Solving the Hardy Cross Problem

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Homework Statement


in this problem , we know that Q (flow rate ) has power of 2 , hf = k(Q^2) , shouldn't the hf = k(Q^2) has positive value ?

Homework Equations

The Attempt at a Solution


since (554)(-40^2) will become positive , same thing goes to hf / Q , when the hf = positive , the hf / Q will become hf / Q divided by Q(has negative value ) , so the hf / Q will have negative value , am i right ?
 
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  • Diagram of a square pipe loop with corners labelled A, B, C and D, showing inflows of 100 L/s at A and 40 L/s at D, and outflows of 20 L/s at B and 40 L/s at C.
    1.PNG
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  • Two slides: left lists Darcy-Weisbach head loss steps and the formula h_f = λ(L/D)(V²/2g) with K = 10λ... (K formula); right shows calculations H_L = KLQ² with K = 554, and a table of pipes AB, BC, CD, AD with flow Q (L/s), head loss H_L (m) and H_L/Q, the AD row with Q = -40 and H_L = -0.886 circled in red.
    2.PNG
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