Question

A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and pressure are 44oF and 24 psia. A water manometer connected to the Pitot-static tube indicates a reading of 3.0 in. (a) Determine the helium velocity. (b) Is it reasonable to consider the flow as incompressible?

Answer 1

Answer:

V=203 ft/s

Explanation:

answer is available in word document named "attach 1" due to some technical error in maths equation. please find the attached document

Answer 2

Answer:

Part A:

$V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s$

Part B:

Ma=0.0737

Since Ma<0.3, it means the flow is in compressible.

Explanation:

Part A:

According to Bernoulli equation:

$P_1+\frac{\rho_H}{2}V^2_1 =P_{2}+\frac{\rho_H}{2}V^2_2\\ V_2=0,\\P_1+\frac{\rho_H}{2}V^2_1 =P_{2}$

Velocity will become:

$V_1=\sqrt{\frac{2(P_2-P_1)}{\rho_H}}$.........Eq (1)

Now,$P_2-P_1$ can be calculated from the specific weight of water and helium$P_2-P_1$$=(\gamma_{h2}o-\gamma_H)h$

Since the specific weight of helium is much smaller than specific weight of water we can neglect the specific weight of helium.

$P_2-P_1$=$=(\gamma_{h2o})h$

For water,$\gamma_{h2o}=62.43 lb/ft^3$

h=3.0 in

Density of helium:

$\rho_H=\frac{P}{RT}$

T=460+44=504 degree R

R=$1.242*10^4 ft.lb/R.slug$

$\rho_H=\frac{24*12^2}{1.242*10^4*504}\\ \rho_H=5.5210*10^{-4} lb/ft^3$

From Eq (1):

$V_1=\sqrt{\frac{2\gamma_{h20}h}{\rho_H}}\\ V_1=\sqrt{\frac{2*62.43*3}{5.5210*10^{-4}*12}}\\ V_1=237.778 ft/s$

Part B:

Checking Ma:

$Ma=\frac{V}{c}$

c is speed of sound:

k=1.66 for helium, In ideal gases:

$c=\sqrt{kRT}\\ c=\sqrt{1.66*1.242*10^4*504}\\ c=3223.51 ft/s\\Ma=\frac{237.778}{3223.51}\\ Ma=0.0737$

Since Ma<0.3, it means the flow is in compressible.