Many theories have been developed since the early 1900s describing the heat and mass transfer phenomenon which takes place in several types of atmospheric water cooling devices. Most of these theories are based on sound engineering principles. The cooling tower may be considered as a heat exchanger in which water and air are in direct contact with one another. There is no acceptable method for accurately calculating the total contact surface between water and air. Therefore, a "K" factor, or heat transfer coefficient, cannot be determined directly from test data or by known heat transfer theories. The process is further complicated by mass transfer. Experimental tests conducted on the specified equipment designs can be evaluated using accepted and proven theories which have been developed using dimensional analysis techniques. These same basic methods and theories can be used for thermal design and to predict performance at the operating conditions other than the design point.

Many types of heat and mass transfer devices defined the solution by theoretical methods or dimensional analysis. Design data must be obtained by the full-scale tests under the actual operating conditions. Items such as evaporative condensers in which an internal heat load is being applied, along with water and air being circulated over the condenser tubes in indefinable flow patterns, presents a problem which cannot be solved directly by mathematical methods. The boundary conditions have not been adequately defined and the fundamental equations describing the variables have not been written. Other devices such as spray ponds, atmospheric spray towers, and the newer spray canal systems have not been accurately evaluated solely by mathematical means. This type of equipment utilizes mixed flow patterns of water and air. Many attempts have been made to correlate performance using "drop theories", "cooling efficiency", number of transfer units, all without proven results. Accurate design data are best obtained by the actual tests over a wide range of operating conditions with the specified arrangement.

The development of cooling tower theory seems to begin with Fitzgerald. The American Society of Civil Engineers had asked Fitzerald to write a paper on evaporation, and what had appeared to be a simple task resulted in a 2 year investigation. The result, probably in keeping with the time, is more of an essay than a modern technical paper. Since the study of Fitzgerald, many peoples like Mosscrop, Coffey & Horne, Robinson, and Walker, etc. tried to develop the theory.

1) Merkel Theory

The early investigators of cooling tower theory grappled with the problem presented by the dual transfer of heat and mass. The Merkel theory overcomes this by combining the two into a single process based on enthalpy potential. Dr. Frederick Merkel was on the faculty of the Technical College of Dresden in Germany. He died untimely after publishing his cooling tower paper. The theory had attracted little attention outside of Germany until it was discovered in German literature by H.B. Nottage in 1938.

Cooling tower research had been conducted for a number of years at University of California at Berkley under the direction of Professor L.K.M. Boelter. Nottage, a graduate student, was assigned a cooling tower project which he began by making a search of the literature. He found a number of references to Merkel, looked up the paper and was immediately struck by its importance. It was brought to the attention of Mason and London who were also working under Boelter and explains how they were able to use the Merkel theory in their paper.

Dr. Merkel developed a cooling tower theory for the mass (evaporation of a small portion of water) and sensible heat transfer between the air and water in a counter flow cooling tower. The theory considers the flow of mass and energy from the bulk water to an interface, and then from the interface to the surrounding air mass. The flow crosses these two boundaries, each offering resistance resulting in gradients in temperature, enthalpy, and humidity ratio. For the details for the derivation of Merkel theory, refer to Cooling Tower Performance edited by Donald Baker and the brief derivation is introduced here. Merkel demonstrated that the total heat transfer is directly proportional to the difference between the enthalpy of saturated air at the water temperature and the enthalpy of air at the point of contact with water.

Q = K x S x (hw - ha)
where,

  • Q = total heat transfer Btu/h
  • K = overall enthalpy transfer coefficient lb/hr.ft2
  • S = heat transfer surface ft2. S equals to a x V, which "a" means area of transfer surface per unit of tower volume. (ft2/ft3), and V means an effective tower volume (ft3).
  • hw = enthalpy of air-water vapor mixture at the bulk water temperature, Btu/Lb dry air
  • ha = enthalpy of air-water vapor mixture at the wet bulb temperature, Btu/Lb dry air

The water temperature and air enthalpy are being changed along the fill and Merkel relation can only be applied to a small element of heat transfer surface dS.

dQ = d[K x S x (hw - ha)] = K x (hw - ha) x dS
The heat transfer rate from water side is Q = Cw x L x Cooling Range, where Cw = specific heat of water = 1, L = water flow rate. Therefore, dQ = d[Cw x L x (tw2 - tw1)] = Cw x L x dtw. Also, the heat transfer rate from air side is Q = G x (ha2 - ha1), where G = air mass flow rate Therefore, dQ = d[G x (ha2 - ha1)] = G x dha.
Then, the relation of K x (hw - ha) x dS = G x dha or K x (hw - ha) x dS = Cw x L x dtw are established, and these can be rewritten in K x dS = G / (hw - ha) x dha or K x dS / L = Cw / (hw - ha) x dtw. By integration,

This basic heat transfer equation is integrated by the four point Tchebycheff, which uses values of y at predetermined values of x within the interval a to b in numerically evaluating the integral . The sum of these values of y multiplied by a constant times the interval (b - a) gives the desired value of the integral. In its four-point form the values of y so selected are taken at values of x of 0.102673.., 0.406204.., 0.593796.., and 0.897327..of the interval (b - a). For the determination of KaV/L, rounding off these values to the nearest tenth is entirely adequate. The approximate formula becomes:

For the evaluation of KaV/L,

2) Heat Balance

HEATin = HEATout
WATER HEATin + AIR HEATin = WATER HEATout + AIR HEATout
Cw L2 tw2 + G ha1 = Cw L1 tw1 + G ha2 Eq. 2-1
(The difference between L2 (entering water flow rate) and L1 (leaving water flow rate) is a loss of water due to the evaporation in the direct contact of water and air. This evaporation loss is a result of difference in the water vapor content between the inlet air and exit air of cooling tower. Evaporation Loss is expressed in G x (w2 -w1) and is equal to L2 - L1. Therefore, L1 = L2 - G x (w2 -w1) is established.)

Let's replace the term of L1 in the right side of Eq. 2-1 with the equation of L1 = L2 - G x (w2 -w1) and rewrite. Then, Cw L2 tw2 + G ha1 = Cw [L2 - G x (w2 - w1)] x tw1 + G ha2 is obtained. This equation could be rewritten in Cw x L2 x (tw2 - tw1) = G x (ha2 - ha1) - Cw x tw1 x G x (w2 - w1). In general, the 2nd term of right side is ignored to simplify the calculation under the assumption of G x (w2 - w1) = 0.

Finally, the relationship of Cw x L2 x (tw2 - tw1) = G x (ha2 - ha1) is established and this can be expressed to Cw x L x (tw2 - tw1) = G x (ha2 - ha1) again. Therefore, the enthalpy of exit air, ha2 = ha1 + Cw x L / G x (tw2 - tw1) is obtained. The value of specific heat of water is Eq. 2-1 and the term of tw2 (entering water temperature) - tw1 (leaving water temperature) is called the cooling range.

Simply, ha2 = ha1 + L/G x Range Eq. 2-2


 

 

 

 

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 87.8oF, 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 60oF 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 @ 87.8oF, 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: 82.4oF
  • Relative Humidity: 80%
  • Site Altitude: sea level
  • L/G Ratio: 1.4928
  • Entering Water Temperature: 107.6oF
  • Leaving Water Temperature: 89.6oF

(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 82.4oF 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 = 18oF

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)

3) NTU (Number of Transfer Unit) Calculation

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/ (hw - ha) 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: 80.0oF
  • Site Altitude: sea level
  • Hot Water Temperature: 104.0oF
  • Cold Water Temperature: 89.0oF

(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

 

 

 

 


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 81.0oF from 80.0oF.

(Solution)

WATER SIDE

AIR SIDE

ENTH DIFF.

Descriptions

tw (oF)

hw (But/Lb)

Description

ha (Btu/Lb)

1/(hw-ha)

tw1 + 0.1 x R

90.50

56.6478

ha1 + 0.1 x L/G x R

47.2587

0.1065

tw1 + 0.4 x R

95.00

63.3426

ha1 + 0.4 x L/G x R

54.6800

0.1154

tw1 + 0.6 x R

98.00

68.2591

ha1 + 0.6 x L/G x R

59.6276

0.1159

tw1 + 0.9 x R

102.50

76.4013

ha1 + 0.9 x L/G x R

67.0489

0.1069

Sum of 1 / (hw - ha)

0.4447

Total Tower Demand (NTU) = Cooling Range x Sum of 1 / (hw - ha)

1.6677

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 101.0oF from 104oF.

(Solution)

WATER SIDE

AIR SIDE

ENTH DIFF.

Descriptions

tw (oF)

hw (But/Lb)

Description

ha (Btu/Lb)

1/(hw-ha)

tw1 + 0.1 x R

90.20

56.2283

ha1 + 0.1 x L/G x R

45.6697

0.0947

tw1 + 0.4 x R

93.80

61.4808

ha1 + 0.4 x L/G x R

51.6068

0.1013

tw1 + 0.6 x R

96.20

65.2631

ha1 + 0.6 x L/G x R

55.5648

0.1031

tw1 + 0.9 x R

99.80

71.4001

ha1 + 0.9 x L/G x R

61.5019

0.1010

Sum of 1 / (hw - ha)

0.4001

Total Tower Demand (NTU) = Cooling Range x Sum of 1 / (hw - ha)

1.2004

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

COMPARISON TABLE

Descriptions

Exe. 2-4

Exe. 2-5

Exe. 2-6

Exe. 2-7

Range (oF)

15.0

15.0

15.0

12.0

Approach(oF)

9.0

9.0

8.0

9.0

Wet Bulb Temp.(oF)

80.0

80.0

81.0

80.0

L/G Ratio

1.6492

1.2540

1.6492

1.6492

KaV/L

1.4866

1.1677

1.6677

1.2004

Driving Force (BTU/Lb)

10.1031

13.0670

9.0089

10.0073

Order of Cooling Difficulty

2nd

4th

1st

3rd