Environmental Engineering Reference
In-Depth Information
3.1.1 Gradeability
Hybrid power plants must have their heat engines sized to meet sustained perfor-
mance on a grade. As we saw above for the Focus vehicle, a 33% grade consumes
its entire engine output in order to sustain 23mph. If there was some headwind or
more vehicle occupants, this would not be possible in high gear. The conclusions
above have taken into account that the driveline will downshift as appropriate to
move the full engine power to these lower speeds.
3.1.2 Wide open throttle
Analysis of vehicle propulsion under wide open throttle (WOT) conditions is
generally the approach taken to illustrate the best vehicle performance in accel-
eration times and passing. When the vehicle's accelerator pedal is pressed com-
pletely to the floor, the engine controller senses the demand for full performance
and autonomously declutches the vehicle's air conditioner compressor by de-
energizing its electromagnetic clutch. In vehicles with controllable fans and water
pumps, there may be some further gains by restraining the power consumed by
these ancillaries.
Example 3:
Using the 2010 Camaro SS specifications given in Table 3.1:
(a) compute the maximum tractive force at the rear wheels when accelerating
from standstill and (b) compute the ratio of traction limited force to the maximum
engine delivered traction force from part (a).
Given: The wheel base
L
of the Camaro SS is specified as 2,852.4mm and the
weight balance is 52%/48% (front/rear).
Solution:
●
Assuming the Camaro SS launches in first gear and for the specifications listed
in Table 3.1, the maximum engine delivered tractive force to the rear wheels is
F
t
eng
¼
ð
g
1st
g
fd
Þ
r
w
m
e
¼
ð
4
:
03
Þð
3
:
27
Þ
0
:
364
553
¼
2
;
0020
:
6N
From Chapter 1 the front weight balance (
B/L
) = 0.52 and the rear (
A/L
) = 0.48,
where
A
is the moment arm from vehicle centre of gravity (cg) to front axle
centreline and
B
is the moment arm from vehicle cg to rear axle centreline.
Therefore,
A
= 1,369.2mm and
F
Nr
=
g
Mr
=
g
(
A/L
)(
M
v
+
M
pass
) = 9.8066(0.48)
(1,754 + 75.5) = 8,611 N. Given that the maximum slip coefficient is 0.85, the
traction limited force,
F
t
lim
= 0.85(
F
Nr
) = 7,320 N:
●
F
t
lim
F
t
eng
¼
7
;
320
20
;
020
:
6
¼
0
:
365
Therefore, launch in first gear demands only 36.5% of available tractive force.
