF
• Water Flow Through Tower (W₁) = Design Water Flow x ( 1 - % By-Pass / 100) = 12,500 x ( 1 - 3.27 /100) = 12,091.25 GPM
• By-Pass Water Flow without Cooling (W₂) =Design Water Flow x % By-Pass / 100 = 12,500 x 3.27 / 100 = 408.75 GPM
• Cold Water Temperature at Water Basin = (W₁ x Ctemp + W₂ x Btemp) / (W₁ + W₂) = (12,091.25 x 88.126 + 408.75 x 103.633) / (12,091.25 + 408.75) = 88.633^oF

Example 13-3. Check if the relation of HEATin = HEATout is established from the above example

(Solution)
HEATin = Total Heat Removal from Water
= Water Flow Rate in GPM x ( 500 / 60 ) x Cooling Range
= 12,500 x ( 500 / 60 ) x ( 104 - 89 )
= 1,562,500 BTU/Min

or = Water Flow Through Tower in GPM x (500 / 60) x Cooling Range Through Tower
= 12,500 x ( 1 - % By Pass / 100 ) x (500 / 60) x Cooling Range / ( 1 - % By Pass / 100)
= 12,500 x ( 1 - 3.27 / 100 ) x (500 / 60) x 15 / ( 1 -3.27 / 100)
= 1,562,500 BTU/Min

HEATout = Total Heat Gain from Air
= Air Mass Flow in LB/Min x (Exit Air Enthalpy - Inlet Air Enthalpy)
= 69,909.2 x (66.0411 - 43.6907)
= 1,562,500 BTU/Min

Example 13-4. Determine L/G ratio and cold water temperature when the wet bulb temperature was downed to 70^oF from design conditions described in the example 13-1.

(Solution)
First, find a dry bulb temperature for 80% of relative humidity corresponding 70^oF of wet bulb temperature.

Second, find an exit air temperature and air volume of fan until these are ultimately equal.

Water Through Tower in LB/Min = Water Through Tower in GPM x (500 / 60)

Air Mass in LB/Min = Air Volume @ Fan / Specific Volume @ Fan

L/G ratio is obtained from the relation of Water Through Tower in LB/Min / Air Mass in LB/Min

Exit Air Enthalpy = Inlet Air Enthalpy + L/G x Range Through Tower = Inlet Air Enthalpy + {Water Through Tower x (500 / 60) / (Air Volume @ Fan / Specific Volume @ Fan)} x Range Through Tower

Net Fan Power = Motor HP x (1 - Motor Minimum Margin) x Power Transmission Efficiency
= Air Volume @ Fan x Total Pressure / (Fan Efficiency x 6356)

Third, calculate the tower characteristic in accordance with above computed results.
KaV/L = 1.864 x {1 / (L/G)}0.8621 x Fill Air Velocity-0.1902 x Fill Height = 1.864 x (1 / 1.4105)^0.8621 x 578.9^-0.1902 x 40.8764 = 1.3890
Total Kav/L = KaV/L @ Fill / (1 - % of Heat Transfer at Rain & Water Spray Zone / 100) = 1.3890 / (1 - 9.9% / 100) = 1.5416

Fourth, determine the NTU satisfying the value of tower characteristic by the method of iteration with the change of approach figure.

Therefore, Actual Cold Water Temperature = Wet Bulb Temperature + Approach = 70 + 11.891 = 81.89 deg. F