Finding Flow Speed in Smaller Section: A Puzzling Problem

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Homework Help Overview

The problem involves fluid dynamics, specifically the flow of a liquid through two sections of tubing with differing cross-sectional areas. The original poster is tasked with finding the flow speed in the smaller section based on given parameters such as density, flow speed, and pressure in the larger section.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply Bernoulli's equation but encounters difficulty due to having two unknowns: pressure in the second section and the second velocity. Some participants suggest using the continuity equation to relate the flow speeds and areas of the two sections.

Discussion Status

Participants are exploring different equations relevant to the problem, with some guidance offered regarding the use of the continuity equation. There is an acknowledgment of the need for additional equations to resolve the unknowns, but no consensus has been reached on a complete approach.

Contextual Notes

The original poster has noted a potential misunderstanding regarding the number of equations needed to solve for the unknowns, as well as a conversion of units to ensure consistency in calculations.

MichaelC
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Here is a problem don't understand:
A liquid with a density of 1.08*10^3 kg/m^3 flows through two horizontal sections of tubing joined to the end. In the first section the cross sectional areais 10.5 cm^2, the flow speed is 274 cm/s, and the pressure is 1.29*10^5 Pa. In the second section, the cross section area is 2.84 cm^2.
You are supposed to find the flow speed in the smaller section using:

(First Pressure)+ (Density*G*(First velocity)^2)= (pressure two)+ (.5*density*(second veloity)^2)

When I simplify I keep getiing an answer with 2 unknowns (pressure and the second velocity) What am I missing?
 
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If the liquid is incompressible and irrotational or in other words ideal,

You are missin an Equation
[tex]A_{small}v_{small}=A_{large}v_{large}[/tex]

Code:
NOTE: No. of unknown = No. of Equations
 
Here's wat I did:

I changed evrything to meters and i Knew that the density of water was 1000 kg/m^3

129000 Pa + (.5 * 1000kg/m^3 * (.274m/s)^2)=

(second Pressure) + (.5 * 1000 *(2nd Velocity)^2)

So...
(133054.104 - (second Pressure))/540 = (Second Velocity)^2
 
Thank you
 

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