A horizontal venturimeter with inlet diameter 200mm and throat diameter 100mm is employed to measure the flow of water. The reading of the differential manometer connected to the inlet is 180mm of Hg. If the coefficient of discharge is 0.98, determine the rate of flow.
Solution:
Inlet dia of venturimeter, D1= 200mm= 0.2m
Therefore, area of inlet, A1= (π/4)*0.22
Throat dia D2= 100mm= 0.1m
Area of throat, A2= (π/4)*0.12
Reading of differential manometer, x = 180mm
= 0.18m of Hg
Coefficient of discharge, Cd =0.98
Rate of flow, Q
To find the difference of pressure head (h), we have
h = x[(sh/sp)-1]
h = 0.18[(13.6/1) -1]= 2.268 m
To find ‘Q’ using this relation
Q = Cd[a1a2 /√( a1 2 - a2 2 )]* √(2gh)
Q = 0.98[0.0314*0.00785 /√( 0.03142 -0.007852 )]* √(2*9.81*2.268)
Q= (0.000241*6.67)/0.0304
Q = 0.0528 m3 /sec
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