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
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