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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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A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
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Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

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We are given;

Length; L = 5ft = 1.52.

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Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

6 0
3 years ago
Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft/s2 until it reaches a spe
taurus [48]

The distance traveled by car A when they pass each other is 1071m

<h3>Calculations and Parameters</h3>

Given:

t=0

constant acceleration= 6 ft/s^2

speed= 80 ft/s^2

S_A= \frac{at1^2}{2} + V_At\\S_B= V_B (t1 +t)

= 2 * (\frac{27}{2})^{2}/2 + 27 * 32.9

= 1071m

Read more about distance here:

brainly.com/question/2854969

#SPJ1

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