Question 13.8: What are the diameter and height of a hyperbolic natural-dra......
What are the diameter and height of a hyperbolic natural-draught cooling tower handling 4810 kg/s of water with the following temperature conditions:
water entering the tower = 301 K
water leaving the tower = 294 K.
air: dry bulb = 287 K
wet bulb = 284 K
Temperature range for the water, ΔT = (301 – 294) = 7 deg K.
At a mean water temperature of 0.5(301 + 294) = 297.5 K, the enthalpy = 92.6 kJ/kg
At a dry bulb temperature of 287 K, the enthalpy = 49.5 kJ/kg
∴ ΔT’ = (297.5 – 287) = 10.5 deg K
and: ΔH’ = (92.6 – 49.5) = 43.1 kJ/kg
In equation 13.32:
\frac{W_{L}}{D_{t}}=0.00369\frac{\Delta H^{\prime}}{\Delta T}(\Delta T^{\prime}+0.0752\Delta H^{\prime})^{0.5} (13.32)
\frac{4810}{D_{t}}=0.00369\left(\frac{43.1}{7}\right)[10.5+(0.0752\times43.1)]^{0.5}and: D_{t} = 57,110
Taking C_{t} as 5.0 and assuming as a first estimate a tower height of 100 m, then in equation 13.31:
D_{t}=\frac{19.50A_{b}\,\zeta_{t}^{0.5}}{C_{t}^{1.5}} (13.31)
57.110=19.50A_{b}{\frac{100^{0.5}}{5.0^{1.5}}}and: A_{b} = 3274 m²
Thus the internal diameter of the column at sill level = \left({\frac{3274\times4}{\pi}}\right)^{0.5}
= \underline{\underline{64.6\ m}}
Since this gives a height: diameter ratio of (100 : 64.6) ≈ 3 : 2, the design is acceptable.