It has been demonstrated on the graph how to calculate the figures. However, since accuracy is important, the NPF has been worked out by using the equations of the propulsive and resistive curves.
Once the acceleration is calculated, the time taken for achieving the new speed is found by using the classic Newton’s Second Law:
v = u + at
Where v = Final Velocity
u = Initial Velocity
a = Acceleration
t = Acceleration Time
Assume it takes 1 second to pull away and 0.5 second for each gear change. Then the total time taken would be
7.43 + (0.5 x 3) + 1 = 9.93 seconds.
=F3*1000/(1322.5+(4*6.5)+(0.15*(0.91*0.96)^2))
Fuel Consumption
Fuel consumption is to be estimated for the vehicle traveling at 90km/hr and 120km/hr across level ground in still air. It is assumed that the vehicle is cruising in top gear (5th gear) for calculation.
The three equations used for the calculation of the final drive ratio are used, only this time working backwards.
For vehicle traveling at 90 km/hr,
ηwheel = Vehicle Velocity
Circumference
= ((90 x 1000)/60) = 852.62 rpm
2п x 0.28
Ratiototal = Ratiogear x Ratiofinal drive
= 0.8 x 4.42
= 3.536
ηengine = ηwheel x Ratiototal
= 852.62 x 3.536
= 3014 rpm
Reading off the graph for 5th gear, the resistive force at 90 km/hr is found to be 0.493 kN.
Since the vehicle is cruising at constant velocity, the propulsive power at wheels is equal to the resistive power, and can be calculated using:
P = RU/1000
= ((0.493 x 1000) x (90x1000/3600)) / 1000
= 12.32 kW
The propulsive power at engine is then:
BP = 12.32/(0.96 x 0.91)
= 14.10 kW
BMEP = BP
V x N x Z x 10-2
= 14.1 x 1000
100000 x 0.0018 x (3014/60) x 0.5
= 3.12 bar
The Brake Specific Fuel Consumption, BSFC, is given by
BSFC = Mf
BP
From the Engine Performance Curve, the BSFC is estimated to be about 0.321 kg/kWh. Since
Mf = 0.321 x BP = 0.321 x 14.1 = 4.53 kg/hr
The time taken for the vehicle to travel 100km at 90 km/hr is:
t = 100/90 = 1.11 hr
The fuel density is assumed to be 860 kg/m3, so the fuel consumption is:
(4.53 x 1.11 x 1000)/860 = 5.85 L/100 km
For vehicle traveling at 120 km/hr,
ηwheel = ((120 x 1000)/60) = 1136.82 rpm
2п x 0.28
ηengine = ηwheel x Ratiototal
= 1136.82 x 3.536
= 4020 rpm
The resistive force at 120 km/hr is found to be 0.745 kN.
The propulsive power at wheels:
P = ((0.745 x 1000) x (120x1000/3600)) / 1000
= 24.82 kW
The propulsive power at engine is then:
BP = 24.82/(0.96 x 0.91)
= 28.41 kW
BMEP = 28.41 x 1000
100000 x 0.0018 x (4020/60) x 0.5
= 4.71 bar
From the Engine Performance Curve, the BSFC is estimated to be about 0.284 kg/kWh.
Mf = 0.284 x 28.41 = 8.08 kg/hr
The time taken for the vehicle to travel 100km at 120 km/hr is:
t = 100/120 = 0.83 hr
The fuel consumption is then:
(8.08 x 0.83 x 1000)/860 = 7.8 L/100 km
Although these figures are slightly high for a 1.8 L vehicle, the reason behind this selection remains that such fifth gear has the desired power to propel the vehicle forward under less-than-ideal conditions.
