Aerial Refueling The KC-10 Extender tanker aircraft of the United States Air Force is used to refuel other planes in flight. The Extender can carry 365,000 lb of jet fuel, which can be transferred to another aircraft through a boom that temporarily connects the two planes.(a) Express the mass of

Question

Aerial Refueling

The KC-10 Extender tanker aircraft of the United States Air Force is used to refuel other planes in flight. The Extender can carry 365,000 lb of jet fuel, which can be transferred to another aircraft through a boom that temporarily connects the two planes.(a) Express the mass of the fuel in the units of slugs and lbm. (b) Express the mass and weight of the fuel in the SI.

Approach
We will calculate the mass m in terms of the fuel’s weight [latex] w[/latex] and the gravitational acceleration [latex]g = 32.2 ft/s^2[/latex].With [latex] w[/latex] expressed in pounds and [latex]g [/latex] having the units of [latex]ft/s^2[/latex],the expression [latex] w=mg[/latex] is dimensionally consistent when mass has the units of slugs. We will then convert from slugs to lbm using the conversion factor 1 slug 5 32.174 lbm from Equation (3.5).

1 slug = 32.174 lbm 1 lbm = 3.1081 × 10[latex]^{-2}[/latex] slugs (3.5)

In part (b),we convert the fuel’s mass in the USCS to the SI using the conversion factor 1 slug = 14.59 kg from Table 3.6.

Table 3.6 Conversion Factors between Certain Quantities in the USCS and SI

Quantity	Conversion
Length	[latex]1 in. = 25.4 mm[/latex]
	[latex]1 in. = 0.0254 m[/latex]
	[latex]1 ft = 0.3048 m[/latex]
	[latex]1 mi = 1.609 km[/latex]
	[latex]1 mm = 3.9370 ×10^{-2} in.[/latex]
	[latex]1 m = 39.37 in.[/latex]
	[latex]1 m = 3.2808 ft[/latex]
	[latex]1 km = 0.6214 mi[/latex]
Area	[latex]1 in^2 = 645.16 mm^2[/latex]
	[latex]1 ft^2 = 9.2903 × 10^{-2} m^2[/latex]
	[latex]1 mm^2 = 1.5500 × 10^{-3 }in^2[/latex]
	[latex]1 m^2 = 10.7639 ft^2[/latex]
Volume	[latex]1 ft^3 = 2.832 × 10^{-2} m^3[/latex]
	[latex]1 ft^3 = 28.32 L[/latex]
	[latex]1 gal = 3.7854 × 10^{-3} m^3[/latex]
	[latex]1 gal = 3.7854 L[/latex]
	[latex]1 m^3 = 35.32 ft^3[/latex]
	[latex]1 L = 3.532 × 10^{-2} ft^3[/latex]
	[latex]1 m^3 = 264.2 gal[/latex]
	[latex]1 L = 0.2642 gal[/latex]
Mass	[latex]1 slug = 14.5939 kg[/latex]
	[latex]1 lbm = 0.45359 kg[/latex]
	[latex]1 kg = 6.8522 × 10^{-2} slugs[/latex]
	[latex]1 kg = 2.2046 lbm[/latex]
Force	1 lb = 4.4482 N
Force	1 N = 0.22481 lb
Pressure or stress	[latex]1 psi = 6895 Pa[/latex]
	[latex]1 psi = 6.895 kPa[/latex]
	[latex]1 Pa = 1.450 × 10^{-4 }psi[/latex]
	[latex]1 kPa = 0.1450 psi[/latex]
Work, energy, or heat	1 ft · lb = 1.356 J
	1 Btu = 1055 J
	1 J = 0.7376 ft · lb
	[latex]1 J = 9.478 × 10^{-4} Btu[/latex]
Power	1 (ft · lb)/s = 1.356 W
	1 hp = 0.7457 kW
	1 W = 0.7376 (ft · lb)/s
	1 kW = 1.341 hp

Accepted Answer

(a) We first determine the mass of the fuel in the units of slugs:

[latex]m=\frac{3.65 imes 10^{5}lb }{32.2ft/s^{2} } \longleftarrow \left[m=\frac{w }{g} ight] [/latex]

[latex]=1.134 imes 10^{4}\frac{lb.s^{2} }{ft}[/latex]

[latex]=1.134 imes 10^{4} slugs[/latex]

In the last step, we used the definition [latex]1 slug = 1 (lb · s^{2})/ft[/latex]. Since an object that weighs 1 lb has a mass of 1 lbm, the mass of the fuel can alternatively be expressed as 365,000 lbm.
(b) We convert the mass quantity [latex]1.134 × 10^{4}[/latex] slugs into the units of kg in the SI:

[latex]m=(1.134 imes 10^4 slugs)\left(14.59\frac{Kg}{slug} ight)[/latex]

= [latex]1.655 imes 10^5( \cancel{slugs})(\frac{Kg}{\cancel{slugs}} )[/latex]

[latex]= 1.655 imes 10^5 kg [/latex]

Since the numerical value for mass has a large power-of-ten exponent, an SI prefix from Table 3.3

Table 3.3 Order-of-Magnitude Prefixes in the SI

Name Symbol Multiplicative Factor
tera	T	[latex]1,000,000,000,000 = 10^{12}[/latex]
giga	G	[latex]1,000,000,000 = 10^9[/latex]
mega	M	[latex]1,000,000 = 10^6[/latex]
kilo	k	[latex]1000 = 10^3[/latex]
hecto	h	[latex]100 = 10^2[/latex]
deca	da	[latex]10 = 10^1[/latex]
deci	d	[latex]0.1 = 10^{-1}[/latex]
centi	c	[latex]0.01 = 10^{-2}[/latex]
milli	m	[latex]0.001 = 10^{-3}[/latex]
micro	[latex]μ[/latex]	[latex]0.000,001 = 10^{-6}[/latex]
nano	n	[latex]0.000,000,001 = 10^{-9}[/latex]
pico	p	[latex]0.000,000,000,001 = 10^{-12}[/latex]

should be applied. We first write [latex]m = 165.5 × 10^3 kg[/latex] so that the exponent is a multiple of 3. Since the “kilo” prefix
already implies a factor of [latex] 10^3 g[/latex], [latex]m = 165.5 × 10^6 g[/latex] or 165.5 Mg, where “M” denotes the prefix “mega.” The fuel’s weight in the SI is

[latex]w =(1.655 imes 10^5 kg)\left(9.81\frac{m}{s^{2} } ight) \longleftarrow \left[w =mg ight] [/latex]

[latex]=1.62 imes 10^{6}\frac{Kg.m}{s^{2} } [/latex]

[latex]=1.62 imes 10^{6}N [/latex]

Because this quantity also has a large power-of-ten exponent, we use the SI prefix “M” to condense the factor of 1 million. The fuel weighs 1.62 MN.

Discussion

To double-check the calculation of weight in the SI, we note that we can convert the fuel’s 365,000 lb weight directly to the dimensions of newtons. By using the conversion factor 1 lb = 4.448 N from Table 3.6, [latex]w[/latex] becomes

[latex]w=(3.65 imes 10^5 lb) \left(4.448\frac{N}{lb} ight)=1.62 imes 10^6 ( \cancel{lb} )(\frac{N}{\cancel{lb}} )= 1.62 imes 10^6 N[/latex]

or 1.62 MN, confirming our previous answer.

[latex]m=1.134 imes 10^4 slugs[/latex]
[latex]m=365,000 lbm [/latex]
[latex]m =165.5 Mg[/latex]
[latex]w =1.62 MN[/latex]

Quantity	Conversion
Length	$1 in. = 25.4 mm$
	$1 in. = 0.0254 m$
	$1 ft = 0.3048 m$
	$1 mi = 1.609 km$
	$1 mm = 3.9370 ×10^{-2} in.$
	$1 m = 39.37 in.$
	$1 m = 3.2808 ft$
	$1 km = 0.6214 mi$
Area	$1 in^2 = 645.16 mm^2$
	$1 ft^2 = 9.2903 × 10^{-2} m^2$
	$1 mm^2 = 1.5500 × 10^{-3 }in^2$
	$1 m^2 = 10.7639 ft^2$
Volume	$1 ft^3 = 2.832 × 10^{-2} m^3$
	$1 ft^3 = 28.32 L$
	$1 gal = 3.7854 × 10^{-3} m^3$
	$1 gal = 3.7854 L$
	$1 m^3 = 35.32 ft^3$
	$1 L = 3.532 × 10^{-2} ft^3$
	$1 m^3 = 264.2 gal$
	$1 L = 0.2642 gal$
Mass	$1 slug = 14.5939 kg$
	$1 lbm = 0.45359 kg$
	$1 kg = 6.8522 × 10^{-2} slugs$
	$1 kg = 2.2046 lbm$
Force	1 lb = 4.4482 N
Force	1 N = 0.22481 lb
Pressure or stress	$1 psi = 6895 Pa$
	$1 psi = 6.895 kPa$
	$1 Pa = 1.450 × 10^{-4 }psi$
	$1 kPa = 0.1450 psi$
Work, energy, or heat	1 ft · lb = 1.356 J
	1 Btu = 1055 J
	1 J = 0.7376 ft · lb
	$1 J = 9.478 × 10^{-4} Btu$
Power	1 (ft · lb)/s = 1.356 W
	1 hp = 0.7457 kW
	1 W = 0.7376 (ft · lb)/s
	1 kW = 1.341 hp

Question 3.4: Aerial Refueling The KC-10 Extender tanker aircraft of the U...

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