Upon studying this section, you should be familiar with the following:
Why we can express the units of pressure as the length
of a fluid.
The interrelation between gauge pressure to absolute pressure.
Solving manometer and other static pressure problems.
Note on Pressure Units
From physics we learn that Pressure = Force/Area, or P = F/A, and F/A
can also be written as distance*density*gravity.So,
Pliquid = distance * gravity
now, when we recognize that gravity is constant
at sea level, and density is relatively constant
for any liquid, then we can see that,
Pliquid = distance * gravity
* density = distance * constant,
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 Pgauge, 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
Pgauge to Pabsolute is done using the following formula:
Pabs = Pgauge + Patm
note: Let me re-emphasize, when pressure is initially given in gauge,
alway's first convert Pgauge to Pabsolute before
any other calculations.
Perform the following pressure conversions. Assume atmospheric pressure
is 1 atm. Unless other wise stated, the pressures are given in absolute.
275 ft H2O to kPa
20cm Hg to atm (absolute)
25.0 psig to mm Hg (absolute)
35.0 psi to cm of carbon tetrachloride
2600 mm Hg to psi
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
Pi = ρigh
So, taking the 8 oz. cup of water, that is, say 6 inches tall and filled
with water, we can determine the pressure that the water exerts on the
bottom of the cup. We would write our formula the following way:
Pwater = ρwatergh
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:
Ptotal = ρwatergh +
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.
Manometers and Static Pressure Problems
| Interrelations Home
| Dr. Blowers Home
| U of A Home
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.
Ptop,LHS + Pair,LHS
+ Pw,LHS = Ptop,RHS
+ Pair,RHS + PMe,RHS
Ptop,LHS + ρairghair,LHS
= Ptop,RHS + ρairghair,RHS
or, with heights known, and LT representing the total length
of the manometer, we get,
Ptop,LHS + ρairgL1
= Ptop,RHS + ρairgL2
- L2 - L3)
now, we can rearrange the equation, with the total pressure drop written
Ptop,LHS - Ptop,RHS = ρairgL2
- L2 - L3) - ρwg(LT-L1)
DP = ρairg(L2
- L1) + ρwg(L1
- L2 - L3) + ρMegL3
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.
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?
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.
The University of Arizona. All copyrights © reserved.