Lecture P3: Integral Momentum Equation


General comments

A copy of the slides I have been using in lecture is available here (PDF). We completed the discussion of using the integral momentum equation to calculate engine thrust. I began with an example of an eraser falling and discussed solving the momentum equation in a control volume fixed to an inertial reference frame and in a control volume in an eraser-fixed reference frame. In the former, there is no term due to acceleration of the reference frame (it is fixed) but there is a change in momentum of the mass in the control volume as a function of time. In the latter, there is a term due to the acceleration of the reference frame, but no change in momentum of the eraser with respect to the control volume (which is fixed to the eraser). Of course the same answer is obtained either way, the acceleration of the eraser is g. This example is discussed further in the notes. I then discussed the addition of pressure forces to the thrust equation. Remember that differences in pressure around a control volume lead to forces; if the pressure is everywhere equal around the control volume there is no net force. We did two PRS questions (PRS #1, PRS#2). The first was designed to highlight the typical steps required to evaluate momentum flux across a control volume surface. When doing such an evaluation, remember to correctly evaluate the dot product of velocity and the outward unit normal. The seoncd PRS question was designed to provide additional practice in applying the integral momentum equation. Note there are two ways to go about solving a problem like this (both are given under the "Answer" button on the PRS question).

Next lecture we will discuss engine efficiency and aircraft performance and mission analysis. Please read Chapter IV of the notes.


Mud responses

(18 mud cards, 56 students attended class)

1) Are we ever going to do something with an accelerating reference frame? (1 student) See homework problem P2.

2) Why can the imbalance of pressures be added to the net flux of momentum to be the thrust equation? (1 student) The thrust equation relates net flux momentum flux to forces. There are two forces -- the reaction force holding the engine to the wing and the forces due to imbalance of pressures.

3) Shouldn't the resultant force be a vector? How can you add the term (po-pe)Ae to this if it isn't a vector value? The resultant force is a vector, but we have resolved it into the x direction only. The thrust equation was taken as one vector component of the integral momentum equation. And why does the integral momentum equation have (rho-ux-udotn) and not (rho-u-udotn)? (1 student) As noted above, because we were only taking one component of the momentum equation.

4) In the equation (po-pe)Ae is the po the static pressure or does it also include the dynamic pressure? (1 student) Static pressure only.

5) In the first PRS question, how come the vector u was considered in the dot product but only the y component for the other u term? (1 student) Because we were looking for the y-component of the momentum flux. This is equal to the integral over the area of the density time the y component of the velocity times the (scalar) dot product of the velocity vector and the outward unit normal.

6) You drew two engine control volumes with different inlet shapes but said the force due to the pressure imbalance at the exit plane was the same? (1 student) Yes. The net pressure force on a control volume is the sum of each of the pressures acting on each of the individual areas. If the pressure is everywhere the same, there is no net pressure force. If you change the pressure on one small individual area (say the exit area) then the net force is (po-pe)Ae.

7) You should be able to solve the tanker fueling problem with a control volume where the refueling boom comes in from the side instead of the top, right?(1 student). Yes. The answer is independent of where you draw the dotted line. Try it.

8) What would happen if I were to try this (INSERT COMPLICATED PICTURE OF AN ENGINE IN A WING OR A WING IN AN ENGINE)?(1 student). I am not sure, I didn't understand the sketch. Grab me sometime and explain it to me.

9) No mud (8 students). Good.