Example 13-1. Determine the cold water temperature for the following conditions.

First Step: Find a dry bulb temperature at the tower inlet.

Second Step: Find an exit air temperature and air volume of fan. The fan power is determined from the relation of Motor HP x (1 - Motor Margin) x Power Transmission Efficiency. It is to iterate the calculation until the net fan power obtained from this equals to the fan bhp which is formulated with (ACFM x Total Pressure) / (6356 x Fan Efficiency). Two variables in the fan bhp equation are unknown, but can be computed from below relationships.

The main idea is to iterate the calculation until the net fan power equals to the calculated fan bhp varying the air volume, static pressure and tower exit temperature at the fan.

The air mass flow rate through the tower is always constant because the air mass is being considered as dry gas. Then, the air mass flow rate and L/G Ratio can be obtained as below:

Air Mass Flow Rate = Air Volume @ Fan / Specific Volume @ Fan

The L/G Ratio =Water Flow Rate in gpm through Tower x (500/60) / (Air Volume @ Fan / Specific Volume @ Fan)

Water flow rate in gpm through tower in this formula is a net water flow rate considering the % by-pass wall water. That is, Design Water Flow Rate x (1 - % By-Pass Water ).

In order to obtain the specific volume the tower exit temperature should be computed first. The exit temperature requires L/G ratio as the exit air enthalpy (ha₂ = ha₁ + L/G x Range) is a summation of tower inlet air enthalpy and L/G x cooling range. L/G ratio can be given or calculated if the tower dimensions are given and the value of fill characteristic is known.

The air volume at each location of cooling tower is obtained from the relationship of Air Mass Flow Rate x Specific at each location, and then the air velocity at each location is calculated by dividing the air volume by the net area at each location. The total static pressure is a summation of pressure drops obtained from each location.

Third Step: Calculate the tower characteristic in accordance with above results. To calculate this the performance data of fill manufacturer is required.

Fourth Step: Determine NTU(=KaV/L) satisfying the value of tower characteristic by the method of iteration with changing the approach.

Download the example file, Version ID-THERMAL/TOWER (idthermal.zip)
This file covers the example no. 13-4, too.

Therefore, Actual Cold Water Temperature = Wet Bulb Temperature + Approach = 80.00 + 8.633 = 88.63^oF

Above the enthalpies for water side were computed at the below temperatures.

• Water Temperature @ 0.1 x Range = Wet Bulb Temperature + Approach + Range - New Range + 0.1 x New Range
• Water Temperature @ 0.4 x Range = Wet Bulb Temperature + Approach + Range - New Range + 0.4 x New Range
• Water Temperature @ 0.6 x Range = Wet Bulb Temperature + Approach + Range - New Range + 0.6 x New Range
• Water Temperature @ 0.9 x Range = Wet Bulb Temperature + Approach + Range - New Range + 0.9 x New Range

Also, the enthalpies for air side were based on the followings.

• Enthalpy @ 0.1 x Range = Inlet Air Enthalpy + L/G x 0.1 x New Range
• Enthalpy @ 0.4 x Range = Inlet Air Enthalpy + L/G x 0.4 x New Range
• Enthalpy @ 0.6 x Range = Inlet Air Enthalpy + L/G x 0.6 x New Range
• Enthalpy @ 0.9 x Range = Inlet Air Enthalpy + L/G x 0.9 x New Range
   

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