Science for technicians question? urgent help needed!?

Question

An electric trolley bus has the following specification:

1) Gross laden weight= 17Tonnes

2) Constant frictional force= 2Kn

3) Overhead line voltage= 600V

4) Motor and drive line efficiency=87%

For a fully laden trolleybus, calculate the following:

a) Acceleration at 16km/hr

b) The time taken to accelerate from 0 to 28Km/hr

c) The distance covered in accelerating 20 28 28km/hr

e) the electrical power supplied to the motor at 28Km/hr

f) the current drawn from the overhead power lines at 28kn/hr

Any one know how to do this question?

Answer 1

you are missing some info, like motor power, gear ratios, torque.

Answer 2

This is a cool multidisciplinary problem requiring setting up and solving an integration, and is a problem that few engineers could solve quickly, or even at all. It's quite a challenge!

This problem uses several basic formulae:

Power = Force * Velocity

Power = E * I = voltage * current

Force = Mass * Acceleration

Distance = Time * Avg Velocity

Acceleration = deltaV/deltaT

Velocity = deltaX/deltaT

To solve the first parts, you need the motor power, which you are missing.

Part a:

The motor power is used to overcome friction and to accelerate the trolley.

Combining:

Power to overcome friction is F = F*V = Ffriction * V

and power to overcome inertia (mass) = Faccel * Velocity,

Power = (Ffriction + Faccel) * Velocity

Assuming a 30kW (output after derating by efficiency)  motor as an example,

30kW/(16km/hr)*3600sec/hr = (2kN + Faccel)

So, Faccel = 4750N

Using F=ma,

4750N/17Mg = 0.28m/s^2.

Part b and c:

We normally use uniformly accelerated motion. This acceleration is a linear function of velocity (as derived above, but leave V as a function). So, solve for acceleration, then integrate. Hint: I believe integral of V^-1 = ln V. Then, integrate again for distance.

Part d: missing....

Part e:

So, at constant speed at 28km/hr,

output P = F*V = 2kN * 28km/hr * hr/3600 sec = 15.55kW

Considering efficiency,

Input power = 15.55/0.87 = 17.88kW

Part f:

At 600V, since power = E*I,

P/E = I = 29.8Amps