Which formula gives head loss for this problem (f given)?

Darcy-Weisbach h_f = 8fLQ^2/(pi^2 g D^5) = kQ^2

What condition fixes the split between BCD and BFD?

Head loss B-C-D equals head loss B-F-D

2076 Asoj · 8 marks

Velocity, head loss, discharge in loop pipes BC, CD, BF, FD

Short answer

Using Hardy-Cross, the flows are: BC 2.025 m³/s (v=1.79 m/s, hf=0.408 m); CD 1.025 m³/s (2.04 m/s, 0.794 m); BF 1.975 m³/s (2.51 m/s, 0.967 m); FD 0.975 m³/s (1.24 m/s, 0.236 m).

Determine velocity, head loss and discharge in the pipes BC, CD, BF and FD. If diameter of pipes BC, CD, BF and FD are 1200 mm, 800 mm, 1000 mm, and 1000 mm respectively. Assume length of pipes are 100 m and coefficient of friction, f = 0.03.

[Figure: 4 m $^3$ /s enters at A and flows to B; the loop B-C-D-F has C and F as the upper and lower corners; outflows 1 m $^3$ /s at C, 1 m $^3$ /s at F and 2 m $^3$ /s at D towards E.]

The 4 m³/s entering at A reaches B and splits between the upper path B–C–D and the lower path B–F–D; the split is found by Hardy-Cross.

Formulas

Darcy-Weisbach: $h_f = \dfrac{8 f L Q^2}{\pi^2 g D^5} = kQ^2$ , velocity $v = Q/(\pi D^2/4)$ . Hardy-Cross correction for the loop:

\Delta Q = -\frac{\sum kQ|Q|}{2\sum k|Q|}

Flow distribution

Let BC carry $x$ m³/s. Continuity gives CD = $x - 1$ (1 m³/s leaves at C), BF = $4 - x$ and FD = $3 - x$ (1 m³/s leaves at F). At D: CD + FD = 2 m³/s ✓. The loop condition is head loss B→C→D = head loss B→F→D. Trial: BC = 2, CD = 1, BF = 2, FD = 1 (the inflow at B is split equally between the two branches).

Constants k for each pipe ( $f = 0.03$ )

Pipe	L (m)	D (m)	k = 8fL/(π²gD⁵)
BC	100	1.2	0.0996
CD	100	0.8	0.7565
BF	100	1.0	0.2479
FD	100	1.0	0.2479

Hardy-Cross iterations (clockwise positive; flows in m³/s)

Iteration	ΔQ	BC	CD	BF	FD
1	+0.0249	2.0249	1.0249	1.9751	0.9751
2	-0.0001	2.0248	1.0248	1.9752	0.9752
3	-0.0000	2.0248	1.0248	1.9752	0.9752

Results

Pipe	Discharge (m³/s)	Velocity (m/s)	Head loss (m)
BC	2.0248	1.790	0.4084
CD	1.0248	2.039	0.7944
BF	1.9752	2.515	0.9671
FD	0.9752	1.242	0.2357

Check: upper path = 0.4084 + 0.7944 = 1.2028 m; lower path = 0.9671 + 0.2357 = 1.2028 m ✓

Answer: BC = 2.025 m³/s (v = 1.79 m/s, h_f = 0.408 m); CD = 1.025 m³/s (2.04 m/s, 0.794 m); BF = 1.975 m³/s (2.51 m/s, 0.967 m); FD = 0.975 m³/s (1.24 m/s, 0.236 m).

✓ Answer checked by Claude Opus 5.5 (high effort)

Velocity, head loss, discharge in pipes BC, CD, BF, FD (0.1 m) · Water Supply Engineering, chapter 7
Head loss and velocity in loop pipes AB, BC, AC, CD · Water Supply Engineering, chapter 7

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