How is f found for the hose?

Colebrook-White with Re = 24000 and e/D = 0.01

f = 64/Re since the velocity is low

From the relative roughness alone with no Re

What does the '1' in (1 + K + fL/D) represent?

Exit kinetic energy V^2/2g at the free outlet

Exit loss coefficient of the pool

2071 Bhadra · 5 marks

Pool drained by hose, find water depth

Short answer

Colebrook friction factor f=0.0403. Energy equation including entrance loss, friction, and velocity head gives d + 3 = 4.543 m, so water depth above hose inlet d = 1.54 m.

Small swimming pool is drained with velocity of 1.2 m/sec using a pipe with hose diameter 20 mm, length 30 m, and absolute roughness e = 0.2 mm. Find the water depth d at instant shown in figure below considering minor head loss coefficient at entrance K = 0.5. [Figure: pool with water depth d above the hose inlet; the hose runs over the pool rim and discharges 3 m below the inlet level]

Data. $V = 1.2$ m/s, $D = 0.02$ m, $L = 30$ m, $e = 0.2$ mm, entrance loss $K = 0.5$ , water $\nu = 1.0\times10^{-6}$ m²/s. The hose outlet (open to the atmosphere) is 3 m below the hose inlet level. Let $d$ be the depth of water above the hose inlet.

Friction factor

Re = \frac{VD}{\nu} = \frac{1.2(0.02)}{10^{-6}} = 24000, \qquad \frac{e}{D} = 0.010

Colebrook-White (iteration) gives $f = 0.0403$ .

Energy equation

Between the pool surface (atmospheric, velocity zero, elevation $d + 3$ above the outlet) and the free outlet (velocity $V$ ):

d + 3 = \frac{V^2}{2g}\left(1 + K + f\frac{L}{D}\right), \qquad \frac{V^2}{2g} = \frac{1.2^2}{2(9.81)} = 0.0734\ \text{m}

f\frac{L}{D} = 0.0403\left(\frac{30}{0.02}\right) = 60.40

d + 3 = 0.0734\left(1 + 0.5 + 60.40\right) = 0.0734(61.90) = 4.543\ \text{m}

d = 4.543 - 3 = 1.54\ \text{m}

Answer: The water depth above the hose inlet is about $1.54$ m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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