Question

a formula is given for the volumetric flow rate (picture), Q, as a function of density, the pressure difference, and the area ratio, for a venturi meter. Derive a similar equation for a case when the venturi meter is inclined at an angle θ, where the distance between points 1 and 2 in the meter is given by d.

1 1 P1 + 5 poi + pgyi = P2 + z poz + Pg42. M-es= 32 - I 1 1 h Get p1 - p2 from hydrostatics Pi P2 = pi + pghi. 2 P1 – P2 2 P1 – P2 Q = A1 = A2 2 2 р р (4) A1 A2 – 1 1-) A2 A1

Answer 1

Hello there. You may be surprised because we are getting same equation in this inclined case also. It is happening because the effect of theta will be reflected in h (differential head of venturimeter). So don't worry about this. When we will try to find h with some method then the effect of inclination will automatically considered.

Please find the attached solution.

Solution Az, V2 P2 e ه) d Z₂ 요 거 Reference for Datum Applying Berncullie's equ between section 0-0 and 2 Pi V,² Ve Pe sg { Assuming. No + f of + Z, 89 29 ag head losses oo V2>V, and Pr>P

V2 ²v, ² ag Picha Pi-pe 3g + 2,-22) V₂²-v, 2 ГР. Pi 8g 다 3g +22) J - Piesometric head of ☺ Piesomekic heed at @ V2²V, 2 c wo this is differential head of Venturimeter. ag V2²V, ²= Qgh • Qtu = AzV2 = ANI 2 V, ata A2 Ve=alta Az alta Ai J Itu AI =2gh here the only unknown ish Ota = Aida Jagh A,2-A₂2 2 There are so many methods of calculeting ho Venturimeter with Piezometer. Venturimeter with monometer. in actual anolysis ( head loss will come into bicture)