physics
posted by Andy on .
What is the pressure change in water going from a 3.0 cmdiameter pipe to a 1.8 cmdiameter pipe if the velocity in the smaller pipe is 3.0 m/s?
Im using the equation pv2^2(A2^2A1^2)/2A1^2 and get (1000)(3)^2(1.8^23^20/(2(3)^2) and get 2880 Pa but I'm getting wrong? Thanks

I have no idea what you are doing.
pstatic1+rho*V1^2/2=pstatic2+rhoV2^2/2
you want pstatic1pstatic1
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(V2^2V1^2)/2
=1000*(1.08^23^3)/2
check my thinking