PUMPS

BACKGROUND INFORMATION:

Engineers use pumps to deliver a liquid or a gas to a specific place at a required rate. Liquids and gases are known as fluids. A pump is used to change the elevation, velocity, or pressure of a fluid. In nature, a fluid will flow naturally from a high place to a low place (like a waterfall) or from a high pressure to a low pressure (from inside a balloon to the outside air). Its velocity is often a function of geometry (its shape) and the conditions around it. When a fluid is needed uphill from a source like a lake or pond, at higher pressure than a source, or at a higher rate, an engineer will consider a pump to accomplish the task.

Figure 1

This lesson will be broken into 5 basic sections: basic terms, fluid statics, flow rates, the Bernoulli equation, and a form of the energy equation for pumps. As it progresses, we will move from basic definitions and terms to finding the power of an idealized pump. Some of the terms we need to start with are: gravity (or the gravitational constant), density, and pressure.

Basic Terms:

Gravity is often mistakenly labeled a force when it is actually an acceleration. We talk about how much something weighs; that is a force caused by the gravitational attraction of a mass to the earth's surface. We use Newton's 2nd law, F = m*a, to compute the weight. Here, the force F is the weight, m is the mass, and a is equal to the gravitational pull, g. In mechanics, we often call g the gravitational constant, since it doesn't vary until you are far above the earth. In metric units, g is equal to 9.81 m/s2 (meters per second squared), and in English units, g is equal to 32.2 ft/s2 (feet per second squared).

The density of a fluid is an important factor in our calculations. The density is a measure of the amount of mass in a given space, or volume. Denser fluids behave in a different manner than less dense fluids (water is denser than air, for example). When we talk about pumps, density is important because pumps are used to deliver liquids. Liquids are fluids with a constant density. If you had a small air-tight container with air in it, and transferred that air to another, larger, empty container, the air would fill up the new container - even though the first one was smaller! That's because the density changed (decreased) in the bigger container. The air molecules expanded to fill the new volume. If you took a small glass of water, however, and poured it into a larger glass, it would NOT expand to fill the new volume. The density is fixed, or constant, so the amount of molecules stays fixed and so does the volume. Since we can hold the density constant for liquids, we can simplify our equations when we analyze our pump problems.

The units for density in the metric system are kg/m3 (kilograms per meter cubed). In the English system, the units are pounds (mass) per feet cubed (lbm/ft3). However, we have to be very careful about mass units (lbm) and force units (lb>f), so we will use a converted density in pounds(force) second squared per feet to the fourth. Water has a density of 62.4 lbm/ft3, but we will use the converted density of 1.94 lbf s2/ft4. Engineers use the Greek letter r (rho) as the symbol for density.

Pressure is the force of a fluid on a surface. It is measured in units of force per area. Air at sea level, for example, is 14.7 lbf/in2 (pounds per square inch, sometimes abbreviated as psi). When we use this pressure in our equations, we often have to convert it to lbf/ft2, to match with our other units in the equation. To get the pressure of the air at sea level (called atmospheric pressure), take 14.7 lbf/in2, multiply by 144 in2/ft2 (12 inches per foot, squared). The result is patm = 2116.8 lbf/ft2.

Fluid Statics:

When a fluid is sitting still, meaning there is no motion, we can make certain simplifications of our equations. This is called the study of fluid statics, or hydrostatics. We can compute the pressure at any depth in a container as a function of the density, the gravitational constant, atmospheric pressure, and the depth. The equation for the absolute pressure is:

pa= r gh + patm

Absolute pressure is the pressure with the atmospheric pressure included. Since the atmosphere surrounds pretty much everything, it must be included in our calculations. Pressure gauges usually measure only the actual pressure in a container, or the first term in our equation. That pressure is known as the gauge pressure, pg. The units for the gauge pressure are always written as psig, while the units for the absolute pressure maybe written as psi or psia. Unless we are told otherwise in a problem, we assume psi means absolute pressure.

Figure 2

Let's do an example. Look at Figure 2. It shows a container filled with a liquid. The funny upside down symbol with lines under it is the universal engineering symbol for surface of the fluid. We can measure the depth of the fluid. In this figure, that is 6 inches from the bottom. Let's assume the liquid is water, so the density is 1.94 lbf s2/ft4. The absolute pressure at the bottom of the container is

pa = (1.94 lbfs2/ft4) * (32.2 ft/s2) * (6 in * 1/12 ft/in) + 2116.8 lbf/ft2 ,

pa = 2148 lbf/ft2 or 14.9 lbf/in2

(2148/144 to convert ft2 to in2)

Notice how the units in the first expression cancel to make lbf/ft2. It is very important to make sure your units are consistent and cancel properly. We would have computed an incorrect answer if we had not switched the 6 inches to .5 ft, for example.

The size around and the shape of the container are not factors in this pressure calculation. The pressure at the bottom of a glass with 6 inches of water is the same as the pressure 6 inches down in a swimming pool! It is only a function of the height or depth of the liquid.

Flow Rate:

Now we need to talk about when the fluid is moving or flowing. We can't measure pressure as just a function of the height of the fluid. We need to define a measurement for the motion of the fluid. We use the velocity of the fluid as a starting point. The velocity is the measure of the distance traveled in a given time. The units are typically feet per second (ft/s) or meters per second (m/s). We will use the velocity of the moving fluid to help us analyze the action of a pump.

Remember, one of the functions of a pump is to deliver a required amount of fluid in a given time. The amount is usually measured as a volume, and the time is defined as seconds, minutes, or hours. The volume flow rate is a measure of how much fluid is delivered in a specific time. The engineering symbol for volume flow rate is Q, and the common units are feet cubed per second (ft3/s) or meters cubed per second (m3/s). One formula for the volume flow rate is

Q=V/t

where V is the volume and t is the time. One of the interesting things about the flow rate is that it can be measured two different ways. You can collect a known amount of fluid in a container and time how long it took to fill it (as per our equation above), or you can use the velocity of the flow and the cross-sectional area of the opening the fluid is coming out. That is also the flow rate! That formula is

Q=v*A

where v is the velocity of the fluid and A is the cross-sectional area of the opening. If you think about it, this makes sense; at the same velocity, you will get more fluid coming out of a large hole than a small hole. You will often use both equations when solving a pump problem. The flow rate along a path (like a pipe system) will stay the same, but the velocity in an individual section may change as a function of the area of the pipe.

Bernoulli Equation:

When we analyze fluids in motion, we will often do our calculations for "idealized" flows. This means we make assumptions about the flow to simplify our equations of fluid motion to algebraic equations. We can get a good idea of the fluid behavior for these idealized conditions. Usually, the negatives we are ignoring are small and don't affect the answer too much.

When we can define our fluid flow along fixed lines (along a pipe, for example) and with no energy changes, we can use the Bernoulli equation. By energy changes we mean something that takes an effort, work, or chemical catalyst to change. For example, at the top of this lesson we talked about how things flow in nature. It takes energy, or some type effort, to go against nature. When we are following nature and not using any energy, we can use the Bernoulli equation to compute values in our flow.

Bernoulli's equation is derived from the energy equation of thermodynamics. Thermodynamics is the study of energy and how the properties and behavior of solids, liquids, and gases change when forces, heating, cooling, or chemical reactions are applied to a situation.

So, when we have a situation where the motion occurs naturally, like water flowing downhill, and it occurs along a known path, we can apply the Bernoulli equation. It states that the sum of expressions of pressure, velocity, and elevation at one point along the known path must be equal to the sum of the same expressions at a second point. An individual value may change; the velocity might be larger at point B than it was at point A. But another value must then also change. If the velocity increases, and the elevation is the same at both points, then the pressure must decrease at point B so that the total sum is the same at both points. There are several different forms of the Bernoulli equation depending on which forms of the individual expressions are used. For pump flows and fluid flows the following expression is the most common:

p1/r + 1/2(v12) + gh1 = p2/r + 1/2(v22) + gh2

The heights, h1 and h2, in this equation must be measured from the same place, usually considered ground zero. If they are not measured with respect to the same spot, the calculation will be incorrect.

Figure 3

Figure 3 shows a tank with water in it and with a small opening near the bottom. The water flows out at a velocity of 22.5 ft/s. We would like to find the height of the water in the tank and the flow rate if d2 = 6 inches. For the first calculation, the height of the water, state 1 is defined at the surface of the water in the tank; state 2 is at the opening of the pipe where the water is coming out. We start by defining all the known values of the terms in our expression:

p1 = patm, v1 = 0, h1 = h, p2 = patm, v2 = 22.5 ft/s, h2 = 0

The density is 1.94 lbfs2/ft4 and g=32.2 ft/s2. When you put all of these values in our equation, however, you see that the pressure terms are the same on both sides of the equation. They cancel out, and we can drop them from our expression. Removing the zero terms and the pressure terms, our expression looks like:

(32.2 ft/s2) * h = 1/2(22.5 ft/s)2 .

To get h, we divide through by 32.2 ft/s2, and our answer is h = 7.86 ft. To get the flow rate, we need to multiply the velocity by the area of the opening. We'll assume the pipe is circular, so the area of a circle is p r2 or p (d/2)2, since we were given the diameter. The expression for the flow rate is:

Q = v*A = (22.5 ft/s)*(p )*(.5 ft)2/4,

Q = 4.42 ft3/s .

If we wanted to, we could use this example to illustrate how we got the expression for the fluid static pressure. If the pipe were plugged, the velocity would be zero, and we would have still water. If we chose state 2' to be at the same level as before, but inside the tank, we could calculate the pressure at that point from our Bernoulli equation. The pressure at state 1 is atmospheric, and the pressure at state 2' is the unknown variable. Our equation is (throwing out all zero terms):

patm/r + gh = p2/r .

If we multiply through by the density to clear it out of the denominator, and switch sides to put the unknown on the left (the typical way engineers write problems) and the known values on the right, we have the expression we used earlier for the pressure:

p2 = r gh + patm !

Energy Equation:

When we want to deliver a liquid uphill, or at a different velocity than would occur naturally, we need to use a pump. Since these actions go against nature, we need to expend energy to cause them to happen. It will take work, a form of energy, to make them go. We will still be working with idealized flows, because we only want to have an idea of how powerful a pump we will need. We can go back later with more accurate calculations to get the finer details.

The energy equation says that the sum of our properties at one state, plus or minus the work, must be equal to the properties at a second state. In our case, with pumps, we subtract the work term. It takes work to pump the fluid uphill, so we must subtract it in our equation. An engineer would say that the work was done ON the fluid, rather than BY the fluid. We commonly use the energy equation as a set of differences. Our terms are similar to the Bernoulli equation:

(p2 - p1)/r + 1/2(v22 - v12) + g(h2 - h1) -wp = 0

While we call wp the work term, it isn't quite the actual work; the units (ft2/s2 or m2/s2) are incorrect. We use it as a temporary value to use to compute the power required to run the pump. The equation for power is the density times the flow rate times the work term:

P = r Qwp .

The units for power as defined here are ft lbf/sec or kg m2/s3 (kilogram meters squared per second cubed). However, pump power is usually given in horsepower (hp) or watts (W or kW - kilowatts). The conversions are:

1 hp = 550 ft lbf/sec and

1 W = 1 kg m2/s3 .

Figure 1

As an example, consider the pump system in Figure 1 again. We are told the pump delivers water (r = 1.94 lbf s2/ft4) at a flow rate of 3 ft3/s to a machine (state 2) that is 20 ft above the reservoir surface (state 1). The diameter of the pipe at state 2 is 3 inches, and the pressure at state 2 is 10 lbf/in2 (absolute). We are asked to find the horsepower of the pump to accomplish this. The known values for this problem are:

p1 = 14.7 lbf/in2, p2 = 10 lbf/in2, h1 = 0 ft, h2 = 20 ft, v1 = 0 ft/s

The velocity at state 2 is not yet known, but we can calculate it from the flow rate and the diameter of the pipe at state 2:

v2 = Q/A = (3 ft3/s)/[(p /4)*(3/12 ft)2] , so

v2 = 61.1 ft/s .

Now all the values are know and can be plugged into our energy equation to solve for the pump work term:

(10*144 lbf/ft2 - 14.7*144 lbf/ft2)/(1.94 lbfs2/ft4) + 1/2*(61.1 ft/s)2 + (32.2 ft/s2)*(20 ft - 0 ft) - wp = 0 .

Solving this gives us wp = - 2161.7 ft2/s2, where the negative sign tells us that the work is done ON the fluid. Now we need to compute the power using the work term (we drop the negative sign when talking about power):

P = (1.94 lbfs2/ft4)* (3 ft3/s)*(2161.7 ft2/s2) = 12581.3 ft lbf/s

P = (12581.3 ft lbf/s)/(550 ft lbf/s/hp) = 22.9 hp .

So, the power required to drive the pump is 22.9 hp.

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