Question 13.4: Suppose the cooling tower of Example 13-3 must handle 1000 g......
Suppose the cooling tower of Example 13-3 must handle 1000 gpm of water. It has been determined that an air mass velocity of 1500 lbma/(hr-ft²) is acceptable without excessive water carry-over (drift). The overall mass-transfer coefficient per unit volume U_{i} a_{m} is estimated to be 120 lbm/(hr-ft³) for the type of tower to be used. Estimate the tower dimensions for the required duty.
The transfer units N required for the tower were found to be 1.1067 in Example 13-3.
Then the total volume of the tower is given by Eq. 13-30 as
V = \frac{N \dot{m}_{l} c_{l}}{U_{i} a_{m}} (13-30)
V = \frac{1.1067(1000) 8.33 (60) (1.0)}{120} = 4609 ft^{3}
The cross-sectional area of the tower may be determined from Eq. 13-31 using the mass velocity of the air and the water-to-air ratio:
A_{c} = \frac{\dot{m}_{a}}{G_{a}} = \frac{\dot{m}_{l}}{G_{l}} (13-31)
A_{c} = \frac{\dot{m}_{a}}{G_{a}} = \left\lgroup\frac{\dot{m}_{l}}{\dot{m}_{a}} \right\rgroup \frac{\dot{m}_{a}}{G_{a}} = \frac{\dot{m}_{l}}{G_{a}}
A_{c} = \frac{1000 (8.33) (60)}{1500} = 333 ft^{2}
which is approximately equal to an 18 × 18 ft cross section. Then from Eq. 13-32,
L = \frac{V}{A_{c}} (13-32)
L = \frac{V}{A_{c}} = \frac{4609}{333} = 13.8 ft