What competitive strategy was used by Regal Marine?
1 answer:
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Answer:
A) 0.03382 kg/s
B) 7.0372 Kw
C) 4.3982
D) 0.7396 kw
Explanation:
Given data:
Evaporator at 60 C
Space temperature = 25 C
power consumed by compressor = 1.6 kw
T1( evaporator temperature ) = 12°C
attached below is the detailed solution
Answer: power required to maintain level flight=82.20hp
Explanation:
Given
Area = 200 ft^2
Weight = 2000 lb
Cl( Lift coefficient)= 0.39
Cd( Drag coefficient) = 0.06
The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3
For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that
Lift force = Weight of aircraft
(1/2)ρAU²Cl= W
1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000
U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39
U=
Velocity, U= 146.7892ft/s
Drag force of the velocity can be deduced from the formulae
Cd= Drag force(D) /1/2 ρU²A
Drag force=1/2 ρU²ACd
D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06
D=307.69
Drag force= 308lb
power required to maintain level flight is given as
P = Drag force x Velocity = D x U
=308lb X 146.7892ft/s
=45,211.0736lb.ft/s
Changing to hp we have that
1 Horsepower, hp = 550 ft lbf/s
??=45,211.0736lb.ft/s
45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp