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The Graphs in Fig 3A4-6. What does the vertical part of the characteristic mean?

There is a problem with these graphs. What is it and how should it be corrected?
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  • David


    t = t1 + ( (t2 - t1) * ((I2 - I1) / (I2 - I1)) )  should have been t = t1 + ( (t2 - t1) * ((Ix - I1) / (I2 - I1)) ) 


    The error is that I put I2 where it should have been Ix.  The result is now the same as my application produces.  This application also graphs all of the equations and by inspection the given result is probably correct.


    Putting the values in for the 315A fuse


    t = t1 + ( (t2 - t1) * ((Ix - I1) / (I2 - I1)) ) 


    5 + (( 1 - 5) * ((2500 - 2000)/(3050 - 2000)) )


    5 + ( (-4)*((500)/(1050)) ) = 3.1 seconds as required


    t1 = 5 sec

    t2 = 1 sec

    I2 = 3050A (current that gives 1 sec)

    I1 = 2000A (current that gives 5 sec)

    Ix = 2500A (current used for my example)


    IMO for most practical engineering use this is as accurate as it needs to be.  Cables only come in a limited number of sizes!


    Regards

    Geoff Blackwell
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  • David


    t = t1 + ( (t2 - t1) * ((I2 - I1) / (I2 - I1)) )  should have been t = t1 + ( (t2 - t1) * ((Ix - I1) / (I2 - I1)) ) 


    The error is that I put I2 where it should have been Ix.  The result is now the same as my application produces.  This application also graphs all of the equations and by inspection the given result is probably correct.


    Putting the values in for the 315A fuse


    t = t1 + ( (t2 - t1) * ((Ix - I1) / (I2 - I1)) ) 


    5 + (( 1 - 5) * ((2500 - 2000)/(3050 - 2000)) )


    5 + ( (-4)*((500)/(1050)) ) = 3.1 seconds as required


    t1 = 5 sec

    t2 = 1 sec

    I2 = 3050A (current that gives 1 sec)

    I1 = 2000A (current that gives 5 sec)

    Ix = 2500A (current used for my example)


    IMO for most practical engineering use this is as accurate as it needs to be.  Cables only come in a limited number of sizes!


    Regards

    Geoff Blackwell
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