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Calibration Studies
© by Ed Seykota, 1999
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In order to test the theory on the bench, I gather data from actual
laboratory experiments and correlate the data with model predictions. To
calibrate the equipment, I first run some tests on very simple devices.
The first device is a simple round orifice normal to a flat wall in a
plenum. The second device is an extension of the first device that also
includes an additional plate, close up against the orifice so as to
restrict flow. |
Device #1
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Device #1: Air escapes from a plenum through a round orifice normal
to a flat wall.

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Air flows through the flow meter, then past the pressure meter, and
finally into the plenum and out through the orifice. I run this test with
orifice sizes of 1/16" and 1/8". I vary the plenum pressure by
adjusting the pump (not shown) and I measure the mass flux through the
orifices. Interestingly, basic Fluid Mechanics texts do not give formulas
or tables for this particular setup. |
Details of Device #1
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Detail of Plenum and Orifice
Air enters the plenum through the hose and exits through the orifice.
The plenum is a cylinder about an inch long and 1/4" in diameter. The
orifice diameters, [D] for the two tests are 1/8" and 1/16". The
pressure varies from zero to about 60 psig. The device is smooth
acrylic plastic. |
The Data
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Column |
Description |
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1 |
Gauge psig (pounds per square inch, gauge) - reading directly from
pressure gauge. |
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2 |
Gauge k-pa (kilo-pascals) - gauge psig times 6895 k-pa/psi. |
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3 |
Density - 1.2 kg/m3 times (11.6/14.6 + gauge k-pa / 78.91) |
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4 |
1/16" diam. Gauge - numerical reading directly from flow meter for
1/16" diameter orifice. |
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5 |
1/18" diam. Gauge - see directly above. |
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6 |
Adjustor Fraction - from MEM Flow Products Company. To adjust gauge
readings for different ambient temperature and pressure. The meter is
calibrated for 70 degrees F and 100 psig. I ran the tests at 50 degrees F
and 11.6 psig (Lake Tahoe). The formula is for the conversion factor:
f = sqrt(Pg / Ps * Ts / Tg)
Where Pg is the operating pressure + 11.6 psi, Ps is the scale pressure
(100 psi) + 11.6 psi, Ts is the scale temperature (70F) + 460F and Tg is
the operating Temperature, (50F) + 460F. Example, the factor for 5 psig at
50F, f = sqrt(16.6/111.6 * 530/510) = .39316. |
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7 |
1/16" Std. Flux (scfm) - gauge reading times adjustor gives the
flow at standard conditions (70 deg. F and 14.6 psi) |
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8 |
1/8" Std Flux (scfm) - see directly above. |
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9 |
1/16" Flux (kg/sec) - std flux (scfm) times 4.5 nt/lb times .07849
lb/scfm / 9.81 m/s2 / 60 s/min. |
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10 |
1/8" Flux (kg/sec) - see directly above. |
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Problem #2: A circular plate over the orifice creates a ring-shaped
"valve" at the center and changes the flows.  |
Given:
0 < P < 5 atm; P0 = 0 atm.; T = 70 F; T0 =
50 F; D = 1/16", h ~ .001 m.
Find the flows at the exit:
Find M-dot; Mo-dot; and Te-dot. |
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Problem #3: Air leaving the valve area continues to experience
skin-friction drag.  |
Given:
0 < P < 5 atm; P0 = 0 atm.; T = 70 F; T0 =
50 F; D = 1/16", E0 ~ .0000003 m.
Find drag from skin friction:
Derive U (velocity), p (pressure) and d (density) from the satates as
functions of M, Mo, and Te.
Find drag as a function of U, p and d. |
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Problem #4: Difference Equations:
To do: Create control volumes from concentric rings of thickness dR,
radiating out from the orifice between the plates, like nested bicycle
tires. (The rings in this panel appear on their sides.) During the
traverse across dR, within a control volume, Mo and Te change slightly as
a function of drag and pressure gradient across dR. For control volume[r],
write equations for these small changes. |
Continued ... |
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... continuation.
There is (1) a pressure gradient from the inside to the outside and (2)
some skin friction across the top and the bottom. These act to change the
momentum and thermal energy of the air passing through. |

Very small section of a ring showing direction of mass flux. Width (w)
is actually curved slightly and continues around as the circumference of
the ring. |
Definitions, Symbols and Units:
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States |
Symbol |
Units |
Notes |
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Mass |
M |
Kg |
Air: d ~ 1.2 kg/m3 |
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Momentum |
Mo |
Kg-m/s |
Mo = M * U |
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Thermal Energy |
Te |
Kg-m2/s2 |
Te = PV = NRT |
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Rates and Flows |
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Mass Flux |
M-dot |
Kg/s |
M-dot = dM/dt |
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Momentum Flux |
Mo-dot |
Kg-m/s2 |
Mo-dot = dMo/dt |
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Thermal Energy Flux |
Te-dot |
Kg-m2/s3 |
Te-dot = dTe/dt |
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