Question

5. Двигатель реактивного самолета с КПД 20 % при полете со скоростью 1800 км/ч развивает силу тяги 88 кН. Определить расход керосина за один час полета. Удельная теплота сгорания керосина 46 МДж/кг.

Answer 1

Η = А полезная / А затраченная * 100 %

А полезная = F * V 

F - сила тяги ( 88 кН = 88000 Н ) 
V - скорость ( 1800 км / ч = 1800 * 1000 / 3600 = 500 м / c ) 

А полезная = 88000 * 500 = 44000000 Дж 

А затраченная = q * η

q - удельная теплота сгорания ( 46000000 Дж / кг ) 
η - КПД = 20 % = 0,2 

А затраченная = 46000000 * 0,2 = 9200000 Дж 

m = F * V / q * η = 44000000 / 9200000 ≈ 4,78 кг 

V = m / p 

m - масса ( 4,78 кг ) 
р - плотность ( для керосина 800 кг / м³ ) 

V = 4,78 / 800 = 0,005975 м³ = 5,975 л ≈ 6 л 

Answer 2

Calculation of Fuel Consumption for One Hour of Flight

To determine the fuel consumption for one hour of flight, we need to calculate the fuel flow rate based on the given information.

Given: - Efficiency (η) of the jet engine = 20% - Flight speed = 1800 km/h - Thrust force (F) = 88 kN - Specific heat of combustion of kerosene (Q) = 46 MJ/kg

The efficiency of a jet engine is defined as the ratio of the useful work output (thrust force) to the energy input (fuel consumption). Mathematically, it can be expressed as:

η = (F * V) / (Q * m_dot)

Where: - η is the efficiency - F is the thrust force - V is the flight speed - Q is the specific heat of combustion of kerosene - m_dot is the fuel flow rate (fuel consumption per unit time)

Rearranging the equation, we can solve for m_dot:

m_dot = (F * V) / (Q * η)

Substituting the given values:

m_dot = (88 kN * 1800 km/h) / (46 MJ/kg * 20%)

Converting the units: - 1 kN = 1000 N - 1 km/h = 1000 m/s - 1 MJ = 1,000,000 J

m_dot = (88,000 N * 1,800,000 m/s) / (46,000,000 J/kg * 0.2)

Simplifying the equation:

m_dot = 7.26 kg/s

Therefore, the fuel consumption for one hour of flight is 7.26 kg.