Closed tank and inverted manometer question

[razerangel]

Hey guy this is my first post here. I am looking for a little bit of help with a past paper I've been looking over. I've had a look at it and I am drawing a blank due to being off ill when this was covered in class, if anyone would be so kind as to show me how to get started on this question I would be very greatful.

Homework Statement

A closed cylindrical tank filled with water (pwater= 1000kg/m3) has a hemispherical dome and is connected to an inverted piping system. The liquid in the top of the piping system has a relative density of 0.8 and the remaning parts of the system are filled with water. Given that the pressure gauge reading at A is 60kPa, determine:

i)the pressure in the pipe at B
ii) the equivalent pressure head, in mm of mercury (pmercury = 13,600 kg/m3) at the top of the dome

Thanks in advance if anyone can help me out here :)!

 

[redargon]

Hi and welcome to PF.

Before we can get started, I think a sketch would be helpful so we know where A and B are and any other useful information.

 

[Mene]

Hello,

I'm happy to help you with this question. Let's start by breaking down the problem into smaller parts.

First, we need to understand the setup of the system. We have a closed cylindrical tank filled with water and connected to an inverted piping system. This means that the water in the tank is connected to the water in the piping system, but the piping system is inverted, meaning that the water is above the tank.

Next, we are given some information about the liquids in the system. The water in the tank has a density of 1000kg/m3, while the liquid in the top of the piping system has a relative density of 0.8. This means that the liquid in the piping system is lighter than water.

Now, we are asked to determine the pressure in the pipe at B and the equivalent pressure head at the top of the dome. To do this, we need to use the concept of pressure and the properties of liquids.

Pressure is defined as force per unit area. In this case, we have a closed system, so the pressure is the same everywhere. This means that the pressure at point A is the same as the pressure at point B and the top of the dome.

To find the pressure at B, we can use the manometer equation: P = pgh, where P is the pressure, p is the density of the liquid, g is the acceleration due to gravity, and h is the height of the liquid column. In this case, the liquid at point B is water, so we can plug in the values and solve for P.

For the equivalent pressure head at the top of the dome, we can use the same equation, but we need to use the density of mercury instead of water. We are given the density of mercury as 13,600 kg/m3, so we can plug in the values and solve for h.

I hope this helps you get started on the problem. Let me know if you need any further clarification or assistance. Good luck!