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valentina_108 [34]
3 years ago
15

Which of the following can easily reverse motion and are better at varying speeds than electrical motors?

Engineering
2 answers:
yan [13]3 years ago
6 0

Answer:

fluid power system

Explanation:

satela [25.4K]3 years ago
4 0
Fluid power system

Example
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Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
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