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MechanicalFluid MechanicsmediumFE ExamCourseworkPE / Advanced

Friction Head Loss in a Pipe

Water flows at V=2.0 m/sV = 2.0\ \text{m/s} through a straight pipe of internal diameter D=50 mmD = 50\ \text{mm} and length L=100 mL = 100\ \text{m}.

The Darcy friction factor is f=0.02f = 0.02. Take g=9.81 m/s2g = 9.81\ \text{m/s}^2.

Find the head loss due to friction over the 100 m length.

Given
V2 m/sMean velocity
D50 mmInternal diameter
L100 mPipe length
f0.02 -Darcy friction factor
Hint 1

Darcy-Weisbach: h_f = f(L/D)(V²/2g).

Hint 2

L/D must be dimensionless — check both are in the same unit before dividing.

Hint 3

L/D = 100/0.05 = 2000, not 2.

Worked solution — try the problem first

The Darcy-Weisbach equation:

hf=f LD V22gh_f = f\,\frac{L}{D}\,\frac{V^2}{2g}

Convert the diameter to metres so it is consistent with LL:

D=50 mm=0.050 mD = 50\ \text{mm} = 0.050\ \text{m}

Length-to-diameter ratio (dimensionless — a common place to lose a factor of 1000):

LD=1000.050=2000\frac{L}{D} = \frac{100}{0.050} = 2000

Velocity head:

V22g=2.022(9.81)=419.62=0.2039 m\frac{V^2}{2g} = \frac{2.0^2}{2(9.81)} = \frac{4}{19.62} = 0.2039\ \text{m}

Head loss:

hf=0.02×2000×0.2039=8.155 mh_f = 0.02 \times 2000 \times 0.2039 = 8.155\ \text{m}

hf=8.15 m of water\boxed{h_f = 8.15\ \text{m of water}}

Worth checking the flow regime. With ν≈1.0×10−6 m2/s\nu \approx 1.0\times10^{-6}\ \text{m}^2/\text{s}:

Re=VDν=2.0×0.0501.0×10−6=1.0×105Re = \frac{VD}{\nu} = \frac{2.0 \times 0.050}{1.0\times10^{-6}} = 1.0\times10^{5}

Turbulent, as the given f=0.02f = 0.02 implies. If ReRe had come out below 2300 we would have expected f=64/Ref = 64/Re instead.

Concepts:Darcy-WeisbachHead lossReynolds numberPipe flow

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