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Solutions Manual For Fundamentals of Heat and Mass Transfer 8th Edition By Bergman, Lavine, Incropera, DeWitt (All Chapters, 100% Original Verified, A+ Grade) R539,14
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Solutions Manual For Fundamentals of Heat and Mass Transfer 8th Edition By Bergman, Lavine, Incropera, DeWitt (All Chapters, 100% Original Verified, A+ Grade)

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This Is The Original 8th Edition Of The Solution Manual From The Original Author All Other Files In The Market Are Fake/Old Editions. Other Sellers Have Changed The Old Edition Number To The New But The Solution Manual Is An Old Edition. Solutions Manual For Fundamentals of Heat and Mass Transf...

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  • September 16, 2024
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By: erichenryolsen • 2 months ago

This is not the complete solutions manual it is only chapters 14 and 13. scam artists

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By: tutorsection • 2 months ago

There is a download link for all chapters on page 17 and the last page. plz copy that download link and paste it for all chapters download. plz make sure there is no cut/break in the link text.

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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. <

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