introduction to chemical engineering thermodynamics 6th edition in si units solutions manual
introduction to chemical engineering thermodynamics 6th edition in si units solutions manual
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, 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 A = 0.023 in
4
ft F mass = 1000.7 lbm
F = PA g = 32.174 mass = Ans.
2 g
sec
, gm m
1.10 Assume the following: = 13.5 g = 9.8
3 2
cm
P
P = 400bar h =
g h = 302.3 m Ans.
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 = massg = Kx mass = 0.40kg g = 9.81 x = 1.08cm
2
F N
F = massg F = 3.924 N Ks = Ks = 363.333
x m
On Mars:
−3
x = 0.40cm FMars = Kx FMars = 4 10 mK
FMars mK
g=
Mars gMars = 0.01 Ans.
mass kg
d MP d MP
1.12 Given: P = −g and: = Substituting: P = − g
dz RT dz RT
P z
Denver 1 Denver Mg
Separating variables and integrating: dP = − dz
P
P 0 RT
sea
PDenver
After integrating: ln = −Mg zDenver
Psea RT
Taking the exponential of both sides − Mg z
RT Denver
and rearranging: PDenver = Pseae
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