Bernoulli-Change in Pressure

  • #1
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


[tex]A_{1}v_{1}=A_{2}v_{2}=flow-rate[/tex]

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


The Attempt at a Solution


[tex]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}[/tex]

[tex]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}[/tex]

[tex]\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[/tex]

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?
 

Answers and Replies

  • #2
stewartcs
Science Advisor
2,177
3


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


[tex]A_{1}v_{1}=A_{2}v_{2}=flow-rate[/tex]

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


The Attempt at a Solution


[tex]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}[/tex]

[tex]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}[/tex]

[tex]\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[/tex]

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
 
  • #3


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.
 
  • #4


Anymore help out there? :smile:
 

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