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how can i set the load condition to 20% of full load

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i have to design a simulink model for a shunt DC motor under the given load condition
where I am asked to
set the simulation time to 10 sec
set the load condition to 20% of full load
what is meant by 20% of full load?
please help me out with this
TIA

Accepted Answer

Pratyush Roy
Pratyush Roy on 11 May 2021
Hi Ayesha,
Every DC Machine block in Simulink comes with several preset models. Each of these models specify certain values for rated armature voltage, output power, rotational velocity and field voltage. Consider the image below:
This configuration suggests a DC Motor with output power of 5 HP and rotational velocity of 1750 rpm. It also considers armature voltage of 240 V and field voltage of 300 V. From these rated values we can calculate the rated load torque using the following script:
Nrpm = 1750;
w = 2*pi*Nrpm/60; %w = velocity in rad/s
P = 5;
P_watts = 746*5; %P_watt = power in watts
T_L = P_watts/w;
This is the full load torque. When we say that 20% of full load is available, this means the external load torque is 20% of the the calculated rated load torque T_L. We can introduce this torque to the Simulink model through a step block as illustrated below:
Here we can change the final value of the step function to the 20% of load torque and then run the simulation.
The attached Simulink(.slx) file might be helpful to understand the working of the DC shunt motor.
Hope this helps!
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