The characteristic curve serves as a measure of the capability of the cooling tower to which it applies. It relates the familiar design term of KaV/L and L/G, and is of the form;

KaV/L = C L/G-m

C and m are constant for a given cooling tower and are determined by the characteristics of the fill, while m is determined by end effects. The characteristic curve is used in conjunction with a KaV/L vs L/G relationship to determine performance. This curve may be termed "Design requirement" curve, since it is a measure of the degree of difficult of the design requirements, and has nothing to to with the physical characteristics of the tower. It is constructed by assuming values of L/G and computing the corresponding values of KaV/L using the following equations.

---------------- Eq. 21.1

where, hw is the enthalpy of air-water vapor mixture at the bulk water temperature and ha is the enthalpy of air-water vapor mixture at the equilibrium wet bulb temperature.

The intersection of the characteristic and design requirements curves locates the design point. The manufacturer predicts that, when operating at the L/G value so located and at design water circulation rate, inlet water temperature, and wet bulb temperature, design outlet water temperature will be attained.

The test value of L/G is determined from Eq. 22-1, 22-10 & 22-12 which were derived in the chapter 22. From these equations, L/G dsn is the L/G value at the intersection of the characteristic and design requirement curves. The corresponding value of KaV/L is computed from above Eq. 63-1 using the test wet bulb and water temperatures. This point is then plotted, and a line a parallel to the characteristic curve is drawn through it. The intersection of this line and design requirements curve locates the L/G capability at the design conditions.

Example 21-1: Determine the tower capability using the characteristic curve for the initial design conditions and below test records.

(Solution)
First, let's compute the dry bulb temperature for the design and wet bulb temperature for the test conditions.

Exit Enthalpy @ Design = Inlet Enthalpy @ Design + L/G design x Range @ Design = 43.6907 + 0.8600 x (115 - 85) = 67.4907 BTU/LB

Exit Enthalpy @ Test = Inlet Enthalpy @ Test + L/G test x Range @ Test

The L/G test is calculated from the below formula.

L/G test = L/G design x (Water Flow test / Water Flow design) x (Fan BHP design / Fan BHP test)1/3 x (Exit Air Density test / Exit Air Density design)1/3 x (Exit Air Specific Volume test / Exit Air Specific Volume design)

Derivation details of L/G Test are as below:

Fan BHP

= VOL x TP / (6356 x Fan Effi.)
= VOL x (VP + SP) / (6356 x Fan Effi.)
= VOL x (1/2g x Density x Vel2 + K x 1/2g x Density x Vel2) / (6356 x Fan Effi.)
= VOL x Density x Vel2 x (1 + K) / 1/2g / (6356 x Fan Effi.)
= VOL x Density x Vel2 x (Area2 / Area2) x (1 + K) /1/2g / (6356 x Fan Effi.)
= VOL x Density x VOL2 x 1 / Area2 x (1 + K) /1/2g / (6356 x Fan Effi.)
(The term of 1 / Area2 x (1 + K) / 1/2g / (6357 x Fan Effi.) could be considered as a constant under the assumption that the fan efficiency at the design conditions is equal to the fan efficiency at the test conditions.)

Then, above equation could be expressed to Constant = Fan BHP / (VOL3 x Density). Therefore, the following relationship is established.

Constant = BHP dsn / (VOL dsn3 x Density dsn) = BHP test / (VOL test3 x Density test)
Let's rewrite this relationship for the term of Vol test.
VOL test3 = (BHP test / BHP dsn) x (Density dsn / Density test) x VOL dsn3

VOL test

= (BHP test / BHP dsn)1/3 x (Density dsn / Density test)1/3 x VOL dsn
= (BHP test / BHP dsn)1/3 x (Density dsn / Density test)1/3 x L dsn / (L/G dsn) x SV dsn (VOL dsn = L dsn / (L/G dsn) x SV dsn)

L/G test

= L test / G test
= L test / (VOL test x SV test)
= L test x SV test / VOL test
= L test x SV test / ( (BHP test / BHP dsn)1/3 x (Density dsn / Density test)1/3 x L dsn / (L/G dsn) x SV dsn)
= L/G dsn x (L test / L dsn) x (BHP dsn / BHP test)1/3 x (Density test / Density dsn)1/3 x (SV test /SV dsn)

Then,

L/G test

= 0.860 x (9,150 / 10,000) x (240.0 / 216.0)1/3 x (Exit Air Density test / 0.0693)1/3 x (Exit Air Specific Volume test / 15.0327)

 

 

 

 

The two factors in the right side of above equation are unknown and these must be computed by the method of try and error. The approximation of air temperature at the first step is an average temperature of test hot and cold water temperatures. Density and S/Volume are computed per the approximated exit air temperature, and L/G test is calculated as per the above equation which is including two unknown factors. The enthalpy under the approximation is obtained and then the new exit air temperature is computed when to iterate until the approximated enthalpy equals to the equivalent enthalpy with varying the temperature.

The L/G test calculated by the try & error method is 0.79823.
Then,

Exit Enthalpy @ Test

= 36.8485 + (104.7 - 79.3) x L/G test
= 36.8485 + 25.4 x 0.79823
= 57.1235

Second, calculate of NTU at Design and Test respectively.

Third, find a value of L/G which the test tower characteristic curve intersects with the design NTU curve. The test C value of tower characteristic is obtained from the relationship of NTUtest = C x L/G-m.

Test C Value

= NTUtest / L/Gtest-m
= NTUtest x L/Gtestm
= 2.3071 x 0.79820.600
= 2.0153

It is to find the value of L/G intersection with the iteration until the test NTU equals to Design NTU varying L/C Intersection.

Test NTU = Test C x L/C Intersection-m
Design NTU at L/G Intersection must be calculated using the below NTU calculation table. The L/G in Air Side of NTU calculation table is to use L/C Intersection.

Finally, determine the tower capability as per the equation below.

Tower Capability

= (L/G intersection / L/G design) x 100 (%)
= (0.8209 / 0.8600) x 100 = 95.46%

The below curve represents the tower capability using Design NTU, Test NTU and Design Characteristic curves.

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