It is very convenient to use the below tool in calculating NTU.

Download the example file(exe2_4.zip)

Example 2-5. Compare NTU at the same given conditions as above example 2-4 excepting L/G = 1.2540.

(Solution)

Download the example file(exe2_5.zip)
(This file covers the examples of 2-6 & 2-7.)

The NTU at L/G = 1.2540 is smaller than NTU at L/G = 1.6492 under the same design conditions. What the L/G is reduced to 1.2540 from 1.6492 under the same water flow rate means that the air mass is increased. In other word, the decrease of L/G for the same water flow rate means the decrease of enthalpy in the air side and a value of 1 / (hw - ha) is consequently decreased. Also, the exit enthalpy per pound dry air is decreased and the temperature of exit air is reduced.

In the actual cooling tower, what the water is evenly distributed on the entire top of fill is very rare. If the temperature is measured onto the top of drift eliminator, the temperature at the area where the water is smaller than other locations is always lower than the water is larger. This is because the air at the area where the water is small can go easily up due to less pressure drop with the water loading.

Example 2-6. Compare NTU at the same given conditions as above example 2-4 excepting that the ambient wet bulb temperature has been changed to 81.0^oF from 80.0^oF.

(Solution)

Through this example, the higher ambient wet bulb temperature (approach is smaller), the larger NTU. That is, the enthalpy driving force (hw - ha) is reduced as the ambient wet bulb temperature is increased. This means that less driving force requires more heat transfer area or more air. (Sometimes, NTU calls "Degree of Difficulty".)

Example 2-7. Compare NTU at the same given conditions as above example 2-4 excepting that the entering water temperature has been changed to 101.0^oF from 104^oF.

(Solution)

This example presents that the smaller range under the same approach, the smaller NTU.