Set 4 2008
4.2
(a)
Relative to the gun barrel,
K
.
E
.
=
mV
2
2
=
0.02 lbm
⋅
2000
ft
s
2
2
⋅
lbfs
2
32.2lbmft
=
1.242
⋅
10
3
ft lbf =1.684 kJ
(b)
V
=
V
0

gt
;
V
=
0 when
t
=
V
0
g
=
2000
ft
s
32.2
ft
s
2
=
62.11s
z
=
Vdt
∫
=
V
0
t

1
2
gt
2
=
2000
ft
s
⋅
62.1s

1
2
32.2
ft
s
2
⋅
62.1s
(
)
2
=
6.21
⋅
10
4
ft
(c)
kJ
684
.
1
lbf
ft
1242
ft
lbm
32.2
s
lbf
62130
2
.
32
02
.
0
2
2
=
=
⋅
⋅
⋅
⋅
=
=
ft
s
ft
lbm
mgz
PE
The fact that the initial kinetic energy and the potential energy at the top of the trajectory
are the same is not an accident.
Air resistance complicates this (see problem 6.94).
4.3*
W
=
Fdx
=
mg
Δ
z
=
2.0lbm
⋅
6
ft
s
2
⋅
10ft
⋅
lbfs
2
32.2 lbmft
=
3.73 ft lbf = 5.05 J
4.7*
System the pump, steady flow.
Start with the enthalpy form of the energy balance (Eqn 4.13) and note
dm
in
=
dm
out
= dm
(steady flow).
Develop the “steady flow, open system” form by dividing through by
dm
[Example 4.5, Eqn 4.15]. Eliminate the terms that are 0 you are left with:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Smith
 Energy, Joule, Eqn, ft lbf, 2 10ft

Click to edit the document details