In a proposed design for an electric automobile, the shaft of a four-pole three-phase induction motor is connected directly to the drive axle; in other words, there is no gear train. The outside diameter of the tires is 20 inches. Instead of a transmission, an electronic converter produces variable-frequency three-phase ac from a 48-V battery. Assuming negligible slip, find the range of frequencies needed for speeds ranging from 5 to
70mph
. The vehicle, including batteries and occupants, has a mass of
1000 kg
. The power efficiency of the dc-to-ac converter is 85 percent, and the power efficiency of the motor is 89 percent. a. Find the current taken from the battery as a function of time while accelerating from 0 to 40 mph uniformly (i.e., acceleration is constant) in 10 seconds. Neglect wind load and road friction. b. Repeat assuming that the vehicle is accelerated with constant power.

The current taken from the battery as a function of time while accelerating from 0 to 40 mph uniformly (i.e., acceleration is constant) in 10 seconds is; 88t A

How to find the current in the car battery?

From the problem description, we are given;

Outside diameter of tire; d_t =20 inches,

Thus; radius; r_t = 10 inches,

DC Voltage; V_DC−in = 48 V

Initial velocity; v₁ = 5 mph

Final velocity; v₂ = 70 mph

Power efficiency of converter; η_dc = 85%

Power efficiency of motor; η_m = 89%

mass; m = 1000 kg

Now, we want to find the current taken from the battery as a function of time while accelerating from 0 to 40 mph uniformly (i.e., acceleration is constant) in 10 seconds.

The considered speeds will be;

Initial speed; v₁ = 0 mph = 0 m/s

Final speed; v₂ = 40 mph = 17.88 m/s

Thus;

Acceleration is gotten from the formula;

a = (v₂ - v₁)/t

a = (17.88 - 0)/10

a = 1.788 m/s²

The force that the motor has to develop for the given acceleration is;

F = ma

F = 1000 * 1.788

F = 1788 N

The output power would be generated from the formula;

P_out = F * a * t

P_out = 1788 * 1.788

P_out = 3196.944t W

The formula for the power input is;

P_in = P_out/(η_dc * η_m)

P_in = 3196.944t/(0.85 * 0.89)

P_in = 4226t W

Formula for the current taken from the battery is;

I_dc_in = P_in/(V_dc_in)

Thus;

I_dc_in = 4226t/48

I_dc_in = 88t A

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