Question

This is problem 7-3 from El-Wakil’s Powerplant Technology book -- a natural-draft cooling tower is 450 ft high. Air enters the tower at 14.696 psia, 50°F, and 50 percent relative humidity and leaves in a saturated condition. The pressure drop in the tower is 0.015 psi. Calculate (a) the air exit temperature, in degrees Fahrenheit, and (b) the makeup due to evaporation, in pound mass per pound mass of dry air.

Answer 1

Ans.

A cooling tower is a huge chimeny like structure. It is basically a heat rejection device that rejects waste heat to atmosphere, so that the cooling off water could be achieved and also the temperature of water stream should be lowered.

Natural drafts cooling tower cools the temperature of water naturally. There is no Mechanical device like fan or exhaust is used for cooling purpose. Whole mechanisms is done naturally under the influence of buoyancy.

Now moving towards the numerical, we have ;

Cooling tower height (H) = 450 ft

Pressure at entrance (P1) = 14.696 psia

Temperature at entrance (T1) = 50°F

Relative Humidity (Rh) = 50 %

Pressure drop (Pd) = 0.015 psi.

Pressure at exit (P2) = 14.696 - 0.015

P2= 14.681 psi.

(a) air exit temperature in degree Fahrenheit is;

P1/P2 = T1/T2

14.696/14.681 = 50/T2

T2 = T1×P2/P1

T2 = 50×14.681/14.696

T2 = 48°F

(b) As we know,

M = E + D + B

Where,

M is makeup

E is evaporation

D Draft

B blow down

Now, E = C × R × Cp/Ch

C is cycle of concentration ( 3 - 7)

R is range i.e. Hotwater temperature - cold water temperature

Cp is specific heat i.e. 4.184 kJ/Kg/C°

Ch is latent heat i.e. 2260 kJ/kg

E = 5×2×4.184/2260

E = 0.01851 m ³ / hr.

Now, Blow down,

B = E/(coc-1)

E is evaporation

Coc is cycle of concentration

B = 0.01851/(5-1)

B = 0.00462 m ³/hr.

And D = 0.3 - 1 ( for natural draft)

Therefore

M = E + D + B

M = 0.0185 + 0.00462 + 0.8

M = 0.82312 m ³/hr.

Hence, was the makeup due to evaporation.