Question

A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 55oF and 28 psia. A water manometer connected to the Pitot-static tube indicates a reading of 2.3 in. (a) Determine the helium velocity. (b) Is it reasonable to consider the flow as incompressible

Answer 1

Answer:

a) V=203 ft/s

b) Since the Mach number is 0.063 < 0.3, we can conclude that the flow

    is incompressible!  

Explanation:

solution:

Start by deriving the expression for the fluid velocity, from the Bernoulli equation:  

$p_0=p_{static} +\frac{1}{2} p_{fluid}V^2\\V=\sqrt{2(p_0-p_{static})/p_{fluid}}$

Next, we need to express the pressure difference between the stagnation pressure $p_{0}$ and the static pressure. This can be done using the relation between the pressures and the specific weights:  

$p_{0}-p_{static}=(r_{manometer-r_{fluid}) } h$

Since $r_{manometer} =rh20$ and it is much greater than $r_{fluid} =r_{HE}$ we can   obtain from the previous equation:

$p_{0}$-p_static=$r_{manometer}$h

V=√2*$r_{manometer}$h/r_fluid

Using the ideal gas formula we can obtain the neccesarry density of Helium:  

r_fluid=p/R.T

          =5.80*10^-4 sl/ft^3

Using the value for water specific weight from the table B.1, we return to the expression for fluid velocity:  

V=√2*$r_{manometer}$h/r_fluid

V=203 ft/s

Next, we need to obtain the Mach number for the fluid flow velocity, so that we can find out if the flow is in compressible or not. The Mach number formula is:  

M=V/c

  =V/√kRT

Substituting the known values for the real gas constant R, the given temperature T and the adiabatic index k, we obtain:  

M=V/c

  =0.063

Since the Mach number is 0.063 < 0.3, we can conclude that the flow is incompressible!