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Question

A Pitot-static tube used to measure air velocity is connected to a differential pressure gage. If the air temperature is 20C^{\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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Answered 2 years ago
Answered 2 years ago
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Given data\textbf{Given data}

T=200c=20+273=2930K\text{T} = 20^0\text{c} = 20 + 273 =293^0\text{K}

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

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

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

Using Pitot static tube formula,

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

To calculate ρ\rho , using ideal gas equation,

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

Where R = gas constant = 0.287KJkg.k0.287 \frac{\text{KJ}}{\text{kg.k}}

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

101.3=ρ×0.287×293101.3 = \rho \times 0.287 \times 293

ρ=101.30.287×293\rho = \frac{101.3}{0.287 \times293}

ρ=1.205kgm3\rho = 1.205\frac{\text{kg}}{{\text{m}}^{3}}

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

V=2×20001.205\text{V} = \sqrt{\frac{2 \times 2000}{1.205}}

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

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