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Using corrected value for hardy cross method Watch

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    (Original post by wilson dang)
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    Spreadsheet attached. Ignore V3. You should be able to figure it out from this. If not I'm not sure I can help...
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  1. File Type: xlsx hardycross.xlsx (28.9 KB, 39 views)
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    (Original post by natninja)
    Spreadsheet attached. Ignore V3. You should be able to figure it out from this. If not I'm not sure I can help...
    Did you upload the wrong file ? i didnt see it it is related to
    http://i.imgur.com/RWZ5odq.png this case , can you change the RQQ to k(Q^2) , i am not comfortable with RQQ and dir ? what does dir mean?
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    (Original post by wilson dang)
    Did you upload the wrong file ? i didnt see it it is related to
    http://i.imgur.com/RWZ5odq.png this case , can you change the RQQ to k(Q^2) , i am not comfortable with RQQ and dir ? what does dir mean?
    It's related to the other example. Ignore dir, that was for my own use to decide whether to add or subtract. I'm afraid I have my own work to do and have already spent too much time on this.
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    (Original post by natninja)
    It's related to the other example. Ignore dir, that was for my own use to decide whether to add or subtract. I'm afraid I have my own work to do and have already spent too much time on this.
    which example do you refer to ?
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    (Original post by wilson dang)
    which example do you refer to ?
    http://nptel.ac.in/courses/112104118...examp_36_1.htm
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    (Original post by natninja)
    Spreadsheet attached. Ignore V3. You should be able to figure it out from this. If not I'm not sure I can help...
    you have combined 2 loop together ? can you explain what do you mean by v1 and v2 ?
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    (Original post by natninja)
    Just use the final values from your previous iteration.

    In fact, just use uncorrected values. It will make your life easier.
    so ,, based on your working , if we split up the network into 2 loops , then we would 2 values of flow rate at junction BD , right ? the first is the corrected value in the first loop which is to be used in the first iteration in second loop , then we would also get the corrected value from the second loop ,
    so we should get the corrected value from the second loop to do the ieration for the second trial at first loop ?
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    (Original post by wilson dang)
    so ,, based on your working , if we split up the network into 2 loops , then we would 2 values of flow rate at junction BD , right ? the first is the corrected value in the first loop which is to be used in the first iteration in second loop , then we would also get the corrected value from the second loop ,
    so we should get the corrected value from the second loop to do the ieration for the second trial at first loop ?
    If I'm following you right then yes, that is correct.
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    (Original post by natninja)
    If I'm following you right then yes, that is correct.
    is it possible to have the value of delta Q become larger and then become smaller again in the iteration ?
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    (Original post by wilson dang)
    is it possible to have the value of delta Q become larger and then become smaller again in the iteration ?
    Yes, it can converge in an oscillatory manner. As long as it doesn't continue to diverge it is fine.
 
 
 
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