Finding Flow Speed in Smaller Section: A Puzzling Problem

AI Thread Summary
To find the flow speed in the smaller section of tubing, the continuity equation A_small * v_small = A_large * v_large is essential, as it relates the areas and velocities of the two sections. The problem involves a liquid with a density of 1.08*10^3 kg/m^3, flowing from a larger section with a known speed and pressure to a smaller section with an unknown speed and pressure. The user attempted to apply Bernoulli's equation but encountered two unknowns: the second pressure and the second velocity. By correctly applying both Bernoulli's equation and the continuity equation, the user can solve for the unknowns. Understanding the relationship between flow speed and cross-sectional area is crucial for resolving this fluid dynamics problem.
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
A_{small}v_{small}=A_{large}v_{large}

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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