In this closed system, a dog has sprinted to the top of a cliff. The dog (mass of 25kg) had been sprinting with a velocity of 13
1 answer:
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Answer:
- the effective lift area for the aircraft is 8.30 m²
- the required engine thrust is 1275 N
- required power is 79.7 kW
Explanation:
Given the data in the question;
Speed V = 225 km/hr = 62.5 m/s
The lift coefficient CL = 0.45
drag coefficient CD = 0.065
mass = 900 kg
g = 9.81 m/s²
a) the effective lift area for the aircraft
we know that for a steady level flight, weight = lift and thrust = drag
Using the equation for the lift force
F = C
ρV²A = W
we substitute
0.45 × × 1.21 × ( 62.5 )² × A = ( 900 × 9.81 )
1081.05 × A = 8829
A = 8829 / 1081.05
A = 8.30 m²
Therefore, the effective lift area for the aircraft is 8.30 m²
b) the required engine thrust and power to maintain level flight.
we use the expression for drag force
F = T = C
ρV²A
we substitute
= 0.065 × × 1.21 × ( 62.5 )² × 8.30
T = 1275 N
Since drag and thrust force are the same,
Therefore, the required engine thrust is 1275 N
Power required;
P = TV
p = 1275 × 62.5
p = 79687.5 W
p = ( 79687.5 / 1000 )kW
p = 79.7 kW
Therefore, required power is 79.7 kW