13-27
13-53 The volume fractions of components of a gas mixture are given. This mixture is heated while flowing through a tube
at constant pressure. The heat transfer to the mixture per unit mass of the mixture is to be determined.
Assumptions All gases will be modeled as ideal gases with constant specific heats.
Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 kg/kmol, respectively (Table A-1). The
constant-pressure specific heats of these gases at room temperature are 0.918, 1.039, 0.846, and 2.2537 kJ/kgK,
respectively (Table A-2a).
Analysis We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of each
component are
m O2  N O2 M O2  (30 kmol)(32 kg/kmol)  960 kg
m N2  N N2 M N2  (40 kmol)(28 kg/kmol)  1120 kg
m CO2  N CO2 M CO2  (10 kmol)(44 kg/kmol)  440 kg
m CH4  N CH4 M CH4  (20 kmol)(16 kg/kmol)  320 kg
The total mass is
qin
m m  m O2  m N2  m CO2  m CH4
 960  1120  440  320
 2840 kg
150 kPa
20C
30% O2, 40% N2
10% CO2, 20% CH4
(by volume)
150 kPa
200C
Then the mass fractions are
mf O2 
m O2
960 kg

 0.3380
2840 kg
mm
mf N2 
m N2 1120 kg

 0.3944
2840 kg
mm
mf CO2 
m CO2
440 kg

 0.1549
2840 kg
mm
mf CH4 
m CH4
320 kg

 0.1127
mm
2840 kg
The constant-pressure specific heat of the mixture is determined from
c p  mf O2 c p ,O2  mf N2 c p , N2  mf CO2 c p ,CO2  mf CH4 c p ,CH4
 0.3380  0.918  0.3944  1.039  0.1549  0.846  0.1127  2.2537
 1.1051 kJ/kg  K
An energy balance on the tube gives
q in  c p (T2  T1 )  (1.1051 kJ/kg  K )(200  20) K  199 kJ/kg
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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at constant pressure. The heat transfer to the mixture per... 13-53