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Fluid Dynamics and the Navier-Stokes Equations

Lift for Aeroplane and Helicopter

The lifting force for aeroplane can be derived straight from Eq.(2) of the Navier-Stokes Equations with the condition for irrotational flow xu = 0 (and (u)u = (1/2)(u2) - ux(xu)). By neglecting the external force term F, the viscosity and the time variation, the resulting formula can be reduced to a very simple form if we further assume that the air flow velocity in the z-direction uz is much smaller than that in the x-direction ux (uy = 0 for this particular choice of coordinates):

ux [d(ux)/dz] = - (1/) dp/dz ---------- (10)

Integrating Eq.(10) yields:

{ [(ux1)2 - (ux2)2]} / 2 = (p2 - p1) ---------- (11)

Aeroplane where the subscripts 1 and 2 respectively denote the flow velocity ux, and the pressure p to the upper and lower layers of the air flow. It shows that if the structure of the wings are designed to create a higher flow velocity in the upper layer, then a net pressure in the

Figure 05 Lift for Aeroplane
[view large image]

upward direction is created to lift the aeroplane up into the air provided such force can overcome the weight [see Figure 05, the length of the arrows is proportional to the magnitude of ux (blue) and p (red)]. The force to lift up the plane is just f = P x Sref ,
Ref. Area & Lift Coeff where P = p2 - p1 is the net upward pressure, and Sref (Reference Area) is the effective area acted upon by P. Figure 06 shows the Sref for an aircraft and a helicopter. The actual force F on these objects is determined further by the formula F = f x CL, where CL is the Coefficient of Lift. It depends on the type of aircraft as shown in Figure 06, where the angle of attack is

Figure 06 Ref. Area & Lift Coeff. [view large image]

defined as the angle between the wing and the direction of the airstream. Most aircrafts will behave similarly to the Cessna 172 while high-speed planes with short wingspans, like fighters, will more closely resemble the Lightning data.
Helicopter The lift for helicopter can be derived from Eq.(2) of the Navier-Stokes Equations in a similar way (as for the aeroplane) with the role of ux and uz interchanged because the airflow pattern is different (see Figure 07). Thus for uz >> ux , Eq.(10) becomes:

uz [d(uz)/dz] = - (1/) dp/dz ---------- (12)

Integrating Eq.(12) yields:

{ [(-uz2)2 - (-uz1)2]} / 2 = (p2 - p1) ---------- (13)

Figure 07 Helicopter Airflow [view large image]

where uz is negative as it is pointing toward the negative z direction. It has a form similar to Eq.(11) for the aeroplane. Computation for the lifting force follows exactly the same line as developed previously.
Helicopter In stationary position, the helicopter's engine provides only the lifting force. According to the principle of angular momentum conservation, the body would turn in the opposite direction of the rotating blades. To stabilize the helicopter, a tail rotate is installed to counteract this trend. By applying more or less pitch angle to the tail rotor blades, it can

Figure 08 Helicopter
[view large image]

also be used to make the helicopter turn left or right. Forward motion is achieved by tilting the spinning rotor in the direction of the flight (see Figure 08). There are many factors related to the rotor blades to limit the maximum speed of a helicopter at about 400 km/h.
Propeller By rotating the rotor blades and the z-axis 90o, Eq. (13) is applicable to calculate the forward pressure (or thrust) of a propeller on a ship or airplane (Figure 09). Since the density for water is about 1000 times higher than air, the formula shows that the propeller can generate more thrust with lower flow speed on a ship or boat. It also explains why the prop plane cannot fly high up into the rare atmosphere, where the air density is very low.

Figure 09 A Prop Plane [view large image]

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