Total pressure and Static Pressure and V

AI Thread Summary
In a wind tunnel experiment, confusion arises regarding total pressure (Pt) and static pressure (Ps) readings, with Pt recorded at 11.4mm and Ps at 16.8mm, leading to a negative dynamic pressure calculation. Total pressure should always be greater than or equal to static pressure, indicating a potential misinterpretation of the pressure ports. The discussion suggests the possibility of using a pitot-static tube, which combines measurements, to clarify the readings. It is emphasized that if the flow direction is not reversed, dynamic pressure cannot be negative. Accurate pressure difference calculations are crucial for determining freestream velocity.
Ma77h3w
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Confusion... I'm probably being stupid but...
Aerofoil in a wind tunnel... there are lots of pressure tappings including Pt and Ps (static and total pressure freestream) these are currently in mm on a manometer..
my Pt is 11.4mm and Ps is 16.8mm.. (SG=0.8)
1st question: Should total pressure be less than static? because i can't solve the equation to find free stream velocity.. rearranging: Pt - Ps = 1/2 p V^2 (p = density)
as Pt - Ps is negative.. hence finding V is impossible.. can't root a negative. Have i mixed the two up?
however my Cp against x/c graph looks good.. and swapping Pt and Ps messes it up..

Thanks
 
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I'm not sure, I don't have access to the experiment anymore. What difference would using a pito static tube make?
Can the Dynamic pressure (Pt - Ps) be negative in this situation? And am I allowed to swap them around (Ps - Pt) to work out the freestream velocity?
All the other pressure tappings on the aerofoil itself at 0 incidence are between 15 and 19mm. It's only the so called Pt that's as low as 11.4mm.. Thought: maybe I'm being silly and the lower the (mm manometer fluid) the greater the pressure...? Confusion reigns
for information manometer (table?) is at 20 degrees and fluid = 800 kg/m3
 
Ma77h3w said:
What difference would using a pito static tube make?
The static and total ports are linked together to do the subtraction for you, providing one number for your data: the velocity pressure.
Can the Dynamic pressure (Pt - Ps) be negative in this situation? And am I allowed to swap them around (Ps - Pt) to work out the freestream velocity?
Not unless the flow direction was backwards.
All the other pressure tappings on the aerofoil itself at 0 incidence are between 15 and 19mm. It's only the so called Pt that's as low as 11.4mm.. Thought: maybe I'm being silly and the lower the (mm manometer fluid) the greater the pressure...? Confusion reigns
for information manometer (table?) is at 20 degrees and fluid = 800 kg/m3
Without knowing what the port that is labeled "Pt" really is, it is tough to help. But total pressure can't be less than static pressure...

...or are these other ports that are labeled as static pressure acutally pressure ports along the profile of the airfoil? In that case, they aren't freestream static pressure.
 
Total P is always greater than or equal to static pressure

I guess u r wrongly calculating the pressure difference... let me guess..
u would have connected the pitot tube(measures tot P) to a manometer (A) at the top
Static tube(measures Static P) to another manometer (B) at the top...
Condition: both manometers exposed on the other tube to same atmospheric conditions

Now u have less pressure acting on the water column at B.. so the water level will be higher, say h2; on the other hand u ve more pressure acting on the water column at A.. so its level will be lower, say h1..

This implies, higher P keeps the water column lower... so u would have calculated the pressure diff as rho*g*(h2-h1) which is actually (Tot P-Static P)... Now u cud get the vel right
 
manicks said:
Total P is always greater than or equal to static pressure

I guess u r wrongly calculating the pressure difference... let me guess..
u would have connected the pitot tube(measures tot P) to a manometer (A) at the top
Static tube(measures Static P) to another manometer (B) at the top...
Condition: both manometers exposed on the other tube to same atmospheric conditions

Now u have less pressure acting on the water column at B.. so the water level will be higher, say h2; on the other hand u ve more pressure acting on the water column at A.. so its level will be lower, say h1..

This implies, higher P keeps the water column lower... so u would have calculated the pressure diff as rho*g*(h2-h1) which is actually (Tot P-Static P)... Now u cud get the vel right

Can you not use text speak please.
 
Cyrus said:
Can you not use text speak please.

okay...hereafter i will be conscious about that.. thank you
 
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