Question

The air velocity in the duct of a heating system is to be measured by a Pitot-static probe inserted into the duct parallel to the flow. If the differential height between the water columns connected to the two outlets of the probe is 2.4 cm, determine

(a) the flow velocity and
(b) the pressure rise at the tip of the probe. The air temperature and pressure in the duct are 45°C and 98 kPa, respectively.

Answer 1

20.94 m/s, 235.44 Pa

Step-by-step explanation:

Acceleration due to gravity g =
`9.81 m/s^2`

height h = 0.024 m

From density of air = P/RT

= (98000)/(287 * 318.14) =
`1.073 kg/m3`

Using Bernoulli equation

`(P/density*g) + (V^2/2g) + z = constant`

`(P1/density*g) + (V1^2/2g) + z1 = (P2/density*g) + (V2^2/2g) + z2`

Here z1 = z2 (since the outlets have the same differential height) and V2 = 0 (no velocity at the tip)

Solving and making V1 subject of the formula

`V1 = √((P2 - P1)/density of air)`

`V1 = √((2*density of water* g*h)/density of air)`

`V1 = √((2*1000*9.81*0.024/1.073)`

= 20.94 m/s

Change in pressure P2 - P1= density of water * g * height

= 1000*9.81*0.024

=235.44 Pa