Question

If a dynamic pressure of 250 Pa was measured in a water tunnel using a pitot static tube and a mercury (denisty of 13650 kg.m ^ -3) manometer inckunded at 30 degrees, what is the velocity being measured in the water tunnel?

Answer 1

OPTION = D

Pressure measured = 250 Pa

Mercury density = 13650 Kg/m³

Pressure = P = Density*gravity*height.

It is said that manometer is inclined at 30⁰

vertical raised height = h*sin 30⁰

P = 250 Pa = 13650*9.81*h*sin 30⁰.

Mercury raise in manometer = h = 0.003733 m.

In the pitot-static tube, all pressure head is converted into velocity head.

Pressure head = P/w

where P = pressure = 250 pa

w = specific weight = 9.81*1000 = 9810 Kg/m³

pressure head = P/w = 250 / 9810 = 0.02548 m

velocity head = V² / 2g

where V = velocity in water tunnel

Lets us equate pressure head and velocity head.

P/w = V² / 2g

0.02548 = V² / 2*9.81

velocity in watre tunnel = V = 0.7070 m/sec.