for the successive portions, and let l, d, v and i be corresponding quantities for the equivalent uniform main B. The total loss of head in A due to friction is
| h = i1l1 + i2l2 + ... = ζ (v12 · 4l1/2gd1) + ζ (v22 · 4l2/2gd2) + ... |
and in the uniform main
il = ζ (v2 · 4l/2gd).
If the mains are equivalent, as defined above,
ζ (v2 · 4l/2gd) = ζ (v12 · 4l1/2gd1) + ζ (v22 · 4l2/2gd2) + ...
But, since the discharge is the same for all portions,
| 1⁄4πd2v = 1⁄4πd12v1 = 1⁄4πd22v2 = ... v1 = vd2/d12; v2 = vd2/d22 ... |
Also suppose that ζ may be treated as constant for all the pipes. Then
| l/d = (d4/d14) (l1/d1) + (d4/d24) (l2/d2) + ... l = (d5/d15) l1 + (d5/d25) l2 + ... |