Question 6.2: A 1/2-in-diameter water pipe is 60 ft long and delivers wate...

A \frac{1}{2}-in-diameter water pipe is 60 ft long and delivers water at 5 gal/min at 20°C. What fraction of this pipe is taken up by the entrance region?

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Convert

Q = (5 gal/min)\frac{0.00223  ft^3/s}{1  gal/min} = 0.0111  ft^3/s

The average velocity is

V = \frac{Q}{A} = \frac{0.0111  ft^3/s}{(\pi/4) [(\frac{1}{2}/12) ft]^2} = 8.17 ft/s

From Table 1.4 read for water \nu = 1.01 \times 10^{-6}  m^2/s = 1.09 \times 10^{-5}  ft^2/s. Then the pipe Reynolds number is

Re_d = \frac{Vd}{\nu} = \frac{(8.17  ft/s)[(\frac{1}{2}/12) ft]}{1.09 \times 10^{-5}  ft^2/s} = 31,300
Table 1.4 Viscosity and Kinematic Viscosity of Eight Fluids at 1 atm and 20°C
Fluid µ,
kg/(m·s)^†
Ratio
µ/µ(H_2)
ρ,
kg/m^3
v, m^2/s^† Ratio
v/v(Hg)
Hydrogen 9.0 E-6 1.0 0.084 1.05 E-4 910
Air 1.8 E-5 2.1 1.20 1.50 E-5 130
Gasoline 2.9 E-4 33 680 4.22 E-7 3.7
Water 1.0 E-3 114 998 1.01 E-6 8.7
Ethyl alcohol 1.2 E-3 135 789 1.52 E-6 13
Mercury 1.5 E-3 170 13,550 1.16 E-7 1.0
SAE 30 oil 0.29 33,000 891 3.25 E-4 2,850
Glycerin 1.5 170,000 1,260 1.18 E-3 10,300

^†1 kg/(m·s) = 0.0209 slug/(ft·s); 1 m^2/s = 10.76 ft^2/s.

This is greater than 4000; hence the flow is fully turbulent, and Eq. (6.6) applies for entrance length:

\frac{L_e}{d} \approx 1.6 Re_d^{1/4}       for       Re_d \leq 10^7                   (6.6)

\frac{L_e}{d} \approx 1.6 Re_d^{1/4} = (1.6)(31,300)^{1/4} = 21

The actual pipe has L/d = (60 ft)/[(\frac{1}{2}/12)ft] = 1440. Hence the entrance region takes up the fraction

\frac{L_e}{L} = \frac{21}{1440} = 0.015 = 1.5%

This is a very small percentage, so that we can reasonably treat this pipe flow as essentially fully developed.

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