Tutorium Mixture Loesung

Jabatan Kejuruteraan Loji dan AutomotifFakulti Kejuruteraan Mekanikal dan PembuatanUniversiti Tun Hussein Onn Malaysia

SolutionTutorial 6/ ws

BDA 3043

Tutorial 6 SolutionMixtures

Winardi Sani

A vessel of volume 0.4 m3 contains 0.45 kg CO and 1 kg of air, at 15 C.Calculate:

(a) The partial pressure of each component

(b) The total pressure in the vessel

The gravimetric analysis of air is to be taken as 23.3 % O2 and 76.7 % of N2.

Component Chem. symbolAnalysis Molar mass [M]

vol. (%) grav. (%) [kg/kmol]

AirOxygen O2 21 23.3 32.0

Nitrogen N2 79 76.7 28.0

Carbon Monoxide CO 28.0

CO N2 O2

N2

COO2

O2N2

N2

p

p

p

p

p p

p

p

p

p Ideal gas equation:

pV = mRT (1)

R =RM

(2)

pV = mMRT

pi = miMi V RT (partial pressure) (3)

Given: V = 0.4 m3; T = 15 +173 = 288 K and R = 8.3145 kJ/kmol K, mCO = 0.45 kgPartial pressure of each component:

pi =miMi

RTV

(4)

Component O2:

pO2 =mO2MO2

RTV

; MO2 = 32.0 kg/kmol; mO2 =23.3100 1 kg = 0.233 kg

RTV

=8.3145 kJ/kmol K 288 K

0.4m3= 5, 986.44

k Pakmol

(5)

pO2 =0.233 kg

32.0 kg/kmol 5, 986.44 k Pa

kmol= 43.59 kPa

pO2 = 0.4359 bar; 1 bar = 100 kPa (6)

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BDA 3043


Component N2:

MN2 = 28.0 kg/kmol; mN2 =76.7100 1 kg = 0.767 kg

pN2 = 96.21 kPa = 0.9621 bar (7)

Component CO:

MCO = 28.0 kg/kmol; mCO = 0.45 kg

pCO = 163.99 kPa = 1.6399 bar (8)

Total pressure in the vessel:

p =3i=1

= pO2 + pN2 + pCO = 3.0379 bar (9)

A mixture of 1 kmol CO2 and 3.5 kmol of air is contained in a vessel at 1 barand 15 C. The volumetric analysis of air can be taken as 21 % O2 and 79% N2.Calculate for the mixture:

(a) The masses of CO2, O2, N2, and the total mass

(b) The percentage carbon content by mass

(c) The molar mass and the specific gas constant for the mixture

(d) The specific volume of the mixture

(MC = 12, MO2 = 32 , MN2 = 28 kg/kmol, R = 8.3145 kJ/kmol K)(a) The masses of CO2, O2, N2, and the total mass

(i) Mass of CO2

m = nM = 1 kmol 44 kg/kmol; (M = 12 + 32) mCO2 = 44 kg (10)

(ii) Mass of O2

pip=nin=ViV

nO2 = n VO2V

= 3.5 kmol 0.21 = 0.735 kmol mO2 = 0.735 32 = 23.55 kg (11)

(iii) Mass of N2

nN2 = n VN2V

= 3.5 kmol 0.79 = 2.765 kmol mN2 = 2.765 28 = 77.5 kg (12)

The total mass of the mixture:

m = mCO2 +mO2 +mN2 = 145.05 kg (13)

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BDA 3043

(b) The percentage carbon content by mass

mCm

=nC MC

m=

1 kmol 12 kg/kmol145.05 kg

100% = 8.27% (14)

(c) The molar mass and the specific gas constant for the mixture

(i) Molar mass for the mixture

R =RM

M =1ni

ni Mi (15)

n =i

ni = nCO2 + nN2 + nO2

= 1 + 2.765 + 0.735 = 4.5 kmol (16)i

ni Mi = 1 44 + 2.765 28 + 0.735 32 = 144.94 kg (17)

M = 32.2 kg/kmol (18)

(ii) The specific gas constant of the mixture:

R =RM

=8.3145 kJ/kmol K32.2 kg/kmol

= 0.2581 kJ/kg K (19)

(d) The specific volume of the mixture

p =m

VRT =

RT

v

v = RTp

=0.2581 kJ/kg K 288 K

1 bar 1 bar105 N/m2

103 Nm1 kJ

v = 0.7435m3/kg (20)

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BDA 3043


Moist air enters a duct at 10 C, 80 % RH, and a volumetric flow rate of 150m3/min. The mixture is heated as it flows through the duct and exits at 30 C. Nomoisture is added or removed, and the mixture pressure remains approximatelyconstant at 1 bar.

(a) Sketch on T s diagram the heating process, and determine

1T = 10 Co

s

p (T2)g

p (T1)g

30 Co

T

RH = 80%

1 bar

pv

2

1

o10 C

(b) The rate of heat transfer, in kJ/min

Q.

H.

2H.

1

2

1

V=150 m /min3

.

control volume

mo

i st

ai r

Mass balance:

ma1 = ma2 = ma; (dry air) (21)

mv1 = mv2 = mv; (water vapor or H2O) (22)

Energy balance:

Q = H2 H1 (23)H = Ha + Hv (24)

H = maha + mvhv = ma(ha +mvma hv)

H = ma(ha + hv) with = mvma

=mvma

(25)

Q = ma2(ha2 + 2hv2) ma1(ha1 + 1hv1)

= ma

[(ha2 ha1) + (hv2 hv1)

]; with 1 = 2 = . (26)

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BDA 3043

Calculation of mass flow rate:

ma =V1va1

(27)

pa1va1 = RaT1 va1 = RaT1pa1

(28)

Calculation of the partial pressure for dry air at inlet:

pa1 = p pv1; and pv1 = 1 pg1 = 1 pg(T1) (29)selected saturated water temperature

T [C] psat [kPa]Enthalpy

hf hg

10 1.2281 =pg1 42.022 2519.2 = hv130 4.2469 =pg2 125.74 2555.6 = hv2

With = 0.8 the partial vapor pressure at inlet is:

pv1 = 1 pg1 = 0.8 1.2281 = 0.9825 kPa (30)

with total pressure, p = 100 kPa, the partial pressure, pa1:

pa1 = p pv1 = 100 0.9825 = 99.0175 kPa (31)

the specific volume of the dry air at inlet:

va1 =RaT1pa1

=0.287 kJ/kg K 283 K

99.0175 kPa 1 Pa1 N/m2

1 Nm1 J

= 0.8202m3/kg

The mass flow rate required is:

= ma = V1va1

=150m3/min0.8348m3/kg

= 182.87 kg/min (32)

Calculation of the enthalpy of dry air:

ha2 ha1 = cp,a (T2 T1)= 1.005 kJ/kg K(30 10) K = 20.1 kJ/kg (33)

The same result is also found, if you use the ideal gas table for air.

Calculation of the enthalpy of water vapor air:

= 0.622 pv1pa1

= 0.622 0.982599.0175

= 0.0062kg (vapor)kg (d.a.)

(34)

hv2 hv1 = 2555.6 2519.2 = 34.60 kJ/kg (35)

Q = ma[(ha2 ha1) + (hv2 hv1)

]= 182.87 (20.1 + 0.0062 34.60)

= Q = 3715 kJ/min (36)

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BDA 3043


(c) The relative humidity at the exit.The partial vapor pressure at inlet equals to that at the exit, see on T sdiagram.

2 =pv2pg2

=pv1pg2

(37)

=0.98254.2469

= 0.2313

= 2 = 23.13% (38)

(d) Compare your result with the psychrometric chart analysis

1

DBT

=

0.80

=

0.82

= 23

%

= 80%

3010

21

2

h = 25.5

h = 46.0

= 0.0062

a1 = 0.815 m3/kg d.a.

ma1 =V1a1

=150m3/min0.815m3/kg

= 184.05 kg/min

Q = ma(h2 h1) (39)= Q = 184.05 (46.0 25.5) = 3773.01 kJ/min (40)

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BDA 3043

Moist air at 30 Cand 50% RH enters a dehumidifier operating at steady statewith a volumetric flow rate of 280 m3/min. The moist air passes over a coolingcoil and water vapor condenses. Condensate exits the humidifier saturated at10 C. Saturated moist air exits in a separate stream at the same temperature.There is no a significant loss of energy by heat transfer to the surroundings andpressure remains constant at 1.013 bar.

2

2

= 100%

2

1

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