Hardy-Cross method is used for the solution of the pipe network flow distribution using the eq as shown below. where, CW - clock-wise (taken +ve), CCW- counter clock-wise (laken -ve). As a 1st guess, you have assumed the following flow distribution fulfiling the continuity condition. 14er In the given loop, if AQ = - 0.1 m'is obtained, what will be the next flow distribution? 14 O None of the distribution are correct

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Chapter2: Loads On Structures
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Hardy-Cross method is used for the solution of the pipe network flow distribution using the eq as shown below.
where, CW - clock-wise (taken +ve), CCW- counter clock-wise (laken -ve). As a 1st guess, you have assumed the following flow distribution fulfiling the
continuity condition.
14er
In the given loop, if AQ = - 0.1 m'is obtained, what will be the next flow distribution?
14
O None of the distribution are correct
Transcribed Image Text:Hardy-Cross method is used for the solution of the pipe network flow distribution using the eq as shown below. where, CW - clock-wise (taken +ve), CCW- counter clock-wise (laken -ve). As a 1st guess, you have assumed the following flow distribution fulfiling the continuity condition. 14er In the given loop, if AQ = - 0.1 m'is obtained, what will be the next flow distribution? 14 O None of the distribution are correct
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