Mechanical Engineering - Pumps

7/29/2019 Mechanical Engineering - Pumps

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PUMPS

BACKGROUND INFORMATION:

Engineers use pumps to deliver a liquid or a gas to a specific place at a required rate. Liquids andgases are known as fluids. A pump is used to change the elevation, velocity, or pressure of afluid. In nature, a fluid will flow naturally from a high place to a low place (like a waterfall) orfrom a high pressure to a low pressure (from inside a balloon to the outside air). Its velocity isoften a function of geometry (its shape) and the conditions around it. When a fluid is neededuphill from a source like a lake or pond, at higher pressure than a source, or at a higher rate, anengineer 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, theBernoulli equation, and a form of the energy equation for pumps. As it progresses, we will movefrom 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 howmuch something weighs; that is a force caused by the gravitational attraction of a mass to theearth's surface. We use Newton's 2nd law, F =m*a, to compute the weight. Here, the force F isthe weight, m is the mass, and a is equal to the gravitational pull, g. In mechanics, we often call gthe gravitational constant, since it doesn't vary until you are far above the earth. In metric units, g

nical Engineering - Pumps http://www.swe.org/iac/lp/pumps_04.html

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is equal to 9.81 m/s 2 (meters per second squared), and in English units, g is equal to 32.2 ft/s 2

(feet per second squared).

The density of a fluid is an important factor in our calculations. The density is a measure of theamount of mass in a given space, or volume. Denser fluids behave in a different manner than lessdense fluids (water is denser than air, for example). When we talk about pumps, density isimportant 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, emptycontainer, the air would fill up the new container - even though the first one was smaller! That'sbecause the density changed (decreased) in the bigger container. The air molecules expanded tofill 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/m 3 (kilograms per meter cubed). In the Englishsystem, the units are pounds (mass) per feet cubed (lbm/ft 3). However, we have to be verycareful 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/ft 3, but wewill use the converted density of 1.94 lb f s

2/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 lb f /in2

(pounds per square inch, sometimes abbreviated as psi). Whenwe use this pressure in our equations, we often have to convert it to lb f /ft

2, to match with our

other units in the equation. To get the pressure of the air at sea level (called atmosphericpressure), take 14.7 lb f /in

2, multiply by 144 in 2/ft2 (12 inches per foot, squared). The result is

patm =2116.8 lb f /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 thepressure 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 +p atm

Absolute pressure is the pressure with the atmospheric pressure included. Since the atmosphere


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surrounds pretty much everything, it must be included in our calculations. Pressure gaugesusually measure only the actual pressure in a container, or the first term in our equation. Thatpressure is known as the gauge pressure, p g. 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 toldotherwise 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 upsidedown symbol with lines under it is the universal engineering symbol for surface of the fluid. Wecan measure the depth of the fluid. In this figure, that is 6 inches from the bottom. Let's assumethe liquid is water, so the density is 1.94 lb f s

2/ft4. The absolute pressure at the bottom of the

container is

pa =(1.94 lb f s2/ft4) * (32.2 ft/s 2) * (6 in * 1/12 ft/in) +2116.8 lb f /ft

2 ,

pa =2148 lb f /ft2 or 14.9 lb f /in

2

(2148/144 to convert ft 2 to in 2)

Notice how the units in the first expression cancel to make lb f /ft2. It is very important to make

sure your units are consistent and cancel properly. We would have computed an incorrect answerif 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. Thepressure at the bottom of a glass with 6 inches of water is the same as the pressure 6 inches downin a swimming pool! It is only a function of the height or depth of the liquid.

Flow Rate:


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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 thefluid. We use the velocity of the fluid as a starting point. The velocity is the measure of thedistance traveled in a given time. The units are typically feet per second (ft/s) or meters persecond (m/s). We will use the velocity of the moving fluid to help us analyze the action of apump.

Remember, one of the functions of a pump is to deliver a required amount of fluid in a giventime. 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 persecond (ft 3/ s) or meters cubed per second (m 3/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 andtime how long it took to fill it (as per our equation above), or you can use the velocity of the flowand 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 thinkabout it, this makes sense; at the same velocity, you will get more fluid coming out of a large holethan a small hole. You will often use both equations when solving a pump problem. The flow ratealong a path (like a pipe system) will stay the same, but the velocity in an individual section maychange 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. Thismeans we make assumptions about the flow to simplify our equations of fluid motion toalgebraic 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 noenergy changes, we can use the Bernoulli equation. By energy changes we mean something thattakes an effort, work, or chemical catalyst to change. For example, at the top of this lesson wetalked 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 tocompute values in our flow.

Bernoulli's equation is derived from the energy equation of thermodynamics. Thermodynamics is


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the study of energy and how the properties and behavior of solids, liquids, and gases changewhen 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, andit 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 equalto the sum of the same expressions at a second point. An individual value may change; thevelocity might be larger at point B than it was at point A. But another value must then alsochange. If the velocity increases, and the elevation is the same at both points, then the pressuremust decrease at point B so that the total sum is the same at both points. There are severaldifferent forms of the Bernoulli equation depending on which forms of the individual expressionsare used. For pump flows and fluid flows the following expression is the most common:

p1/r +1/2(v 12) +gh 1 =p 2/r +1/2(v 2

2) +gh 2

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 calculationwill be incorrect.

Figure 3

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

p1 =p atm , v 1 =0, h 1 =h, p 2 =p atm , v 2 =22.5 ft/s, h 2 =0

The density is 1.94 lb f s2/ft4 and g=32.2 ft/s 2. 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


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out, and we can drop them from our expression. Removing the zero terms and the pressure terms,our expression looks like:

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

To get h, we divide through by 32.2 ft/s 2, 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 thearea of a circle is p r 2 or p (d/2) 2, since we were given the diameter. The expression for the flowrate is:

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

Q =4.42 ft 3/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 stillwater. If we chose state 2' to be at the same level as before, but inside the tank, we couldcalculate the pressure at that point from our Bernoulli equation. The pressure at state 1 isatmospheric, and the pressure at state 2' is the unknown variable. Our equation is (throwing outall zero terms):

patm /r +gh =p 2/r .

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

p2 =r gh +p atm !

Energy Equation:

When we want to deliver a liquid uphill, or at a different velocity than would occur naturally, weneed to use a pump. Since these actions go against nature, we need to expend energy to causethem to happen. It will take work, a form of energy, to make them go. We will still be workingwith idealized flows, because we only want to have an idea of how powerful a pump we willneed. 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 workterm. It takes work to pump the fluid uphill, so we must subtract it in our equation. An engineerwould say that the work was done ON the fluid, rather than BY the fluid. We commonly use theenergy equation as a set of differences. Our terms are similar to the Bernoulli equation:


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(p2 - p1)/r +1/2(v 22 - v1

2) +g(h 2 - h1) - wp =0

While we call wp the work term, it isn't quite the actual work; the units (ft 2/s2 or m 2/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 Q wp .

The units for power as defined here are ft lb f /sec or kg m 2/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 lb f /sec and

1 W =1 kg m 2/s3 .

Figure 1

As an example, consider the pump system in Figure 1 again. We are told the pump delivers water(r =1.94 lb f s

2/ft4) at a flow rate of 3 ft 3/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 10lbf /in

2 (absolute). We are asked to find the horsepower of the pump to accomplish this. The

known values for this problem are:

p1 =14.7 lb f /in2, p 2 =10 lb f /in

2, h 1 =0 ft, h 2 =20 ft, v 1 =0 ft/s

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


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v2 =Q/A =(3 ft 3/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 pumpwork term:

(10*144 lb f /ft2 - 14.7*144 lb f /ft

2)/(1.94 lb f s2/ft4) +1/2*(61.1 ft/s) 2 +(32.2 ft/s 2)*(20 ft - 0 ft) -w

p =0 .

Solving this gives us wp =- 2161.7 ft 2/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 negativesign when talking about power):

P =(1.94 lb f s2/ft4)* (3 ft 3/s)*(2161.7 ft 2/s2) =12581.3 ft lb f /s

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

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

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Mechanical Engineering - Pumps