Try answering the question using three different approaches:

(1) Newton's second law

(2) Conservation of mass using a control volume encompassing only the stream of water

(3) Conservation of mass using a control volume encompassing both the solid object and the stream of water

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The mechanism of lift can be analyzed in similar ways.

It is convenient to relate lift to the motion of the fluid through the conservation of momentum. On the other hand, the conservation of energy is often not affected by the magnitude of lift. This is because lift, being orthogonal to the direction of relative motion, does zero work when taking either the frame of reference of the solid object or the frame of reference of the freestream. Drag, however, is aligned with the direction of relative motion, and contributes to the conservation of energy.

Parasite Drag

In this video Visualisation of the flow behind a cylinder in a water tunnel - YouTube , for example, the cylinder drags a column of water behind it. Suppose the column of water has the same cross-sectional area as the frontal area of the cylinder LaTeX: A and travels at the same speed LaTeX: V as the cylinder does. Then one can calculate the drag on the cylinder as follows:

(1) The volume of water dragged by the cylinder per second is $LaTeX: A\,V$ , which has a mass of $LaTeX: \rho\,A\,V$ with a unit of kg per second.

(2) The kinetic energy in the wake is $LaTeX: \frac12 V^2$ per kg. So the increment in the total kinetic energy of the wake is $LaTeX: \frac12 \rho\,V^3\,A$ in Watts.

(3) If the cylinder moves at a uniform speed, i.e., its kinetic energy does not change, then conservation of energy requires an external force to propel the cylinder. The power of the propulsive force has to be exactly $LaTeX: \frac12 \rho\,V^3\,A$ in Watts.

(4) Since the propulsive force pushes the cylinder at a speed of LaTeX: V , its magnitude must be $LaTeX: \frac12 \rho\,V^2\,A$ in Newtons. Since we are assuming that the cylinder moves at a uniform speed, the propulsive force balances the drag, which then also must be $LaTeX: \frac12 \rho\,V^2\,A$ in Newtons.

(5) The drag coefficient, $LaTeX: C_D = \frac{D}{\frac12 \rho\,V^2\,A}$ , then is 1.

One can thus relate the drag coefficient of a non-lifting body to the size of its wake, i.e., how much fluid the body drag with it. More aerodynamically-shaped bodies have thinner wakes; less aerodynamic ones have wider wakes. Drag caused by moving air in the direction of travel is generally known as parasite drag.

For very aerodynamically shaped bodies, e.g., wings, its parasite drag can be calculated by integrating the skin friction on its surface:

$LaTeX: C_{D\,parasite} = \frac{1}{A_{ref}\, V^2} \int_S c_f\, u^2\,ds$

where $LaTeX: A_{ref}$ is the reference area, LaTeX: V is the speed of the body, the integral is over all exposed surfaces. LaTeX: u is the local flow velocity on the surface, and LaTeX: c_f is the skin friction coefficient. Usually, $LaTeX: u \sim V$ , and for objects of scales of airplanes and cars, $LaTeX: c_f \approx 0.003$ . A good estimate is thus

$LaTeX: C_{D\,parasite} \approx 0.003\frac{S}{A_{ref}}$

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Induced Drag

Another source of drag is lift-induced drag. To generate lift, an airplane needs to set in motion surrounding air in a direction orthogonal to its direction of travel. Suppose it sets in motion a tube of air with cross section LaTeX: b^2 , where is the wing span. The mass of air it sets in motion per second is $LaTeX: \rho\,b^2 V$ . Further suppose the air is set in motion with a speed of LaTeX: w , then using conservation of momentum, the airplane would generate a lift force of magnitude $LaTeX: L = \rho\,b^2 V\,w$ . If we know how the lift force, then we can calculate $LaTeX: w = \frac{L}{\rho\,b^2 V}$

How much kinetic energy is in the tube of air set in motion per second? Since the mass of air is already calculated to be $LaTeX: \rho\,b^2 V$ , the kinetic energy per second is $LaTeX: \frac12 \rho\,b^2 V\,w^2$ . If the airplane is flying horizontally at a constant speed, then conservation of energy requires that a thrust force must inject energy with a power of $LaTeX: \frac12 \rho\,b^2 V\,w^2$ . Since the airplane is flying at a speed of LaTeX: V , the thrust force, which balances the drag force induced by the lift, must be $LaTeX: D_i = \frac12 \rho\,b^2 w^2$ .

If we substitute $LaTeX: w = \frac{L}{\rho\,b^2 V}$ , we obtain that $LaTeX: D_i = \frac{L^2}{2 \rho b^2 V^2}$ . This induced drag formula is correct up to a constant -- the exact formula is $LaTeX: D_i = \frac{2\,L^2}{\pi\,e\, \rho b^2 V^2}$ , where LaTeX: e is the span efficiency factor that typically ranges from about 0.7 to about 1.

How to be aerodynamic?

1. Being aerodynamically effective often means generating a large lift.

2. Being aerodynamically efficient usually means consuming or losing as little energy as possible, which means a low drag.

Being aerodynamic, therefore, often means having a high lift-to-drag ratio. In this subject, we focus on the aerodynamic lift and drag from the perspective of the fluid (air). To generate a high lift, one needs to either move a massive amount of air per second and/or move air at a high speed, in an orthogonal direction. The low drag requirement, however, forbids moving air at a high speed because that implies a high energy consumption. Therefore, an aerodynamic body would have to move a massive amount of air per second, which means a high span. The low drag requirement also means a small viscous wake, which forbids flow separation and requires a small exposed surface area.

To summarize:

(a) High span

(b) No flow separation

(c) Small expoesd surface