Homework Help: physics

Posted by Andy on Monday, November 30, 2009 at 4:28pm.

What is the pressure change in water going from a 3.0- cm-diameter pipe to a 1.8- cm-diameter pipe if the velocity in the smaller pipe is 3.0 m/s?

Im using the equation pv2^2(A2^2-A1^2)/2A1^2 and get (1000)(3)^2(1.8^2-3^20/(2(3)^2) and get -2880 Pa but I'm getting wrong? Thanks

physics - bobpursley, Monday, November 30, 2009 at 4:41pm
I have no idea what you are doing.

pstatic1+rho*V1^2/2=pstatic2+rhoV2^2/2
you want pstatic1-pstatic1
So you need V2: Using the law of continuity,
Area1*V1=area2*V2 or V2=V1(1.8/3.0)^2
or V2=1.08m/s

pressure difference=rho(V2^2-V1^2)/2
=1000*(1.08^2-3^3)/2
check my thinking

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