Pressure Topics

From physics we learn that Pressure = Force/Area, or P = F/A, and F/A can also be written as distance*density*gravity.So,

now, when we recognize that gravity is constant at sea level, and density is relatively constant for any liquid, then we can see that,

This shows that pressure is a direct function only of the distance (vertical distance) of a liquid between points, so because the gravity and the density of liquid are constant, pressure can be expressed with units of liquid height, such as ft H₂O, or mm Hg.

Gauge pressure, or P_gauge, doesn't take into account the atmospheric pressure in its readings. In chemical engineering, unless otherwise stated, we want to use absolute pressure in calculations. Converting from P_gauge to P_absolute is done using the following formula:

note: Let me re-emphasize, when pressure is initially given in gauge, alway's first convert P_gauge to P_absolute before any other calculations.

Example 3.a.1:

Perform the following pressure conversions. Assume atmospheric pressure is 1 atm. Unless other wise stated, the pressures are given in absolute.

1. 2600 mm Hg to psi
50.3 psi.
1.34 x 10⁵ psi.
5.03 psi.
1.34 x 10⁴ psi.

2. 275 ft H₂O to kPa
0.822 kPa.
92 kPa.
822 kPa.
0.092 kPa.

3. 20cm Hg to atm (absolute)
0.026 atm.
2.63 atm
26.3 atm
0.263 atm

4. 25.0 psig to mm Hg (absolute)
1293 mmHg.
2052 mmHg.
129.3 mmHg.
205.2 mmHg.

5. 35.0 psi to cm of carbon tetrachloride
2.67 x 10⁷ lb_m
2.67 x 10¹ lb_m
2.67 x 10⁶ lb_m
2.67 x 10³ lb_m

The purpose of this section is to introduce the interrelation that relates the pressure exerted by a fluid or solid to its density. For some fluid or solid, designated i, at height h above a referrence point, we have the following interrelation.

P_water = ρ_watergh

Now, just for fun, I pose the following question. Is the pressure exerted on the bottom of the flask completely given by the water in the flask? That is, is there anything else exerting pressure on the bottom of the flask? The answer is yes, there is air exerting pressure on top of the water. So, while we have just written an expression for the pressure exerted on the bottom of the cup due to water, if we want to find to total pressure exerted on the bottom of the cup, we must account for the atmospheric pressure being exerted on the water. So our expression for the total pressure exerted on the bottom of the cup would include the atmospheric pressure:

This pressure is called the hydrostatic pressure. I wanted to formally introduce this interrelation here, and a practical application of this interrelation is given in the next section.

Typical applications to the interrelations given above are in manometer problems. A typical manometer could be exposed to different pressures at it's openings and is filled with one or more different fluids. A typical manometer problem could ask you to find the pressure drop across the two openings. Here, given the following diagram, we derive an expression for the pressure drop.

Here, LHS and RHS represent the left and right hand side of the manometer. We begin with writing the pressure representation of each component, then we gradually simplify the expression to get it into a reduced form. 

or

or, with heights known, and L_T representing the total length of the manometer, we get,

now, we can rearrange the equation, with the total pressure drop written as DP

or, 

This is our expression, given or looking up the quantities on the right side of the equation allows for a determination of the pressure drop. 

Example 3.a.2:

The great Boston molasses flood occurred on January 15, 1919. In it, 2.3 million gallons of crude molasses flowed from a 30-foot high storage tank that ruptured, killing 21 people and injuring 150. The estimated specific gravity of crude molasses is 1.4 (so the density of the molasses was 1.4 g/ml). What was the mass of molasses in the tank?

Example 3.a.3:

What was the pressure at the bottom of the molasses tank prior to the tank rupture? Assume that the tank was vented to the atmosphere.