Bernoulli's Equation for Pressure Drop in a Constricted Venturi Tube

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Homework Help Overview

The discussion revolves around a problem involving Bernoulli's equation applied to a venturi tube, specifically focusing on calculating the pressure drop in a constricted section of the tube due to airflow. The subject area includes fluid dynamics and principles related to pressure changes in varying cross-sectional areas.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of flow rate and velocity in different sections of the venturi tube. There are questions about the correct application of Bernoulli's equation and the interpretation of the results, particularly regarding the sign and magnitude of the pressure drop. Some participants also inquire about the appropriate symbols for flow rate.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the calculations. Some have pointed out potential errors in the application of the equation, while others seek clarification on the correct approach and notation. There is no explicit consensus yet, as participants are still analyzing the problem.

Contextual Notes

Participants note discrepancies in the expected results, with one mentioning that their calculated answer was significantly smaller than anticipated. There is also a reference to the multiple-choice nature of the problem, which may influence the discussion dynamics.

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Bernoulli--Change in Pressure

Homework Statement


A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 cm/s in the 2.0-cm section? ( = 1.2 kg/m3.)


Homework Equations


A_{1}v_{1}=A_{2}v_{2}=flow-rate

\triangle P=\frac{1}{2}\rho\left(v_{1}^{2}-v_{2}^{2}\right) (I knew how to derive this from Bernoulli's)


The Attempt at a Solution


flow-rate=A_{1}v_{1}=\frac{1}{4}\pi d_{1}^{2}=\frac{1}{4}\pi\left(0.02m\right)^{2}\left(0.03\frac{m}{s}\right)=9.42\times10^{-6}\frac{m^{3}}{s}

v_{2}=\frac{flow-rate}{A_{2}}=\frac{9.42\times10^{-6}\frac{m^{3}}{s}}{\frac{1}{4}\pi\left(0.01m\right)^{2}}=0.12\frac{m}{s}

\triangle P=\frac{1}{2}\rho\left(v_{1}^{2}-v_{2}^{2}\right)=\frac{1}{2}\left(1.2\frac{kg}{m^{3}}\right)\left[\left(0.03\frac{m}{s}\right)^{2}-\left(0.12\frac{m}{s}\right)^{2}\right]=-8.1\times10^{-3}Pa

All the answers were in Pascals, but somehow my answer was 1,000 times smaller! Since it was multiple-choice, I got the answer right anyways, but I'd like someone to point out where I went wrong. Thanks!

btw, what symbol should I use for the flow rate?
 
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wrongusername said:

Homework Statement


A venturi tube may be used as the inlet to an automobile carburetor. If the 2.0-cm diameter pipe narrows to a 1.0-cm diameter, what is the pressure drop in the constricted section for an airflow of 3.0 cm/s in the 2.0-cm section? ( = 1.2 kg/m3.)

Homework Equations


A_{1}v_{1}=A_{2}v_{2}=flow-rate

\triangle P=\frac{1}{2}\rho\left(v_{1}^{2}-v_{2}^{2}\right) (I knew how to derive this from Bernoulli's)

The Attempt at a Solution


flow-rate=A_{1}v_{1}=\frac{1}{4}\pi d_{1}^{2}=\frac{1}{4}\pi\left(0.02m\right)^{2}\left(0.03\frac{m}{s}\right)=9.42\times10^{-6}\frac{m^{3}}{s}

v_{2}=\frac{flow-rate}{A_{2}}=\frac{9.42\times10^{-6}\frac{m^{3}}{s}}{\frac{1}{4}\pi\left(0.01m\right)^{2}}=0.12\frac{m}{s}

\triangle P=\frac{1}{2}\rho\left(v_{1}^{2}-v_{2}^{2}\right)=\frac{1}{2}\left(1.2\frac{kg}{m^{3}}\right)\left[\left(0.03\frac{m}{s}\right)^{2}-\left(0.12\frac{m}{s}\right)^{2}\right]=-8.1\times10^{-3}Pa

All the answers were in Pascals, but somehow my answer was 1,000 times smaller! Since it was multiple-choice, I got the answer right anyways, but I'd like someone to point out where I went wrong. Thanks!

btw, what symbol should I use for the flow rate?

What did you get for the correct answer?

Note: that your equation should be V2^2 - V1^2 and not V1^2-V2^2...

CS
 


stewartcs said:
What did you get for the correct answer?

Note: that your equation should be V2^2 - V1^2 and not V1^2-V2^2...

CS

Correct answer was 81 Pa. So my answer is actually 10,000 times smaller.

Thanks. I correct that.
 


Anymore help out there? :smile:
 

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