A jet airplane in level filight has a mass of 865 x 10 kg an

A jet airplane in level filight has a mass of 8.65 x 10* kg, and the twe wings have an estimated total (a) What is the pressure difference between the lower and upper surfeces of the wings? (b) If the speed of air under the wings is 230 m/s, what is the speed of the air over the wings? Assume air has a (e) Explain why all aircraft have a \"celing\"a maximum operational altitude air has a density of 1.29 kg/m m/s with increasing height, resting in a )pressum dmerence, Beyond the masinum operate al attude, the preen d wa am pressure difference can no longer support the aircraft. Noed Help? R A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point 14.4 m below the water level. If the rate of fow frem following the leak is 3.00 x 10-3 m2/min, determine the (a) Determine the speed at which the water leaves the hole (b) Determine the diameter of the hole

Solution

a)

the net force due to pressure on the wing must counter the weight of the plane.

The weight of the plane is W=m*g. = 8.65*10^4*9.8 = 84.77*10^4 N

Now, pressure is force spread over an area, P=F/A.

the force must be equal to the weight of the plane,

so P=W/A = 84.77*10^4/95 = 8923.157 pa

b)

Bernoulli\'s equation:

p1 + 1/2*rho*v1^2 = p2 + 1/2*rho*v2^2

for points 1 and 2 along the same path in steady, incompressible flow.

We assume that the air on top and bottom of the wing starts upstream at the same conditions.

That leads us to pt + 1/2*rho*vt^2 = pb + 1/2*rho*vb^2 (t=top, b=bottom).

1/2*rho*vt^2 - 1/2*rho*vb^2 = pb - pt.

Here, (pb - pt) is the pressure difference

0.5*1.29*vt^2 - 0.5*1.29*230^2 = 8923.157

Vt = 258.3299 m/s

c)

As you can see, density plays a very significant role in the lift equations.

If density is too low (high altitude) the airplane may not be able to get enough of a pressure difference to offset its

weight.

2a)

special case of Bernoulli\'s equation

v = (2gh) = (2*14.4 9.8) = 16.8 m/s.

2b)

Fluid flow Q is given by Av, where A stands for the (cross-sectional) area of flow tube, and v is the velocity (speed)

of fluid.

3.00*10^-3 m^3/min = 3.0*10^-3/ 60 s = 5*10^-5 m^3/s.

Since both Q and v are known,

A = Q/v = 5*10^-5/16.8 = 3.125*10^-6 m^2.

A = 3.125 mm^2.

Now, A = pi*d^2/4

d = (4A/pi) = 1.9952 mm.