Question

Figure shows the schematic for the measurement of velocity of air (density =1.2 kg/$m^3$) through a constant-area duct using a pitot tube and a water tube manometer. The differential head of water (density=1000 kg/$m^3$) in the two columns of the manometer is 10 mm. Take acceleration due to gravity as 9.8 m/$s^2$. The velocity of air in m/s is

• A. 6.4
• B. 9.0
  
• C. 12.8
• D. 25.6

Answer 1

The correct option is C 12.8

Given data:

Density of air,

${\rho }_a=1.2kg/m^3$
Density of water,
${\rho }_w=1000kg/m^3$
Differential head in manometer
h= 10mm of water
=0.01 m of water
This reading is the dynamic pressure head. Hence dynamic pressure.

${\rho }_{dyn}=(\rho gh{)}_{water}={\rho }_{w}gh$

$1000\times 9.81\times 0.01=98.1N/m^2$

$also{\rho }_{dyn}={\left(\frac{1}{2}\rho V^2\right)}_{air}=\frac{1}{2}{\rho }_a{V}^2$

$\therefore 98.1=\frac{1}{2}\times 1.2\times V^2$

$or{V}^2=163.5$

$V=12.78m/s$

Alternatively

$V=\sqrt{2gh}$

$h=x(\frac{{\rho }_m}{\rho }-1)$
where
x = Reading of differential manometer
$=10\times {10}^{-3}m$
${\rho }_m$ =Density of manometric fluid
$=1000kg/m^3$
$\rho$ = Density of following fluid = 1.2 kg/$m^3$
$\therefore V=\sqrt{2\times 9.81\times 10\times {10}^{-3}(\frac{1000}{1.2}-1)}$

$V=12.779m/s=12.8m/s$