(Solution)
Range, R₁ = HWT - CWT = tw₂ - tw₁ = 104 - 89 = 15^oF
Water Flow Rate in Pound, L₁ = L₂ = Water Flow Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Air Mass Flow Rate, G₁ = G₂ = 80,848 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = L/G₂ = 133,333.3 / 80,848 = 1.6492
Range, R₂ = 20^oF
The value of NTU, and "C" at the design conditions is same as a value calculated in the example 4-1. Also, the new tower characteristic for even a increased cooling range is same as the example 4-1. Iterate until the value of new characteristic is equal to the new NTU varying the value of approach. (New Cold Water Temperature = Wet Bulb Temperature + New Approach)

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 80 + 10.65 = 90.65^oF
New HWT = CWT + Range = 90.65 + 20 = 110.65^oF

Example 4-5. Assume the existing mechanical-draft cooling tower is operating at the initial conditions of example 4-1. Determine the cold & hot water temperature if the air mass flow rate is reduced to 53,900 lb/min by the adjustment of the fan pitch angle and/or fan speed.

(Solution)
Range, R₁ = HWT - CWT = tw₂ - tw₁ = 104 - 89 = 15^oF
Water Flow Rate in Pound, L₁ = L₂ = Water Flow Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Air Mass Flow Rate, G₁ = 80,848 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = 133,333.3 / 80,848 = 1.6492
Air Mass Flow Rate, G₂ = 53,900 lb/min
Liquid to Gas Ratio, L/G₂ = L₂ / G₂ = 133,333.3 / 53,900 = 2.4737

The value of NTU, and "C" at the design conditions is same as a value calculated in the example 4-1.Calculate a new tower characteristic for the decreased air mass flow.

New Tower Characteristic = C x (L/G)^-m = 2.21825 x (2.4737)^-0.8 = 1.0748
Iterate until the value of new characteristic is equal to the new NTU varying the value of approach. New Cold Water Temperature = Wet Bulb Temperature + New Approach

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 80 + 14.85 = 94.85^oF
New HWT = CWT + Range = 94.85 + 15 = 109.85^oF

Example 4-6. Assume that the cold & hot water temperature at the conditions where the wet bulb temperature is decreased to 77^oF from 80^oF and the air mass flow is changed to 53,900 lb/min. Others remain unchanged from example 4-1.

(Solution)
Range, R₁ = R₂ = HWT - CWT = tw₂ - tw₁ = 104 - 89 = 15^oF
Water Flow Rate in Pound, L₁ = Water Flow Rate x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Air Mass Flow Rate, G₁ = 80,848 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = L₂ = 133,333.3 / 80,848 = 1.6492
Air Mass Flow Rate, G₂ = 53,900 lb/min
Liquid to Gas Ratio, L/G₂ = L₂ / G₂ = 133,333.3 / 53,900 = 2.4737

The value of NTU, and "C" at the design conditions is same as a value calculated in the example 4-1. Calculate a new tower characteristic for the decreased air mass flow.

New Tower Characteristic = C x (L/G)^-m = 2.21825 x (2.4737)^-0.8 = 1.0748

Forth Step: Iterate until the value of new characteristic is equal to the new NTU varying the value of approach. New Cold Water Temperature = Wet Bulb Temperature + New Approach

Fifth Step: Compute the cold water temperature with the result of iteration as follows;

New CWT = WBT + New Approach = 77.0 + 16.25 = 93.25^oF
New HWT = CWT + Range = 93.25 + 15 = 108.25^oF

Download the example file(exe4_6.zip)