P
3000atm
:=
D
0.17in
:=
A
π
4
D
2
⋅
:=
A
0.023in
2
=
FP
A
⋅
:=
g
32.174
ft
sec
2
=
mass
F
g
:=
mass
1000.7lb
m
=
Ans.
1.7
P
abs
ρ
g
⋅
h
⋅
P
atm
+
=
ρ
13.535
gm
cm
3
⋅
:=
g
9.832
m
s
2
⋅
:=
h
56.38cm
:=
P
atm
101.78kPa
:=
P
abs
ρ
g
⋅
h
⋅
P
atm
+
:=
P
abs
176.808kPa
=
Ans.
1.8
ρ
13.535
gm
cm
3
⋅
:=
g
32.243
ft
s
2
⋅
:=
h
25.62in
:=
P
atm
29.86in_Hg
:=
P
abs
ρ
g
⋅
h
⋅
P
atm
+
:=
P
abs
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:
t0
:=
Given
t
1.8t
32
+
=
Find t
()
40
−
=
Ans.
1.5 By definition:
P
F
A
=
F
mass g
⋅
=
Note: Pressures are in
gauge pressure.
P
3000bar
:=
D
4mm
:=
A
π
4
D
2
⋅
:=
A
12.566mm
2
=
F
PA
⋅
:=
g
9.807
m
s
2
=
mass
F
g
:=
mass
384.4kg
=
Ans.
1.6 By definition:
P
F
A
=
F
mass g
⋅
=
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentF
Mars
Kx
⋅
:=
F
Mars
41
0
3
−
×
mK
=
g
Mars
F
Mars
mass
:=
g
Mars
0.01
kg
=
Ans.
1.12 Given:
z
P
d
d
ρ
−
g
⋅
=
and:
ρ
MP
⋅
RT
⋅
=
Substituting:
z
P
d
d
⋅
⋅
−
g
⋅
=
Separating variables and integrating:
P
sea
P
Denver
P
1
P
⌠
⎮
⎮
⌡
d
0
z
Denver
z
Mg
⋅
⋅
⎛
⎝
⎞
⎠
−
⌠
⎮
⎮
⌡
d
=
After integrating:
ln
P
Denver
P
sea
⎛
⎜
⎝
⎞
⎠
M
−
g
⋅
⋅
z
Denver
⋅
=
Taking the exponential of both sides
and rearranging:
P
Denver
P
sea
e
M
−
g
⋅
⋅
z
Denver
⋅
⎛
⎝
⎞
⎠
⋅
=
P
sea
1atm
:=
M2
9
gm
mol
:=
g
9.8
m
s
2
:=
1.10
Assume the following:
ρ
13.5
gm
cm
3
:=
g
9.8
m
s
2
:=
P
400bar
:=
h
P
ρ
g
⋅
:=
h
302.3m
=
Ans.
1.11
The force on a spring is described by: F = K
s
x where K
s
is the spring
constant.
First calculate K based on the earth measurement then g
Mars
based on spring measurement on Mars.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Ku
 Derivative, Trigraph, Pallavolo Modena, Sisley Volley Treviso

Click to edit the document details