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Jlenok [28]
3 years ago
14

A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0

.06. Determine the power required to maintain level flight.
Engineering
1 answer:
Gnoma [55]3 years ago
6 0

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=\sqrt{21,547.08}

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

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Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo
Rom4ik [11]

Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

X=quality=1

T=-26C

density 1=α1=5.27kg/m^3  

entalpy1=h1=234.7KJ/kg

state 2

T2=70

P2=8bar=800kPa

density 2=α2=31.91kg/m^3  

entalpy2=h2=306.9KJ/kg

Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

(Q1)(α1)=(Q2)(α2)

\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s

the volumetric flow rate at the exit is 0.0297M^3/s

To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

W=m(h2-h1)

m=Qα

W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

4 0
3 years ago
3–102 One of the common procedures in fitness programs is to determine the fat-to-muscle ratio of the body. This is based on the
gayaneshka [121]

Answer:

x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

Explanation:

Given:

- The total volume of body = V

- The average density of the body = p_avg

- The density of muscle = p_muscle

- The density of fat = p_fat

Find:

Obtain a relation for the volume fraction of body fat x_fat

Solution:

- The volume of the fat is given by:

                          V_fat = x_fat*V

- The volume of the muscle is given by:

                          V_muscle = V - V_fat

                                            = V - x_fat*V

                                            =V*( 1 - x_fat )

- We will use the conservation of mass for the body related as:

                         mass_fat + mass_muscle = Total average mass

                         p_fat*V_fat + p_muscle*V_muscle = p_avg*V

                         p_fat*x_fat*V + p_muscle*V*( 1 - x_fat ) = p_avg*V

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = p_avg

- To determine p_1 we weigh the body in air:

                         Weight reading (Wsa) = m = p_1*V

                         p_1 = Wsa / V*g

- To determine p_2 we weigh the body in water:

                         Weight reading (Wsw) = m - p_w*V= p_1*V - p_w*V

                         Weight reading (Wsw) = V*(p_1 - p_w) = V*(p_2)

                         Where, p_2 = p_1 - p_water

                         p_2 = Wsw / V

- The average density p_avg:

                         p_avg = 0.5*(p_1 + p_2)  

                         p_avg = 0.5*(Wsa / V + Wsw / V)  

                         p_avg = 0.5*(Wsa + Wsw) / V                      

- Plug in the mass equation:

                         p_fat*x_fat + p_muscle*( 1 - x_fat ) = 0.5*(Wsa + Wsw) / V

                         x_fat*( p_fat - p_muscle ) = 0.5*(Wsa + Wsw) / V - p_muscle

                   x_fat = [ 0.5*(Wsa + Wsw) -  p_muscle*V ] / V*( p_fat - p_muscle )

                         

6 0
3 years ago
Why is it reasonable to say that no system is 100% efficient?​
Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

5 0
3 years ago
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

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Answer:

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