Q1 = cB (H2 − H1) √ {g (H2 + H1) }.
From (8) the discharge is
Q2 = 2⁄3 cB √2g (H23/2 − H13/2).
Hence, for the same value of c in the two cases,
Q2/Q1 = 2⁄3 (H23/2 − H13/2) / [ (H2 − H1) √ { (H2 + H1)/2} ].
Let H1/H2 = σ, then
Q2/Q1 = 0.9427 (1 − σ3/2) / {1 − σ √ (1 + σ) }.
(9)
If H1 varies from 0 to ∞, σ( = H1/H2) varies from 0 to 1. The following table gives values of the two estimates of the discharge for different values of σ:—
| H1/H2 = σ. | Q2/Q1. | H1/H2 = σ. | Q2/Q1. |
| 0.0 | .943 | 0.8 | .999 |
| 0.2 | .979 | 0.9 | .999 |
| 0.5 | .995 | 1.0 | 1.000 |
| 0.7 | .998 |