OPERAÇÕES UNTÁRIAS C - EXERCÍCIOS TORRES DE RESFRIAMENTO

C

O C

M R

E

U D

E

N N

I D C

I A

A D

D O

E P

E E

L

V A O

D

N E

G

C

R

É E

T

L

I C

O

D

A E

L 2

6

/ 0

OPERAÇÕES UNTÁRIAS C AULA 5 – Exercícios de fixação sobre torres de resfriamento PROFESSOR: Nazareno Braga

1. Uma água quente gerada no resfriamento de uma planta de energia nuclear entra em uma torre de resfriamento a 40 ºC e 90 kg/s. A água é resfriada para 25 ºC na torre de resfriamento por um ar que entra na torre a 1 atm, 23 ºC e 60% de umidade relativa e deixa a torre saturado a 32 ºC. Determine: (a) o fluxo de massa do ar na entrada da torre e (b) taxa de escoamento da água de reposição requerida.  Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air  remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kineti kineticc and potent potential ial energy energy change changess are neglig negligibl ible. e. 4 The The cool coolin ing g towe towerr is adiabatic.  Analysis (a) The The mass mass flow flow rate rate of dry dry air air thro throug ugh h the the towe towerr rema remain inss cons consta tant nt  a1 = m a 2 = m a ) , but the mass flow rate of liquid water decreases by an amount equal to (m the amount of water that vaporizes in the tower during the cooling process. The water  lost lost thro throug ugh h evap evapor orat atio ion n must must be made made up late laterr in the the cycle cycle to main mainta tain in stead steady y operation. Applying the the mass and energy balances yields yields

 Dry Air Mass Balance: Balance: 2

 a,i = ∑m  a,e     a1 = m  a2 = m a ∑m  → m

32°C 100%

Water Mass Balance: Balance:  w,i = ∑m  w,e     3 +m  a1ω  1 = m  4 +m  a 2ω  2 ∑m  → m  3 −m  4 =m  a (ω  2 − ω  1 ) = m  makeup m

3

WATER  40°C 90 kg/s

 Energy Balance: Balance: 0 (steady)  −  E    = ∆ =0  E   E  in out system

System  boundary 1

 =  E    E  in out  i hi = ∑ m  e he ∑m

 =0 since Q = W 

 e he − ∑ m  i hi 0 = ∑m  a 2 h2 + m  4 h4 − m  a1h1 − m  3 h3 0=m

4 25°C

Makeup water 

 a (h2 − h1 ) + (m 3 − m  makeup )h4 − m  3 h3 0=m

 , Solving for  m a

a = m

 3 (h3 − h4 ) m (h2 − h1 ) − (ω 2 − ω 1 )h4

From the psychometric chart (Figure A-33),

AIR 

1 atm 23°C 60%

C

O C

M R

E

U D

E

N N

I D C

I A

A D

D O

E P

E E

L

V A O

D

N E

C

G R

É E

T

L

I C

O

D

A E

L 2

6

h1 = 50 .3 kJ/kg dry air  ω   1

= 0.0106 kg H 2 O/kg dry air 

v1 = 0.854 m 3 /kg dry air 

and h2 = 110 .8 kJ/kg dry air  ω   2

= 0.0307 kg H 2O/kg dry air 

From Table A-4, h3

≅ h  f   @ 40 °C =167

.57 kJ/kg H 2 O

h4

≅ h  f   @ 25 °C =104

.89 kJ/kg H 2 O

Substituting,

a = m

(90 kg/s)(167. 57 − 104.89)kJ/ kg (110 .8 − 50 .3) kJ/kg − (0.0307 − 0.0106 )(104 .89 ) kJ/kg

= 96.6 kg/s

Then the volume flow rate of air into the cooling tower becomes

1 V 

 a v1 = (96.6 =m

kg/s)(0.85 4 m 3 / kg ) = 82.5

3

m /s

(b) The mass flow rate of the required makeup water is determined from

 makeup = m  a (ω  2 − ω  1 ) = (96 .6 kg/s)(0.03 07 − 0.0106) = 1.942 kg/s m

2. Uma água quente proveniente de uma planta de resfriamento entre em uma torre de resfriamento a 43 ºC a uma velocidade de alimentação de 45 kg/s. A água é resfriada a 27 ºC por um ar que entra na torre a 1 atm, 24 ºC e 60% de umidade relativa e deixa

/ 0

C

O C

M R

E

U D

E

N N

I D C

A

I A

D

D O

E P

E E

L

V A O

N

D

E

G

C

R

É E

T

L

I C

O

D

A E

L 2

6

/ 0

saturado a 35 ºC. Determine: (a) a vazão volumétrica do ar que entra na torre e (b) vazão mássica requerida para a água de reposição.  Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air  remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic.  Analysis (a) The mass flow rate of dry air through the tower remains constant  a1 = m  a2 = m  a ) , but the mass flow rate of liquid water decreases by an amount equal (m to the amount of water that vaporizes in the tower during the cooling process. The water lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass balance and the energy balance equations yields

Dry Air Mass Balance:  a,i = ∑m  a, e     a1 ∑m  → m

 a2 = m a =m

Water Mass Balance:  w,i = ∑m  w,e     3 +m  a1ω  1 = m 4 ∑m  → m  3 −m  4 =m  a (ω  2 − ω  1 ) = m  makeup m

 a 2ω  2 +m

Energy Balance:   E  i n

  E  − o u t 

  E  = ∆ s y s t e m

  E  i n  i hi m ∑

s t e a d y ) 0 (

= 0

  E  = o u t   e he m = ∑

 e he m ∑ 0 =

  Q = W   = 0)

( s i n c e  i hi m − ∑

 a 2 h2 m 0 =

 4 h4 + m

 a1h1 − m

 a ( h2 m 0 =

 3 − h1 ) + (m

 3 h3 − m

 m − m a k e u p

) h4

 3 h3 − m

 , Solving for  m a

a = m

 3 (h3 − h4 ) m

2

(h2 − h1 ) − (ω 2 − ω 1 ) h4 From the psychometric chart (Figure A-33), h1 = 30 .9 Btu/lbm dry air  ω   1

= 0.0115 lbm H 2 O/lbm dry air 

v1 = 13 .76 ft 3 /lbm dry air 

95°F 100%

3 WATER  110°F 100 lbm/s

and h2 = 63 .2 Btu/lbm dry air  ω   2

= 0.0366 lbm H 2 O/lbm dry air 

System  boundary 1

From Table A-4E, h3

≅ h  f   @ 110 °F = 78 .02

h4

≅ h  f   @ 80 °F = 48 .09

Btu/lbm H 2 O Btu/lbm H 2 O

Substituting,

a = m

(100 lbm/s)(78. 02 − 48 .09 )Btu/lbm

4

1 atm 76°F 60%

80°F

Makeup water 

(63 .2 − 30 .9) Btu/lbm − (0.0366 − 0.0115 )( 48 .09 ) Btu/lbm

Then the volume flow rate of air into the cooling tower becomes

AIR 

= 96.3 lbm/s

C

O C

M R

E

U D

E

N N

I D C

I A

A D

D O

E P

E E

L

V A O

D

N E

C

G R

É E

T

L

I C

O

D

A E

L 2

6

/ 0

1 = m  a v1 = (96 .3 lbm/s)(13. 76 ft /lbm ) = 1325 ft /s V  3

3

(b) The mass flow rate of the required makeup water is determined from

 makeup = m  a (ω  2 − ω  1 ) = (96 .3 lbm/s)(0.0 366 − 0.0115) = 2.42 lbm/s m

3. Uma torre está resfriamento uma corrente de água quente de 40 ºC para 25 ºC onde a  pressão atmosférica é 96 kPa. Ar entra na torre a 20 ºC e 70% de umidade e deixa saturado a 35 ºC. Determine: a vazão volumétrica do ar que entra na torre e (b) vazão mássica da água de reposição.  Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air  remains constant during the entire process. 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy changes are negligible. 4 The cooling tower is adiabatic.  Analysis (a) The mass flow rate of dry air through the tower remains constant  a1 = m a 2 = m a ) , but the mass flow rate of liquid water decreases by an amount equal to (m the amount of water that vaporizes in the tower during the cooling process. The water  lost through evaporation must be made up later in the cycle to maintain steady operation. Applying the mass and energy balances yields

 Dry Air Mass Balance:  a,i = ∑m  a, e     a1 = m  a2 = m a ∑m  → m

Water Mass Balance:

 w,i = ∑ m w,e     3 + m a1ω 1 = m 4 + m a 2ω 2 ∑m  → m  3 − m 4 = m a (ω 2 − ω 1 ) = m makeup m

2

35°C 100%

3 WATER  40°C 50 kg/s

System  boundary 1 4 25°C

Makeup water 

AIR 

96 kPa 20 ºC 70%

C

O C

M R

E

U D

N

E

I D

N

C

A

I A

D

D

E

O

P

E E

L

V A O

N

D

E

G

C

É

R

E

T

L

I C

O

D

A

L

E

2

6

/ 0

 Energy Balance:   E  i n

  E  − o u t 

0 ( s t e a d y )

  E  = ∆ s y s t e m

  E  i n

= 0

 =  E  o u t 

 i hi ∑ m

 e he = ∑ m

  Q = W   = 0)

( s i n c e

 e he m ∑ 0 =

 i hi m − ∑

 a 2 h2 m 0 =

 4 h4 m +

 a1h1 m −

 a ( h2 m 0 =

 3 h1 ) + − (m

 3 h3 m −

 m m − a k e u p

 3 h3 m −

) h4

Solving for  m a ,  3 (h3 − h4 ) m

a = m

(h2 − h1 ) − (ω 2 − ω 1 ) h4

The properties of air at the inlet and the exit of the tower are calculated to be   P  1 v

1  P    g 1

1  P  s a t 

(0 .7 0

) (

2.3 3 9

k P a )

1.6 3 7

k P a

φ  φ   = = = = ° − = − = ⋅ ⋅ ω  = = =° − − ⋅° ω  + = φ  φ  = = = = ° ω  = = − ⋅° ° =− + ω =

  P  a 1 v 1

  P  1

2 0

C

1.6 3 7

( 0.2 8 7

m

  P  1 a

0.6 2 2

  P  v 1

  P  1

.5

d r y

.3 6 3

/  k g

k P a

K ) ( 2 9 3

K )

0.8 9 1

k P a

k P a )

1.6 3 7

( 1.0 0 5

1h   g 1

k J / k g

3

.3 6 3

( 1.6 3 7

( 9 6

  P  v 1

C       pT     1 4 7

9 4

k P a

9 4

0.6 2 2

1

h 1

@

9 6

  P  v 1

  RaT     1

)

k P a

k J / k g

0 . 0 1 0 8

k g

C ) ( 2 0

C )

H

+

m3

/ k g 2O

/  k g

d r y

( 0 . 0 1 0 8 ) ( 2

d r y

a i r 

a i r 

5 3 8 . 1

k J / k g )

a i r 

and

 P  v2

2  P   g  2

2

0.6 2 2

2

 P  2

h2

 P  v2

 P  v2

C      pT    2

1 3 4

 P  s a t 

3 5

2

d r y

C

(1.0 0

(5.6 2 8

( 9 6

2 h g 

.5 k J / k g

@

0.6 2 2

5.6 2 8

(1.0 0 5

) (

5.6 2 8

k P a )

)

k P a

k J / k g

k P a )

0 . 0 3 8 7

C ) ( 3 5

5.6 2 8

k g

C )

a i r 

From Table A-4, h3

≅

h  f   @ 40°C

=

167 .57 kJ/kg H 2 O

h4

≅

h  f   @ 25°C

=

104 .89 kJ/kg H 2 O

Substituting,

a = m

(50 kg/s)(167. 57 − 104.89)kJ/ kg (134 .5 − 47 .5) kJ/kg − (0.0387 − 0.0108 )(104 .89 ) kJ/kg

= 37 .3 kg/s

Then the volume flow rate of air into the cooling tower becomes

1 = m  a v1 = (37 .3 kg/s)(0.89 1 m3 / kg ) = 33.2 m 3 /s V  (b) The mass flow rate of the required makeup water is determined from

 makeup = m  a (ω  2 − ω  1 ) = (33 .2 kg/s)(0.03 87 − 0.0108) = 1.04 kg/s m

H

+

/ k g 2O

( 0 . 0 3 8 7 ) ( 2

k P a d r y