Verify suggested discharge distribution (Hardy Cross)

Data. Suggested discharges (units): B to A 58, B to C 42, A to C 20, A to D 18, C to D 32 (these satisfy continuity at every node with 100 in at B, 20 out at A, 30 out at C, 50 out at D). $r$: AB 2, BC 4, AC 1, AD 5, CD 1; $h_f = rQ^2$ ($n = 2$).

Loop I = A-C-B-A (AC +, CB $-$, BA +). Loop II = A-D-C-A (AD +, DC $-$, CA $-$).

Test of the suggested distribution (first iteration)

Loop I: $\sum rQ|Q| = 1(20)^2 - 4(42)^2 + 2(58)^2 = 400 - 7056 + 6728 = 72$, $2\sum r|Q| = 2(20 + 168 + 116) = 608$, so $\Delta Q_I = -\dfrac{72}{608} = -0.118$.

Loop II: $\sum rQ|Q| = 5(18)^2 - 1(32)^2 - 1(20)^2 = 1620 - 1024 - 400 = 196$, $2\sum r|Q| = 2(90 + 32 + 20) = 284$, so $\Delta Q_{II} = -\dfrac{196}{284} = -0.690$.

The loop head-loss sums are not zero (72 and 196), so strictly the suggested distribution is not exact, although the corrections are small (under 1.2% of the pipe discharges). Continuing the Hardy-Cross iterations:

| Iter | $\sum rQ|Q|$ (I) | $2\sum r|Q|$ (I) | $\Delta Q_I$ | $\sum rQ|Q|$ (II) | $2\sum r|Q|$ (II) | $\Delta Q_{II}$ | |---|---|---|---|---|---|---| | 1 | 72.00 | 608.00 | -0.118 | 196.00 | 284.00 | -0.690 | | 2 | 27.90 | 609.62 | -0.046 | 6.32 | 279.62 | -0.023 | | 3 | 0.93 | 609.75 | -0.002 | 1.88 | 279.40 | -0.007 |

Corrected distribution

(Checks: $57.83 + 42.17 = 100$; at D, $17.28 + 32.72 = 50$.) The suggested values are therefore correct to within about 4%; the correction mainly shifts about 0.7 unit of flow from pipe AD to the route A-C-D.

Pressure at D

Piezometric head at B: $H_B = 50 + 40 = 90$ m. Head loss from B to D via A: $h_{BD} = h_{BA} + h_{AD} = 2(57.83)^2 + 5(17.28)^2 = 6689.5 + 1493.0 = 8182.5$ (the same through C: $7111.9 + 1070.6$). Then

$$H_D = H_B - h_{BD}, \qquad \frac{p_D}{\gamma} = H_D - z_D = 90 - h_{BD} - 40 = 50 - h_{BD}$$

With $r$ and $Q$ used exactly as given, $h_{BD} = 8182.5$, so $p_D/\gamma = 50 - 8182.5 = -8132.5$, which is physically impossible. This shows that, with the "units" of discharge in the data, $rQ^2$ does not come out in metres of head. The method is still the one above: the pressure head at D is $50 - h_{BD}$ (with $h_{BD}$ in metres), and the pressure is $p_D = \gamma\,(50 - h_{BD})$.

Answer: The suggested distribution is nearly, but not exactly, satisfactory; the proper values are BA = 57.8, BC = 42.2, AC = 20.6, AD = 17.3, CD = 32.7. Pressure head at D: $p_D/\gamma = 50 - h_{BD}$; with the data taken literally, $h_{BD} = 8182.5$ and the result is negative ($-8132.5$), so the given data do not give a physically meaningful pressure in metres.