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).
t := 0
Guess solution:
Given
t = 1.8t + 32
1.5 By definition:
P=
Find ( t) = −40
F
A
F = mass⋅ g
P := 3000bar
D := 4mm
F := P⋅ A
g = 9.807
π 2
⋅D
4
F
mass :=
g
A :=
m
2
s
1.6 By definition:
P=
F
A
Ans.
Note: Pressures are in
gauge pressure.
A = 12.566 mm
2
mass = 384.4 kg
Ans.
F = mass⋅ g
P := 3000atm
D := 0.17in
F := P⋅ A
g = 32.174
π 2
⋅D
4
F
mass :=
g
A :=
ft
2
sec
2
A = 0.023 in
mass = 1000.7 lbm
Ans.
1.7 Pabs = ρ ⋅ g⋅ h + Patm
ρ := 13.535⋅
gm
3
g := 9.832⋅
cm
Patm := 101.78kPa
1.8
ρ := 13.535⋅
gm
3
h := 56.38cm
2
s
Pabs := ρ ⋅ g⋅ h + Patm
g := 32.243⋅
cm
Patm := 29.86in_Hg
m
ft
2
Pabs = 176.808 kPa Ans.
h := 25.62in
s
Pabs := ρ ⋅ g⋅ h + Patm
Pabs = 27.22 psia
Ans.
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1.10
Assume the following: ρ := 13.5
gm
g := 9.8
3
cm
P := 400bar
1.11
h :=
m
2
s
P
h = 302.3 m
ρ⋅g
Ans.
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
g := 9.81
m
x := 1.08cm
2
s
F := mass⋅ g
F
x
F = 3.924 N
Ks :=
Ks = 363.333
FMars := K⋅ x
FMars = 4 × 10
N
m
On Mars:
x := 0.40cm
gMars :=
1.12 Given:
FMars
mass
d
P = −ρ ⋅ g
dz
gMars = 0.01
and:
−3
mK
kg
ρ=
mK
Ans.
M⋅ P
R⋅ T
Substituting:
P
M⋅ P
d
P= −
⋅g
R⋅ T
dz
z
⌠ Denver 1
⌠ Denver ⎛ M⋅ g ⎞
Separating variables and integrating: ⎮
dP = ⎮
−⎜
dz
⎮
P
⎮
R⋅ T ⎠
⎝
⌡0
⌡P
sea
⎛ PDenver ⎞
ln ⎜
After integrating:
⎝ Psea ⎠
Taking the exponential of both sides
and rearranging:
Psea := 1atm
M := 29
=
PDenver = Psea
gm
mol
−M⋅ g
⋅ zDenver
R⋅ T
⎞
⎛ − M⋅ g ⋅ z
⎜
Denver
R⋅ T
⎠
⋅ e⎝
g := 9.8
m
2
s
2
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3
R := 82.06
cm ⋅ atm
mol⋅ K
T := ( 10 + 273.15)K
zDenver := 1⋅ mi
M⋅ g
⋅ zDenver = 0.194
R⋅ T
PDenver := Psea
⎞
⎛ − M⋅ g ⋅ z
⎜
Denver
R⋅ T
⎠
⋅ e⎝
PDenver = 0.823 atm
Ans.
PDenver = 0.834 bar
Ans.
1.13 The same proportionality applies as in Pb. 1.11.
gearth := 32.186⋅
ft
gmoon := 5.32⋅
2
s
∆learth := ∆lmoon⋅
1.14
∆lmoon := 18.76
2
s
gearth
gmoon
∆learth = 113.498
M := ∆learth⋅ lbm
M = 113.498 lbm
Ans.
wmoon := M⋅ gmoon
wmoon = 18.767 lbf
Ans.
costbulb :=
hr
5.00dollars
⋅ 10
1000hr
day
costbulb = 18.262
costelec :=
dollars
yr
costtotal := costbulb + costelec
1.15
ft
D := 1.25ft
hr
0.1dollars
⋅ 10
⋅ 70W
kW⋅ hr
day
costelec = 25.567
dollars
yr
costtotal = 43.829
dollars
yr Ans.
mass := 250lbm
g := 32.169
ft
2
s
3
PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted
only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
Patm := 30.12in_Hg
A :=
(c)
F
A
∆l := 1.7ft
F = 2.8642 × 10 lbf
Ans.
Pabs = 16.208 psia
Ans.
3
Work = 4.8691 × 10 ft⋅ lbf Ans.
∆PE := mass⋅ g⋅ ∆l
∆PE = 424.9 ft⋅ lbf
mass := 150kg
Patm := 101.57kPa
A :=
(c)
1.18
F
A
∆l := 0.83m
mass := 1250kg
EK :=
1
2
mass⋅ u
2
Work := EK
1.19
Wdot =
g := 9.813
Ans.
m
2
s
π 2
⋅D
4
2
A = 0.173 m
4
(a) F := Patm⋅ A + mass⋅ g
(b) Pabs :=
2
Work := F⋅ ∆l
D := 0.47m
1.16
A = 1.227 ft
3
(a) F := Patm⋅ A + mass⋅ g
(b) Pabs :=
π 2
⋅D
4
F = 1.909 × 10 N
Ans.
Pabs = 110.054 kPa
Ans.
Work := F⋅ ∆l
Work = 15.848 kJ Ans.
∆EP := mass⋅ g⋅ ∆l
∆EP = 1.222 kJ
u := 40
Ans.
m
s
EK = 1000 kJ
Ans.
Work = 1000 kJ
Ans.
mass⋅ g⋅ ∆h
⋅ 0.91⋅ 0.92
time
Wdot := 200W
g := 9.8
m
2
∆h := 50m
s
4
PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted
only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
mdot :=
1.22
a) cost_coal :=
Wdot
mdot = 0.488
g⋅ ∆h⋅ 0.91⋅ 0.92
25.00
ton
cost_coal = 0.95 GJ
MJ
29⋅
kg
2.00
gal
cost_gasoline :=
37⋅
GJ
kg
s
Ans.
−1
cost_gasoline = 14.28 GJ
−1
3
m
cost_electricity :=
0.1000
kW⋅ hr
cost_electricity = 27.778 GJ
−1
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
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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:
⎛ A− B ⎞
⎜
T+C ⎠
f ( T , A , B , C) := e⎝
First derivative of the function with respect to parameter A
B ⎞
d
f ( T , A , B , C) → exp ⎛⎜ A −
T + C⎠
dA
⎝
First derivative of the function with respect to parameter B
−1
B ⎞
d
f ( T , A , B , C) →
⋅ exp ⎛⎜ A −
T+C
T + C⎠
dB
⎝
First derivative of the function with respect to parameter C
B
B ⎞
d
f ( T , A , B , C) →
⋅ exp ⎛⎜ A −
2
T + C⎠
dC
⎝
( T + C)
⎛ −18.5 ⎞
⎜
⎜ −9.5 ⎟
⎜ 0.2 ⎟
⎟
⎜
⎜ 11.8 ⎟
⎜ 23.1 ⎟
t := ⎜
⎟
32.7
⎟
⎜
⎜ 44.4 ⎟
⎟
⎜
52.1
⎟
⎜
⎜ 63.3 ⎟
⎜
⎝ 75.5 ⎠
⎛ 3.18 ⎞
⎜
⎜ 5.48 ⎟
⎜ 9.45 ⎟
⎟
⎜
⎜ 16.9 ⎟
⎜ 28.2 ⎟
Psat := ⎜
⎟
41.9
⎟
⎜
⎜ 66.6 ⎟
⎟
⎜
89.5
⎟
⎜
⎜ 129 ⎟
⎜
⎝ 187 ⎠
6
PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted
only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.
T := t + 273.15
lnPsat := ln ( Psat)
Array of functions used by Mathcad. In this case, a0 = A, a1 = B and a2 = C.
a1 ⎞
⎡
⎛
⎤
exp
a
−
⎜
0
⎢
⎥
T + a2 ⎠
⎝
⎢
⎥
a1 ⎞
⎢
⎥
⎛
exp ⎜ a0 −
⎢
⎥
T + a2 ⎠
⎝
⎥
F ( T , a) := ⎢
a
⎢ −1
⎛
1 ⎞ ⎥
⋅
exp
a
−
⎜
0
⎢ T + a2
T + a2 ⎠ ⎥
⎝
⎢
⎥
a1 ⎞ ⎥
⎢ a1
⎛
⋅
exp
a
−
⎜
0
⎢
2
T + a2 ⎠ ⎥
⎝
⎣ ( T + a2)
⎦
Guess values of parameters
⎛⎜ 15 ⎞
guess := ⎜ 3000 ⎟
⎜ −50
⎝
⎠
Apply the genfit function
⎛⎜ A ⎞ ⎛⎜ 13.421 ⎞
⎜ B ⎟ = ⎜ 2.29 × 103 ⎟
⎜C
⎝ ⎠ ⎜⎝ −69.053 ⎠
⎛⎜ A ⎞
⎜ B ⎟ := genfit ( T , Psat , guess , F)
⎜C
⎝ ⎠
Ans.
Compare fit with data.
200
150
Psat
f ( T , A , B , C)
100
50
0
240
260
280
300
320
340
360
T
To find the normal boiling point, find the value of T for which Psat = 1 atm.
7
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Tnb := ⎛⎜
Psat := 1atm
B
Psat ⎞
⎜ A − ln ⎛⎜
⎝
⎝ kPa ⎠
− C⎞ ⋅ K
⎠
Ans.
Tnb − 273.15K = 56.004 degC
1.25
a) t1 := 1970
t2 := 2000
C2 := C1⋅ ( 1 + i)
t2 − t1
Tnb = 329.154 K
dollars
gal
dollars
C2 = 1.513
gal
C1 := 0.35
i := 5%
The increase in price of gasoline over this period kept pace with the rate of
inflation.
b) t1 := 1970
Given
t2 := 2000
C2
C1
= ( 1 + i)
C1 := 16000
t2 − t1
dollars
yr
i := Find ( i)
C2 := 80000
dollars
yr
i = 5.511 %
The salary of a Ph. D. engineer over this period increased at a rate of 5.5%,
slightly higher than the rate of inflation.
c) 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.
8
PROPRIETARY MATERIAL. © 2005 The McGraw-Hill Companies, Inc. All rights reserved. Limited distribution permitted
only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.