Example2

A conical vessel is connected to a U-tube having 

mercury and water as shown in the Figure. When 

the vessel is empty the manometer reads 0.25 m. 

find the reading in manometer, when the vessel is 

full of water. 

إضافة شرح

Solution:

P1 = P2

P1 = (0.25 + H) ρw g + Po 

P2 = 0.25 ρm g +Po

⇒ (0.25 + H) ρw g + Po = 0.25 ρm g +Po

⇒ H = 0.25 (ρm – ρw)/ ρw

= 0.25 (12600 /1000) = 3.15 m 

When the vessel is full of water, let the mercury 

level in the left limp go down by (x) meter and the mercury level in the right limp go to 

up by the same amount (x) meter. 

i.e. the reading manometer = (0.25 + 2x) 

 P1 = P2

P1 = (0.25 + x +H + 3.5) ρw g + Po 

P2 = (0.25 + 2x) ρm g +Po

⇒ (0.25 + x +H + 3.5) ρw g + Po = (0.25 + 2x) ρm g +Po

⇒ 6.9 + x = (0.25 + 2x) (ρm/ ρw) ⇒ x = 0.1431 m

The manometer reading = 0.25 + 2 (0.1431) = 0.536 m 

Example -3.7-

The following Figure shows a 

compound manometer connected to the 

pipeline containing oil of sp.gr. 0.8. 

Calculate Pa. 

Solution:

ρa = 0.8 (1000) = 800 kg/m3

Pa + 0.4 ρa g – 0.3 ρm g + 0.3 ρa g – 

0.3 ρm g – Po = 0 

⇒ Pa = Po + 0.7 ρa g – 0.6 ρm g

 = 1.01325 x 105

 – 0.7 (800) 

9.81 + 0.6 (13600) 9.81 

 = 1.75881 x 105

 Pa