(a) Explanation Given: The differential height ( h ) between the water columns is 2.4 cm. The air temperature in the duct is 45 ° C . The pressure in the duct is 98 kPa. Calculation: Consider the density ( ρ ) of water as 1 , 000 kg / m 3 Consider the gas constant ( R ) of air is 0.287 kPa ⋅ m 3 / kg ⋅ K . Consider the point 1 as the side of the probe which is parallel to the flow and it is connected to the Pitot - static tube and the point 2 as the tip of the probe which is normal to the flow and it is connected to the dynamic arm of the Pitot - static probe. The point 2 is stagnation , so V 2 = 0 . Consider the elevation at point 1 and point 2 is equal. i.e., z 1 = z 2 . Consider the acceleration due to gravity ( g ) as 9.81 m/s 2 . Find the pressure rise at the top ( P 2 − P 1 ) of the Pitot-static probe using the relation: P 2 − P 1 = ρ g h Substitute ρ = 1 , 000 kg / m 3 , g = 9.81 m/s 2 , and h = 2.4 cm in the above Equation: P 2 − P 1 = 1 , 000 × 9.81 × 2.4 cm × 1 m 100 cm = 235.4 N/m 2 = 235.4 Pa Find the density of air ( ρ air ) using the relation: ρ air = P R T = 98 kPa 0.287 kPa .m 3 / kg ⋅ K × ( 45 + 273 ) K = 1 (b) To determine The pressure rise at the tip of the probe.

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Chapter 12, Problem 17P

(a)

To determine

The flow velocity.

(b)

To determine

The pressure rise at the tip of the probe.

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Chapter 12, Problem 16P

Chapter 12, Problem 18P

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