Question 17.5: Cooling Tower Design and Part-Load Operation Use the effecti......

First, the tower heat rejection is found. It is the sum of the heat removed from the space and the compressor power:

= 7,500 kBtu/h

The tower water flow rate can be found since the tower water range is known (105°F − 85°F = 20°F):

Equations 17.14 through 17.18 are now used to find the peak-load performance of the tower. Since the airflow rate appears implicitly in these equations, an iterative solution is needed. We set up the equations, substituting all known numerical values, and then we present the results of the iterative solution. The iteration is started with an estimate of the tower effectiveness, which is refined at each iteration until convergence is reached.

\varepsilon_{tower} = \frac{\dot{Q}}{\dot{m}_{a} (h_{a,sat,i}  –  h_{a,i})}            (17.14)

NTU = a \left\lgroup \frac{\dot{m}_{w}}{\dot{m}_{a}} \right\rgroup^{m}                  (17.18)

The effective specific heat of saturated air c_{p,a,sat} is given by Equation 17.17:

c_{p,a,sat} = \frac{h_{a,sat,i}  –  h_{a,sat,o}}{T_{w,i}  –  T_{w,o}}           (17.17)

The coupled equations governing tower performance are solved in the following order. First, Equation 17.14 is used to find the first estimate of the tower airflow rate. If we use an initial estimate of effectiveness halfway between the thermodynamic limits of 0.0 and 1.0,

\varepsilon_{tower} = 0.500                (17.21)

Then, the airflow rate by rearranging Equation 17.14 is

Second, the capacitance ratio R can be found from Equation 17.16:

R \equiv \frac{\dot{m}_{a} c_{p,a,sat}}{\dot{m}_{w} c_{p,w}}             (17.16)

Third, the NTU value is found from Equation 17.18 as follows:

Finally, the new value of the tower effectiveness can be found by inserting these values of R and NTU into Equation 17.15. If it is sufficiently close to that given in Equation 17.21, the iteration is complete.
If the difference between the two values is less than the acceptable tolerance, say, ±0.1%, the iteration is at an end. If not, the iteration is repeated, beginning from Equation 17.21 and using the new value of the effectiveness found:

This value of \varepsilon_{tower} is too far from the estimate to be considered accurate, so the iteration is repeated.
The final results for this problem after several iterations ^{†} are

NTU = 2.131,      R = 1.288,

\dot{m}_{a} = 303,700  lb_{m}/h(38.35  kg/s),  and  \varepsilon_{tower} = 0.614

The fan airflow rate is the key result of the calculation. The designer will specify this along with the cooling tower heat rate, range, and approach.
At part load, the tower performance can be found as the aforementioned by using a new value of the airflow rate and finding the corresponding heat rejection rate. At reduced heat rejection rates at part load, the tower exit water temperature is higher than the design point of 85°F; this has a small effect on c_{p,a,sat}. Table 17.5 summarizes the performance of this tower at flow rates ranging from slightly above the design ow rate to 50,000  lb_{m}/h, representing about 15% of full load.
Comments
The effectiveness increases as the airflow decreases at part load because the residence time of air within the tower is longer. However, the heat rate decreases with airflow since Equation 17.16 shows that the heat rate decreases linearly with decreasing flow, whereas the effectiveness increases with decreasing flow but at a less than linear rate.
Problem 17.5 requires the reader to calculate the entries in Table 17.5.

^{†} This simple iteration method converges, but it converges slowly. Better methods that converge more quickly include the Newton–Raphson method. Computer software packages, such as Mathematica, exist specifically for this purpose.