ME-EM 3220 ENERGY LABORATORY
Air Flow Measurements
Pitot Static Tube
A slender tube aligned with the flow can measure local velocity by means of pressure differences. It
has sidewall holes to measure the static pressure ps in the moving stream and a hole in the front to
measure the stagnation pressure po, where the stream is decelerated to zero velocity. The pitot static tube
may also be of the modified ellipsoidal-nose type. The tube has a very small diameter compared to that of
the duct diameter, but the resultant error caused by the additional blockage effect is considered minimal for
this investigation.
Instead of measuring po and ps separately, it is customary to measure their difference with, say, a
transducer, as in Figure 1.
Fluid flow
V
ps
po
ps
ps
po
Figure 1
po
po
Pitot Tube Configuration
Pitot tubes are affected by Reynolds number at low fluid velocities. The minimum Reynolds number for
total pressure measurements is approximately 30. This is the point where the characteristic length of the
pitot tube is equal to the diameter of the impact hole. Below this value, the indicated impact pressure is
higher than the stream impact pressure due to viscosity effects. For air at standard atmospheric conditions,
the error due to low Reynolds number is only apparent for air velocities less than 12 ft./sec. (3.66 m/sec.).
And this is for pitot tubes with impact hole diameters of 0.010 inches (0.2543 mm) or less. For low-velocity
for example, U = 1 ft./sec. in standard air, p0-p equal to only 0.001 lbf/ft2 (0.048 Pa). This is beyond the
resolution of most pressure gages.
The accuracy of pitot tubes is also affected if the sensor head is not parallel to the fluid. The total and
static pressure measurement error due to yaw and pitch angles increase rapidly above angles of 5 o.
Fortunately, they cancel each other out so velocity pressure measurements are 2% accurate up to angles
of attack of 30o.
1
The measurements of static pressure is also sensitive to the presence of fluid boundaries. The
presence of a pitot tube in a pipe also affects the static pressure. The pitot tube partially blocks the flow
passage which increases the flow velocity in the vicinity of the device. This results in an indicated static
pressure which is less than the actual static pressure.
The speed of response of pitot tubes is also geometry dependent. The diameter of the air passage
within the probe, the diameter and length of the interconnecting tubes, and the displacement volume of the
manometer determines the time constant. For tubes diameters greater than 1/8 inches (3.175 mm) and
ordinary manometer connections, the time constant is very short. However the time constant increases
rapidly for smaller diameter tubes, with a response time of approximately 15 to 60 seconds for tubes
having a 1/16 inches (1.59 mm) diameter. Because of the slow response of the fluid-filled tubes leading to
the pressure sensors, it is not useful for unsteady-flow measurements. One common problem with pitot
tubes that have very small diameters is that they tend to choke up easily if there is fine dirt in the fluid.
The pitot static tube is useful in liquids and gases; for gases a compressibility correction is necessary if
the stream Mach number is high.
If ReD > 1000, where D is the probe diameter, the flow around the probe is nearly frictionless and
Bernoulli’s relation applies with good accuracy.
p2 – p1 1 2
2
----------------+ --- ( V 2 – V 1 ) + g ( z 2 – z 1 ) = 0
2
ρ
[1]
p
p
1 2
1 2
-----1 + --- V 1 + gz 1 = -----2 + --- V 2 + gz 2 = const
ρ 2
ρ 2
[2]
or
For an incompressible flow
1 2
1
2
p s + --- ρV + ρgz s ≈ p 0 + --- ρ ( 0 ) + ρgz 0
2
2
Assuming that the elevation pressure difference
V th =
[3]
ρg ( z s – z 0 ) is negligible, this reduces to
2 ( p0 – ps )
-----------------------ρ
[4]
Here ρ is the air density in kg/m3;
p
ρ = -------s ,
RT
[5]
p s is the static pressure in Pascal, R is the gas constant and the value is 287 m2/(s2.K), T is the absolute
temperature in Kelvin.
The velocity measured by the pitot tube needs to be corrected due to geometry and flow interactions. It
is done as follows
V = αV th
where
[6]
α = f ( Re D ) , and the Reynolds number,
ρVD
Re D = -----------µ
[7]
2
Table below gives the range of α as a function of Reynolds numbers.,
Re D
3 × 10
α
0.986
4
1 × 10
0.988
5
3 × 10
0.990
5
1 × 10
6
0.991
Since V is not known, process is one which requires iteration, as described next;
•
STEP 1: - guess your α values anywhere from 0.986 to 0.991
Hint: Start from the lowest value.
•
STEP 2: - plug in the α values and calculate V
•
STEP 4: - Evaluate Re D
•
STEP 5: - Look up your guess α values whether it matches with the calculated Re D or not
•
STEP 6: - If the assumed α values is not in the calculated Re D range, repeat Steps 1 - 5.
NOTE:
Different sensors will require a calibration data generated for that particular case. If such information is not
available use the table above as valid for the pitot tube you are using. Depending on the application you
have, you may also be advised to delete velocity correction altogether, α = 1.
For channel flow where an average velocity is required, ( p 0 – p s ) can also be determined by
evaluating the effective pressure (or average) for the channel after multiple measurements as follows:
∆P eff
1
= ---N
2
j=N
∑
0.5
∆P j
[8]
j=1
where N is the number of measurements.
3
Calculating the Flow Rate from the Velocity Profile
The volumetric airflow rate can be directly determined from the velocity profile across the duct. Recall
that a fluid of velocity u passing across an infinitesimal area dA with outward unit normal vector n̂ .
u
n
dA
Figure 2
Velocity Profile
Denoting the cross sectional area of the duct as A, the volume flow rate through any given cross
section of the duct can be found from integration
Q =
∫ –( u • n̂ ) d A
[9]
A
For a flow which is axisymmetric, the velocity is only a function of radial distance from the tube axis
(centerline) as it is in the case of circular cross-section but it will have also have an azimuthal angle
dependency otherwise.
What follows is a discussion for a channel of circular cross-section area only. In this instance,
u = u(r )
[10]
and the infinitesimal area element
dA = 2πrdr
[11]
where r is the radial distance from centerline, which varies from r=0 to r=R (the tube wall). The vector dotproduct can be simplified by recognizing the flow velocity is always perpendicular to the cross section of
the tube. Thus
u ( r ) • n̂ = – u ( r )
[12]
where u is the magnitude of the velocity. With these simplifications, the volumetric flow-rate in the circular
tube is given by
r=R
Q =
∫
2πu ( r )r dr
[13]
r=0
Knowing the velocity profile u(r) then allows us to calculate the volumetric flow rate by integration.
Since the data set is usually limited to a finite number of values u(r), we cannot perform the exact analytic
integral. We can, however, use a numerical estimate to approximately evaluate the flow rate. This
numerical estimate is best understood by recognizing that integration is the process of calculating the area
underneath a curve.
The curve in Figure 3 represents the true function u(r), for which the discrete values u 1 = u ( r 1 ) ,
u 2 = u ( r 2 ) , etc. are available. We can approximate the integration by summing the areas represented
by the shaded rectangles. In this case, the flow rate can be estimated from
4
u(r)
u1
u2
dr1 dr2
r1
r2
r3
r4
r5
r6=R
Figure 3 Approximating an Analytic Integral With Discrete Numeric Values
6
Q ≈ 2π
∑ un r n dr n
[14]
n=1
Q ≈ 2π ( u 1 r 1 dr 1 + u 2 r 2 dr 2 + u 3 r 3 dr 3 + u 4 r 4 dr 4 + u 5 r 5 dr 5 + u 6 r 6 dr 6 )
[15]
Mass flow rate can be evaluated as
ṁ = ρQ
[16]
An analysis similar to above can be performed for rectangular cross-sections as well.
5
Air Flow Measurements
Bernoulli Obstruction Theory
Consider the generalized flow obstruction shown in Figure 4.
Figure 4
Velocity and Pressure Change through a Generalized Bernoulli Obstruction
Meter
The energy grade line (EGL) shows the height of the total Bernoulli constant
2
p V
h 0 = z + --- + -----γ 2g
[17]
The hydraulic grade line (HGL) shows
2
V
-----2g
[18]
where the height corresponding to elevation and pressure head
6
p
z + --γ
[19]
that is, the EGL minus the velocity head. The HGL is the height to which liquid would rise in a piezometer
tube attached to the flow. In an open-channel flow the HGL is identical to the free surface of the water.
The flow in the basic duct of diameter D is forced through an obstruction of diameter d; the β is a key
parameter of the device,
d
β = ---D
[20]
After leaving the obstruction, the flow may neck down even more through a vena contracta of diameter
D2 < d, as shown. Apply the continuity and Bernoulli equations for “incompressible steady frictionless flow”
to estimate the pressure change:
Continuity:
π 2
2
Q = --- D V 1 = D 2 V 2
4
[21]
Bernoulli
1 2
1 2
p 0 = p 1 + --- ρV 1 = p + --- ρV 2
2
2
Eliminating
[22]
V 1 , we solve these for V 2 or Q in terms of the pressure change ( p 1 – p 2 ) :
2 ( p1 – p2 )
Q
------ = V 2 ≈ ---------------------------------4
4
A2
ρ ( 1 – D2 ⁄ D )
1⁄2
[23]
But this is surely inaccurate because we neglected friction in a duct flow, where we know friction will be
very important. Nor do we want to get into the business of measuring vena contracta ratios D2/d for use in
Equation 23. We assume that D 2 ⁄ D ≈ β and then calibrate the device to fit the relation.
2 ( p1 – p2 )
Q = A t V t ≈ C d A t ------------------------4
ρ(1 – β )
1⁄2
[24]
where subscript t denotes the throat of the obstruction. The dimensionless discharge coefficient Cd
accounts for the discrepancies in the approximate analysis. By dimensional analysis for a given design we
expect
C d = f ( β, Re D )
[25]
ρV 1 D
Re D = -------------µ
[26]
where
The geometric factor involving
β in Equation 24 is called the velocity-of-approach factor
4 –1 ⁄ 2
E = (1 – β )
[27]
One can also group Cd and E in Equation 24 to form the dimensionless flow coefficient
Cd
α = C d E = -------------------------4 1⁄2
(1 – β )
α
[28]
7
Thus,
2 ( p1 – p2 )
Q = α A t -------------------------ρ
1⁄2
[29]
Obviously the flow coefficient is correlated in the same manner:
α = f ( β, Re D )
[30]
Occasionally one uses the throat Reynolds number instead of the approach Reynolds number
ρV t d
Re
= ---------DRe D = -----------β
µ
[31]
Since the design parameters are assumed known, the correlation of α or of C d is the desired solution
to the fluid-metering problem.
Figure 5 shows three basic devices recommended for use by the International Organization for
Standardization (ISO): the orifice, nozzle, and venturi tube.
Figure 5
Orifice, Nozzle, and Venturi Tube Configurations
8
Thin-Plate Orifice.
An orifice plate is nothing but a flat plate with a hole in it. Once placed in the duct, it restricts flow and
causes an increase in velocity similar to a venturi. Directly behind the orifice plate an area of low pressure
exists. By measuring the difference in pressure from this point to the-free flowing duct, the volumetric flow
rate can be found using the following equation:
2
πd 2∆P 3
Q = αε --------- ----------- m ⁄ s
4
ρu
[32]
where α flow coefficient, ε expansibility factor, d diameter of orifice, m, ∆P pressure drop over orifice
plate, Pa, ρ u density upstream of the device (i.e. at atmospheric pressure) kg/m3.
The values of α for the orifice plates are as follows:
65 mm orifice: α = 0.599
95 mm orifice: α = 0.596
The value of ε for an inlet orifice is given by the following expression
∆P
ε = 0.42 ------Pu
where
[33]
P u = pressure upstream of the device (atmospheric), Pa.
Nozzle.
The flow nozzle, with its smooth rounded entrance convergence, practically eliminates the vena
contracta and gives discharge coefficients near unity. The volumetric flow rate is determined from the
following expression;
2
πd 2∆P 3
Q = αε --------- ----------- m ⁄ s
4
ρu
[34]
The values of the dimensionless compound coefficient is given by the expression
αε = 0.986 – ( 0.0055 × 10
–3
× ∆P )
[35]
Nozzle Inlet Venturi Flow-Rate Measuring Device.
A venturi is just a gradual constriction in the duct. Since the mass flow rate is constant, the velocity
must increase as the area decreases. A change in static pressure occurs, and that change in pressure can
be used to find the flow rate using the following formula and dimensions (same as the nozzle).
2
πd V 2∆P 3
Q = αε ---------- ----------- m ⁄ s
4
ρu
[36]
The values of the dimensionless compound coefficient is given by the expression
αε = 0.986 – ( 0.0055 × 10
–3
× ∆P )
[37]
9
Conical Inlet Venturi Flow-Rate Measuring Device.
This device is mounted on the inlet side of the fan ducting. The volume flow
the expressions
Q relationship is given by
2
πd V 2∆P 3
Q = αε ---------- ----------- m ⁄ s
4
ρu
[38]
where
– 0.2
αε = 1.0 – 0.5Re d
4
when 2 × 10 < Re d < 30 × 10
4
[39]
and
αε = 0.960 when Re d ≥ 30 × 10
4
[40]
Note that conical inlet flow measurement should not be used when Red<2x104. Here, d diameter of constant diameter duct section (meters) = 0.095, ρI is density of air upstream device - (kg/m3), ∆P differential
pressure measured in pascals [kg/(m.s2)].
Flow Rate Measurements: Cusson Wind Tunnel
On/Off switch
large and small
manometers
Start button
Fan outlet
valve
venturi
Off button
Pitot tube
Fan
Place the
orifice or
nozzle
here
nozzle
conical inlet
65 mm orifice
95 mm nozzle
95 mm orifice
10
Experimental setups and Apparatus. (Cusson - For orifice plate)
Apparatus
Air Flow Bench, Manometers (small & large scale), Orifice (65 mm & 95 mm), Nozzle (95 mm), Conical
Inlet cone, Venturi, Pitot tube
Procedure: (Emperiments I, and III)
1.
Couple the 1 meter long ducting to the flow straightening section positioned at the inlet of the fan
using the toggle catches
2.
Attached the orifice inlet adapter housing to that 1 meter duct
3.
Support the overhanging section of the assembly to a suitable height using the stand
4.
Ensure that the flow straightening honeycomb disc is positioned squarely within the orifice inlet
adapter housing
5.
Insert the 65 mm orifice plate into the inlet adaptor housing. (The orifice plate is positioned with
the counter sunk side downstream from the inlet, i.e. facing into the housing)
6.
Fit the pitot static tube and scale to the 1 meter ductwork at any of the three radial positions.
(Blank off the pitot static tube tappings if not in use)
7.
Connect the pitot static tube to the smaller manometer (Total Pressure to the back of the small
manometer inlet) (Static Pressure to the limb of the smaller adapter
8.
Connect the orifice adapter housing tapping to the larger manometer limb.
9.
Set each manometer limb in the upright position, level and zero the manometer
10. Fully close the fan outlet valve and then switch on the fan
11. Set the orifice plate manometer limb to the most sensitive position possible and re-zero it again
(after disconnecting the pressure taping tube). Record the reading in kPa
Note: The readings should be multiplied with the multiplier written on that manometer.
Experimental setups and Apparatus. (Hampden)
Apparatus
Standard test section, pitot-static probe, probe positioner, Manometers
Procedure: (Experiment II)
1.
Install the standard test section in the wind tunnel
2.
Install a pitot-static probe, in the probe positioner and through the duct access hole in the test
section
3.
Connect pressure tubing from the static and total pressure taps on the pitot-static tube to one of
the appropriate manometers
4.
Using the variable frequency drive control to adjust fan speed and the pitot tube positioner to
locate the pitot tube vertical location in the duct, read and record velocity pressures at various locations
Note: The lab instructor will show how to turn on/off the drive control and monitor the speed.
Area of the inside duct (standard test section) is 0.444 ft2 (0.0413 m2) inside dimensions.
11
EXPERIMENT I
Pitot-Tube Based Velocity Profile and Flow Rate Measurements
Cusson Wind Tunnel (D = 147 mm)
Trial #
Pitot Tube Locations
Almost closed
∆p
M=1
+ 6 cm
M=2
+ 5 cm
M=3
+ 4 cm
M=4
+ 3 cm
M=5
+ 2 cm
M=6
+ 1 cm
M=7
CENTER: 0 cm
Effective
∆p
Average
Velocity
Volume Flow
Rate
Mass Flow
rate
V
Valve Outlet Positions
Center
∆p
V
Almost open
∆p
V
∆P eff = Eq ( 8 ) (Pascal)
V av (m/s)
Q = AV av
ṁ = ρQ (kg/s)
For air at 20oC and 1 atm; ρ = 1.20 kg/m3, µ = 1.8 E-5 kg/(m.s), ν = 1.51 E-5 m2/sec
1.
Traverse the pitot static tube across the diameter of the ductwork, noting the manometer reading
at each position and record them in Data Sheet. (The manometer should be in the most sensitive position
possible, bearing in mind that the maximum values will be obtained at the central positions)
2.
Repeat the experiments again by adjusting the fan outlet to the “center” and almost open” position and record your data in Data Sheet.
Assignments:
•
Complete the table above.
•
Plot the velocity across the diameter.
•
Calculate the entrance length required to establish a fully developed boundary layer at inlet (refer
to Chapter 6 of your text book).
•
Discuss the velocity profile within the light of the entrance length.
12
EXPERIMENT II
Pitot-Tube Based Velocity Profile and Flow Rate Measurements
Hampden Wind Tunnel (Test section Inside area 0.0413 m2)
Trial #
Pitot Tube Locations
N = 800 rpm
∆p
M=5
+ 8 cm
M=4
+ 6 cm
M=3
+ 4 cm
M=2
+ 2 cm
M=1
CENTER: 0 cm
Effective
∆p
Average
Velocity
Volume Flow
Rate
Mass Flow
Rate
V
Valve Outlet Positions
N = 1600 rpm
∆p
V
N = 2400 rpm
∆p
V
∆P eff = Eq ( 8 ) (Pascal)
V av (m/s)
Q = AV av
ṁ = ρQ (kg/s)
For air at 20oC and 1 atm; ρ = 1.20 kg/m3, µ = 1.8 E-5 kg/(m.s), ν = 1.51 E-5 m2/sec
1.
Traverse the pitot static tube across the cross section of the test chamber, noting the manometer
reading at each position and record them in Data Sheet. (The manometer should be in the most sensitive
position possible, bearing in mind that the maximum values will be obtained at the central positions)
2.
Repeat the experiments again by adjusting the fan outlet to the “center” and almost open” position and record your data in Data Sheet.
Assignments:
•
Complete the table above.
•
Plot the velocity across the plane.
•
Calculate the entrance length required to establish a fully developed boundary layer at inlet (refer
to Chapter 6 of your text book).
•
Discuss the velocity profile within the light of the entrance length.
13
EXPERIMENT III
Cusson Wind Tunnel (D = 147 mm)
Orifice Plate For Flow Rate Measurements
Valve Positions)
∆P kPa
αε
Q m3/s
αε
Q m3/s
Almost closed
Center
Almost Open
Nozzle For Flow Rate Measurements
Valve Positions)
∆P kPa
Almost closed
Center
Almost Open
Conical Inlet Venturi For Flow Rate Measurements
Valve Positions)
∆P kPa
αε
Q m3/s
Almost closed
Center
Almost Open
Nozzle Inlet Venturi For Flow Rate Measurements
Valve Positions)
∆P kPa
αε
Almost closed
Center
Almost Open
14
Q m3/s
Bernoulli Equation Demonstrator
Background
Pressure Measurements In Moving Fluids
Pressure measurements in moving fluids deserve special considerations. Consider the flow over the
bluff body shown in Figure 1.
Figure 6
Streamlines over a bluff body
Assume that the upstream flow is uniform and steady. Points along the two streamlines labeled as A
are to be studied. Along streamline A, the flow moves with a velocity, V 1 , such as at point 1 upstream of
the body. As the flow approaches point 2 it must slow down and finally stop at the front end of the body.
Point 2 is known as the stagnation point and streamline A the stagnation streamline for this flow. Along
streamline B, the velocity at point 3 will be V 3 and because the upstream flow is considered to be uniform
it follows that V 1 = V 3 . As the flow along B approaches the body, it is deflected around the body. From
conservation of mass principles, V 4 > V 3 . Application of conservation of energy between points 1 and 2
and between 3 and 4 yields
2
2
2
2
p 1 + ( ρV 1 ) ⁄ ( 2g ) = p 2 + ( ρV 2 ) ⁄ ( 2g )
[41]
p 3 + ( ρV 3 ) ⁄ ( 2g ) = p 4 + ( ρV 4 ) ⁄ ( 2g )
[42]
However, because point 2 is the stagnation point,
V 2 = 0 , and
2
p 2 = p total = p 0 = p 1 + ( ρV 1 ) ⁄ ( 2g )
[43]
2
Hence, it follows that p 2 > p 1 by an amount equal to V 2 ⁄ 2g , an amount equivalent to the kinetic
energy per unit mass of the flow as it moves along the streamline. If the flow is brought to rest in an
isentropic manner (i.e., no energy lost through irreversible processes such as through a transfer of heat1),
15
this translational kinetic energy will be transferred completely into p2 is known as the stagnation or total
pressure and will be noted as p0. The “total pressure” can be determined by bringing the flow to rest at a
point in an isentropic manner.
The flow at 1, 3, and 4 are known as the stream or static pressures2 of the flow. Because the flow is
uniform, V 1 = V 3 , so that p 3 = p 1 . The static pressure and velocity at points 1 and 3 are known as the
freestream pressure and freestream velocity. However, as the flow accelerates around the body its velocity
increases such that, from Eq (1), p 4 ≠ p 3 . The pressure, such as at point 4, is called a local static
pressure. The “static pressure” is that pressure sensed by a fluid particle as it is moves with the same
velocity as the local flow.
Background
Closely related to the steady-flow energy equation is relation between pressure, velocity, and elevation
in a frictionless flow, now called the “Bernoulli equation”. The Bernoulli equation is very famous and very
widely used, but one should be wary of its restrictions - all fluids are viscous and thus all flows have friction
to some extent. to use the Bernoulli equation correctly, one must confine it to regions of the flow which are
nearly frictionless.
For an incompressible fluid, the Bernoulli equation is
( p2 – p1 ) 1 2
2
---------------------- + --- ( V 2 – V 1 ) + g ( z 2 – z 1 ) = 0
[44]
2
ρ 2
2
V
p
V
p
-----1 + -----1- + gz 1 = -----2 + -----2- + gz 2 = const
[45]
ρ
22
ρ 22
V
p
V
p
-----1 + -----1- + z 1 = -----2 + -----2- + z 2 = const
[46]
γ 2g
ρ 2g 2
p
V
where --- is the static pressure head; ------ is the velocity pressure head; and z is the potential energy head.
γ pressure head is equal to the2gsum of the static and velocity pressure heads. This is the Bernoulli
The total
equation for steady frictionless incompressible flow along a streamline.
A venturi tube can be used to demonstrate the Bernoulli equation as shown in Figure 3.
Figure 3: Venturi
AIR FLOW
DIRECTION
to
It is readily apparent that the potential energy head is zero ( z 1
2
= z 2 , the Bernoulli equation reduces
2
p
V
p
V
-----1 + -----1- = -----2 + -----2- = const
γ 2g
ρ 2g
[47]
Using the conservation of mass, the mass flow rate at points 1 and 2 must be the same, or
ρV 1 A 1 = ρV 2 A 2
[48]
16
MEEM 3220 ENERGY LABORATORY
EXPERIMENT IV
(HAMPDEN WINDTUNNEL)
Objective
To investigate the Bernoulli equation as it relates to pressure and velocity of a fluid along a streamline.
Experimental Setups
3.
Attached the venturi section with convergent and divergent sections oriented as shown in Figure
4. Make sure that the fastening screws are tightened to provide a tight seal.
Figure 7
Schematic Diagram
4.
Install a pitot-static probe into the probe positioner and one access hole in the duct. Align the
probe head with the center line of the convergent-divergent section.
5.
Connect the total and static pressure taps of the pitot-static tube to the appropriate manometers
as shown in Figure 4. please note that the choice of which inclined manometer is used depends on the
magnitude of the static pressure; and this changes along the length of the venturi tube. Two universal tee
connectors are required.
17
Experimental Procedure
6.
Turn ON the fan and use the variable frequency drive control to adjust fan speed. Always run the
fan in forward for proper air flow direction.
STEP 1
Press “JOG” button.
STEP 2
Press “LOCAL”.
STEP 3
Press “FWD”.
STEP 4
Press “ARROW BUTTON UP” to increase the speed or Press “ARROW BUTTON
DOWN” to decrease the speed.
STEP 5
Press “STOP” to turn OFF the fan
STEP 6
Directly Press “FWD” to choose the same speed as chosen before.
7.
Measure and record the total, static, and velocity pressures (inches of H2O) at various points
along the cross section of the venturi tube at each location.
Equations:
P total = P static + P dynamic
[49]
∆P = P total – P static = P dynamic = P velocity ; P dynamic is also known as P velocity
[50]
P = ρgh static = γ h static
[51]
( P static ) abs
ρ = -------------------------RT
[52]
2
where P static in absolute pressure (Pa or N ⁄ m ) ( P static ) abs
2
2
constant (287 m ⁄ ( s • K ) T is the absolute temperature (K)
2
2∆P
----------- in m/s; where ∆P in Pascal or N ⁄ m
ρ
V =
= P atm + ( P static ) gauge R is the gas
[53]
Conversions:
2
–4
1cm = 1 × 10 m
1Pa = 1N ⁄ m
2
[54]
2
[55]
1inH 2 O = 249.1Pa
[56]
1atm = 101325Pa
[57]
1inHg = 3372.2Pa
[58]
1mmHg = 133.32Pa
[59]
Assignments
Plot ρAV and Bernoulli Constant along the venturi.
(Refer to Eq (3) for Bernoulli Constant)
18
DATA SHEET IV
Fan = _____rpm Patm = ___ inH2O = ____N/m2 T = ________ K
(Pressure)gauge
Test
#
Types
inH2O
N/m2
A
V
ρAV
P/γ
V2/2g
Bernoulli
Constant
m2
m/s
kg/s
meter
meter
meter
Pstatic
1
Ptotal
∆P
Pstatic
2
Ptotal
∆P
Pstatic
3
Ptotal
∆P
Pstatic
4
Ptotal
∆P
Pstatic
5
Ptotal
∆P
Pstatic
6
Ptotal
∆P
Pstatic
7
Ptotal
∆P
Pstatic
8
Ptotal
∆P
Pstatic
9
Ptotal
∆P
Pstatic
10
ρ = _______ kg/m3
Ptotal
∆P
19