Pumps and Turbines Continued

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En provenance du cours de Georgia Institute of Technology

Fundamentals of Engineering Exam Review

203 notes

Georgia Institute of Technology

Fundamentals of Engineering Exam Review

203 notes

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

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Hydraulics and Hydrologic Systems

This module applies basic principles of fluid mechanics to practical problems in hydraulics, hydrology, and groundwater flow. We first discuss the generalized and one-dimensional momentum theorem and apply it to various typical problems. Flow in pipes and non-circular conduits is discussed beginning with the Bernoulli equation accounting for energy losses and gains. Calculation of head loss due to friction and minor losses due to valves and other accoutrements are presented. Friction losses are calculated for laminar Poiseuille flow and turbulent flow using the Moody chart; examples include computation of pressure drop in laminar pipe flow and turbulent water flow. Methods to calculate flow in pipe networks consisting of multiple connecting pipes and other fittings is then discussed with examples for parallel pipes. Pipes and turbines are then discussed along with their basic equations and definitions. Characteristic curves, especially of centrifugal pumps, are presented and it is shown how to match a pump to a system head.Flow in open channels are discussed including classification of flow types and prediction of uniform flow by the Manning equation. The use of specific energy concepts to solve gradually varying flows, and the importance of the Froude number and sub and supercritical flows are presented. Predictions of hydraulic jumps and flow over weirs are given.Hydrological principles include predictions of surface runoff by the curve number method and peak runoff by the rational formula. Groundwater principles include Darcy’s law for flow through porous media and prediction of drawdown by wells in confined and unconfined aquifers by the Dupuit and Thiem equations.In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions.Time: Approximately 3 hours | Difficulty Level: Medium

Momentum Theorem9:26

Momentum Theorem Continued15:26

Flowin Pipesand Conducts12:44

Flow in Pipes10:56

Flow in Pipes Continued9:57

Pumps and Turbines6:51

Pumps and Turbines Continued11:38

Flow in Open Channels10:04

Flow in Open Channels Continued13:56

Groundwater5:56

Groundwater Continued7:20

Rencontrer les enseignants

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

Now that we've covered the basic theory of pumps and

turbines, we'll do some examples.

Next we have to consider if we have a pump and

we know its characteristics, and we have a particular system.

For example, in this case we have a pump.

We're pumping water from a lower reservoir to an upper reservoir.

And in this system we have friction losses.

We have minor losses for a valve here.

How do we match the system to the characteristics of the pump?

I applied Bernoulli equation between these two free surfaces here, surface one,

the lower reservoir, and surface two, the upper reservoir and

here is our full Bernoulli equation, which looks like this.

But as usual, we can make a lot of simplifications.

The pressure in the lower reservoir is at the surface is atmospheric,

which is zero, so that goes out.

Similarly, the pressure at the upper reservoir is zero, so that goes out.

If the reservoirs are large, the velocities are very small, so V one and

V two go out.

We don't have a turbine in the system, so that goes out.

Re-arranging that equation, I can write it like this.

hP, the head added by the pump

is equal to the elevation difference between the two free surfaces,

z two minus z one plus the summation of all of the head losses in the system.

I can rewrite this equation in this form, hF is f L over D, V squared over two g.

The minor losses are terms which look like this,

loss coefficient times V squared over two g.

Combining those two equations I get hP is z two over z one plus K times Q squared

because each of these terms here are proportional to velocity squared.

However, the velocity squared is proportional to the flow rate, so

this is proportional to Q squared.

So all of those last terms I can replace by this single term here, KQ squared,

where K accounts for the friction loss in the pipe and the minor losses.

If I plot that equation out, I get a curve which looks like this.

As the flow rate through the system goes to zero,

the K Q squared term goes to zero, and I get to this point here,

which is the elevation static head Z two minus Z one.

Then, as the flow rate increases,

the head loss increases by Q squared and I get a graph which looks like this.

If the pump characteristic curve of flow versus head looks like this,

then my operating point occurs where these two graphs intersect right here.

And that is the operating point for this system.

If something changes, for example, the, the head characteristics change,

then I might get a second curve here, and my operating point could move to here.

And in this case, my energy grade lines for the system look like this.

The energy grade line starts off coincident with the free surface here.

If there's no energy loss up to the pump, we get to this point.

The pump increases the head and

the flow, so it kicks up the head by an amount equal to hP.

Then we have some friction loss here, a sudden drop across the valve.

And then, another loss back to the free surface here.

And the geometry of this energy grade line is exactly same as

these equations here, because it, the, the energy grade

line is simply a graphical representation of the ener, of the Bernoulli equation.

Questions are not likely to be very complicated.

This might be a typical one.

We're given that the characteristic curve of flow and head for

two centrifugal pumps are as shown.

At a flow rate of six hundred cubic feet per minute

which of the following statements is correct?

Power delivered by pump A is 34 horsepower etcetera.

So to solve this here is our basic equation of power in USCS units.

Power is gamma QhP divided by 550 horsepower.

And in this case we're given that the flow is 600 cubic feet per minute, or

ten cubic feet per second.

For pump A operating at 600 feet per minute.

Here's 600 feet per minute.

The head is 30 feet, so the power substituting in is 62.4

specific weight of water, etcetera, is 34 horsepower.

Similarly, pump B is at this point here,

the head is approximately 35 feet.

So substituting in we have the, the power for pump B is 40 horsepower.

So, out of those solutions, the correct one is A.

Another one.

Characteristic curves of head and efficiency for

a centrifugal pump are shown.

At a water flow rate of 700 feet, cubic feet per minute,

the power required by the pump is most nearly which of these?

So in this case we get, from the graph at 700 cubic feet per minute,

which is converted to 11.6 cubic feet per second.

Reading off here we have 700 cubic feet per minute right here.

The head,

hP is 20 feet and the efficiency, which is this curve

right over here, is approximately 85% or 0.85.

For a pump, efficiency is defined this way, eta is gamma Q hP divided by W dot.

And in this case we're asked for the power required by the pump.

In other words the power required by the shaft to drive it.

So that is given by W dot is gamma Q hP divided by eta and now I can substitute

in the numbers 62.4 times 11.7 times 20 divided by the efficiency which

is 0.85 is equal to 17 thousand 180 foot pounds per second.

Divide by 550 is 31.2 horsepower.

So the closest answer is D.

Another one, in this case we're pumping water from a canal

at the bottom here to a reservoir at a higher elevation.

The pump is 0.1 meters in diameter, has a total length of 60 meters and

a friction factor is given as 0.02.

If the flow velocity is three meters per second, and minor loss is inevitable,

the head added by the pump is most nearly which of these?

So the starting point is the usual Bernoulli equation.

So the Bernoulli equation from the lower reservoir at one

to the upper reservoir at two is given here.

And, usual approximations, the pressures in the reservoirs are zero,

the velocities are negligible, so those terms go out.

We don't have a turbine, so that goes out.

We're told to neglect minor losses, th, so that term goes out.

So we're left with hP,

the head added by the pump is equal to z two minus z one plus hf.

Which physically states that the head required by the pump

is the head required to lift water from this elevation to this elevation.

In other words, z two minus z one, plus overcome the energy losses due to friction

between the two points, which is hf, is the physical meaning of that equation.

So hf is f L over DV squared over two g.

Now we can plug in the numbers z two minus z one, 140 minus 100.

Friction factor we're given is 0.02 and the other variables R is given.

And the answer is 40 plus 5.5 or 5.5 meters.

The answer is 45.5 and the closest answer is B.

One last example.

We have two reservoirs connected by a pipeline as shown.

But in this case, this device here is a pump or a turbine.

So it can operate either as a pump

if we're pumping water from the lower level to the upper level, or

a turbine if the flow is running from the upper reservoir to the lower reservoir.

It's a reversible pump turbine.

So, question is, when it's functioned at, at a,

as a turbine at a certain flow rate, the head removed by the turbine is 70 meters.

When functioning as a pump, the head that should be added

to pump water at the same flow rate, is most nearly which of these?

So here's our Bernoulli equation.

Starting point is the same as usual.

So firstly, when this is operating as a turbine,

in other words, the flow is from the upper reservoir from the lower reservoir,

the equation from the same approximations that we've made previously, sums to that.

ht, the head extracted,

is z one minus z two minus the summation of all the head losses in the system.

So, the energy grade line then looks like this.

It starts off here, and we have a head loss here in the pipeline,

let's say to the turbine, then a sudden drop across the turbine and

we move to the lower reservoir.

So, the head drop here, the head extracted by the turbine is that.

And this we'll say is the summation of the head losses in the system.

So, in this case we have the summation of the head losses

is equal to the elevation difference minus the head extracted by the turbine.

And the elevations are 220 and 140 meters.

We've given that the head removed is 70 meters so

therefore the summation of the head losses due to friction etcetera is ten meters.

Now, if we reverse this, so

that the flow is from the lower reservoir to the upper reservoir.

In other words, this is acting as a pump.

Then our equation becomes this.

hP is equal to z one minus z two plus hL.

And the grade line in this case starts off here, but now the head is added here.

The head goes up and then drops down by that amount.

It looks something like that.

So here is the head added by the pump and

here are the the energy losses summation hL.

So it looks like the same equation except now we have plus hL here.

So that is equal to z one minus z two is 220 minus 40 and

the head losses we're assuming is, are the same

as when it acts as a turbine because the flow and everything is the same.

So therefore that is equal to 90 meters and the closest answer is D.

So this finishes our discussion of pumps and turbines.

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