OPERAÇÕES UNTÁRIAS C - EXERCÍCIOS TORRES DE RESFRIAMENTO

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C RR EE DD EE NN CC II AA DD OO PP EE LL OO DD EE CC RR EE TT OO DD EE 2 2 66 // 00

OPERAÇÕES UNTÁRIAS C

AULA 5 –

Exercícios de fixação sobre torres de resfriamentoExercícios de fixação sobre torres de resfriamento PROFESSOR: Nazareno Braga

PROFESSOR: Nazareno Braga

1. Uma água quente gerada no resfriamento de uma planta de energia nuclear entra em 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 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 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 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.

e (b) taxa de escoamento da água de reposição requerida. Assumptions

Assumptions

11

Steady operating conditions exist and thus mass flow rate of dry air Steady operating conditions exist and thus mass flow rate of dry air remains constant during the entire process.

remains constant during the entire process.

22

Dry air and water vapor are ideal gases.Dry air and water vapor are ideal gases.

33

The

The kinkinetietic c and and potpotentential ial eneenergy rgy chachangenges s are are negnegligligiblible.e.

44

ThThe e cocoololining g totowewer r isis adiabatic.

adiabatic. Analysis

Analysis ((aa) ) ThThe e mamass ss floflow w ratrate e of of drdry y aiair r ththrorougugh h ththe e totowewer r reremamainins s coconsnstatantnt

((mm  _a_a₁₁= = mm  _a_a₂₂ ==mm_a_a))_{, but the mass flow rate of liquid water decreases by an amount equal to}_{, but the mass flow rate of liquid water decreases by an amount equal to}

the amount of

the amount of water that vaporizes in water that vaporizes in the tower during the tower during the cooling process. the cooling process. The water The water lo

lost st ththrorougugh h evevapapororatatioion n mumust st be be mamade de up up lalateter r in in ththe e cycycle cle to to mamainintatain in ststeadeadyy operation.

operation. Applying thApplying the mass and e mass and energy balances yieldenergy balances yieldss Dry Air Mass Balance

Dry Air Mass Balance::

a a a a a a e e a a ii a a mm mm mm mm m m_ ==∑∑ _    →→ _ == _ == _ ∑ ∑ _,, _,, ₁₁ ₂₂

Water Mass Balance Water Mass Balance::

makeup makeup 1 1 2 2 4 4 3 3 2 2 2 2 4 4 1 1 1 1 3 3 ,, ,, )) (( mm m m m m m m m m m m m m m m m m m m a a a a a a e e w w ii w w           = = − − = = − − + + = = + +  →→   ∑ ∑ = = ∑ ∑ ω ω ω ω ω ω ω ω Energy Balance Energy Balance:: 3 3 3 3 4 4 makeup makeup 3 3 1 1 2 2 3 3 3 3 1 1 1 1 4 4 4 4 2 2 2 2 out out in in (steady) (steady) 0 0 system system out out in in

))

((

))

((

0

0 =

=

since

0

0 h

h

m

h

m

h

m

h

m

h

m

h

m

h

m

h

m

h

m

W

Q

h

m

h

m

E

a a a a a a ii ii ee ee ee ee ii ii



−

+

−

=

−

+

=

∑

−

∑

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∑

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∆

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 Solving for Solving for mm



aa,, 4 4 1 1 2 2 1 1 2 2 4 4 3 3 3 3

))

((

))

((

))

((

h

m

_a_a ω ω ω ω

−

=



From the psychometric chart (Figure A-33), From the psychometric chart (Figure A-33),

System System boundary boundary 1 1 2 2 4 4 32 32°°CC 100% 100% WATER WATER 40 40°°CC 90 kg/s 90 kg/s 1 atm 1 atm 23 23°°CC 60% 60% AIR AIR 3 3 Makeup water Makeup water 25 25°°CC

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air dry /kg m 854 . 0 air dry O/kg H kg 0106 . 0 air dry kJ/kg 3 . 50 3 1 2 1 1 = = = v h ω and air dry O/kg H kg 0307 . 0 air dry kJ/kg 8 . 110 2 2 2 = = ω h

From Table A-4,

O H kJ/kg 89 . 104 O H kJ/kg 57 . 167 2 C 25 @ 4 2 C 40 @ 3 = ≅ = ≅ ° ° f f h h h h Substituting, kg/s 96.6 kJ/kg ) 89 . 104 )( 0106 . 0 0307 . 0 ( kJ/kg ) 3 . 50 8 . 110 ( kg 104.89)kJ/ 57 kg/s)(167. 90 ( = − − − − = a m

_

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

/s m 82.5 3 = = = v₁ (96.6kg/s)(0.85 4m3/kg) 1 ma V



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

kg/s 1.942 = − = − = ( ₂ ₁) (96.6kg/s)(0.03 07 0.0106) makeup ma ω ω 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

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C R E D E N C I A D O P E L O D E C R E T O D E 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 )

(m_a₁ =m_a₂ =m_a _{, but the mass flow rate of liquid water decreases by an amount equal} 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 a a e a i a m m m m m_ =∑_  → _ = _ = _ ∑ _, _, ₁ ₂

Water Mass Balance:

makeup 1 2 4 3 2 2 4 1 1 3 , , ) ( m m m m m m m m m m a a a e w i w           = − = − + = + →  ∑ = ∑ ω ω ω ω Energy Balance: 3 3 4 makeup 3 1 2 3 3 1 1 4 4 2 2 out in (steady) 0 system out in ) ( ) ( 0 0 0 ) 0 = = (since 0 h m h m m h h m h m h m h m h m h m h m W Q h m h m E E E E E a a a i i e e e e i i                    − − + − = − − + = ∑ − ∑ = ∑ = ∑ = = ∆ = −  Solving for m



a, 4 1 2 1 2 4 3 3 ) ( ) ( ) ( h h h h h m m_a ω ω − − − − =  

From the psychometric chart (Figure A-33),

air dry /lbm ft 76 . 13 air dry O/lbm H lbm 0115 . 0 air dry Btu/lbm 9 . 30 3 1 2 1 1 = = = v h ω and air dry O/lbm H lbm 0366 . 0 air dry Btu/lbm 2 . 63 2 2 2 = = ω h

From Table A-4E,

O H Btu/lbm 09 . 48 O H Btu/lbm 02 . 78 2 F 80 @ 4 2 F 110 @ 3 = ≅ = ≅ ° ° f f h h h h Substituting, lbm/s 96.3 Btu/lbm ) 09 . 48 )( 0115 . 0 0366 . 0 ( Btu/lbm ) 9 . 30 2 . 63 ( )Btu/lbm 09 . 48 02 lbm/s)(78. 100 ( = − − − − = a m

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

System boundary 1 2 4 95°F 100% WATER 110°F 100 lbm/s 1 atm 76°F 60% AIR 3 Makeup water 80°F

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/s ft 1325 3 = = = ₁ ₍₉₆_.₃ _lbm/s)(13. ₇₆_ft3_/lbm₎ 1 m v V



_a

lbm/s 2.42 = − = − = ( ₂ ₁) (96.3lbm/s)(0.0 366 0.0115) makeup ma ω ω 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

(m _a₁= m _a₂ =m_a)_{, but the mass flow rate of liquid water decreases by an amount equal 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 and energy balances yields

Dry Air Mass Balance:

a a a e a i a m m m m m_ =∑_  → _ = _ = _ ∑ _, _, ₁ ₂

Water Mass Balance:

makeup 1 2 4 3 2 2 4 1 1 3 , ,

)

(

m

a a a e w i w



=

−

=

−

+

=

+

→



∑

=

∑

ω ω ω ω System boundary 1 2 4 35°C 100% WATER 40°C 50 kg/s 96 kPa 20 ºC 70% AIR 3 Makeup water 25°C

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C R E D E N C I A D O P E L O D E C R E T O D E 2 6 / 0 Energy Balance: 3 3 4 makeup 3 1 2 3 3 1 1 4 4 2 2 out in (steady) 0 system out in ) ( ) ( 0 0 0 ) 0 = = (since 0 h m h m m h h m h m h m h m h m h m h m W Q h m h m E E E E E a a a i i e e e e i i                    − − + − = − − + = ∑ − ∑ = ∑ = ∑ = = ∆ = −  Solving for m_a _, 4 1 2 1 2 4 3 3 ) ( ) ( ) ( h h h h h m m_a ω ω − − − − =  

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

air dry kJ/kg 5 . 47 kJ/kg) 538.1 (0.0108)(2 + C) C)(20 kJ/kg 005 . 1 ( air dry O/kg H kg 0.0108 kPa ) 637 . 1 (96 kPa) 637 . 1 ( 622 . 0 622 . 0 air dry kg / m 891 . 0 kPa 363 . 94 K) K)(293 kg / m kPa 287 . 0 ( kPa 363 . 94 637 . 1 96 kPa 637 . 1 kPa) 339 . 2 )( 70 . 0 ( 1 1 1 1 2 1 1 1 1 3 3 1 1 1 1 1 1 C 20 @ sat 1 1 1 1

=

+

=

⋅

°

=

−

=

−

=

⋅

=

−

=

−

=

°

g p v v a a v a g v h T C h P P P P T R v P P P P P P

ω

φ

and air dry kJ/kg 5 . 134 (0.0387)(2 + C) C)(35 kJ/kg 005 . 1 ( dry O/kg H kg 0.0387 kPa ) 628 . 5 (96 kPa) 628 . 5 ( 622 . 0 622 . 0 kPa 628 . 5 kPa) 628 . 5 )( 00 . 1 ( 2 2 2 2 2 2 2 2 2 C 35 @ sat 2 2 2 2

=

+

=

⋅

°

=

−

=

−

=

°

g p v v g v h T C h P P P P P P

ω

φ

From Table A-4,

O H kJ/kg 89 . 104 O H kJ/kg 57 . 167 2 C 25 @ 4 2 C 40 @ 3 = ≅ = ≅ ° ° f f h h h h Substituting, kg/s .3 37 kJ/kg ) 89 . 104 )( 0108 . 0 0387 . 0 ( kJ/kg ) 5 . 47 5 . 134 ( kg 104.89)kJ/ 57 kg/s)(167. 50 ( = − − − − = a m

_

Then the volume flow rate of air into the cooling tower becomes /s m 33.2 3 = = = ₁ (37.3kg/s)(0.89 1m3/kg) 1 m v V



_a

kg/s 1.04 = − = − = ( ₂ ₁) (33.2kg/s)(0.03 87 0.0108) makeup ma ω ω m 