Question

21) The following figures apply to a certain car running in top gear on a level road: Road speed (km/h) 20 40 60 70 80 Power required to overcome resistance to motion (kW) 2.5 10 27 40 59 Power developed by engine (kW) 20 40 52 50 37 22) Draw the curves of power required and power developed on a km/h base and determine: a) the maximum speed cm on the level b) the maximum power available for acceleration: c) the speed at which (b) occurs; . d) the maximum power of the engine. 23) The following results were obtained for a vehicle during a series of road speed Road speed (km/h) 15 30 45 60 75 Tractive resistance (N) 270 330 430 570 750 Tractive effort (N) 1200 1500 1625 1650 1600 Plot these values to a base of toad speeds and use the curves to determine (a) the road speed at which the acceleration is a maximum. (b) the value of this maximum acceleration if the total mass of the vehicle

Answer 1

Speed (km/hr)	Resistance Power (kW)	Engine Power (kW)
20	2.5	20
40	10	40
60	27	52
70	40	50
80	59	37

70 60 50 . Resistance Power • Engine Power 40 kW 30 20 10 . 0 0 20 40 80 100 60 km/hr

Fig.1: Engine power and Resistance Power Vs speed graph

(a) The maximum speed is 80 km/hr.

(b) The maximum power available for acceleration is given by:

$P_{max}=40-10=30kW$

(c) The maximum power occur at 40 km/hr.

(d) The maximum power of the engine is :

50 ᎬᏉ

Speed (km/h)	Tractive Resistance (N)	Tractive effort (N)
15	270	1200
30	330	1500
45	430	1625
60	570	1650
75	750	1600

1800 1600 1400 1200 Tractive Resistance (N) Tractive effort (N) 1000 z 800 600 400 . 200 0 20 40 60 80 km/hr

Fig.2: Tractive resistance and effort vs speed of the vehicle.

Speed (km/h)	Tractive Resistance (N)	Tractive effort (N)	Diff.=Tractive effort (N)-Tractive Resistance (N)
15	270	1200	930
30	330	1500	1170
45	430	1625	1195
60	570	1650	1080
75	750	1600	850

(a) The road speed at which the acceleration is maximum is corresponding to the speed having maximum available effort.

Hence the speed is

(b) The mass is not given. Let the mass be 'M' kg. Then the maximum acceleration is given by:

$a_{max}=\frac{F_{max}}{M} \ m/s^2$

$\Rightarrow a_{max}=\frac{1195}{M} \ m/s^2$

if M=1000 kg (say) then

$\Rightarrow a_{max}=\frac{1195}{1000}=1.195\ m/s^2$