This factor accounts for the amount of water which unavoidably bypasses the fill along the outside and partition walls, internal columns, internal risers etc. This water is not cooled as much as the water passing through the fill. This effect is well known and recognized as the WALL EFFECT but there is no precise theory on how to predict and account for it. This is not a reason for neglecting it in the calculations. It may be very large particularly in a small tower where it can be as big as 20%. Even large towers can have 2% to 5% on the walls. The approach to this problem is very simple. The by-pass wall water is assumed to be only half cooled.

How to estimate the by-pass wall water? Through an example, the estimation can be discussed. 
A 36 x 36 ft tower cell has 144 nozzles. 40 nozzles are near to the four walls each projecting 10 % of their water onto those walls. 40 x 10% / 144 = 2.78 %. 4 nozzles are in the corners and project 20% of their water into the wall. 4 x 20% / 144 = 0.56 %. There are 25 internal columns. Each column receives 5% of the water from 4 adjacent nozzles 25 x 4 x 5% / 144 = 3.47 %. Then total by-pass is 6.81% and the water amount for being half cooled is 6.81 / 2 = 3.4%. This means that 3.4% of total water flow is passing through the wall under not being cooled. This is not an exaggerated number. Obviously this evaluation largely depends on water distribution design and the type of nozzles used. A lot of precautions can be taken to minimize this value but it must be kept in mind that that it is better to have a little water on the walls than leaving dry spots with no water at all. Many cooling tower fills do not redistribute the water very well and air will rush through a dry spot where there is less resistance.

If the tower was 18 x 18 ft the same type of evaluation would give: 

16 x 10% / 36 = 4.4 % 
4 x 20% / 36 = 2.2 % 
4 x 4 x 5% / 36 = 2.2 % 
Total = 8.8 %

This means that the total 4.4% of water flow is being passed through the cooling tower without the heat exchange.

Example 5-1: Let's assume that the % by-pass wall water was 4% and compare the tower demand using the example 4-1.

(Solution) 
Since the 4% of water flow rate is considered not to be completely cooled, the cooling tower has to remove the heat for the original heat load duty and reduced water flow rate. Therefore, it is nature that the cooling range is increased and the tower demand must be based on these new cooling range and cold water temperature.

Original Range, R₁ = 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 
Heat Load, D₁ = L₁ x R₁ = 133,333.3 x 15 = 2,000,000 BTU/min 
Heat Load, D₂ = D₁ = 2,000,000 BTU/min 
Tower Water Flow Rate, L₂ = Water Circulation Rate x (1 - % By-Pass Wall Water / 100) x (500 / 60) 
= 16,000 x (1 - 4 / 100) x (500 / 60) = 128,000.0 lb/min 
Range, R₂ = D₂ / L₂ = 2,000,000 / 128,000 = 15.625^oF 
= L₁ x R₁ / {L₁ x (1 - %By-Pass Wall Water / 100)} 
= R1 / (1 - % By-Pass Wall Water / 100) = (104 - 89) / (1 - 4 / 100) = 15.625^oF 

Tower Cold Water Temp., CWT₂ = CWT₁ + R₁ - R₂ = 89 + 15 - 15.625 = 88.375^oF 

Air Mass Flow Rate, G₂ = G₁ = 80,848 lb/min, Liquid to Gas Ratio, L/G₂ = L₂ / G₂ = 128,000.0 / 80,848 = 1.5832

This example shows that the tower demand is increased by about 9.44% when the by-pass wall
water is considered. That is, the degree of cooling difficulty with the consideration of by-pass wall water is higher than the degree with the ignorance of by-pass wall water.

Example 5-2. The example number 18 was based on the assumption that the heat load is constant for the increase of water flow rate. Estimate the cold & hot water temperature under the assumption that the cooling range is constant for the increase of water flow rate to 20,000 from 16,000 GPM, and the assumption of 4% of total water is being by-passed without the heat removal through the tower.

(Solution)
Range, R₁ = tw₂ - tw₁ = 104 - 89 = 15.0^oF
Tower Water Flow Rate in Pound, L₁ = Water Flow Rate  x (500 / 60) = 16,000 x (500 / 60) = 133,333.3 lb/min
Liquid to Gas Ratio, L/G₁ = L₁ / G₁ = 133,333.3 / 80,848 = 1.6492
Air Mass Flow Rate, G₁ = G₂ = 80,848 lb/min
Tower Water Flow Rate in Pound, L₂ = Water Flow Rate x (1 - % By-Pass Wall Water / 100) x (500 / 60)
= 20,000 x (1 - 4 / 100) x (500 / 60) = 160,000.0 lb/min
Liquid to Gas Ratio, L/G₂ = L₂ / G₂ = 160,000.0 / 80,848 = 1.9790
Range, R₂ = (tw₂ - tw₁) / (1 - % By-Pass Water / 100) = (104 - 89) / 0.96 = 15.625^oF

The value of NTU is same as a value calculated in the example 4-1.
Calculate a value of "C" of tower characteristic for the design conditions as follows;

C = KaV/L / (L/G)^-m = KaV/L x (L/G)^m = 1.4866 x (1.6492)^0.8 = 2.21825

Calculate a new tower characteristic for the increased water flow.
New Tower Characteristic = C x (L/G)^-m = 2.21825 x (1.9790)^-0.8 = 1.2848
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

Finally, compute the cold water temperature with the result of iteration as follows;

New CWT through Tower = WBT + New Approach + Design Range - Actual Range = 80 + 12.331 + 15 - 15.625 = 91.706
Final CWT = (New CWT through Tower x Water Flow through Tower + New HWT x By-Pass Wall Water Flow) / Total Water Flow Rate = (19,200 x 91.706 + 800 x 107.331) / 20,000 = 92.331^oF
Water Flow Rate through Tower = Alternative Water Flow x (1 - % By-Pass) = 20,000 x (1 - 0.04) = 19,200 GPM
By-Pass Wall Water Flow = Alternative Water Flow x % By-Pass = 20,000 x 0.04 = 800 GPM

Final HWT = Final CWT + Heat Build Up from Heat Exchanger (Range) = 92.331 + 15.0 = 107.331^oF

Or, Final HWT = New CWT through Tower + New Range through Tower = 91.706 + 15.625 = 107.331^oF

Therefore, the hot water temperature when to consider the by-pass wall water is higher than example no. 4-3 by 0.321^oF.

Download the example file (exe5_2.zip)