Cooling Tower Design Calculations - Height of Packing & Air Flow Rate

Problem Statement & Given Data:

Warm water at 45°C is to be cooled to 30°C by countercurrent contact with air in a tower packed with wood slats. The inlet air has a dry-bulb temperature of 31°C and a wet-bulb temperature of 22°C. The mass flow rate of water is 6000 kg/m².h and that of air is 1.4 times the minimum. The individual gas-phase mass transfer coefficient is k_Y’a = 6000 kg/m³.h.∆Y’. The volumetric water-side heat transfer coefficient is given by h_La = 0.059 x L^0.51 x G_S, in kcal/m³.h.K, where L and G_S are mass low rates of water and air (dry basis). Determine (a) the dry air flow rate to be used, (b) the height of packing.

Solution:

(a)                                              Inlet air: T_G = 31°C; T_W = 22°C = T_as 

                            Humidity (from Psychrometric chart), Y’₁ = 0.01295.

       

Enthalpy, H’ = [2500 x Y’₁ + (1.005 + 1.88 x Y’₁)] x (31 – 0) = 64.3 kg/kg dry air.

                                  

                                        Exit water temperature, T_L1 = 30°C

Draw the saturation curve (the equilibrium line) from the calculated values of saturation enthalpies at different temperatures as shown in Figure 1. For example, take T_L = 35°C (= 308 K). Vapor pressure of water at this temperature, P^v = 0.05521 bar, and

          Y’ = [(0.05521/(1.013 – 0.05521)] x (18.02/28.97) = 0.02 kg moisture/kg dry air

                           Enthalpy of saturated air at 35°C [ref. temp = 0°C]

                 H’ = [(2500 x 0.02) + (1.005 + 1.88 x 0.02)] x (35 – 0) = 128 kJ/kg dry air

Figure 1

Table 1 Equilibrium line data

Locate the point Q (30°C; 64.3 kJ/kg) on the T_L – H’ plane (Q is the lower terminal of the operating line). In order to determine the minimum air rate, draw the tangent to the equilibrium curve from the point Q shown in Figure 1.

                                 Slope of the tangent = 10.76 = L C_WL / G_S,min

                                                         C_WL = 4.187 kJ/kg.K

                                                            L = 6000 kg/m².h

                                                        G_S,min = 2335 kg/m².h

          Actural air rate to be used,

                                                 G_S = 1.4 x G_S,min = 1.4 x 2335

                                                G_S = 3270 kg/m².h (dry basis)     

(b)  Given, feed water temperature, T_L2 = 45°C. Determine H’₂ (enthalpy of the exit air stream) from the equation of the operating line given below.

L x C_WL x (T_L2 – T_L1) = G x (H’₂ – H’₁) = 6000 x 4.187 x (45 – 30) = 3270 x (H’₂ – 64.3)

                                                     H’₂ = 179.6 kJ/kg

Locate the point P (45°C, 179.6 kJ/kg) which is upper terminal of the operating line. Join PQ.

 Calculate the liquid phase heat transfer coefficient from the given correlation.

                            h_La = 0.059 x L^0.51 x G_S = 0.059 x (6000)^0.51 x 3270 

                                   

                                   = 16,300 kcal/h.m³.°C = 68,260 kJ/h.m³.°C

                        Slope of the ‘tie lines’ = - h_La/k_Y’a = -68,260/6000 = -11.4

A set of tie lines of this slope (-11.4) is drawn from several points on the operating line (including the terminal points) as shown in Figure 2. Any such line meets the equilibrium line at (T_Li, H’_i). The set of points (H’, H’_i) thus obtained are given below in Table 2. The values of 1/(H’_i – H’) are plotted against H’ (Figure 3) and the integral in Eq.1 is evaluated numerically.

Equation 1

Figure 2 Equilibrium and operating lines

Table 2

Figure 3 

 Use Trapezoidal Rule or any other numerical integration technique to determine N_tG ,

            

              Area under the curve = 7.004 = N_tG = number of gas-phase transfer units

                               Height of transfer unit, H_tG = G_S/k_Y’a = 3270/6000 = 0.545

                                 Packed height, Z = N_tG x H_tG = 7.004 x 0.545 = 3.82 m

Z = 3.82 m