Smith e Van Ness   Introdução a engenharia Química   Termodinâmica   7ª ed   Soluções
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Smith e Van Ness Introdução a engenharia Química Termodinâmica 7ª ed Soluções


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P 3000atm\ufffd D 0.17in\ufffd A S
4
D
2˜\ufffd A 0.023 in2 
F P A˜\ufffd g 32.174 ft
sec
2
 mass F
g
\ufffd mass 1000.7 lbm Ans.
1.7 Pabs U g˜ h˜ Patm\ufffd=
U 13.535 gm
cm
3
˜\ufffd g 9.832 m
s
2
˜\ufffd h 56.38cm\ufffd 
Patm 101.78kPa\ufffd Pabs U g˜ h˜ Patm\ufffd\ufffd Pabs 176.808kPa Ans.
1.8 U 13.535 gm
cm
3
˜\ufffd g 32.243 ft
s
2
˜\ufffd h 25.62in\ufffd 
Patm 29.86in_Hg\ufffd Pabs U g˜ h˜ Patm\ufffd\ufffd Pabs 27.22psia Ans.
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\ufffd 
Given t 1.8t 32\ufffd= Find t( ) 40\ufffd Ans.
1.5 By definition: P
F
A
= F mass g˜= Note: Pressures are in
gauge pressure.
P 3000bar\ufffd D 4mm\ufffd A S
4
D
2˜\ufffd A 12.566mm2 
F P A˜\ufffd g 9.807 m
s
2
 mass F
g
\ufffd mass 384.4kg Ans.
1.6 By definition: P
F
A
= F mass g˜=
1
FMars K x˜\ufffd FMars 4 10 3\ufffdu mK 
gMars
FMars
mass
\ufffd gMars 0.01 mK
kg
 Ans.
1.12 Given:
z
P
d
d
U\ufffd g˜= and: U M P˜
R T˜= Substituting: zP
d
d
M P˜
R T˜\ufffd g˜=
Separating variables and integrating:
Psea
PDenver
P
1
P
µ´
µ¶
d
0
zDenver
z
M g˜
R T˜
§¨
©
·
¹\ufffd
µ´
µ¶
d=
After integrating: ln
PDenver
Psea
§¨
©
·
¹
M\ufffd g˜
R T˜ zDenver˜=
Taking the exponential of both sides
and rearranging: PDenver Psea e
M\ufffd g˜
R T˜ zDenver˜
§¨
©
·
¹˜=
Psea 1atm\ufffd M 29 gm
mol
\ufffd g 9.8 m
s
2
\ufffd 
1.10 Assume the following: U 13.5 gm
cm
3
\ufffd g 9.8 m
s
2
\ufffd 
P 400bar\ufffd h PU g˜\ufffd h 302.3m 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:
F mass g˜= K x˜= mass 0.40kg\ufffd g 9.81 m
s
2
\ufffd x 1.08cm\ufffd 
F mass g˜\ufffd F 3.924N Ks F
x
\ufffd Ks 363.333 N
m
 
On Mars:
x 0.40cm\ufffd 
2
Ans.
wmoon M gmoon˜\ufffd wmoon 18.767 lbf Ans.
1.14 costbulb
5.00dollars
1000hr
10˜ hr
day
\ufffd costelec 0.1dollars
kW hr˜ 10˜
hr
day
70˜ W\ufffd 
costbulb 18.262
dollars
yr
 costelec 25.567 dollars
yr
 
costtotal costbulb costelec\ufffd\ufffd costtotal 43.829 dollars
yr
 Ans.
1.15 D 1.25ft\ufffd mass 250lbm\ufffd g 32.169 ft
s
2
\ufffd 
R 82.06
cm
3
atm˜
mol K˜\ufffd T 10 273.15\ufffd( )K\ufffd zDenver 1 mi˜\ufffd 
M g˜
R T˜ zDenver˜ 0.194 
PDenver Psea e
M\ufffd g˜
R T˜ zDenver˜
§¨
©
·
¹˜\ufffd PDenver 0.823atm Ans.
PDenver 0.834bar Ans.
1.13 The same proportionality applies as in Pb. 1.11.
gearth 32.186
ft
s
2
˜\ufffd gmoon 5.32 ft
s
2
˜\ufffd 'lmoon 18.76\ufffd 
'learth 'lmoon
gearth
gmoon
˜\ufffd 'learth 113.498 
M 'learth lbm˜\ufffd M 113.498 lbm 
3
Ans.
(b) Pabs
F
A
\ufffd Pabs 110.054kPa Ans.
(c) 'l 0.83m\ufffd Work F 'l˜\ufffd Work 15.848kJ Ans.
'EP mass g˜' l˜\ufffd 'EP 1.222kJ Ans.
1.18 mass 1250kg\ufffd u 40 m
s
\ufffd 
EK
1
2
mass u
2˜\ufffd EK 1000kJ Ans.
Work EK\ufffd Work 1000kJ Ans.
1.19 Wdot
mass g˜' h˜
time
0.91˜ 0.92˜=
Wdot 200W\ufffd g 9.8 m
s
2
\ufffd 'h 50m\ufffd 
Patm 30.12in_Hg\ufffd A S
4
D
2˜\ufffd A 1.227 ft2 
(a) F Patm A˜ mass g˜\ufffd\ufffd F 2.8642 103u lbf Ans.
(b) Pabs
F
A
\ufffd Pabs 16.208psia Ans.
(c) 'l 1.7ft\ufffd Work F 'l˜\ufffd Work 4.8691 103u ft lbf˜ Ans.
'PE mass g˜' l˜\ufffd 'PE 424.9 ft lbf˜ Ans.
1.16 D 0.47m\ufffd mass 150kg\ufffd g 9.813 m
s
2
\ufffd 
Patm 101.57kPa\ufffd A S
4
D
2˜\ufffd A 0.173m2 
(a) F Patm A˜ mass g˜\ufffd\ufffd F 1.909 104u N 
4
mdot
Wdot
g 'h˜ 0.91˜ 0.92˜\ufffd mdot 0.488
kg
s
 Ans.
1.22
a) cost_coal
25.00
ton
29
MJ
kg
˜
\ufffd 
cost_coal 0.95GJ
1\ufffd 
cost_gasoline
2.00
gal
37
GJ
m
3
˜
\ufffd 
cost_gasoline 14.28GJ
1\ufffd 
cost_electricity
0.1000
kW hr˜\ufffd cost_electricity 27.778GJ 1\ufffd 
b)The electrical energy can directly be converted to other forms of energy
whereas the coal and gasoline would typically need to be converted to heat
and then into some other form of energy before being useful.
The obvious advantage of coal is that it is cheap if it is used as a heat
source. Otherwise it is messy to handle and bulky for tranport and
storage.
Gasoline is an important transportation fuel. It is more convenient to
transport and store than coal. It can be used to generate electricity by
burning it but the efficiency is limited. However, fuel cells are currently
being developed which will allow for the conversion of gasoline to electricity
by chemical means, a more efficient process.
Electricity has the most uses though it is expensive. It is easy to transport
but expensive to store. As a transportation fuel it is clean but batteries to
store it on-board have limited capacity and are heavy.
5
1.24 Use the Matcad genfit function to fit the data to Antoine's equation.
The genfit function requires the first derivatives of the function with
respect to the parameters being fitted.
Function being fit: f T A\ufffd B\ufffd C\ufffd( ) e
A
B
T C\ufffd\ufffd
§¨
©
·
¹\ufffd 
First derivative of the function with respect to parameter A
A
f T A\ufffd B\ufffd C\ufffd( )d
d
exp A
B
T C\ufffd\ufffd
§¨
©
·
¹o
First derivative of the function with respect to parameter B
B
f T A\ufffd B\ufffd C\ufffd( )d
d
1\ufffd
T C\ufffd exp A
B
T C\ufffd\ufffd
§¨
©
·
¹˜o
First derivative of the function with respect to parameter C
C
f T A\ufffd B\ufffd C\ufffd( )d
d
B
T C\ufffd( )2
exp A
B
T C\ufffd\ufffd
§¨
©
·
¹˜o
t
18.5\ufffd
9.5\ufffd
0.2
11.8
23.1
32.7
44.4
52.1
63.3
75.5
§¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
©¨
·
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¹
\ufffd Psat
3.18
5.48
9.45
16.9
28.2
41.9
66.6
89.5
129
187
§¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
©¨
·
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¹
\ufffd 
6
T t 273.15\ufffd\ufffd lnPsat ln Psat( )\ufffd 
Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.
Guess values of parameters
F T a\ufffd( )
exp a0
a1
T a2\ufffd
\ufffd§¨©
·
¹
exp a0
a1
T a2\ufffd
\ufffd§¨©
·
¹
1\ufffd
T a2\ufffd
exp a0
a1
T a2\ufffd
\ufffd§¨©
·
¹˜
a1
T a2\ufffd\ufffd\ufffd 2 exp a0
a1
T a2\ufffd
\ufffd§¨©
·
¹˜
ª«
«
«
«
«
«
«
«
«
«¬
º»
»
»
»
»
»
»
»
»
»¼
\ufffd guess
15
3000
50\ufffd
§¨
¨
©¨
·
¸
¹
\ufffd 
Apply the genfit function
A
B
C
§¨
¨
©¨
·
¸
¹
genfit T Psat\ufffd guess\ufffd F\ufffd( )\ufffd 
A
B
C
§¨
¨
©¨
·
¸
¹
13.421
2.29 10
3u
69.053\ufffd
§¨
¨
©¨
·
¸
¹
 Ans.
Compare fit with data.
240 260 280 300 320 340 360
0
50
100
150
200
Psat
f T A\ufffd B\ufffd C\ufffd( )
T
To find the normal boiling point, find the value of T for which Psat = 1 atm.
7
This is an open-ended problem. The strategy depends on age of the child,
and on such unpredictable items as possible financial aid, monies earned
by the child, and length of time spent in earning a degree.
c)
The salary of a Ph. D. engineer over this period increased at a rate of 5.5%,
slightly higher than the rate of inflation. 
i 5.511% i Find i( )\ufffd C2
C1
1 i\ufffd( )t2 t1\ufffd=Given
C2 80000
dollars
yr
\ufffd C1 16000 dollars
yr
\ufffd t2 2000\ufffd t1 1970\ufffd b)
The increase in price of gasoline over this period kept pace with the rate of
inflation.
C2 1.513
dollars
gal
 C2 C1 1 i\ufffd( )t2 t1\ufffd˜\ufffd 
i 5%\ufffd C1 0.35 dollars
gal
\ufffd t2 2000\ufffd t1 1970\ufffd a)
1.25
Tnb 273.15K\ufffd 56.004degC Ans.
Tnb 329.154K Tnb B
A ln
Psat
kPa
§¨
©
·
¹\ufffd
C\ufffd§¨
©¨
·
¹
K˜\ufffd Psat 1atm\ufffd 
8
t1 20 degC˜\ufffd CP 4.18 kJ
kg degC˜˜\ufffd MH2O 30 kg˜\ufffd 
t2 t1
'Utotal
MH2O CP˜
\ufffd\ufffd t2 20.014degC Ans.
(d) For the restoration process, the change in internal energy is equal but of
opposite sign to that of the initial process. Thus 
Q 'Utotal\ufffd\ufffd Q 1.715\ufffd kJ Ans.Ans.
(e) In all cases the total internal energy change of the universe is zero.
2.2 Similar to Pb. 2.1 with mass of water = 30 kg.
Answers are: (a) W = 1.715 kJ
(b) Internal energy change of
 the water = 1.429 kJ
(c) Final temp. = 20.014