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Total? Dynamic? Static? Are you feeling it?

Freemasm

Well Known Member
Happy Friday. Interesting Engine Cooling thread has me thinking about an often misunderstood application of a relatively simple relationship; Total, dynamic, and static pressure. Hopefully the Mods will allow this as it is applicable to our endeavors.

The first picture is the test set-up. A fan is drawing air through the test chamber. Like our flying aircraft, it is outside or basically an infinitely large room. Water is the fluid in the manometer tube. For us nerds that want to overcomplicate things, don't; Flow is not turbulent, no compressibility, ignore friction effects.

The second picture has the relevant parameters (?)

If VAF is similar to the undergrad engineering population; one third will get the answer, another third will be wrong initially and then get the correct solution, another third will tell me I'm an idiot.

Feel free to make assumptions. Are there enough listed parameters to determine water column height X? If so, what is it?
 

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Is this going to be on the final? It takes me back to engineering classes I took in the last century...oops, i think I had a little vomit in my mouth for just a second there.
 
My answers: Pd = 3.78 psi. Ptotal can't be defined because you didn't give an altitude or temp. Assuming sea level on a standard day, that would make P static = 14.7 psi and P total equal to 18.5 psi. Manometer height is 10.7 inches (5.35 up, 5.35 down).

Other assumption: V amb = 0, so that Pamb = Pstatic
 

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All I know

One thing certain is if pie are square, corn bread are round, or vice versa!
 
No correct answer yet

Maybe question is below most VAFer’s intellect to be worthy of their thoughts.

Maybe most are afraid they might be wrong to post an answer.

Maybe I’m the idiot and the whole question is just dumb.

The right answer will surprise (guessing) 2/3s of anyone that actually cares.

Anyone else?
 
I don't think that there is enough information presented. Eg. diameter of the entry and exit points and the diameter of the narrowest point. The real question should be how much Jack Daniels was consumed before publishing this question.
 
Bad assumption

“VAF is similar to undergrad engineering students.”

My impression - we’re closer to pilots at happy hour.
 
Cumulo is correct

The exercise is conceptual.

There were plenty of parameters available if needed. No energy is added to the system until the fan., so Pt is constant until that point.

The configuration in the pseudo test set up was chosen for its similarity to GA aircraft application

That’s all I’ll say for now. Concepts can’t be taught, only presented; a statement of mine that used to pi$$ off my math teacher wife until she reconciled it.

If you don’t agree, that’s fine. If you have a question and don’t want to post, PM me.

So, for those ~600 to now VAFers that bothered to read this (I realized most could care less), you know what third you fall into. I’ll say again, this concept applies to our EAB/GapA endeavors.

Cheers boys.
 
The exercise is conceptual....Concepts can’t be taught, only presented

So why present an illustration showing a positive pressure water column?

Here's a useful conceptual illustration, credited to Fred Moreno:
.
 

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Maybe question is below most VAFer’s intellect to be worthy of their thoughts.

Maybe most are afraid they might be wrong to post an answer.

Maybe I’m the idiot and the whole question is just dumb.

The right answer will surprise (guessing) 2/3s of anyone that actually cares.

Anyone else?

The exercise is conceptual.

There were plenty of parameters available if needed. No energy is added to the system until the fan., so Pt is constant until that point.

The configuration in the pseudo test set up was chosen for its similarity to GA aircraft application

That’s all I’ll say for now. Concepts can’t be taught, only presented; a statement of mine that used to pi$$ off my math teacher wife until she reconciled it.

If you don’t agree, that’s fine. If you have a question and don’t want to post, PM me.

So, for those ~600 to now VAFers that bothered to read this (I realized most could care less), you know what third you fall into. I’ll say again, this concept applies to our EAB/GapA endeavors.

Cheers boys.
My question is:
Why are you tainting your posts with “insults” to your reader’s intellect?

If you really want to have meaningful dialogue about any particular subject matter I don’t think it wise to openly insult your audience. That is a sure fire way to soon not have an audience at all!
 
So why present an illustration showing a positive pressure water column?

Here's a useful conceptual illustration, credited to Fred Moreno:
.

In the other thread. I asked how/where Pd was measured (engine cooling flow). You said it was calculated. Fine. Measuring anything other than Ps in that set-up is probably a waste of time as an array of pitots adjustable pitots would be needed to begin to understand the related flow fields. I started thinking about very common misunderstandings/mistakes regarding such experimentation/data gathering and stated I would start this thread. Nothing makes me think these general assumptions were wrong.

How should I have represented the head differential? Friction losses would have rendered x negative. Sure, anything can be improved upon but I'm not feeling any related guilt.

I'll agree that your attachment is a good illustration of a process but not of a concept.

The intent here was the Ps/Pd/Ps relationship; not necessarily related to engine cooling. See below.

My question is:
Why are you tainting your posts with “insults” to your reader’s intellect?

If you really want to have meaningful dialogue about any particular subject matter I don’t think it wise to openly insult your audience. That is a sure fire way to soon not have an audience at all!

Taunts? People hear what they want to hear; believe what they want to believe.

Self taught knowledge always sticks better than that handed to someone. Hats off to the ones that laid their's on the table and offered an answer. Not many did. I'll guess they fall into the aforementioned second group of "Man, I should have known/seen that." Happens to all of us. Anyway, if the lower side of the sense line would have been tapped into any straight wall of the test chamber, the provided parameters and equations could have been useful. Pt is constant unless you add energy to a system. I'll state again, the provided illustration is a very useful corollary for our aviation applications.

If you asked someone what the pitot tube on their aircraft measures, most would probably say airspeed. If you asked again and clarified "measures", the most common, different reply would probably be Dynamic Pressure. Again, this is not correct. Any such device can only measure Total Pressure; assumptions made for proper orientation. It has to be referenced to static pressure reference to derive Pd (and thus indirectly airspeed).

Sure, I could have made simple statements about known mathematical relationships or a common misconception about an aircraft pitot system. It would then becomes a matter or trivia or memorization for someone not skilled in such art.

A real understanding of systems can save someone a lot of time, anguish, or even more. The size of an audience that reads anything I write is not important to me. Don't want to read any replies by me, no worries here. If it helps a single person to a better understanding/saves them troubleshooting efforts/etc. I'm good with that.

Cheers boys,
 
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Static “Vent???”

This thread reminds me of a discussion I had a long time ago with a young CFII who asked me to explain how the Pitot-static system on our rental airplane worked. After I correctly explained it, he said I was wrong because the static port was clearly labeled “Static Vent” which meant the air flowed into the Pitot tube and came out the “Static Vent.”

What??? :confused: :eek: :confused:

I found a new instructor.
 
This thread reminds me of a discussion I had a long time ago with a young CFII who asked me to explain how the Pitot-static system on our rental airplane worked. After I correctly explained it, he said I was wrong because the static port was clearly labeled “Static Vent” which meant the air flowed into the Pitot tube and came out the “Static Vent.”

What??? :confused: :eek: :confused:

I found a new instructor.

Another post that makes me wish we had a "like" button
 
Here you go:

May I see if I can help with this?

First, the mass flow thru the venturi is the same at station 1 as station 2. Assuming incompressible flow (good assumption up to about Mach 0.4)

So (rho)(V)(A)_1 = (rho)(V)(A)_2

(1.2)(20)(10) = (1.2)(V_2)(3). Solve for V_2 = 66.666 M/s

My apologies - the rest of what i did a minute ago was wrong -- I didn't see that the design had a pitot tube in the throat, or that the other end of the manometer was open to the ambient.

This is actually a really good trick problem, because I jumped to a tasty contusion that the manometer was measuring the differential between the static pressures at station 1 and station 2, which is a common case. In fact, it is measuring the differential between the ambient static pressure (which is at stagnation pressure) and the pitot pressure at station 2. In the absence of any losses, those are the same. X= 0.


And cudo's for Cumulo who got it straight away. I should have looked at the drawing more closely before I dove in the first time.
 
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I'll agree that your attachment is a good illustration of a process but not of a concept.

Take another look. The concept of a Ps + Pd = Pt trade is right there. And the illustrated process is real. It's happening at the nose of every RV in the fleet.

Yes, I realize the example treats ambient static as zero.
.
 

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May I see if I can help with this?

First, the mass flow thru the venturi is the same at station 1 as station 2. Assuming incompressible flow (good assumption up to about Mach 0.4)

So (rho)(V)(A)_1 = (rho)(V)(A)_2

(1.2)(20)(10) = (1.2)(V_2)(3). Solve for V_2 = 66.666 M/s

My apologies - the rest of what i did a minute ago was wrong -- I didn't see that the design had a pitot tube in the throat, or that the other end of the manometer was open to the ambient.

This is actually a really good trick problem, because I jumped to a tasty contusion that the manometer was measuring the differential between the static pressures at station 1 and station 2, which is a common case. In fact, it is measuring the differential between the ambient static pressure (which is at stagnation pressure) and the pitot pressure at station 2. In the absence of any losses, those are the same. X= 0.


And cudo's for Cumulo who got it straight away. I should have looked at the drawing more closely before I dove in the first time.

Thanks for posting Steve, I reasoned it was zero, but did not prove it in the formulas, so did not post a guess.

I am not sure how a clever gotcha helps train people. In a classroom we captured paid for the change to sit there and not exit the door. I chose my professors carefully (still do) when I could. A funny story about an electrical prof and vibrations prof each teaching the others fundamentals by referring to the other. I called the both to the same meeting and told them what was happening - they laughed profusely and changed their approaches and did the math. I was happy and then understood both.
 
Thanks for posting Steve, I reasoned it was zero, but did not prove it in the formulas, so did not post a guess.

I am not sure how a clever gotcha helps train people.

Well, it trains you to look closely at the problem before diving in to answer the problem you think you have. I'm a bit out of practice apparently, because I just glanced quickly and set about to solve for the pressure difference and related X value if the manometer was sensing the static pressure at stations 1,2, which would be a very common installation. X=9.74" in that case. Then I looked more closely at the picture and saw what the real set-up of the manometer was.
 
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