Pitot Stagnation Tubes
Under Laminar
Conditions
*P7.24
Jace Benoit
February 15, 2007
Problem Statement
Air at 20°C and 1 atm flows past the flat plate
in Fig. P7.24 under laminar conditions. There
are two equally spaced pitot stagnation tubes,
each placed 2 mm from the wall. The
manometer fluid is water at 20°C. If
U=15m/s and L=50cm, determine the values
of the monometer readings h1 and h2.
Determine these values if the manometer fluid
is replaced with mercury.
Sketch
u1
u2
Fig P7.24
Assumptions
Laminar Flat-Plate Flow
 Incompressible Flow
 Constant Temperature
 No Mixing at Liquid-Air Interface
 Frictionless Flow
 Steady Flow

Solution
The following gives a summary of the densities and
kinematic velocity from tables A.1, A.2, and A.3.
Density (ρ)
Kinematic
Velocity (v)
Air
Water
Mercury
1.20 kg/m3
998 kg/m3
13,550 kg/m3
1.50 E-5 m2/s Not Required Not Required
Solution
Since this is laminar flow along a boundary layer, the
Blasius equation (7.21) can be used to determine η.
1/ 2
U 
  y

 vx 


15m / s
1  0.002m

5
2
 (1.50 10 m / s)(0.5m) 


15m / s
 2  0.002m 

5
2
 (1.50 10 m / s)(1.0m) 
1/ 2
 2.828
1/ 2
 2.000
Solution
Now that η is known, Table 7.1 on page 462 can be
used to approximate u/U.
For η = 2.8, u/U = 0.81152 and for η = 3.0, u/U = 0.84605, so
for η1 = 2.828:
3.0  2.8
3.0  2.828

0.84605  0.81152 0.84605  u
U
u1
 0.8164
U
Fortunately, the u/U value corresponding to η = 2.0 can be read
directly from Table 7.1 to be 0.62997.
Solution
Plugging U = 15 m/s into the u/U expression yields
the velocities at the entrances of each manometer.
u1  (15m / s)  (0.8164)  12.25m / s
u2  (15m / s)  (0.6298)  9.447m / s
Solution
With the inlet velocities known, Bernoulli’s equation
(3.77) can be used to solve for the inlet pressure
assuming atmospheric pressure at the outlet of the
manometer.
P


1 2
2
 u2  u1  g z2  z1   0
 2
Solution
Rearranging this equations with the appropriate
cancellations yields the following:
 u2 
Pinlet  Poutlet    
 2
2


(
12
.
25
m
/
s
)
2
Pi ,1  101325Pa  (1.20kg / m )  
  101235Pa
2


 (9.447m / s) 2 
Pi , 2  101325Pa  (1.20kg / m )  
  101271Pa
2


2
Solution
Now that the inlet and outlet pressures are known,
this problem can now be solved as a manometer
problem.
Pi   water gh  Po
Po  Pi
h
 water g
Solution
Finally, the manometer heights for water are as
follows:
(101325  101235) Pa
h1 
 9.2mm
3
2
(998kg / m )(9.81m / s )
(101325  101271) Pa
h2 
 5.5mm
3
2
(998kg / m )(9.81m / s )
Solution
All that needs to be done to determine the height if
mercury is the manometer fluid is changed the
density in the previous equation to 13,550 kg/m3.
(101325  101235) Pa
h1 
 0.68mm
3
2
(13,550kg / m )(9.81m / s )
(101325  101271) Pa
h2 
 0.41mm
3
2
(13,550kg / m )(9.81m / s )
Application to BME
The manometer with one end being submerged
in water and the other in another medium can
mimic a catheter which is used to measure
various biological effects such as cardiac output
and blood pressure. The catheter is filled with a
saline solution similar to how this manometer is
filled with water and mercury, and pressure is
exerted at each ends of the catheter.
Reference
White, Frank M. Fluid Mechanics. 5th Ed.
McGraw-Hill Companies, Inc.: New York. 2003.