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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
P+\frac{1}{2}\rho v^2 +\rho g h = constant
vA=constant
P=P_{0} + \rho g h
The Attempt at a Solution
P_{i}= pressure in the pipe where the diameter is d_{i}
v_{i}= speed of water where the pressure is P_{i}
d_{1} = 0.019
d_{2} = 0.0064
\rho_{w} is the density of water
\rho_{oil} is the density of oil
H=0.014 is the height difference of oil
Firstly I neglected the potential energy component to obtain
P_{1} + \frac{1}{2} \rho_{w} v_{1}^2 = P_{2} + \frac{1}{2} \rho_{w} v_{2}^2
v_{1}A_{1} = v_{2} A_{2}
where
A_{i} = \frac{1}{4}\pi d_{i}^2
and
P_{1}-P_{2}=\rho_{oil} g H
popping this into the mix gets
\rho_{oil} g H + \frac{1}{2} \rho_{w} v_{1}^2 = \frac{1}{2} \rho_{w} v_{2}^2
\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
\frac{1}{2}\rho_{w}v_{2}^2(1-\frac{A_{2}^2}{A_{1}^2})=\rho_{oil} g H
v_{2}=\sqrt{\frac{\rho_{oil}}{\rho_{w}}\frac{2gH}{(1-\frac{d_{2}^4}{d_{1}^4})}}
v_{2} = 0.4774
then the volume flow rate is just
v_{2}A_{2} = 0.4774 * \frac{1}{4}\pi 0.0064^2 = 1.5 \times 10^{-5} m^3 / sec
The book gives an answer of 7.2 cm^3 /sec. Where did I go wrong?