SOLUTION MANUAL
1
,1) Introduction
2) The First Law and Other Basic Concepts
3) Volumetric Properties of Pure Fluids
4) Heat Effects
5) The Second Law of Thermodynamics
6) Thermodynamic Properties of Fluids
7) Applications of Thermodynamics to Flow Processes
8) Production of Power from Heat
9) Refrigeration and Liquefaction
10) The Framework of Solution Thermodynamics
11) Mixing Processes
12) Phase Equilibrium: Introduction
13) Thermodynamic Formulations for Vapor/Liquid Equilibrium
14) Chemical-Reaction Equilibria
15) Topics in Phase Equilibria
16) Thermodynamic Analysis of Processes
2
, Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
Guess solution: t = 0
Given t = 1.8t + 32 Find(t) = −40 Ans.
F
1.5 By definition: P= F = massg Note: Pressures are in
A gauge pressure.
2
P = 3000bar D = 4mm A = D A = 12.566 mm
2
4
m F mass = 384.4 kg
F = PA g = 9.807 mass = Ans.
2 g
s
F
1.6 By definition: P= F = massg
A
2
P = 3000atm D = 0.17in A = D 2
4
ft F mass = 1000.7 lbm
F = PA g = 32.174 mass = Ans.
2 g
sec
1.7 Pabs = gh + Patm
gm m
= 13.535 g = 9.832 h = 56.38cm
3 2
cm s
Patm = 101.78kPa Pabs = gh + Patm Pabs = 176.808 kPa Ans.
3
, gm m
1.10 Assume the following: U
13.5 g
9.8
3 2
cm s
P
P
400bar h
h 302.3 m Ans.
Ug
1.11 The force on a spring is described by: F = Ks x where Ks is the spring
constant. First calculate K based on the earth measurement then gMars
based on spring measurement on Mars.
On Earth:
m
F = mass g = K x mass
0.40kg g
9.81 x
1.08cm
2
s
F N
F
mass g F 3.924 N Ks
Ks 363.333
x m
On Mars:
3
x
0.40cm FMars
K x FMars 4 u 10 mK
FMars mK
gMars
gMars 0.01 Ans.
mass kg
d M P d M P
1.12 Given: P = U g and: U= Substituting: P= g
dz R T dz R T
P zDenver
´ Denver 1 ´ § M g · dz
Separating variables and integrating: µ dP = µ ¨
µ P µ © R T ¹
¶P ¶0
sea
§ PDenver · M g
After integrating: ln ¨ = zDenver
© Psea ¹ R T
Taking the exponential of both sides § M g z ·
¨ Denver
e© ¹
and rearranging: R T
PDenver = Psea
gm m
Psea
1atm M
29 g
9.8
mol 2
s
2
1
,1) Introduction
2) The First Law and Other Basic Concepts
3) Volumetric Properties of Pure Fluids
4) Heat Effects
5) The Second Law of Thermodynamics
6) Thermodynamic Properties of Fluids
7) Applications of Thermodynamics to Flow Processes
8) Production of Power from Heat
9) Refrigeration and Liquefaction
10) The Framework of Solution Thermodynamics
11) Mixing Processes
12) Phase Equilibrium: Introduction
13) Thermodynamic Formulations for Vapor/Liquid Equilibrium
14) Chemical-Reaction Equilibria
15) Topics in Phase Equilibria
16) Thermodynamic Analysis of Processes
2
, Chapter 1 - Section A - Mathcad Solutions
1.4 The equation that relates deg F to deg C is: t(F) = 1.8 t(C) + 32. Solve this
equation by setting t(F) = t(C).
Guess solution: t = 0
Given t = 1.8t + 32 Find(t) = −40 Ans.
F
1.5 By definition: P= F = massg Note: Pressures are in
A gauge pressure.
2
P = 3000bar D = 4mm A = D A = 12.566 mm
2
4
m F mass = 384.4 kg
F = PA g = 9.807 mass = Ans.
2 g
s
F
1.6 By definition: P= F = massg
A
2
P = 3000atm D = 0.17in A = D 2
4
ft F mass = 1000.7 lbm
F = PA g = 32.174 mass = Ans.
2 g
sec
1.7 Pabs = gh + Patm
gm m
= 13.535 g = 9.832 h = 56.38cm
3 2
cm s
Patm = 101.78kPa Pabs = gh + Patm Pabs = 176.808 kPa Ans.
3
, gm m
1.10 Assume the following: U
13.5 g
9.8
3 2
cm s
P
P
400bar h
h 302.3 m Ans.
Ug
1.11 The force on a spring is described by: F = Ks x where Ks is the spring
constant. First calculate K based on the earth measurement then gMars
based on spring measurement on Mars.
On Earth:
m
F = mass g = K x mass
0.40kg g
9.81 x
1.08cm
2
s
F N
F
mass g F 3.924 N Ks
Ks 363.333
x m
On Mars:
3
x
0.40cm FMars
K x FMars 4 u 10 mK
FMars mK
gMars
gMars 0.01 Ans.
mass kg
d M P d M P
1.12 Given: P = U g and: U= Substituting: P= g
dz R T dz R T
P zDenver
´ Denver 1 ´ § M g · dz
Separating variables and integrating: µ dP = µ ¨
µ P µ © R T ¹
¶P ¶0
sea
§ PDenver · M g
After integrating: ln ¨ = zDenver
© Psea ¹ R T
Taking the exponential of both sides § M g z ·
¨ Denver
e© ¹
and rearranging: R T
PDenver = Psea
gm m
Psea
1atm M
29 g
9.8
mol 2
s
2
