Cooling Tower

COOLING TOWERS 1.

A mechanical-draft cooling tower receives 115 m3 per second of atmospheric air at 103 kPa, 32 C dry bulb temperature, 55% RH and discharges the air saturated at 36 C. If the tower receives 200 kg/s of water at 40 C, what will be the exit temperature of the cooled water? Solution:

 3 = 200 kg s m t3 = 40 C

tdb2 = 36 C saturated V1 =115 m3 s tdb1 = 32 C

φ1 = 55% at 1, for tdb1 = 32 C pd = 4.799 kPa hg1 =2559.9 kJ kg

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COOLING TOWERS ps = φ1 pd = 0.55( 4.799 ) = 2.639 kPa 0.622 ps 0.622( 2.639 ) W1 = = = 0.0164 kg kg pt − ps 103 − 2.639

h1 = cptdb1 + W1hg = (1.0062)( 32 ) + ( 0.0164)( 2559.9 ) = 74.2 kJ kg RT ( 0.287 )( 32 + 273) = 0.8722 m3 kg = pt − ps 103 − 2.639 V 115 a= 1 = m =131.85 kg s v1 0.8722

v1 =

at 2, t db2 = 36 C , saturated ps = pd = 5.979 kPa hg 2 = 2567.1 kJ kg

W2 =

0.622ps 0.622( 5.979 ) = = 0.0383 kg kg pt − ps 103 − 5.979

h2 = cptdb2 + W2 hg = (1.0062 )( 36) + ( 0.0383)( 2567.1) = 134.5 kJ kg At 3, t3 = 40 C h3 =hf at 40 C =167.57 kJ kg

 4: To solve for m By Mass Balance:  3 +m  a + W1m  a =m  4 +m  a + W2 m a m

 4 =m  3 + ( W1 − W2 ) m  a = 200 + ( 0.0164 − 0.0383)(131.85) = 197.1 kg s m

To solve for h4 :  ah1 + m  3 h3 = m  4 h4 + m  ah2 m

(131.85)( 74.2) + ( 200)(167.57 ) = (197.1) ( h4 ) + (131.85)(134.5)

h4 =129.70 kJ kg ∴ t4 = 31.9 C - exit water temperature.

2.

In a cooling tower water enters at 52 C and leaves at 27 C. Air at 29 C and 47% RH also enters the cooling tower and leaves at 46 C fully saturated with moisture. It is desired to determine (a) the volume and mass of air necessary to cool 1 kg of water, and (b) the quantity of water that can be cooled with 142 cu m per minute of atmospheric air.

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COOLING TOWERS Solution:

t3 = 52 C tdb2 = 46 C sat

tdb1 = 29 C

φ1 = 47% t4 = 27 C From psychrometric chart At 1, tdb1 = 29 C , φ1 = 47% h1 =59 kJ kg W1 = 0.0116 kg kg v1 = 0.873 m3 kg

at 2, tdb2 = 46 C , saturated ps = pd =10.144 kPa hg2 =2585.0 kJ kg

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COOLING TOWERS W2 =

0.622 ps 0.622(10.144 ) = = 0.0692 kg kg pt − ps 101.325 − 10.144

h2 = cptdb2 + W2 hg = (1.0062 )( 46) + ( 0.0692)( 2585) = 225.2 kJ kg At 3, t3 = 52 C h3 =hf at 52 C =217.69 kJ kg

At 4, t4 = 27 C h4 =hf at 27 C =113.25 kJ kg

(a)

Volume of air necessary to cool 1 kg of water = V1 = mav1 To solve for ma when m3 =1 kg By energy balance: Eq. (1) mah1 + m3 h3 = mah2 + m4 h4 By mass balance

m3 − m4 = ma (W2 − W1 ) Eq. (2) m4 = m3 − ma (W2 − W1 ) Substitute in (1)

= mah2 + [ m3 − ma ( W2 − W1 ) ] h4 ( ma ) ( 59) + (1)( 217.69) = ( ma ) ( 225.2) + [1 − ( ma ) ( 0.0692 − 0.0116) ] (113.25)

Eq. (3) mah1 + m3 h3 ma = 0.654 kg

Volume of air = V1 = mav1 = ( 0.654 )( 0.873) = 0.5709 m3 Mass of air required = ma = 0.654 kg (b)

 3 = quantity of water. Let m V 142 a= 1 = m =162.66 kg min v1 0.873

 a , m3 → m 3 Use Eq. (3), change ma → m

 ah1 + m  3 h3 = m  ah2 + [ m  3 −m  a ( W2 − W1 ) ] h4 m

 3 ) ( 217.69) = (162.66)( 225.2 ) + [ m  3 − (162.66)( 0.0692 − 0.0116 ) ](113.25) (162.66)( 59) + ( m  3 =36,631 +113.25m  3 −1061.1 9597 + 217.69m  3 = 248.7 kg min m

∴ Quantity of water =

 3 = 248.7 kg min m

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COOLING TOWERS

3. A cooling tower receives 6 kg/s of water of 60 C. Air enters the tower at 32 C dry bulb and 27 C wet bulb temperatures and leaves at 50 C and 90% relative humidity. The cooling efficiency is 60.6%. Determine (a) the mass flow rate of air entering, and (b) the quantity of make-up water required. Solution:

 3 = 6 kg s , t3 = 60 C m tdb2 = 50 C

φ 2 = 90% tdb1 = 32 C twb1 = 27 C Cooling tower efficiency = 60.6% To solve for t 4 : Efficiency =

0.606 =

t3 − t 4 t3 − twb1

60 − t4 60 − 27

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COOLING TOWERS t3 = 40 C at 1, tdb1 = 32 C , twb1 = 27 C h1 =85 kJ kg W1 = 0.0208 kg kg at 2, tdb2 = 50 C , φ 2 = 90% pd =12.349 kPa ps = φ2 pd = ( 0.90 )(12.349) = 11.114 kPa hg2 =2592.1 kJ kg

W2 =

0.622 ps 0.622(11.114 ) = = 0.0766 kg kg pt − ps 101.325 −11.114

h2 = cptdb2 + W2 hg = (1.0062 )( 50) + ( 0.0766 )( 2592.1) = 248.9 kJ kg At 3, t3 = 60 C h3 =hf at 60 C =251.13 kJ kg

At 4, t4 = 40 C h4 =hf at 40 C =167.57 kJ kg

By mass balance

m3 − m4 = ma (W2 − W1 ) m4 = m3 − ma (W2 − W1 )

By energy valance  a h1 + m  3h3 = m  a h2 + m  4 h4 m

 ah1 + m  3 h3 = m  ah2 + [ m  3 −m  a ( W2 − W1 ) ] h4 m ( m a ) ( 85) + ( 6)( 251.13) = ( m a ) ( 248.9) + [ 6 − ( m a ) ( 0.0766 − 0.0208) ](167.57 )

 a +1506.8 = 248.9m  a +1005.4 −9.35m a 85m  a = 3.244 kg s m  a = 3.244 kg s (a) Mass flow rate of air = m

(b) Make Up Water = ma (W2 − W1 ) = ( 3.244 )( 0.0766 − 0.0208) = 0.1810 kg s

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