Question

A 1800 kg hybrid vehicle operates on ethanol and is equipped with a multipurpose motorgenerator-flywheel. When the vehicle slows or stops, 50% of the kinetic energy is recovered as electrical energy in the battery. When the IC engine is used to recharge the battery, there is a 25% efficiency of converting chemical energy in the fuel to electrical energy stored in the battery. The vehicle slows from 70 miles per hour to 20 miles per hour. Calculate:

Answer 1

This question is incomplete, the complete question is;

A 1800 kg hybrid vehicle operates on ethanol and is equipped with a multipurpose motorgenerator-flywheel. When the vehicle slows or stops, 50% of the kinetic energy is recovered as electrical energy in the battery. When the IC engine is used to recharge the battery, there is a 25% efficiency of converting chemical energy in the fuel to electrical energy stored in the battery. The vehicle slows from 70 miles per hour to 20 miles per hour. Calculate: (A) Electrical energy recovered in the battery in [kJ] (B) Mass of fuel needed to store same amount of energy in the battery in [kg]

Answer:

a) Electrical energy recovered in the battery is 404.6895 kJ

b) Mass of fuel needed to store same amount of energy in the battery is 0.0606 kg

Step-by-step explanation:

Given that;

Initial speed of the vehicle V = 70 miles per hour = 31.293 m/s

Final speed of the vehicle u = 20 miles per hour = 8.941 m/s

mass of vehicle m = 1800 kg

Noe, change in kinetic energy of the vehicle will be;

`E_(kinetic)` =
`(1)/(2)`m( v² - u² )

we substitute

=
`(1)/(2)` × 1800( (31.293)² - (8.941)² )

= 900( 979.2518 - 79.9414)

= 900 × 899.3104

= 809379.36 J

`E_(kinetic)` = 809.379 kJ

now, Electrical energy recovered in the battery when the vehicle slows will be;

`E_(battery)` = 50% ×
`E_(kinetic)`

`E_(battery)` = 50/100 × 809.379 kJ

`E_(battery)` = 404.6895 kJ

Therefore, Electrical energy recovered in the battery is 404.6895 kJ

b)

For this electrical energy to be obtained from fuel, the chemical energy required will be;

`E_(chemical)` =
`E_(battery)` / 25%

`E_(chemical)` = 404.6895 kJ / 0.25

`E_(chemical)` = 1618.758 kJ

Heat energy released per mass of ethanol combustion

(Lower heating value of ethanol) is 26.7kJ/g

Now, the mass of fuel needed to generate 1618.758 kJ will be;

`m_(fuel)` = 1618.758 kJ / 26.7kJ/g

`m_(fuel)` = 60.63 g

`m_(fuel)` = 0.0606 kg

Therefore, Mass of fuel needed to store same amount of energy in the battery is 0.0606 kg