I reviewed the material we discussed in the last two lectures starting with the steady flow energy equation. This is a form of the first law written for an open system (with the same mass flow coming in and out). The only assumption I invoked was that I neglected changes in potential energy. That got me to and . The common difficulty with these two equations is understanding the physical difference between flow work (p2v2-p1v1) and shaft (or other external) work.
We then looked at the special case of an adiabatic process with no external work (q=0 and ws=0). This is an excellent model for the acceleration or deceleration of many fluids. For this case the stagnation enthalpy of the flow (hsubT = cpTsubT for an ideal gas) is a constant. If the flow is accelerated (c increasing), the enthalpy decreases (and the internal energy and the temperature). And vice versa if the flow is decelerated. If we deecelerate the flow to zero speed (c=0) then the temperature = the stagnation temperature and the enthalpy = the stagnation enthalpy. We also noted that if the process was also quasi-static, then pv^g=constant and we can define a stagnation pressure as the pressure one would reach if the flow were decelerated to zero speed in a particular reference frame via an adiabatic, quasi-static process with no external work. The stagnation quantities are dependent on the speed of the reference frame in which we assume the flow is stagnating ( a faster moving vehicle has a higher skin temperature). Also note that you can only apply the steady flow energy equation in a reference frame that is STEADY. So you must put yourself on the moving blade or on the moving aircraft (so you have a steady flow coming towards you), then determine what the speed of the flow is that is moving towards you, allow all of this kinetic energy (c^2/2) to be converted to enthapy (cpT) via an adiabatic process with no external work , arriving at the stagnation enthalpy (=cpTsubT for an ideal gas).(Add quasi-static as a requirement for the process if you want to determine the stagnation pressure).
We reviewed the PRS question we ended the last lecture with, and introduced one new PRS question.
Responses to 'Muddiest Part of the Lecture Cards'
(20 respondents out of 66 students)
1)Unclear on the frame dependence of stagnation temperature and related questions. (16 students) All of the mud cards focused on the same thing. So I will make a few general comments rather than answering each separately. First, we will have more time to go over this in the recitation on Tuesday. Also, you will have practice on the homework.
Now, I will review the rotor blade and supersonic airplane examples (they are the same case--instead of blades spinning around, think of tiny supersonic airplanes spinning around in the engine). There are two ways to think about the problems. First, in the reference frame of the blade or the airplane. They see flow coming at them with kinetic energy associated with the moving flow, and with an internal energy (or more appropriately enthalpy) associated with its temperature far away in the atmosphere. When the flow is stagnated on the blade, both of these sum into the stagnation enthalpy. Thus the temperature is higher than the ambient temperature. The second way to look at it is in the fixed reference frame (not moving with respect to the atmosphere). In this case the flow has some stagnation temperature as it comes in the inlet which is equal to the ambient atmospheric temperature. Then the blade or airplane comes along and the fluid sticks to the blade or airplane (work is done on the fluid particle) gaining kinetic energy. Thus the total or stagnation temperature on the surface of the blade is greater than the stagnation temperature the flow had when coming in the inlet. But this is an unsteady problem (the blades and the airplane are moving), so you can only think about it this way conceptually -- you can't apply the steady flow energy equation in this frame. You need to apply the steady flow energy equation in a reference frame where the flow is steady.
Now the engine sitting on the ground example The flow starts out stagnant someplace (in the atmosphere far away) and moves to a new location with no heat or external work where it stagnates again on a surface -- the wall of the engine inlet. The two reference frames are the same with no relative motion, therefore the stagnation temperature is the same. Any time the flow is moving relative to the fixed reference frame (like in the inlet of the engine), the (static) temperature is lower. With the stagnation temperature constant, enthalpy (cpT) is just traded for kinetic energy (c^2/2) as the flow accelerates and decelerates. So as the flow moves faster, its static temperature drops. As the flow moves slower, its static temperature increases.
2) No mud (4 students). Good.