An oil of specific gravity 0.8 is flowing through a venturimeter having inlet diameter 20cm and throat diameter 10cm. The oil(so = 0.8)-mercury differential manometer shows a reading of 25cm. Calculate the discharge of oil through the horizontal venturimeter. Take Cd=0.98.

 An oil of specific gravity 0.8 is flowing through a venturimeter having inlet diameter 20cm and throat diameter 10cm. The oil(so = 0.8)-mercury differential manometer shows a reading of 25cm. Calculate the discharge of oil through the horizontal venturimeter. Take Cd=0.98.

 Solution: 

Given: 

Specific gravity of oil, so =0.8 

Specific gravity of mercury sh =13.6 

Reading of differential manometer x =25cm 

Therefore difference of pressure head, h =x [(sh/so) -1] 

 =25[(13.6/0.8) -1] cm of oil = 25[17-1] =400 cm of oil. 

Dia at inlet, d1=20cm 

 Area at inlet, a1 = (πd1 2 )/4 = (π202 )/4 =314.16cm2 

 Similarly at throat, d2 =10cm 

 a2 =(π102 )/4 =78.54cm2 

 Cd = 0.98 (given) 

Therefore discharge Q is given by 

Q = Cd*(a1a2/(√a1 2 -a2 2 ))*(√2gh) 

 =0.98*(314.16*78.54/(√314.162 -78.542 ))*(√2*9.81*400) 

 =21421375.68/ (√98696-6168) 

 =21421375.68/304 cm3 /s   

 =70465cm3 /s                   Q =70.465 lit/s