Question

A 133.7 ft3 volume of liquid hydrogen rocket fuel has a mass of 268 kg. Calculate: the weight of the fuel at standard (Earth) sea level conditions in N and Ibr, the density of the fuel in kg/m3, and the specific volume in ft3/lbm (Ibm=pound-mass, which is not the same as a a lbf or a slug)

Answer 1

Answer:

given,

volume = 133.7 ft³

1 ft³ = 0.0283 m³

133.7 ft³ =  133.7 × 0.0283 = 3.784 m³

mass = 268 kg

1 kg = 2.204 lb

268 kg = 268 × 2.204 = 590.672 lb

weight of the fuel = 268 × 9.8 = 2626.4 N

1 N = 0.2248 lbf

2626.4 N =  2626.4 × 0.2248 = 590.414 lbf

$Density = \dfrac{mass}{volume}$

$\rho= \dfrac{268}{3.784}$

ρ = 70.82 kg/m³

specific volume = $\dfrac{133.7}{590.672}$

                           = 0.226 ft³/lbm