# Still lost

1. Jan 17, 2004

### MichaelC

Here is a problem dont 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?

2. Jan 17, 2004

### himanshu121

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 (Text):

NOTE: No. of unknown = No. of Equations

3. Jan 17, 2004

### MichaelC

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

4. Jan 17, 2004

Thank you