Solving Pressure: A & B in Horizontal Glass Tube

[Ertosthnes]

I don't have a diagram for this, so I'm going to do my best to describe it.

A glass tube lying horizontally has three different cross-sectional areas, A, B and C. Area A is 12 cm^2, B is 5.6 cm^2, and C is 6.0 cm^2. Mercury (density = 13,600 kg/m^3) is being pushed through the tube by a piston open to the atmosphere at the end of area A. The mercury flows through area B and leaves the tube at area C with a velocity of 8.0 m/s.

a) What is the total pressure at point A in area A?
b) What is the total pressure at point B in area B?

Relevant equations:

Pressure = Force/Area
Density = mass/volume
A(1)v(1) = A(2)v(2)
P(1) + (1/2)pv(1)^2 = P(2) + (1/2)pv(2)^2 } part of Bernoulli's equation

^all that comes to mind

My attempt at a solution:

To calculate the velocity of mercury through area A:
A(1)v(1) = A(2)v(2)
.12 m^2 * v(1) = .06 m^2 * 8 m/s
v(1) = 4 m/s

To calculate the velocity of mercury through area B:
A(1)v(1) = A(2)v(2)
.12 m^2 * 4 m/s = .056 m^2 * v(2)
v(2) = 8.57 m/s

I think that the total pressure at point A is just atmospheric pressure (1.013 * 10^5 Pa), but I'm not sure. When I use this value in Bernoulli's equation to calculate the pressure in area B, I get a negative number, which makes me think that my assumption is wrong.

 

[member 553083]

For the pressure at point B in area B, I used Bernoulli's equation:P(1) + (1/2)pv(1)^2 = P(2) + (1/2)pv(2)^21.013 * 10^5 Pa + (1/2)(13,600 kg/m^3)(4 m/s)^2 = P(2) + (1/2)(13,600 kg/m^3)(8.57 m/s)^2P(2) = 1.042 * 10^5 PaIf I'm wrong about any of this, I'd really appreciate it if someone could explain to me where I went wrong. Thank you!