Consequently, the enthalpy of exit air is a summation of the enthalpy of entering air and the addition of enthalpy from water to air (this is a value of L/G x Range).

Example 2-1. Calculate the ratio of water and air rate for the 20,000 gpm of water flow and 1,600,000 acfm of air flow at DBT 87.8^oF, 80% RH, and sea level.

(Solution)
Water Flow Rate = GPM x (500 / 60) lb/min = 20,000 x (500 / 60) =166,666.67 lb/min
(The weight of 1 gallon of water at 60^oF equals to 8.345238 pounds and 500 was obtained from 8.345238 x 60 for simplifying the figure.)

Air Flow Rate = ACFM / Specific Volume =1,600,000 / 14.3309 =111,646.76 lb/min
(Specific Volume @ 87.8^oF, 80% & sea level = 14.3309 ft³/lb)

Ratio of Water to Air = Water Flow Rate / Air Flow Rate =166,666.67 / 111,646.76 =1.4928

Example 2-2. Why is L/G in the equation of ha₂ = ha₁ + L/G x Range called a slope?

(Solution)
This curve is exactly same as a linear function of y = a + b x. The ha₁ corresponds to "a" , L/G corresponds to "b", and the cooling range corresponds to "x". So, L/G is a slope of linear curve.

Example 2-3. Calculate the enthalpy and temperature of exit air for the following cooling tower design conditions.

Given,

• Ambient Wet Bulb Temperature: 82.4^oF
• Relative Humidity: 80%
• Site Altitude: sea level
• L/G Ratio: 1.4928
• Entering Water Temperature: 107.6^oF
• Leaving Water Temperature: 89.6^oF

(Solution)
The enthalpy of exit air is calculated from Eq. 2-2 which was derived above. That is, ha₂ = ha₁ + L/G x Range. The enthalpy of inlet air (ha₁) at 82.4^oF WBT & sea level is 46.3624 Btu/Lb dry air.

The cooling range = Entering Water Temp. - Leaving Water Temp. = (tw₂ - tw₁) = 107.6 - 89.6 = 18^oF

Therefore, the enthalpy of exit air (ha₂) is obtained as below.

ha₂ = ha₁ + L/G x Range = 46.3624 + 1.4928 x (107.6 - 89.6) = 73.2328 BTU/lb

A temperature corresponding to this value of air enthalpy can be obtained from the table published by Cooling Tower Institute or other psychrometric curve. However, this can be computed from the computer program. The procedure of computing a temperature at a given enthalpy is to find a temperature satisfying the same value of enthalpy varying a temperature by means of iteration.

Download the example file(exe2_3.zip)

The right side of the above equations is obviously dimensionless factor. This can be calculated using only the temperature and flows entering the cooling tower. It is totally independent from the tower size and fill configuration and is often called, for lack of another name, NTU. Plotting several values of NTU as a function of L/G gives what is known as the "Demand" curve. So, NTU is called Tower Demand too.

As shown on above, NTU is an area of multiplying the cooling range by the average of 1/ (h_w - h_a) at four points in the x axis (Temp.).

NTU or KaV/L = Cooling Range x [Sum of 1 / (hw - ha)] / 4

Example 2-4. Determine the tower demand (called Number of Transfer Unit) for the below given conditions.

Given,

• Water Circulation Rate: 16000 GPM
• Entering Air Flow Rate: 80848 Lb of dry air / min
• Ambient Wet Bulb Temperature: 80.0^oF
• Site Altitude: sea level
• Hot Water Temperature: 104.0^oF
• Cold Water Temperature: 89.0^oF

(Solution)
Water Flow Rate = 16,000 x (500 / 60) = 133,333 Lb/min
L/G Ratio = Water Flow Rate / Air Flow Rate = 133,333 / 80,848 = 1.6492

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