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happysmiles36
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Homework Statement
Air is blown through a horizontal Venturi tube. The diameters of the narrow and wider sections of the tube are 1.0 cm and 3.0 cm respectively, and the height h of the mercury column is measured to be 1.00 mm. What are the air speeds in the wider and the narrower sections of the tube? The density of mercury is 13,600 kg/m3. (Assume the density of air is 1.29 kg/m3.)
Air is blowing through the wider section to the smaller section.
Homework Equations
p=rho
P1 + (1/2)(p)(v1)^2 + pgy1 = P2 + (1/2)(p)(v2)^2 + pgy2
A1v1=A2v2
P1 = Po + pgh
A(circle)=pi(r)^2
The Attempt at a Solution
Formulas:
y1=y2 so the pgy1 and pgy2 can cancel and v2=A1v1/A2
We can rearrange the equation to give:
P1-P2 = (1/2)(p)(v1)^2((A1/A2)^2-1)
v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)
P1-P2
Po +pgh1 - Po - pgh2
Change in pressure: pg(h1-h2)
v2=A1v1/A2
What we have:
r1= 3.0cm/2/100= 0.015m
r2= 1.0cm/2/100= 0.005m
(P1-P2)= change in pressure
change in height = 1.00mm/1000= 0.001m
p(mercury)= 13600kg/m^3
p(air)= 1.29kg/m^3
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Change in pressure:
pg(h1-h2)
(13600kg/m^3)(9.8m/s^2)(0.001m)= 133.28pa
A1= (0.015m)^2(pi)= 7.068583471x10^-4 m^2
A2= (0.005m)^2(pi)= 7.853981634x10^-5 m^2
v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)
(133.28pa/(1/2)(1.29kg/m^3)((7.068583471x10^-4 m^2/7.853981634x10^-5 m^2)^2-1))^(1/2)
v1= 1.607154546 m/s
v2=A1v1/A2
(7.068583471x10^-4 m^2)(1.607154546 m/s)/(7.853981634x10^-5 m^2)
v2=14.46439092 m/s
I am not sure if I am even doing this right. :(
Sorry if formatting is hard to follow/ugly, I am still getting used to the new changes, don't know where everything is yet.