Question

A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is connected between two sections of the pipe (Fig. 14-55); the cross-sectional area A of the entrance and exit of the meter matches the pipe's cross-sectional area. Between the entrance and exit, the fluid flows from the pipe with speed V and then through a narrow ''throat'' of cross-sectional area a with speed v. A manometer connects the wider portion of the meter to the narrower portion. The change in the fluid speed is accompanied by a change Δp in the fluid's pressure, which causes a height difference h of the liquid in the two arms of the manometer. (Here Δp means pressure in the throat minus pressure in the pipe.) Suppose that the fluid is freshwater, that the cross-sectional areas are 71 cm2 in the pipe and 34 cm2 in the throat, and that the pressure is 58 kPa in the pipe and 43 kPa in the throat. What is the rate of water flow in cubic meters per second?

Answer 1

Bernoulli'sequationstates:

both pressres are static

fluid is fresh water -->ρ= 1000 kg/m³

Δp=-.5*ρ*Δ(v²)= 58-43 kPa

equation of continuity

mass inflow=mass outflow

A₁*V₁= A₂*V₂ israte of water flow

2*Δp/ρ=v₂²-V₁²

2 equations and 2 varibles

A1=71 cm2

A2= 34cm2

2*Δp/ρ=V₁² * [(A₁/A₂)² -1]

V₁ =2.98 m/sec from SI units

rate of water flow=A₁*V₁= 212.13 * 10^-4 m³/sec

Answer 2

(a)
from the theory we know that the equation ofcontinuity as
A1 v1 = a2v2
from the given
?p = p1
= 1.80atm
so we get
v = v{2 ?p / [?((A /a)2 - 1)]}
= v{2 (1.01 x105 Pa) / [(1000 kg / m3)((5a /a)2 - 1)]}
= .......... m / s
(b)
again using equation of continuity we get
V = (A / a) v
= (5a / a) v
= ......... m /s
(c)
the rate of flow is
A v = (p / 4) (5.0 x 10-4m2) (v)
=......... m3 / s

Answer 3

(a)
from the theory we know that the equation ofcontinuity as
A1 v1 = a2v2
from the given
?p = p1
= 1.80atm
so we get
v = v{2 ?p / [?((A /a)2 - 1)]}
= v{2 (1.01 x105 Pa) / [(1000 kg / m3)((5a /a)2 - 1)]}
= 0.02356 m / s
(b)
again using equation of continuity we get
V = (A / a) v
= (5a / a) v
= 0.11783. m /s
(c)
the rate of flow is
A v = (p / 4) (5.0 x 10-4m2) (v)
=9.2546x10^-6 m3 / s