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:

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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]

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

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;

= m( v² - u² )

we substitute

=   × 1800( (31.293)² - (8.941)² )

= 900( 979.2518 - 79.9414)

= 900 × 899.3104

=  809379.36 J

= 809.379 kJ

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

= 50% ×

= 50/100 × 809.379 kJ

=  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;

=   / 25%

=  404.6895 kJ / 0.25

 = 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;

= 1618.758 kJ / 26.7kJ/g

= 60.63 g

= 0.0606 kg

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

need to see circuit to get time constant

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft