¿Cuál es el objetivo de la participación del gobierno en la economía?
1 answer:
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Answer:
mass of LNG: 129501.3388 kg
quality: 0.005048662
Explanation:
Volume occupied by liquid:
400 m^3*0.9 = 360 m^3
Volume occupied by vapor
400 m^3*0.1 = 40 m^3
A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present
For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg
From specific volume definition:
vf = liquid volume/liquid mass
liquid mass = liquid volume/vf
liquid mass = 360 m^3/0.002794 m^3/kg
liquid mass = 128847.5304 kg
vg = vapor volume/vapor mass
vapor mass = liquid volume/vg
vapor mass = 40 m^3/0.06118 m^3/kg
vapor mass = 653.8084341 kg
total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg
Quality is defined as the ratio between vapor mass and total mass
quality = 653.8084341 kg/129501.3388 kg = 0.005048662
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
Answer:
The atmospheric pressure in atm=0.885 atm
Explanation:
Given that
Local pressure (h)= 30 ft of water height ( 1 ft= 0.3048 m)
We know that pressure in given by
P=ρgh
We know that ρ of water is 1000
So pressure
P=1000(9.81)(9.144)
We know that 1000 Pa=0.00986 atm
So P=0.885 atm
The atmospheric pressure in atm=0.885 atm