Question

Figure shows a pilot-static tube to measure air speed. (attached to a manometer)
The manometer shows a difference in levels $9mm$. lt contains water of density 998 $kg/m^3$. The density of air is 1.2 $kg/m^3$. The gravitational field strength is 9.81 m/s${}^2$. The agreement between the measured and calculated values of air speed is not very good so that $V_{calibrated}=C{v}_{air}$ where C is called calibration coefficient. Then:

• A. the pressure difference is $88Pa$ (nearly).
• B. the coefficient C is due to fact that even air is slightly viscous causing some turbulence by the hypodermic needles.
• C. in order to apply Bernoulli equation, the flow need not be streamline non-viscous and incompressible.
• D. if calibrated flow meter records air speed of $8.8m/s$, the factor C is $\mathrm{0.73.}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A the pressure difference is $88Pa$ (nearly).
B the coefficient C is due to fact that even air is slightly viscous causing some turbulence by the hypodermic needles.
D if calibrated flow meter records air speed of $8.8m/s$, the factor C is $\mathrm{0.73.}$

For(A):Pressure difference $={\rho }_{water}g.\left(\Delta h\right)$
$=\left(998\right)\left(9.81\right)(9\times {10}^{-3})$
$=88.11N/m^2=88.11Pa$
$\simeq 88Pa$
For(B): Coefficient C is also called Mach factor and also takes account of compressibility of air.
For(C): Bernoulli equation applies strictly to streamlined, non-viscous, incompressible flow.
For(D): $p_s=$ static pressure above the left hand manometer arm
$p_T$ Total pressure at the head of the right hand manometer arm
$p_T$ $-$$p_s+\frac{1}{2}{\rho }_{air}{v}^2$
$p_T-p_s={\rho }_{water}.g.\Delta h=\frac{1}{2}{\rho }_{air}{v}^2$

$\Rightarrow v=\sqrt{\frac{2{\rho }_{water}.g.\left(\Delta h\right)}{{\rho }_{air}}}$

$=\sqrt{\frac{2\times 88Pa}{1.2kg/m^3}}$

$=12.11m/s$

Coefficient, $C=\frac{v_{calibrated}}{v_{air}}=\frac{8.8m/s}{12.11m/s}$$\simeq 0.73$ (to two significant figure)