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vladimir69
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Homework Statement
The venturi flowmeter is used to measure the flow rate of water in a solar collector system. The flowmeter is inserted in a pipe with diameter 1.9cm; at the venturi of the flowmeter the diameter is reduced to 0.64cm. The manometer tube contains oil with density 0.82 times that of water. If the difference in oil levels on the two sides of the manometer tube is 1.4cm, what is the volume flow rate?
Homework Equations
[tex]P+\frac{1}{2}\rho v^2 +\rho g h = constant[/tex]
[tex]vA=constant[/tex]
[tex]P=P_{0} + \rho g h[/tex]
The Attempt at a Solution
[itex]P_{i}=[/itex] pressure in the pipe where the diameter is [itex]d_{i}[/itex]
[itex]v_{i}=[/itex] speed of water where the pressure is [itex]P_{i}[/itex]
[itex]d_{1} = 0.019[/itex]
[itex]d_{2} = 0.0064[/itex]
[itex]\rho_{w}[/itex] is the density of water
[itex]\rho_{oil}[/itex] is the density of oil
[itex]H=0.014[/itex] is the height difference of oil
Firstly I neglected the potential energy component to obtain
[tex]P_{1} + \frac{1}{2} \rho_{w} v_{1}^2 = P_{2} + \frac{1}{2} \rho_{w} v_{2}^2[/tex]
[tex]v_{1}A_{1} = v_{2} A_{2} [/tex]
where
[tex]A_{i} = \frac{1}{4}\pi d_{i}^2[/tex]
and
[tex]P_{1}-P_{2}=\rho_{oil} g H[/tex]
popping this into the mix gets
[tex]\rho_{oil} g H + \frac{1}{2} \rho_{w} v_{1}^2 = \frac{1}{2} \rho_{w} v_{2}^2[/tex]
[tex]\rho_{oil} g H + \frac{1}{2} \rho_{w} v_{2}^2\frac{A_{2}^2}{A_{1}^2}- \frac{1}{2} \rho_{w} v_{2}^2=0[/tex]
[tex]\frac{1}{2}\rho_{w}v_{2}^2(1-\frac{A_{2}^2}{A_{1}^2})=\rho_{oil} g H[/tex]
[tex]v_{2}=\sqrt{\frac{\rho_{oil}}{\rho_{w}}\frac{2gH}{(1-\frac{d_{2}^4}{d_{1}^4})}}[/tex]
[tex]v_{2} = 0.4774[/tex]
then the volume flow rate is just
[tex]v_{2}A_{2} = 0.4774 * \frac{1}{4}\pi 0.0064^2 = 1.5 \times 10^{-5} m^3 / sec[/tex]
The book gives an answer of 7.2 cm^3 /sec. Where did I go wrong?