Patylulu
Páginas: 451 (112750 palabras)
Publicado: 6 de junio de 2010
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.
1.5 By definition:
P=
F A
F = mass g
Note: Pressures are in gauge pressure.
P
3000bar
D
4mm
A
4
D
2
A
12.566 mm
2F
PA
g
9.807
m s
2
mass
F g
mass
384.4 kg
Ans.
1.6 By definition:
P 3000atm
P=
D
F A
0.17in
F = mass g
A
4
D
2
A
0.023 in
2
F
PA
g
32.174
ft sec
2
mass
F g
mass
1000.7 lbm
Ans.
1.7 Pabs =
gh
Patm
13.535
gm cm
3
g
9.832
m s
2
h
56.38cm
Patm
101.78kPa
Pabs
ghPatm
Pabs
176.808 kPa Ans.
1.8
13.535
gm cm
3
g
32.243
ft s
2
h
25.62in
Patm
29.86in_Hg
Pabs
gh
Patm
Pabs
27.22 psia
Ans.
1
1.10
Assume the following:
13.5
gm cm
3
g
9.8
m s
2
P
400bar
h
P g
h
302.3 m
Ans.
1.11
The force on a spring is described by: F = Ks x where Ks is the springconstant. 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 s
2
x
1.08cm
F
mass g
F
3.924 N
Ks
F x
4 10
Ks
3
363.333
N m
On Mars:
x
0.40cm
FMars mass
FMars
Kx
FMars
mK
gMars
gMars
0.01
mK kg
Ans.
1.12 Given:
d P=dz
g
and:
=
MP RT
Substituting:
d P= dz
MP g RT
PDenver
Separating variables and integrating:
Psea
1 dP = P
zDenver
Mg dz RT
0
After integrating:
ln
PDenver Psea
=
Mg RT
zDenver
Taking the exponential of both sides and rearranging: Psea 1atm
PDenver = Psea e gm mol
Mg zDenver RT
M
29
g
9.8
m s
2
2
R
82.06cm atm mol K
3
T
( 10
273.15)K
zDenver
1 mi
Mg zDenver RT
Mg
0.194
PDenver
Psea e
RT
zDenver
PDenver
0.823 atm
Ans.
PDenver 1.13 The same proportionality applies as in Pb. 1.11.
0.834 bar
Ans.
gearth
32.186
ft s
2
gmoon
5.32
ft s
2
lmoon
18.76
learth
lmoon
gearth gmoon
learth
113.498
M
wmoonlearth lbm
M gmoon
M
113.498 lbm
18.767 lbf
Ans.
wmoon
Ans.
1.14
costbulb
hr 5.00dollars 10 day 1000hr
costelec
hr 0.1dollars 70W 10 day kW hr
costbulb
18.262
dollars yr
costelec
25.567
dollars yr
costtotal
costbulb
costelec
costtotal
43.829
dollars yr Ans.
1.15
D
1.25ft
mass
250lbm
g
32.169
ft s
2
3Patm
30.12in_Hg
A
4
D
2
A
3
1.227 ft
2
(a) F
Patm A
mass g
F
2.8642 10 lbf
Ans.
(b) Pabs
F A
Pabs
16.208 psia
Ans.
3 4.8691 10 ft lbf Ans.
(c)
l
1.7ft
Work
F l
Work
PE
mass g l
PE
424.9 ft lbf
Ans.
1.16
D
0.47m
mass
150kg
g
9.813
m s
2
Patm
101.57kPa
A
4
D
2
A4
0.173 m
2
(a) F
(b) Pabs
Patm A
F A
mass g
F
Pabs
1.909 10 N
110.054 kPa
Ans.
Ans.
(c)
l
0.83m
Work
F l
Work
15.848 kJ Ans.
EP
mass g l
EP
1.222 kJ
Ans.
1.18
mass
1250kg
u
40
m s
EK
1 2
mass u
2
EK
1000 kJ
Ans.
Work
EK
Work
1000 kJ
Ans.
1.19
Wdot =
mass g h 0.91 0.92time
Wdot
200W
g
9.8
m s
2
h
50m
4
mdot
Wdot g h 0.91 0.92
25.00 ton
mdot
0.488
kg s
Ans.
1.22 a) cost_coal
29
MJ kg
2.00 gal
cost_coal
0.95 GJ
1
cost_gasoline
37
GJ m
3
cost_gasoline
14.28 GJ
1
cost_electricity
0.1000 kW hr
cost_electricity
27.778 GJ
1
b) The electrical energy can directly be...
Leer documento completo
Regístrate para leer el documento completo.