## Simulink - Vehicle Acceleration 5

1. Introduction
In this section we will develop the model a little differently using engine rpm and throttle opening as the inputs to a 2d look up table and this table has been given very simplistic parameters to give as the output the air mass flow rate into the engine.
The initial assumption for this model is that the engine has a peak output of 120kW at 5000 rpm and assuming the efficiency is 30%, this means that the fuel burnt must provide 120/0.3 = 400kW.
400 kW requires a fuel mass flow rate of 400000/46000000 = 0.00869 kg/s
This requires an air flow rate of 0.00869 x 14.7 = 0.128 kg/s

2. The Model
The Matlab Simulink model is shown in the diagram below:

The entries in the 2d look up table are forclosed and fully open throttle and for 0 and 5000 rpm. Values entered were [0 1], [0 5000] with outputs estimated to be [0.001 0.005; 0.005 0.128]

The first gain block assumes stoichiometric combustion and multiplies the air flow by 0.068 to give the fuel mass flow rate.
The next gain block, Gain 1, multiplies the fuel mass flow rate by 46000000 to give a value for the power theoretically available from burning fuel at the rate it is being used. This needs to be converted to torque, which can normally be done by dividing by the engine angular velocity. However, this does not work when the engine is starting from rest as a division by zero stops the simulation and at low engine speeds the torque calculated is unrealistically high. To get over this the constant is added to the value of the engine angular velocity that is fed back, before it is inverted with the 'Math Function'. Choosing an appropriate constant is tricky. It is probably reasonable to choose the approximate tick over speed, say 900 rpm (94.2 radians/s) However putting this in the model gives a high torque peak, which although it quickly drops down, is some twice the maximum that the engine can develop. For this reason a value equivalent to letting the clutch out at 1800 rpm (188.5 rad/s) has been used.

Gain 2 contains a value of 0.3, to represent the overall efficiency of the engine.
Gain 3 contains a value of 10, the assumed gear ratio.
Gain 4 contains the value of 1/wheel radius to give the tractive force provided at the tyre periphery.
The summing point subtracts the air resistance computed by the inner feed back loop, the output is then the net force available to accelerate the vehicle.
Gain 5 is 1/equivalent mass of the vehicle, the output is the acceleration. This is fed into the integrator and the output from this is the vehicle velocity.

3. Results
This model shows a 0 - 60 time of about 6 seconds (the engine is by this time theoretically running above its maximum capability as there is only one gear available and it is too low to reach 60 mph).

David J Grieve, 4th November 2004