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.

 

[Nate810]

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!

 

[m2003]

Can you please help me understand this problem better?
Thank you for your question. Let's start by clarifying some of the information given in the problem. The piston is open to the atmosphere, which means that the pressure at point A is indeed atmospheric pressure (1.013 * 10^5 Pa). This is because there is no obstruction to the flow of air, so the pressure at point A is equal to the pressure of the surrounding air.

Now, let's move on to calculating the total pressure at point B in area B. You are correct in using Bernoulli's equation to calculate this. However, the negative value you are getting is most likely due to a mistake in the units used. Since the density of mercury is given in kg/m^3, we need to convert the cross-sectional areas to m^2 as well. This means that area A is 0.0012 m^2, area B is 0.00056 m^2, and area C is 0.0006 m^2.

Using these values, we can calculate the velocity of mercury through area A:
A(1)v(1) = A(2)v(2)
0.0012 m^2 * v(1) = 0.00056 m^2 * 8 m/s
v(1) = 3.33 m/s

And the velocity of mercury through area B:
A(1)v(1) = A(2)v(2)
0.00056 m^2 * 3.33 m/s = 0.0006 m^2 * v(2)
v(2) = 3.11 m/s

Now, using these velocities in Bernoulli's equation, we get:
P(1) + (1/2)pv(1)^2 = P(2) + (1/2)pv(2)^2
(1.013 * 10^5 Pa) + (1/2)(13600 kg/m^3)(3.33 m/s)^2 = P(2) + (1/2)(13600 kg/m^3)(3.11 m/s)^2
P(2) = 1.013 * 10^5 Pa + 1.65 * 10^5 Pa - 1.49 * 10^5 Pa
P(2) = 1.173 *