Reynolds Number Calculate the Reynolds number for the following situations. (a) A Winchester.30–30 bullet leaves the muzzle of a rifle at 2400 ft/s. (b) Freshwater flows through a 1-cm-diameter pipe at the average speed of 0.5 m/s. (c) SAE 30 oil flows under the same conditions as (b). (d) A fast

Question

Reynolds Number

Calculate the Reynolds number for the following situations. (a) A Winchester.30–30 bullet leaves the muzzle of a rifle at 2400 ft/s. (b) Freshwater flows through a 1-cm-diameter pipe at the average speed of 0.5 m/s.(c) SAE 30 oil flows under the same conditions as (b). (d) A fast attack submarine with hull diameter of 33 ft cruises at 15 knots. One knot is equivalent to 1.152 mph.

Approach
To calculate the Reynolds number for each situation, we apply the definition in Equation (6.6),

[latex]R{e}=\frac{\rho vl}{\mu } [/latex] (6.6)

making sure that the numerical quantities are dimensionally consistent. The density and viscosity values for air, freshwater, oil, and seawater are listed in Table 6.1.

Table 6.1 Density and Viscosity Values for Several Gases and Liquids at Room Temperature and Pressure

	Density,ρ		Viscosity, μ
Fluid	[latex]kg/m^3[/latex]	[latex]slug/ft^3[/latex]	kg/(m . s)	slug/(ft . s)
Air	1.2	[latex]2.33×10^{-3}[/latex]	[latex]1.8×10^{-5}[/latex]	[latex]3.8×10^{-7}[/latex]
Helium	0.182	[latex]5.53×10^{-4}[/latex]	[latex]1.9×10^{-5}[/latex]	[latex]4.1×10^{-7}[/latex]
Freshwater	1000	1.94	[latex]1.0×10^{-3}[/latex]	[latex]2.1×10^{-5}[/latex]
Seawater	1026	1.99	[latex]1.2×10^{-3}[/latex]	[latex]2.5×10^{-5}[/latex]
Gasoline	680	1.32	[latex]2.9×10^{-4}[/latex]	[latex]6.1×10^{-6}[/latex]
SAE 30 oil	917	1.78	0.26	[latex]5.4×10^{-3}[/latex]

Accepted Answer

(a) The bullet’s diameter is 0.3 in., which we convert to the consistent dimension of ft (Table 3.5):

Table 3.5 Certain Derived Units in the USCS

Quantity	Derived Unit	Abbreviation	Definition
Length	mil	mil	1 mil = 0.001 in.
	inch	in.	1 in. = 0.0833 ft
	mile	mi	1 mi = 5280 ft
Volume	gallon	gal	[latex] 1 gal = 0.1337 ft^3 [/latex]
Mass	slug	slug	[latex]1 slug = 1 (lb . S^2)/ft[/latex]
Mass	pound-mass	lbm	[latex] 1 lbm = 3.1081 imes 10^{-2}(lb s^{2})/ft [/latex]
Force	ounce	oz	1 oz = 0.0625 lb
Force	ton	ton	1 ton = 2000 lb
Torque, or moment of a force	foot-pound	ft · lb	—
Pressure or stress	[latex]pound/inch^2[/latex]	psi	[latex]1 psi = 1 lb/in^2[/latex]
Energy, work, or heat	foot-pound	ft · lb	—
Energy, work, or heat	British thermal unit	Btu	1 Btu = 778.2 ft . lb
Power	horsepower	hp	1 hp = 550 (ft . lb)/s
Temperature	degree Fahrenheit	[latex]°F[/latex]	[latex]°F = °R-459.67[/latex]
Although a change in temperature of [latex] 1°[/latex] Rankine also equals a change of 1 [latex]°F[/latex], numerical , numerical values are converted using the formula.

[latex]d=(0.3in.)\left(0.0833\frac{ft}{in.} ight) [/latex]

[latex]=0.025(\cancel{in.})\left(\frac{ft}{\cancel{in.}} ight)[/latex]

[latex]=0.025ft[/latex]

The Reynolds number becomes

[latex]R{e}=\frac{(2.33 imes 10^{-3}slug/ft^3)(2400 ft/s)(0.025ft)}{3.8 imes 10^{-7}slug/(ft.s)} \longleftarrow \left[R{e}=\frac{ ho v l}{\mu } ight] [/latex]

[latex]=3.679 imes 10^5\left(\frac{\cancel{slug}}{\cancel{ft^3}} ight)\left(\frac{\cancel{ft}}{\cancel{s}} ight)(\cancel{ft})\left(\frac{\cancel{ft}.\cancel{s}}{\cancel{slug}} ight) [/latex]

[latex]= 3.679 imes 10^5[/latex]

(b) With the numerical values in the SI, the Reynolds number for water
flowing in the pipe is

[latex]R{e}=\frac{(1000kg/m^3)(0.5m/s)(0.01m)}{1.0 imes 10^{-3}kg/(m.s)} \longleftarrow \left[R{e}=\frac{ ho vl}{\mu } ight] [/latex]

[latex]=5000\left(\frac{\cancel{kg}}{\cancel{m^3}} ight)\left(\frac{\cancel{m}}{\cancel{s}} ight)(\cancel{m})\left(\frac{\cancel{m}.\cancel{s}}{\cancel{kg}} ight) [/latex]

[latex]=5000[/latex]

(c) When SAE 30 oil is pumped through the pipe instead of water, the
Reynolds number is reduced to

[latex]R{e}=\frac{(917kg/m^3)(0.5m/s)(0.01m)}{0.26kg/(m.s)} \longleftarrow \left[R{e}=\frac{ ho vl}{\mu } ight] [/latex]

[latex]=17.63\left(\frac{\cancel{kg}}{\cancel{m^3}} ight)\left(\frac{\cancel{m}}{\cancel{s}} ight)(\cancel{m})\left(\frac{\cancel{m}.\cancel{s}}{\cancel{kg}} ight) [/latex]

[latex]=17.63[/latex]

(d) We need to convert the submarine’s speed into consistent dimensions. The first step is to convert from knots to mph:

[latex]v = (15 knots)(1.152\frac{mph}{knot} )[/latex]

[latex]=17.28(\cancel{knot})\left(\frac{mph}{\cancel{knot}} ight) [/latex]

[latex]=17.28 mph [/latex]

and then from mph to ft/s:

[latex]v = (17.28 \frac{mi}{h} )(5280\frac{ft}{mi} )(\frac{1}{3600}\frac{h}{s} )[/latex]

[latex]=25.34\left(\frac{\cancel{mi}}{\cancel{h}} ight)\left(\frac{ft}{\cancel{mi}} ight)\left(\frac{\cancel{h}}{s} ight) [/latex]

[latex]=25.34 \frac{ft}{s} [/latex]

The boat’s Reynolds number becomes

[latex]R{e}=\frac{(1.99 slug/ft^3)(25.34 ft/s)(33ft)}{2.5 imes 10^{-5}slug/(ft.s)} \longleftarrow \left[R{e}=\frac{ ho v l}{\mu } ight] [/latex]

[latex]=6.657 imes 10^7\left(\frac{\cancel{slug}}{\cancel{ft^3}} ight)\left(\frac{\cancel{ft}}{\cancel{s}} ight)\cancel{ft}\left(\frac{\cancel{ft}.\cancel{s}}{\cancel{slug}} ight) [/latex]

[latex]= 6.657 imes 10^7[/latex]

Discussion
As expected for a dimensionless quantity, the units of the numerator in Re exactly cancel those of the denominator. Laboratory measurements have shown that fluid flows through a pipe in a laminar pattern when Re is smaller than approximately 2000. The flow is turbulent for larger values of Re. In (b), we would expect the flow of water in the pipe to be turbulent, while the flow in (c) would certainly be laminar because the oil is so much more viscous than water.

[latex]Re_{bullet} = 3.679 imes 10^5[/latex]
[latex]Re_{water pipeline}= 5000[/latex]
[latex]Re_{oil pipeline} = 17.63[/latex]
[latex]Re_{submarine} = 6.657 imes 10^7[/latex]

Question 6.5: Reynolds Number Calculate the Reynolds number for the follow...

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