chapter 13 gas- vapor mixtures and air-conditioning o o (j ... · chapter 13 gas- vapor mixtures...

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Chapter 13 Gas- Vapor Mixtures and Air-Conditioning

13-14 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative

humidity, and the volume of the tank are to be determined.

Assumptions The air and the water vapor are ideal gases.

Analysis (a) The specific humidity can be determined form its definition,

n1v 0.3 kg O O 4 .(J)=-=-= .1 3 kgH2O/kgdryaIr

ma 21 kg

(b) The saturation pressure of water at 30°C is

Pg = Pgli) :I1C = 4246kPa

Then the relative humidity can be determined from

fj) = wP = (O.OI43)(100 kPa) = 52.9%

(0.622+w)Pg (0.622+0.0143)4.246 kPa

(c) The volume of the tank can be determined from the ideal gas relation for the dry air ,

P" = fj)P 9 = (0529)(4.246 kPa) = 2.246 kPa

Pa = P-P" = 100-2.246=97.754 kPa

V = ~= (21 kg)(0.287 kJ / kg. K)(303 K) = 18.7 mJ

Pa 97.754 kPa ~~ --~

---

13-15 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative

humidity, and the volume of the tank are to be determined.

Assumptions The air and the water vapor are ideal gases.

Analysis (a) The specific humidity can be determined form its definition,

n1v 0.3 kg O O 4 I . (f)=-=-= .1 3kgH2O kgdryalr

ma 21 kg

(b) The saturation pressure of water at 30°C is

Pg = Pg@ '.FC = 5628ld=>a

Then the relative humidity can be determined from

tjJ= (J)P= (QO1~(100kPa)=39.9%(Q~+(J)Pg (Q~+QO1~5m.skPa

( c ) The volume of the tank can be determined from the ideal gas relation for the dry air .

PlJ = tjJPg = (a.E)(5~kPa) = 2.2A6kPa

Pa = P- PlJ = 100- 22A6= 97.754kPa

V=~= (21kg)(0.287kJ/kgoK)(3::SK)= 19.0 m3Pa 97.754kPa

~


1200 kJ/min

13.68 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The

exit temperature, the exit relative humidity of the air, and the exit velocity are to be determined.

Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant duringthe entire process (mal = ma2 = ma) .2 Dry air and water vapor are ideal gases. 3 The kinetic and potential

energy changes are negligible.

Analysis (a) The amount of moisture in the air remains constant ({J) I = {J)2) as it flows through the coolingsection since the process involves no humidification or dehumidification. The inlet state of the air is

completely specified, and the total pressure is 1 atm. The properties of the air at the inlet state aredetermined from the psychometric chart (Figure A-33) to be

hl = 55.0 kJ / kg dry air

{J)J = 0.0089 kg H2O/kg dry air (= {J)2)

vi = 0.877 m3 / kg dry air f-:\ 32°C

\.!/ 30%18 m/s 0

AIRThe mass flow rate of dry air through the cooling section is

ma =~VIAI =- 1-(18m/s)(1l"xO.42/4m2)=258kg/s(0.877 m3 / kg)VI

From the energy balance on air in the cooling section,

-~\t. = irb(~- '1)

-l~ /00 kJ /s = (2.59 kg /s) ( ~ -550) kJ /kg

~ = 47.2 kJ /kg dry air

The exit state of the air is fixed now since we know both h2 and CV2. From the psychometric chart at this

state we read

(b)

T2 = 24.4°C

(12 = 46.6 %

V2 = 0.856 m3 / kg dry air

(c) The exit velocity is determined from the conservation of mass of dry air,

~ V2-=-

~ ~

~=~Ut ~

irbI=iT!t2 --,+ --+

~~~

13-19,

~


~

Heatingcoils

13-73 Air is first heated and then humidified by wet steam. The temperature and relative humidity of air at

the exit of heating section, the rate of heat transfer, and the rate at which water is added to the air are to be

determined.



Properties The inlet and the exit states of the air are completely specified, and the total pressure is 1 atm.The properties of the:air at various states are determined from the psychometric chart (Figure A-33) to be

" -..'0~1rT /kgdry .1 Sat. vapor.\-~!'wJ air, 100°C

ml = Qcx:53~ H2O /kg dry air(= m2) Humidifier

q = Qa:E m3 /kg dry air--~ AIR I

20°C60%

1 aIm

10°C70%

h3 = 42.3 kJ / kg dry air 35 mJ/min

(J)3 = 0.0088 kg H2O / kg dry air (1) ~ 0

Analysis (a) The amount of moisture in the air remains constant it flows through the heating section ((J) 1 =

(J)2), but increases in the humidifying section ((J)3 > (J)2). The mass flow rate of dry air is

35 m3 /n:in~fr1=.!t=

q

= 433 kg /ninQa:e rn3 /kg

Noting that Q = W =0, the energy balance on the humidifying section can be expressed as

E. E.-A £0 "0 (steady) -in -out -tiLsystem -0

Ein = Eout

}::;ni;h; = }::;ni"h" --+ m".h". + ma2h2 = milh3

(W3 -w2)h". +h2 = h3

Solving for h2-

~ = h3 -«(J)3 -(J)2)hK@IOO°C = 42.3- (0.0088- 0.0053)(2676.1) = 32.9 kJ / kg dry air

Thus at the exit of the heating section we have (J) = 0.0053 kg H2O dry air and h2 = 32.9 kJ/kg dry air, which

completely fixes the state. Then from the psychometric chart we read

T2 = 19.4°C

q)2 = 37.8%

(b) The rate of heat transfer to the air in the heating section is

9m = i7h( f2 -1\) = ( 433 ~ /nin) (::r2,.9- 235) kJ /kg = 407 kJ / miD

(c) The amount of water added to the air in the humidifying section is determined from the conservation of

mass equation of water in the humidifying section,

1'r6= irh((J)3-(J)2) = (433kg/~(Q~- QCX6.1 = 0.15 kg /mill

13-23


13-77 Air is first cooled, then dehumidified, and finally heated. The temperature of air before it enters the

heating section, the amount of heat removed in the cooling section, and the amount of heat supplied in theheating section are to be determined.



Analysis (a) The amount of moisture in the air decreases due to dehumidification (fI) 3 < fI) 1), and remains

constant during heating (fI) 3 = fI) 2). The inlet and the exit states of the air are completely specified, and the

total pressure is 1 atm. The intermediate state (state 2) is also known since ~ = 100% and fI) 2 = fI) 3.

Therefore, we can determined the properties of the air at all three states from the psychometric chart (TableA-33) to be

hl = 95.2 kJ / kg dry air

{J)l = 0.0238 kg H2O / kg dry air

and Ti = 34°C

(>I = 70%h3 = 43.1 kJ / kg dry air

(t}3 = 0.0082 kg H2O / kg dry air ( = (t}2 )

Also,

h", = hf@IO°C = 42.0 kJ / kg

~ = 31.8 kJ / kg dry air

T2 = 11.1° C

(b) The amount of heat removed in the cooling section is determined from the energy balance equation

applied to the cooling section,

Eo Eo A z:. 710 (steady)

Oin -out = ilLsystem =

Ein = Eout

Lm;hi = Lmehe + Qout,cOOJing

QOut,cooling = mQ1h1 -(mQ2~ + mwhw ) = mQ (h1 -~ ) -m",hw

or, per unit mass of dry air ,

qout,cooling = (h1 -~)-«(J)1 -(J)2)hw

= (95.2- 31.8) -(0.0238- 0.0082)42.0

= 62.7 k.J Ikg dry air

(c) The amount of heat supplied in the heating section per unit mass of dry air is

q;n,heating = h3 -~ = 43.1- 31.8 = 11.3 k.J I kg dry air

13-27


C!) 3~

~~J,-~~20mJ/min (f'J

2 c , ""p=95kPa t/>J 0

/ AIR T3

25~ Jlmin 12°C

G) 90%c

13-97 Two airstreams are mixed steadily. The specific humidity, the relative humidity, the dry-bulb

temperature, and the volume flow rate of the mixture are to be determined.

Assumptions 1 Steady operating conditions exist 2 Dry air and water vapor are ideal gases. 3 The kinetic

and potential energy changes are negligible. 4 The mixing section is adiabatic.

Analysis The properties of each inlet stream are determined to be

Pvl = tjJlPgl = tjJlP9@32"C = (Q4Q)(48X>kPa) = 192 kPa

l! al = ll- P vl = !=*5- 192 = 9300 kPa

Rn1i ( Q287 kpa. m3 /kg .K)(3:E K)q=-=

P al 9300 kPa

= Q94Q m3 /kg dry air

wl = Q622 Pvl = Q~192 kPa) = 0.0128 kg H2O /kg chy air

ll- Pvl (!=*5-192) kPa

~ = Cp1i + Wlhgl = (l(ffikJ/kg.oC)(';!Z'C) + (0.0128)(~.9kJ /kg)

=64.9kJ/kgdryair

and

p Ifl = 1jJ2P g2 = 1jJ2P Si@ l2"C = ( 09)) ( 14100 kPa) = 1277 kPa

p a2 = ~ -p Ifl = gj:- l277 = 9.3 7'2'3 kPa

U]. = ~ = ( 02S7 kPa .m3 /kg .K) (285 K) = 0873 m3 /kg chy air

p a2 93 7'2'3 kPa

-O~ p Ifl -o~ 1277 kPa) - OrY\Oc: kgH O /kg chy .m2--- .~ 2 8lr

~ -p Ifl (gj:-1277) kPa

f2 = Cp~ + (tJ2~ = (l00;kJ /kg.oC)( 12"C) + (0.<XE5}(2523.4kJ /kg) = 335kJ /kg chy air

Then the mass flow rate of dry air in each stream is

.3irh! = .Y!. = X> m /n:in = 2l3 kg /rrin

q Q94O m3 /kg chy air

.3.V2 25m /rrin 286kg / .TT1:a mn-U]. -Q87J m3 /kg dry air -

From the conservation of mass,

ma3 =mal +ma2 =(21.3+28.6) kg!min =49.9 kg!min

The specific humidity and the enthalpy of the mixture can be determined from

Eqs. 13-24, which are obtained by combining the conservation of mass and

energy equations for the adiabatic mixing of two streams:

!!!EL = {1)2 -{1)3 = ~

ma2 {1)3- {1)\ h3 -hi

~ = 0.0085-{1)3 = 335-h3

28.6 {1)3-0.0]28 h3-64.9

which yields,

{1)3 = 0.0103 kg H2O/kg dry air

h3 = 46.9 kJ I kg dry air

13-43

ica


.1

These two properties fix the state of the mixture. Other properties are determined from

h3 = CpT3 + (J)3hg3 = CpT3 +(J)3(2501.3+ 1.82T3)

46.9 kJ / kg = (1.005 kJ / kg.oC) ~ + (0.0103)(2501.3+ 1.82 I:,\) kJ / kg

T3 = 20.6°C

Q~ P v3 Q~ P u3155 kPa(J)3= ~ QOICX3= ~ Pv3=

~-Pv3 ~-Pv3

p p 155kPafjJ3 = --!:§.. = v3 = = Q635 or 63.5 %

Pg3 Pg@~ 2.44kPa

Finally,

Pa'3 = &- Pll3 = ~-155= ffi45kPa

~ = ~ = ( Q287 kPa .m3 /kg .K) ( 2:}3.6 K) = Q!:XJ2 m3 /kg chy air

P a'3 ffi45 kPa

~= ~~ = (49.9kg/n:in)(0.!:XJ2m3 /kg) = 45.0 m3/min

13-44


AIR ~ 34°CEXIT \2:; 90%

I~IWARM

WATER

40°C

60 kg/s

CD

AAAAAA'

13-106 Water is cooled by air in a cooling tower. The volume flow rate of air and the mass flow rate of the

required makeup water are to be determined.

Assumptions 1 Steady operating conditions exist and thus mass flow rate of dry air remains constant duringthe 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 (mal =ma2 = ma) , 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:

Lma.i = Lma.e ~ mal = ma2 = ma

Water Mass Balance:

L 1'f1ui = L 1'f1ue ~ ~ + iTh1{Ol = 1'14 + fT!O;{02

~ -1'14 = iTh({02 -{OJ = ~

Energy Balance:

~ -~lt = ~~ ~o (,gea:Iy! = 0

~=~lt

L 1'r}~ = L 1'T1~ (Sn::e Q= w = 0)

0= L 1'T1~ -L 1'r}~

0= fT!O;l2 + 1'I4l\ -1'Th1" -r'rsls

0= iTh(l2- ,,) +(~- ~)~- ~Is(j) AIR

"",INLET1 aIm

TUb = 22°C

Twh = 16°C

Solving for ma .

~

@'

~ ,~COOL U2~o;WATER

t Makeup

niJ (hJ -h4)m -a -

.(h2-h,)-(W2-W,)h4

From the psychometric chart (Table A-33),

" = 44. 7 kJ /kg <hy air

w 1 = Q<Xm kg H2O /kg <hy air

Ut = Q849 m3 /kg <hy air

and

~ = 1135 kJ / kg dry air

{i)2 = 0.0309 kg H2O / kg dry air

From Table A-4,

h3 = hf@40°C = 16757 kJ / kg H2O

h4 = hf@26°C = 109.07 kJ / kg H2O

Substituting,

~/s)(16~IO9.07)k~m =a (1135-44.7) kJ/kg-(O.O309-0.0089)(IO9.07) kJ/kg

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

~ = irhq = (529 kg/s)(0.849m3 /kg) = 44.9 mJ Is

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

~ = irh(m2-mJ = (529kg/s)(O.m:9-Qcx:m) = 1.16 kg/s

=52.9 kg/s

13-50

chapter 13 gas- vapor mixtures and air-conditioning o o (j ... · chapter 13 gas- vapor mixtures...

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