**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.

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.