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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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yarga [219]

Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series  circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

5 0
3 years ago
The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter
Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

(p_t)₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ [ α₂/(p_t)₂ ]^{1/2 ------- let this be equation 2

where (p_t)₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2 from equation for (σf)₂  in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ [ α₁/(p_t)₁ ]^{1/2  = 2(6( σ₀ )₁) [ 0.84α₁/(p_t)₂ ]^{1/2  

divide both sides by 2(σ₀)₁

[ α₁/(p_t)₁ ]^{1/2  =  6 [ 0.84α₁/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]^{1/2  =  6 [ 0.84/(p_t)₂ ]^{1/2

[ 1/(p_t)₁ ]  =  36 [ 0.84/(p_t)₂ ]

1 / (p_t)₁ = 30.24 / (p_t)₂

(p_t)₂ = 30.24(p_t)₁

(p_t)₂/(p_t)₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0
3 years ago
Basil is installing a system in Toronto, Ontario where they are wide temperature variations. What is the TD of the system he’s i
marysya [2.9K]
In places with cold winters, space heating systems have a fundamental role in buildings. Without them, indoor temperatures would quickly become unsuitable for human occupancy. The local weather is one of the most important factors when designing a heating system; if two identical buildings are developed in Miami FL and New York City, the heating load will be much higher for the NYC property.

4 0
3 years ago
A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m
Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

\omega=\dfrac{2\pi N}{60}\ rad/s

Now by putting the values in the above equation

\omega=\dfrac{2\pi \times 3000}{60}\ rad/s

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

\omega=\dfrac{ 3000}{60}\ rad/s

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

4 0
3 years ago
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of
yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question:  determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613  ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

we will apply interpolation

attached below is the remaining part of the solution

8 0
2 years ago
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