Problem 2.20
Known:
The drag force on a truck is known as a function of velocity and other parameters.
Find:
Determine the power required by the truck to overcome drag.
Schematic & Given Data:
Power required to overcome drag
Drag force
Velocity
Drag coefficient
Projected frontal area
Air density
d
d
d
W
F
V
C
A
ρ
=
=
=
=
=
=
±
Model and Assumptions:
The truck is the system.
Solution:
“The rate of energy transfer by work is called
power
and is denoted by
W
±
. The rate of energy transfer
by work is equal to the product of the force and the velocity at the point of application of the force.”
WF
V
=⋅
±
(Eq.2.13 in a textbook)
Applying Eq. 2.13, the power required to overcome drag is
dd
V
±
(1)
The drag force,
d
F
, is given by
2
1
2
FC
A
V
=
(2)
Substituting Eq. (2) to Eq. (1),
23
11
22
d
WC
A
V
V
C
A
V
ρρ
⎛⎞
==
⎜⎟
⎝⎠
±
(3)
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()
3
3
2
32
3
1
kg
km
10 m
1h
1N
1kJ
0.65 10m
1.1
110
2
m
h
1km 3600s
1kg m s
10 N m
kJ 1kW
102
s1
k
J
/
s
102kW
d
W
⎛⎞
⎛
⎞
=
⎜⎟
⎜
⎟
⋅⋅
⎝⎠
⎝
⎠
=
=
±
Problem 2.33
Known:
Carbon monoxide gas within a pistoncylinder assembly undergoes three processes in series.
Find:
Sketch the processes in series in a PV diagram and evaluate the work for each.
Schematic & Given Data:
Process 12: Expansion from P
1
=5bar, V
1
=0.2m
3
to V
2
=1m
3
, during which the pressurevolume
relationship is
PV C
=
(constant).
Process 23: Constantvolume heating from state 2 to state 3, where P
3
=5bar.
Process 31: Constantpressure compression to the initial state
Work done (+:by the system, while :on the system )
Pressure
System volume
Constant
W
P
V
C
=
=
=
=
Model and Assumptions:
1.
The gas is the closed system.
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 Fall '08
 TORRANCE
 Thermodynamics, Energy, Work, Heat Transfer, Wcycle Qin

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