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# A Pitot-static tube used to measure air velocity is connected to a differential pressure gage. If the air temperature is 20$^{\circ}C$ at standard atmospheric pressure at sea level, and if the differential gage reads a pressure difference of 2 kPa, what is the air velocity?

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$\textbf{Given data}$

$\text{T} = 20^0\text{c} = 20 + 273 =293^0\text{K}$

Air is at standartd atmospheric pressure i.e. ${\text{ P}}_{atm} = 101.3\text{kPa}$

Differential gage pressure $\delta{\text{P}} = 2 \text{kPa}$

$\textbf{Goal is to find out the air velocity}$

Using Pitot static tube formula,

$\text{V} = \sqrt{\frac{2 \cdot \delta{\text{P}}}{\rho}} \ \ \ \ ... (1)$

To calculate $\rho$ , using ideal gas equation,

$\text{P} = \rho \cdot \text{R} \cdot \text{T} \ \ \ ...(2)$

Where R = gas constant = $0.287 \frac{\text{KJ}}{\text{kg.k}}$

Substituting P, R ,T values in equation (2).

$101.3 = \rho \times 0.287 \times 293$

$\rho = \frac{101.3}{0.287 \times293}$

$\rho = 1.205\frac{\text{kg}}{{\text{m}}^{3}}$

Substituting $\rho$ and $\delta{P}$ value in equation (1)

$\text{V} = \sqrt{\frac{2 \times 2000}{1.205}}$

$\boxed{\text{V} = 57.62 \frac{\text{m}}{\text{s}} }$

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