(20 14? Wuxi has a hybrid car, which is driven by an automobile engine (gasoline engine) and an electric motor. Every gasoline engine

(1) Electric energy stored in the battery:

w = UQ = 200v× 150 a×3600s = 1.08× 108j；

(2)a: When the car is driving at a constant speed ν 1= 108km/h=30m/s, the resistance of the car is equal to the traction, including

P=Wt=Fst=Fv=Ffv

ff = Pv = 54× 103 w30m/s = 1.8× 103n；

B: According to the data in the table, when the speed is 1m/s, E=6.0× 102, and E=m 12v2, and M = 2ev21= 2× 6.0×1002.

V0= 108km/h=30m/s, V-end =72km/h=20m/s,

Let the total work done by the gasoline engine in this process be W, and the kinetic energy theorem is: W=Ff? s = 12mv 20- 12mv 2 end = 12x 1.2x 103kg×[(30m/s)2-(20m/s)2]= 3x 105j。

The power obtained by the battery? E power = w× 50% =1.5×105j;

(3) Rotate 5000 revolutions in 65438 0 minutes, and inhale the mixed gas of gasoline and air for 2500 times.

The useful work done by the engine at the maximum power 1 min;

w has = 69× 103 w×60s = 4. 14× 106j，

The compression ratio is 10 (that is, the ratio of the total cylinder volume to the combustion chamber volume).

∴ The original inhalation gas is the exhaust gas 109,

∴ gasoline accounts for 19,

Mass of gasoline consumed by the engine: m = 7.2×10-4 kg×19×1.8 =1.44×10-4 kg.

Heat released by complete combustion of gasoline in 1 min:

q = 2500× 1.44× 10-4kg×4.6× 107j/kg = 1.656× 107j，

The thermal efficiency of this engine at maximum power.

η=W Useful q×100% = 4.14×106j10.656×107j100% = 25%.

Answer: (1) The power battery of this car can store 1.08× 108J at most;

(2)a: At this time, the resistance of the automobile is1.8×103n; B: The electric energy obtained by the power storage battery is1.5×105j;

(3) At this time, the efficiency of this gasoline engine is 25%.