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a_sh-v [17]
3 years ago
8

Consider a 2D airfoil with a zero lift angle of attack of −4 degrees. The stall angle for this airfoil is 15 degrees, where the

maximum 2D lift coefficient is found to be 2.0. This airfoil is used in an aircraft with a non-elliptic wing of span 20m and planform area of 60 m2 . If ????1 = ????2 = 0.01, the weight of the plane is 105 Newton, density of air is 1 kg/m3 , find out the maximum landing speed that the plane can have
Engineering
1 answer:
zzz [600]3 years ago
4 0

Answer:

Explanation:

Recquired Data

lift coefficient = 2.0

Distance = 20m

Area = 60m²

Force = 105 N

Density = 1kg/m³

we are recquired to find Maximum panding speed

The lift coefficient CL is defined by

CL≡L/qs=L/1/2рμ²S = 2L/рμ²S

where L, is the lift force, S, is the relevant surface area and q, is the fluid dynamic pressure, in turn linked to the fluid density rho ,, and to the flow speed u,

substituting the figure respectively

2.0 = 2 X 105 / 1 X μ² X 60

μ² = 2L /CL X P X S

μ² = 2 X 150 / 2.0 X 1 X 60

μ² = 300 / 120

μ² = 2.5

μ =1.58ms⁻²

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Answer:

The correct answer is "1341.288 W/m".

Explanation:

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⇒       =3.7258 \ l

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⇒ Q=SF\times K(T_2-T_1)

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Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

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(d) X_{C} =49.37 Ω

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Explanation:

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\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

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S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

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The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

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I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

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Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

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I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

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