COOLING TOWERS 1 PDF

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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: m...

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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 t 3 = 40 C t db2 = 36 C saturated V1 = 115 m 3 s t db1 = 32 C

φ1 = 55% at 1, for t db1 = 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 p s 0.622(2.639) W1 = = = 0.0164 kg kg pt − ps 103 − 2.639 h1 = c pt db1 + W1hg = (1.0062)(32) + (0.0164)(2559.9) = 74.2 kJ kg

RT (0.287 )(32 + 273) = 0.8722 m 3 kg = pt − p s 103 − 2.639 V& 115 &a = 1 = m = 131.85 kg s v1 0.8722

v1 =

at 2, tdb2 = 36 C , saturated

ps = pd = 5.979 kPa hg 2 = 2567.1 kJ kg W2 =

0.622p s 0.622(5.979) = = 0.0383 kg kg pt − p s 103 − 5.979

h2 = c p t db2 + W2 hg = (1.0062)(36 ) + (0.0383)(2567.1) = 134.5 kJ kg

At 3, t 3 = 40 C h3 = h f at 40 C = 167.57 kJ kg

&4: To solve for m By Mass Balance: &3 +m & a + W1 m &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 : & a h1 + m & 3 h3 = m & 4 h4 + m & a h2 m

(131.85)(74.2) + (200)(167.57 ) = (197.1)(h4 ) + (131.85)(134.5) h4 = 129.70 kJ kg ∴ t 4 = 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:

t 3 = 52 C t db2 = 46 C sat t db1 = 29 C

φ1 = 47% t 4 = 27 C

From psychrometric chart At 1, t db1 = 29 C , φ1 = 47% h1 = 59 kJ kg W1 = 0.0116 kg kg v1 = 0.873 m 3 kg

at 2, t db2 = 46 C , saturated ps = pd = 10.144 kPa hg 2 = 2585.0 kJ kg

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COOLING TOWERS

W2 =

0.622p s 0.622(10.144) = = 0.0692 kg kg pt − p s 101.325 − 10.144

h2 = c p t db2 + W2 hg = (1.0062)(46 ) + (0.0692)(2585) = 225.2 kJ kg

At 3, t 3 = 52 C h3 = h f at 52 C = 217.69 kJ kg At 4, t 4 = 27 C h4 = h f at 27 C = 113.25 kJ kg (a) Volume of air necessary to cool 1 kg of water = V1 = ma v1 To solve for ma when m3 = 1 kg By energy balance: Eq. (1) ma h1 + m3 h3 = ma h2 + m 4 h4 By mass balance m3 − m4 = ma (W2 − W1 ) Eq. (2) m 4 = m3 − ma (W2 − W1 ) Substitute in (1) Eq. (3) ma h1 + m3 h3 = ma h2 + [m3 − ma (W2 − W1 )]h4 (ma )(59) + (1)(217.69) = (ma )(225.2 ) + [1 − (ma )(0.0692 − 0.0116)](113.25)

ma = 0.654 kg Volume of air = V1 = ma v1 = (0.654)(0.873) = 0.5709 m 3 Mass of air required = ma = 0.654 kg

& 3 = quantity of water. (b) Let m V& 142 &a = 1 = m = 162.66 kg min v1 0.873 & a , m3 → m &3 Use Eq. (3), change ma → m & a h1 + m & 3 h3 = m & a h2 + [m & 3 −m & a (W2 − W1 )]h4 m (162.66)(59 ) + (m& 3 )(217.69) = (162.66 )(225.2) + [m& 3 − (162.66 )(0.0692 − 0.0116)](113.25) & 3 = 36,631 + 113.25m & 3 − 1061.1 9597 + 217.69m & 3 = 248.7 kg min m

∴ Quantity of water = m& 3 = 248.7 kg min 4

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 , t 3 = 60 C m t db2 = 50 C

φ 2 = 90% t db1 = 32 C t wb1 = 27 C

Cooling tower efficiency = 60.6% To solve for t 4 : t −t Efficiency = 3 4 t 3 − t wb1 0.606 =

60 − t 4 60 − 27

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COOLING TOWERS t 3 = 40 C

at 1, t db1 = 32 C , t wb1 = 27 C h1 = 85 kJ kg W1 = 0.0208 kg kg

at 2, t db2 = 50 C , φ 2 = 90% pd = 12.349 kPa ps = φ 2 pd = (0.90 )(12.349) = 11.114 kPa hg 2 = 2592.1 kJ kg W2 =

0.622p s 0.622(11.114) = = 0.0766 kg kg pt − p s 101.325 − 11.114

h2 = c p t db2 + W2 hg = (1.0062)(50) + (0.0766)(2592.1) = 248.9 kJ kg

At 3, t 3 = 60 C h3 = h f at 60 C = 251.13 kJ kg At 4, t 4 = 40 C h4 = h f at 40 C = 167.57 kJ kg By mass balance m3 − m4 = ma (W2 − W1 ) m 4 = m3 − ma (W2 − W1 ) By energy valance

m& a h1 + m& 3h3 = m& a h2 + m& 4h4 & a h1 + m & 3 h3 = m & a h2 + [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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