Fliud mechanic

Examples

A simple manometer is used to measure the 

pressure of oil sp.gr. 0.8 flowing in a pipeline. 

Its right limb is open to atmosphere and the 

left limb is connected to the pipe. The center 

of the pipe is 9.0 cm below the level of the 

mercury in the right limb. If the difference of 

the mercury level in the two limbs is 15 cm, 

determine the absolute and the gauge 

pressures of the oil in the pipe. 

Solution:

ρ = 0.8 (1000) = 800 kg/m3

P1 = P2

P1 = (0.15 –0.0 9)m(800 kg/m3

)9.81m/s2

+ Pa

P2 = (0.15) m (13600 kg/m3

) 9.81 m/s2

 + Po

Pa = 15 (13600) 9.81 + Po + [(15 – 9)cm 

(800 kg/m3

) 9.81 m/s2

] 

 = 1.20866 x 105

 Pa (Absolute pressure) 

The gauge press. = Abs. press. – Atm. Press. 

= 1.20866 x 105

 - 1.0325 x 105

= 1.9541 x 104

 Pa 

Example -3.5-

The following Figure shows a 

manometer connected to the 

pipeline containing oil of sp.gr. 0.8. 

Determine the absolute pressure of 

the oil in the pipe, and the gauge 

pressure. 

Solution:

ρa = 0.8 (1000) = 800 kg/m3

P1 = P2

P1 = Pa – h2 ρa g 

P2 = Po + h1 ρm g 

⇒ Pa = Po + h1 ρm g + h2 ρa g

 = 1.0325 x 105

 + (0.25) m 

(13600 kg/m3

) 9.81 m/s2

 + 

(0.75) m (800 kg/m3

) 9.81 m/s2

 = 1.40565 x 105

 Pa