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 ρ=7×102  kg/m3ρ=7 \times 10^2 \;kg/m^3

Inlet hose radius r1=0.024  mr_1 = 0.024 \;m

Outlet hose radius r2=0.012  mr_2 =0.012 \;m

Pressure difference ΔP=1.2×103  PaΔP=1.2 \times 10^3 \;Pa

A) From equation of continuity

A1v1=A2v2v1=A2A1v2v12=(πr22πr12)2v22=((0.012  m)2(0.024  m)2)2v22=(0.0325)v22A_1v_1=A_2v_2 \\ v_1= \frac{A_2}{A_1}v_2 \\ v_1^2 =(\frac{\pi r_2^2}{\pi r_1^2})^2 v^2_2 \\ = (\frac{(0.012 \;m)^2}{(0.024 \;m)^2})^2 v^2_2 \\ = (0.0325)v^2_2

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

ΔP=12ρ[v22v12]=12ρ[v22(0.0625)v22]=12ρv22[10.0625]v2=2ΔPρ(10.0625)=2×1200  Pa(700  kg/m3)×(10.0625)=1.9  m/sΔP= \frac{1}{2}ρ[v^2_2 -v^2_1] \\ = \frac{1}{2}ρ[v^2_2 -(0.0625)v^2_2] \\ = \frac{1}{2}ρ v^2_2 [1-0.0625] \\ v_2 = \sqrt{ \frac{2ΔP}{ρ(1-0.0625)}} \\ = \sqrt{\frac{2 \times 1200 \;Pa}{(700 \;kg/m^3) \times (1-0.0625)}} \\ =1.9 \;m/s

b) The flow rate is

Q=A2v2=πr22v22=π×(0.012  m)2×(1.9123  m/s)=8.65×104  m3/sQ=A_2v_2 \\ = \pi r^2_2 v^2_2 \\ = \pi \times (0.012 \;m)^2 \times (1.9123 \;m/s) \\ = 8.65 \times 10^{-4} \;m^3/s


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