Fluid motion in venturi flowmeter

[vladimir69]

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?

 

[Redbelly98]

Weird, your calculations seem good to me. I even checked them with an online calculator:

http://www.efunda.com/formulae/fluids/venturi_flowmeter.cfm#calc

Was there any information in the problem statement about loss of flow after the meter? Though I doubt that is the problem, you'd need to lose about 1/2 the velocity or 3/4 of the kinetic energy in order to get the book answer; and venturi tubes are designed to be low-loss.

p.s. if you use the above online calculator, they will let you try it one time, and after that ask for a login. However, I found that deleting the efunda.com cookie on my computer allowed me to use the calculator a 2nd time.

 

[vladimir69]

thanks for checking for me
as for the problem statement, i just checked and its been copied word for word from the book (minus a diagram)

regards,
vladimir