Pressure Drag


So what is pressure drag? In order to understand
that. We have to understand what drag itself is. This occurs when a body is moving through
a fluid, and when we have drag. We assume that there is not only pressure forces, but
there are viscous forces as well. So the forces that oppose movement. Are what is known as
drag, and we have 2 main components of drag. We have friction drag, which is due to a tangential
force we call tau. We also have pressure drag, and pressure drag is due to the pressure on
the object and it is a normal force. So lets take a look at some examples. If we consider
a flat plate, and we are going to incline this plate, and you have some velocity of
the fluid. Normal to the plate you have your pressure forces. Tangential to the plate you
have your friction forces. So lets take a look at a couple of other situations. What
would happen if you had a plate that was parallel to the direction of the flow. Well there you
will only have tangential forces. We are considering this a template. So there will be no normal
forces. The opposite situation is when you have a thin plate that is perpendicular to
the flow. So again we have a u going this way, and now we only have normal forces or
pressure forces. If you look at a sphere for example, and you look at flow pass the sphere.
So here we have a sphere and we have some velocity of a fluid coming in. You will see
that we not only have normal forces that are pressure. Up here we also have these tangential
forces, which are our friction drag. So as you can see by looking at this the greater
the area of the forces that are acted on. In other words the bigger the object the greater
the drag that is on this. So why would be interested in drag. Well it has a great impact
on things such as acceleration, because it is drag that causes the deceleration of a
drag car, or if you have parachute it is the deceleration of this parachute. That makes
sure you hit the ground at a safe speed. So lets take a look at the pressure drag. So
when we are look at the pressure drag. What we are really look at are these forces that
are normal. So this drag is the integral of the pressure times its position in another
words the angle that it hits at. So its orientation times our differential area, and we define
a dimensionless coefficient of pressure drag as this drag, and here we go d of p because
it is the pressure of drag over 1/2 times rho times u^2 divided by the area. You can
see that this coefficient of drag depends on area and velocity. If we rewrite this,
it is this integral times its pressure times the cosine of theta, da divided by our 1/2,
rho, u^2,A and we let this equal our Cp, which is our coefficient of pressure times the cosine
of theta, da, which is in the integral divided by area, where our Cp equals p- some reference
pressure, divided by rho, u^2, divided by 2.This pressure coefficient is actually a
dimensionless form of the pressure. So you might want to know. What about this reference
pressure? Well the level doesn’t influence the drag directly because what you are really
looking for is the net pressure force on the body, and that is 0 if the pressure is constant
on the entire surface. So what happens if the inertial forces are large. What that means
is that we have a large Re number. If that is the case our pressure coefficient is independent
of the Re. However if we have viscous forces that are large. Now our Re number cannot be
neglected, and the drag coefficient is proportional to 1 over Re. What if the viscosity were 0.
Then you would have no pressure drag in a steady flow at all. There would be large pressure
forces on the front portion, but you would have the opposite and equally large forces
on the rear.

5 thoughts on “Pressure Drag

  1. how would you describe the drag if a fluid was flowing over a stationary circular rod? Does drag always happen whenever there is relative motion between the fluid and the solid?

  2. Thanks for the quick response! In my problem, the velocities are so small (of the order 10E-5 ms-1) so is it reasonable to suggest that the effects of both friction drag and pressure drag are sufficiently small that they can be neglected? My problem is basically a flow of a fluid between two plates with two circular rods inserted. The fluid flows because of an imposed pressure difference and I have to analyse the forces affecting the fluid during its flow. I'm having difficulties describing the flow as the fluid encounters the circular rod (where I imagine the friction drag will predominate). I have used a software to numerically compute the velocities at the outlet but describing the theory between the inlet and outlet is proving a bit stubborn…Thanks for the help in advance!

  3. isn't it (Drag coefficient) = Total Drag/(0.5*c*u*u*A) instead of it being (Pressure Drag coefficient) = Pressure Drag/(0.5*c*u*u*A). Or am I missing something here

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