Question

Given that a commercial airliner has a range a range of 9,500 miles using 265 tons of JP-4 fuel. Estimate the range of the aircraft burning either gaseous or liquid hydrogen at_____________.

Answer 1

Here is the full question.

Given that a commercial airliner has a range of 9,500 miles using 265 tonnes of JP-4 fuel. Estimate the range of the aircraft burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4.

The dry mass of the aircraft is 800 tonnes. The fuel properties are as follows:

Fuel                             Density (kg/m³)              Heating value (kJ/kg)

JP4                                 800.0                            45000

H₂ (gaseous, S.T.P)       0.0824                          120900

H₂ (liquid, 1 atm)             70.8                              120900

Answer:

25,514 mi

Explanation:

Let first calculate the value of initial mass of the aircraft $m_1$ when burning JP4 fuel by using the expression:

$m_1 =m_2 + m_{JP4}$

where;

$m_2 = 800t$

$m_{JP4} = 265t$

$m_1 = 800t+265t$

$m_1=1065 t$

We need to employ the use of the range of the airliner which can be expressed by the formula;

$s = \eta__0}(\frac{L}{\delta g})In(\frac{m_1}{m_2})(\frac{Q_R}{g})$

where;

$\eta__0}$ = overall efficiency

L = lift

$\delta g$ = drag force

$m_1 =$ initial mass of the vehicle

$m_2$ = is the final mass of the airliner after the burning of fuel

$Q_R$ = is the heat of the reaction of the fuel burning.

g = gravitational acceleration

Rearranging the above equation; we have:

$\eta__0}(\frac{L}{\delta g}) = \frac{ s}{In(\frac{m_1}{m_2})({Q_R})}$

$\eta__0}(\frac{L}{\delta g}) = \frac{9500 mi}{In(\frac{1065t}{800t})({45000 kJ/kg})}$

$\eta__0}(\frac{L}{\delta g}) = \frac{0.2111}{In(1.33)}$

$\eta__0}(\frac{L}{\delta g}) = 0.74 mi.kg/kJ$

To estimate the range of the aircraft burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4; we have:

$s = \eta__0}(\frac{L}{\delta g})In(\frac{m_1}{m_2})(\frac{Q_R}{g})$

$s= (0.74mi.kg/kJ)(120900kJ/kg)In(\frac{1065t}{800t})$

$s= (89466mi)In(\frac{1065t}{800t})$

$s = (89466mi)In(1.33)$

s = 25,513.82 mi

s ≅ 25,514 mi

Thus, the range of the aircraft when burning either gaseous or liquid hydrogen at the same mass of hydrogen as JP4 is 25,514 mi