Answer to Question #233449 in Mechanics | Relativity for john

Question #233449

The Venturi tube shown in Figure 8 may be used as a fluid flow meter. Suppose

the device is used at a service station to measure the flow rate of gasoline

(ρ = 7.00310

2 kg

m

3

) through a hose having an outlet radius of 1.20 cm. If the

difference in pressure is measured to be P1 − P2

= 1.20kP a and the radius of

the inlet tube to the meter is 2.40cm, find

(a) the speed of the gasoline as it leaves the hose and

(b) the fluid flow rate in cubic meters per second.


1
Expert's answer
2021-09-06T11:35:15-0400

Given that

Density of fluid "\u03c1=7 \\times 10^2 \\;kg\/m^3"

Inlet hose radius "r_1 = 0.024 \\;m"

Outlet hose radius "r_2 =0.012 \\;m"

Pressure difference "\u0394P=1.2 \\times 10^3 \\;Pa"

A) From equation of continuity

"A_1v_1=A_2v_2 \\\\\n\nv_1= \\frac{A_2}{A_1}v_2 \\\\\n\nv_1^2 =(\\frac{\\pi r_2^2}{\\pi r_1^2})^2 v^2_2 \\\\\n\n= (\\frac{(0.012 \\;m)^2}{(0.024 \\;m)^2})^2 v^2_2 \\\\\n\n= (0.0325)v^2_2"

Applying Bernoulli’s theorem for a hose pipe of the same height

"\u0394P= \\frac{1}{2}\u03c1[v^2_2 -v^2_1] \\\\\n\n= \\frac{1}{2}\u03c1[v^2_2 -(0.0625)v^2_2] \\\\\n\n= \\frac{1}{2}\u03c1 v^2_2 [1-0.0625] \\\\\n\nv_2 = \\sqrt{ \\frac{2\u0394P}{\u03c1(1-0.0625)}} \\\\\n\n= \\sqrt{\\frac{2 \\times 1200 \\;Pa}{(700 \\;kg\/m^3) \\times (1-0.0625)}} \\\\\n\n=1.9 \\;m\/s"

b) The flow rate is

"Q=A_2v_2 \\\\\n\n= \\pi r^2_2 v^2_2 \\\\\n\n= \\pi \\times (0.012 \\;m)^2 \\times (1.9123 \\;m\/s) \\\\\n\n= 8.65 \\times 10^{-4} \\;m^3\/s"


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