Solutions Manual For Fundamentals of Heat and Mass Transfer 8th Edition By Bergman, Lavine, Incropera, DeWitt (All Chapters, 100% Original Verified, A+ Grade)

Solu�ons Manual for
Fundamentals of Heat and
Mass Transfer 8th Edition By
Bergman, Lavine, Incropera,
DeWit
(All chapters 100% Original
Verified, A+ Grade)



All Chapters 14 to 1 with
Supplement files download link at the
end of this file.

, PROBLEM 14.1
KNOWN: Mixture of O 2 and N 2 with partial pressures in the ratio 0.21 to 0.79.
FIND: Mass fraction of each species in the mixture.
SCHEMATIC:

pO 2 0.21
=
p N2 0.79

MO = 32.00 kg / kmol
2

M N = 28.01 kg / kmol
2




ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: From the definition of the mass fraction,
ρi ρi
m=
i =
ρ Σρ i
Hence, with
pi pi Mi pi
ρi
= = = .
R iT ( ℜ / M i ) T ℜT
Hence
M i p i / ℜT
mi =
ΣM i p i / ℜT
or, canceling terms and dividing numerator and denominator by the total pressure p,
Mi x i
mi = .
ΣM i x i
With the mole fractions as
0.21
x O2 p=
= O2 / p = 0.21
0.21 + 0.79
x N 2 p=
= N 2 / p 0.79,
find the mass fractions as
32.00 × 0.21
=mO
2
= 0.233 <
32.00 × 0.21 + 28.01× 0.79

m N2 =
1 − mO2 =
0.767. <

, PROBLEM 14.2
KNOWN: Mole fraction (or mass fraction) and molecular weight of each species in a mixture of n
species. Equal mole fractions (or mass fractions) of O 2 , N 2 and CO 2 in a mixture.
FIND: (a) Equation for determining mass fraction of species i from knowledge of mole fraction and
molecular weight of each of n species. Equation for determining mole fraction of species i from
knowledge of mass fraction and molecular weight of each of n species. (b) For mixture containing
equal mole fractions of O 2 , N 2 , and CO 2 , find mass fraction of each species. For mixture containing
equal mass fractions of O 2 , N 2 , and CO 2 , find mole fraction of each species.
SCHEMATIC:

x=
O2 x=
N 2 x CO
= 1/ 3
2
or
m
= O2 m
= N 2 mCO
= 1/ 3
2

MCO = 44.01 kg/kmol
2
=MO 32.00
= kg/kmol, M N 28.01 kg/kmol
2 2
ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: (a) With
ρi ρi pi / R i T p i M i / ℜT
m=
i = = =
ρ ∑ ρi ∑ pi / R i T ∑ p i M i / ℜT
i i i
and dividing numerator and denominator by the total pressure p,
Mi x i
mi = . (1) <
∑ Mi x i
i
Similarly,

xi
= =
pi ρi R i T
=
( ρ i / M i ) ℜT
∑ pi ∑ ρi R i T ∑ ( ρi / M i ) ℜT
i i i
or, dividing numerator and denominator by the total density ρ
m i / Mi
xi = . (2) <
∑ m i / Mi
i
(b) With equal mole fractions of each species, x i = 1/3, using Eq. (1),
MO x O + M N x N + MCO x CO = (32.00 + 28.01 + 44.01) / 3 = 34.7 kg/kmol
2 2 2 2 2 2

=mO2 0.31,
= m N 2 0.27,
= mCO2 0.42. <
With equal mass fractions of each species, m i = 1/3, using Eq. (2),
mO / MO + m N / M N + mCO / M 2.99 ×10−2 kmol/kg
(1/ 32.00 + 1/ 28.01 + 1/ 44.01) / 3 =
=
2 2 2 2 2 CO2

find
=x O2 0.35,
= x N 2 0.40,
= x CO2 0.25. <

, PROBLEM 14.3
KNOWN: Partial pressures and temperature for a mixture of CO 2 and N 2 .
FIND: Molar concentration, mass density, mole fraction and mass fraction of each species.
SCHEMATIC:

A → CO 2 , M A = 44.01 kg / kmol
pA = pB = 0.75 bar B → N2 , MB = 28.01 kg / kmol
T = 318K



ASSUMPTIONS: (1) Ideal gas behavior.
ANALYSIS: From the equation of state for an ideal gas,
pi
Ci = .
ℜT
Hence, with p A = p B ,
0.75 bar
C=
A C=
B
8.314 × 10−2 m3 ⋅ bar / kmol ⋅ K × 318 K

C=
A C=
3
B 0.0284 kmol / m . <
With ρi = Mi Ci , it follows that

ρA =44.01 kg / kmol × 0.0284 kmol / m3 =
1.25 kg / m3 <
ρB =28.01 kg / kmol × 0.0284 kmol / m3 =0.795 kg / m3. <
Also, with
x i Ci / Σi Ci
=
find
x=
A x=
B 0..0568
= 0.5 <
and with
m
= i ρ i / Σρ i
find

A 1.25 / (1.25 + 0.795
m= = ) 0.611 <
m
= B 0.795 / (1.25 + 0.795
= ) 0.389. <