5 replies to this topic

[RJam]

A Pitot tube is used to determine the volumetric flowrate of a gas in a duct of circular cross section of diameter 1.5 m. A Pitot traverse using three data readings is used, presented in Table Q1. If the readings are taken at three radii that simplifies the calculation in which the centroids are of equal area, determine the position of the three readings expressed as r1/D, r2/D and r2/D, and the volumetric flowrate of the gas in the duct.
The volumetric flowrate, Q, is determined from;
Equation contained in attachement.

Q=Aduct 1/n ∑_(i=1)^n▒Vi
where Aduct is the duct cross sectional area, n is the number of points surveyed and vi is the indicated velocity at each measurement point.Table Q1

Position r1/D r2/D r3/D
Diff Pressure (Pa) 441 529 576

The density of the gas may be taken as 2.0 kg.m-3.

State any assumptions.

Can anyone help please! I understand the Pitot traverse concepts, just can' t put this calculation together.

Thanks

P.S sorry about the layout of the equation.

[Shawfield Stallion]

I'll just bump this topic if you don't mind guys.

I have an exam coming up pretty soon and pitot traverses are something that I just can't get my head around. I've emailed my tutor about some help a few times but he hasn't emailed back (surely his new year hangover has gone by now....hasn't it? )

Anyway, if anyone can solve this it'd be a great help!

[Andree]

I'm not sure if this is realistic figure for 1.5m duct, seems it is, but I obtained Q = 40m3/s (this corresponds to velocity 22.667m/s).

solution: the only difficulty is to calculate velocity from dynamic pressure measured by Pitot tube as v_i=sqrt(2*p_i/2) and subbstitute into your formula for Q

[Andree]

I did a mistake, the general formula for velocity is v_i=sqrt(2*p_i/density) (but in our case density was 2, so there is no error)

[newtonsown]

Guess it was just plain luck I came across this thread.

Anyway, has anyone got detailed solutions for the OP's questions? Also, is it correct to assume that V_3 is zero? Because if it is, how can you calculate the position of r_3?

Any help would be greatly appreciated.

[Andree]

From geometrical analysis the radius of the rings which divide the cross section area of the 1.5m dia pipe into 3 equal areas are: 0.866m and 1.225m
Not sure, but IMO the assuming linear change of velocity profile (not the best assumption, even if the flow is turbulent) one gets central position of each area as 0.433m, 1.045m and 1.363m, which gives r1/D 0.289, r2/D 0.697 and r3/D 0.909