The inlet and throat diameter of a horizontal venturimeter are 30cm and 10cm respectively. The liquid flowing through the meter is water. The pressure intensity at inlet is 13.734N/cm3 while the vacuum pressure head at the throat is 37cm of mercury. Find the rate of flow. Assume that 4% of the differential head is lost between the inlet and throat. Find also the value of Cd for the venturimeter.

 The inlet and throat diameter of a horizontal venturimeter are 30cm and 10cm respectively. The liquid flowing through the meter is water. The pressure intensity at inlet is 13.734N/cm3 while the vacuum pressure head at the throat is 37cm of mercury. Find the rate of flow. Assume that 4% of the differential head is lost between the inlet and throat. Find also the value of Cd for the venturimeter. 

Solution: 

Given:

 Dia at inlet, d1=30cm 

Area at inlet, a1=(πd1 2 )/4=(π302 )/4=706.85cm2 

 Dia at throat, d2=10cm 

Area at throat, a2=(π102 )/4=78.54cm2 

 Pressure at entry, p1=13.734N/cm2 =13.734*104N/m2 

 Therefore pressure head, p1/ρg=13.734*104 /9.81*1000 

 =14m of water 

p2/ρg= -37cm of mercury

 = (-37*13.6/100) m of water 

 =-5.032 m of water 

Differential head, h = p1/ρg- p2/ρg=14-(-5.032) =14+5.032

                                                           = 19.032 m of water 

                                                              =1903.2 cm 

Head lost, hf= 4% of h=0.04*19.032=0.7613 m 

Cd=√((h- hf)/h) =√(19.032-0.7613)/19.032 

 =0.98 

Therefore discharge 

Q= Cd*(a1a2/(√a1 2 -a2 2 ))*(√2gh) 

 = 0.98*(706.85*78.54/(√706.852 - 78.542 ))*(√2*9.81*1903.2) 

 = (105132247.8)/√(499636.9-6168)=149692.8cm3 /s 

 Q = 0.14969m3 /s