Display MoreThere is some understandable confusion about the outlet. Here is some information from an unpublished paper. This is translated by me:
"The blower exit is rectangular, 58 × 38 mm. The wind velocity in a square outlet is uneven. It is better to measure it some distance from the fan, so a 200-mm long cover pipe cylinder made of paper (Fig. 1-4) was attached in front of the blower outlet. One end is rectangular to fit the blower, and the other end is circular, 66 mm in diameter. This cylinder can be made of paper or plastic. 2-mm-thick urethane insulation is wrapped around the outside of the cylinder."
The traverse test covers 6 cm, so that leaves out only 6 mm to the edge, 3 mm on each side. The wind speed 3 mm from the wall of the pipe is the same as the wind speed in the middle, to the limits of precision with this anemometer. It is not possible the wind speed is zero 2 mm from the wall, or 1 mm from it. Somewhere at a fraction of a millimeter, the speed is down to zero. That's basic physics. I wouldn't know where, but you can see from the calibrations that the error is not significant.
The Reynolds number is 18,000, which is far about the textbook value of 2300 where turbulent flow begins.
Writing a summary on another thread, and having there been accused of laziness, I realise I need to revisit this.
Are the anenometer air velocity measurements used to calibrate the blower made in the 58X38mm outlet, or in the 66mm tube? I believe the 66mm tube.
Are the traverse tests used determine average velocity made in the 58X38mm outlet or the 66mm tube?
I believe the 66mm tube.
Assuming all is in the 66mm tube we have at 3.5 m/s Re=12,000 (18,000 would be 5.25 m/s)
This is well into turbulent flow but that does not mean velocity profile is flat.
From: https://www.chegg.com/homework…ile-shown-figur-q13459661 or many other places.
we get v = (1 - r/R)^0.2 (power law, Re=10,000, answer is not highly sensitive to Re).
Set R=1 to do the integration of 2*pi*v(r)dr / pi*r*r
average velocity is then integral^{r=1}_{r=0} 2*pi*r(1-r)^0.5 / pi = int (2r*(1-r)^0.2) = 0.76
You can see that looks about right from the picture (noting that larger values, nearer the edge, comprise more of the total area).
This is a 24% difference between calculated and actual heat flow. Worth noting (basically subtract 20% or maybe 24% from the results). It is unclear why this was not seen in the calibration data, maybe this was done at a different time, or on a slightly different system? Or maybe there is some other exactly compensating error.
{ Using, RB please note, online Wolfram math Integrate[2x(1-x)^0.2,{x,0,1}] }
Confirming the ~ 20% difference between middle of tube velocity and average over area velocity.