Introduction to Chemical Engineering Thermodynamics 6th Edition in SI Units Solutions Manual

, 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 = massg 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 = PA g = 9.807 mass = Ans.
2 g
s


F
1.6 By definition: P= F = massg
A

 2
P = 3000atm D = 0.17in A = D A = 0.023 in
4

ft F mass = 1000.7 lbm
F = PA g = 32.174 mass = Ans.
2 g
sec



1.7 Pabs = gh + Patm

gm
 = 13.535 h = 56.38cm
cm

Patm = 101.78kPa Pabs = gh + Patm Pabs = 176.808 kPa Ans.



gm ft
1.8  = 13.535 g = 32.243 h = 25.62in
cm

,Patm = 29.86in_Hg Pabs = gh + Patm Pabs = 27.22 psia Ans.


1

, 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 = massg = Kx mass = 0.40kg g = 9.81 x = 1.08cm
2

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  10 mK
FMars mK
g=
Mars gMars = 0.01 Ans.
mass kg




d MP d MP
1.12 Given: P = −g and: = Substituting: P = − g
dz RT dz RT


P z
 Denver 1  Denver  Mg
Separating variables and integrating:  dP =  − dz
 P 
P 0  RT 
sea



 PDenver
After integrating: ln = −Mg zDenver

 Psea  RT



Taking the exponential of both sides  − Mg z 
 RT Denver

and rearranging: PDenver = Pseae