Before we launch into the more complicated case of moving fluids around the brewery, we will first consider stationary, or “static,” fluids to set some basic definitions and build a knowledge base. Once we have considered the simple static case, we will look at moving fluids, and finally, we will add extra complications such as movingfluids through a grain bed.
7.2.1 Pressure
When considering movementfluids in pipes, we often want to know the force, or the pressure, that afluid exerts on a container. The words pressure and force are frequently confused and used interchangeably. However, these two terms are very different. By definition, pressure is defined as the force that exerted on a surface divided by the area that the force is distributed across,
P¼F
A: ð7:1Þ
In the SI system, with forces measured in Newtons (N) and area in square meters (m2), the pressure is measured in N/m2. This combination of units is known as the Pascal (Pa), named after the French scientist Blaise Pascal (1623–1662). A unit that is sometimes used is known as the bar (bar). One bar is equal to 100,000 Pa.
Another common unit for pressure, particularly influids engineering in the USA, is pounds per square inch (commonly abbreviated psi, or sometimes as lbs/in2).
Let us consider the difference between force and pressure using a simple experiment: Place a thumbtack between your indexfinger and thumb such that the pointy end is pressing on your thumb. Press together gently. You will notice that (ouch!) the pointy end of the tack is digging in and hurting compared to the other, flat end. It is important to note that the force is the same at either end of the tack and thus the same on either finger/thumb. But since each end of the tack has consid- erably different areas, the pressure is significantly different. You should see that for the same force, a smaller area gives a larger pressure. Likewise, a larger area will give a smaller pressure—again with the same force.
7.2.2 Pascal’s Law
Pascal’s law (also Pascal’s principle) is a principle in fluid physics that states that pressure exerted anywhere in a confined and incompressiblefluid (such as water) is transmitted equally in all directions throughout thefluid. This means that, at any given depth of thefluid, the pressure is the same.
Another example here will highlight once more the difference between force and pressure and use Pascal’s principle as well. Consider a hydraulic car lift as shown in Fig.7.1. A car weighting 26.7 kN (about 6000 lbs) is placed on one piston with a radius of 45 cm. The hydraulic fluid is connected to another piston, but with a radius of 1 cm. Since the pressure must be the same in thefluid at identical depths, the required force at the smaller piston to lift the car can be determined from
P1¼P2
F1
pr12¼ F2
pr22
ð7:2Þ
or
F1¼F2
r12
r22¼26;700 N 12 452 F1¼13:2 Nðabout 3 lbsÞ
ð7:3Þ
Of course, for a given amount of travel in the large piston, the smaller piston must travel further. The amount, or volume, of thefluid moved must be the same at either piston,
V1¼V2: ð7:4Þ
So for every centimeter that the large piston moves, the smaller one must move
pr21x¼pr22ð1 cmÞ x¼ ð1 cmÞ452
12 x¼2025 cm
ð7:5Þ
Pascal’s law is actually a statement of the weight of the fluid above a certain point. Looking at Fig.7.2, we now consider a liquid and draw an imaginary cube somewhere in the liquid. The difference in pressure between the top and the bottom of the cube can be determined by finding the weight of this cube of fluid. The weight (a force) of this cube is given as:
W ¼mg¼qVg ð7:6Þ
whereρis the density of thefluid andVis the volume of the cube. So, the pressure at the bottom of the cube is simply the pressure at the top of the cubeplus the pressure–“force”due to the weight of the cube.
Pbottom ¼PtopþqVg
A : ð7:7Þ
Rearranging, Pascal’s law is more generally stated as
DP¼qgDh ð7:8Þ
F1
r1
r2
F2
Fig. 7.1 Example illustrating Pascal’s principle. The pressure at each piston must be the same
with ΔP being the difference between the top and bottom pressures. It just states that the pressure difference between two points depends only on the net difference in height between the two points. So, if two points in afluid are at the same depth, they will have the same pressure.
Consider another example: A tube, bent into a U-shape, is partiallyfilled with water (ρ= 1 g/cm3). We add enough water such the bottom part of the U is completely covered. At this point we expect after equilibrium has been reached that the water will be at the same level on either side of the U. Then, on the left side of tube we add oil (ρ= 0.85 g/cm3) so that the total height of the oil in the tube is 3 cm as shown in Fig.7.3. Then, the question is: What is the difference in the water level,x, on either side of the tube?
The weight of the oil on the left side will push down on the water below it until the pressures at positions aand b are equal. Since the pressures at the top of the fluids due to the atmosphere are the same, we start with Pascal’s law:
DPoil¼DPwater
qoilgDhoil¼qwatergDhwater
ð7:9Þ
The densities are given for bothfluids, and the gravitational constantg cancels on both sides. Then substituting,
0:85 g/cm3
1:0 g/cm3 3 cm¼x x¼2:55 cm
ð7:10Þ
We can use Pascal’s law to illustrate another idea. Let us consider an elevated container of water, such as a hot-liquor tank, as shown in Fig.7.4. The container has 1.5 m of water, and it is elevated by 3 m. The container has a pipe leading down to the floor where we have a closed valve. It is important to note that the valve is shut, and thefluid is not moving. We can use Pascal’s law to calculate the pressure difference between the top of the water and the location of the valve. Since, the density of water is 1000 kg/m3,
∆h Ptop
Pbottom
A
Fig. 7.2 An imaginary cube is drawn around a parcel offluid to illustrate Pascal’s law
DP¼qgDh
DP¼1000 kg/m39:8 m/s24:5 m DP¼44;100 N/m2¼44:1 kPa
ð7:11Þ
where we have used the total height in meters.
In the above example, note that we ignored the atmospheric pressure pressing on the top of the water in the container. Pascal’s law only gives the pressure change due to the weight of thefluid. Since there is about 101.3 kPa (14.7 psi) of atmo- spheric pressure pressing at the top of the water the total, orabsolute pressure,is 44.1 + 101.3 = 145.4 kPa. We will write this asP = 145.4 kPaabsolute. In the past, this would have been expressed as 14.7 psia where the “a” means absolute.
Absolute pressure is measured against a reference pressure of zero, a perfect vacuum.
On the other hand, there is also atmospheric pressure pressing on the outside of the valve. Since it is pressuredifferencesthat will tend to causefluids to move (or vessels and pipes to expand), we are frequently only interested in the pressure relative to the outside, or the atmosphere. Thus, the relative pressure, or gauge pressure, is always measured relative to the local atmospheric pressure. The pressure measured at the valve in the example will be written asP= 44.1 kPgauge. If a pressure measurement does not specify absolute or gauge, then gauge is assumed.
3cm x
b a
Fig. 7.3 A U-tube isfirstfilled with water and then with oil
CHECKPOINT 7.1
A famous experiment is attributed to Pascal in which it is said that he inserted a long tube into a wooden barrel. It is reported that the barrel busted after he filled the tube with water. While the accuracy of this tale has not been verified, it is educational to calculate theforceof the water on the bottom of the barrel. Why calculate force and not just pressure? Because the staves of the barrel must resist this total force to hold the bottom in place, in the same way the piston held up the total weight of the car in Fig.7.1. Consider a standard-sized bourbon barrel; the ends have diameters of about 50 cm and are 1 m tall. It is said that Pascal used a 10-m tube. Calculate the total force on the bottom of the barrel once the tube isfilled with water.
3m 1.5m
Valve
Patm
Patm
Fig. 7.4 An elevated container of water is connected to a valve