# Bernoulli's Equation and liquid density

1. Aug 17, 2011

### physkid1

1. The density of the liquid flowing through the horizontal pipe in the drawing is 1200 kg/m3. The speed of the fluid at point A is 7.5 m/s while at point B it is 11 m/s. What is the difference in pressure, PB – PA, between points B and A?

2. Bernoulli's Equation P1 + 1/2 pv1^2 + pgy1 = P2 + 1/2 pv2^2

3. ok to find P is it possible to make P1 = 1/2 pv1^2 + pg (y1 can be neglected) then do the same with P2 then subtract the two values to work out the pressure ? or am i way off ?

2. Aug 17, 2011

### Staff: Mentor

While you would arrive at the correct final answer, your working would not be quite right.
You would be saying that P1=0.5 p v12 but that's not right. You are disregarding the constant in Bernoulli's Equation; in effect, equating it to zero.

It is better to rearrange Bernoulli's Equation to give the pressure difference, viz., P2 - P1 and that way the constant vanishes and your working will be correct all the way through and no more complicated.