What is the velocity and pressure of water in a narrowed pipe?

[Foghorn]

Homework Statement

Water with pressure of 120 kPa is flowing through a rectangular pipe with velocity 1.9 m/s. At one point, the pipe narrows to one half of its original diameter. What is the velocity of the water at this narrower point? What is the pressure?

Homework Equations

v1A1 = v2A2?
P = F/A

The Attempt at a Solution

I figured the pressure was irrelevant to figuring out the velocity in this equation, since water is basically incompressible. So, I just established v2 as the unknown variable...

1.9 m/s * A = v2 * A/2

(1.9 m/s * A)/(A/2) = v2

v2 = 3.8 m/s

For the pressure, I said that no change in energy is going to occur, so there's not going to be any change in force either...

120 = F/A
P2 = (F/.5A)

.5P2 = F/A
.5P2 = 120
P2 = 60 kPa

 

[tomwilliam]

I think your answer to the first part is right. However, as the pipe narrows, the area halves, and the energy remains the same. Pressure is the same as energy per unit volume (J/m^3), so if energy remains the same, I don't think the force can be the same.
I would use Bernouilli's equation:
P_1 + 0.5p*v_1^2 = P_2 + 0.5p*v_2^2
where p is the density of water (about 1000 kg/m^3)
I may be wrong on this, but this is what I think.