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Lec 13: Machines (except heat exchangers)

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• For next time:– Read: § 5-4– HW7 due Oct. 15, 2003

• Outline:– Diffusers and nozzles– Turbines– Pumps and compressors

• Important points:– Know the standard assumptions that go

with each device– Know how to simplify the governing

equations using these assumptions– Consider what each device would be used

for in real-world applications

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Applications to some steady state systems

• Start simple– nozzles– diffusers– valves

• Include systems with power in/out– turbines– compressors/pumps

• Finish with multiple inlet/outlet devices– heat exchangers– mixers

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We will need everything we have covered

• Conservation of mass• Conservation of energy• Property relationships• Ideal gas equation of state• Property tables• Systematic analysis approach

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Nozzles and Diffusers

• Nozzle--a device which accelerates a fluid as the pressure is decreased.

V1, p1V2, p2

This configuration is for subsonic flow.

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Nozzles and Diffusers

• Diffuser--a device which decelerates a fluid and increases the pressure.

V1, p1V2, p2

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Nozzles

For supersonic flow, the shape of the nozzle is reversed.

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General shapes of nozzles and diffusers

Subsonic Flow

Nozzle Diffuser

Nozzle Diffuser

Supersonic Flow

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Common assumptions for nozzles and diffusers

• Steady state, steady flow.

• Nozzles and diffusers do no work and use no work.

• Potential energy changes are usually small.

• Sometimes adiabatic.

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TEAMPLAYTEAMPLAY

• For nozzles, diffusers and other machines--just how important is PE?

• The energy in the head of a kitchen match is reportedly about 1 Btu.

• How far does 1 lbm have to fall in a standard earth gravity field to “match” this much energy?

• Example 5-12 on p. 175 has an enthalpy change h1 - h2 less than 20 Btu. What does your result mean physically for a nozzle or diffuser?

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We start our analysis of diffusers and nozzles with the conservation

of mass

dt

dmCV 0

mmm 21

If we have steady state, steady flow, then:

And

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We continue with conservation of energy

dt

dECV 0)]zz(g2

)hh[(mWQ ei

2e

2i

eiCV

VV

)zz(g2

)hh(wq 12

21

22

12

VV

We can simplify by dividing by mass flow:

Applying the definition that w=0 and using some other assumptions...

0 0 0

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We can rearrange to get a much simpler expression:

With a nozzle or diffuser, we are converting flow energy and internal energy, represented by Dh into kinetic energy, or vice-versa.

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Sample Problem

An adiabatic diffuser is employed to reduce thevelocity of a stream of air from 250 m/s to 35 m/s.The inlet pressure is 100 kPa and the inlettemperature is 300°C. Determine the requredoutlet area in cm2 if the mass flow rate is 7 kg/sand the final pressure is 167 kPa.

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Sample Problem:Assumptions

• SSSF (Steady state, steady flow) - no time dependent terms

• adiabatic• no work• potential energy change is zero• air is ideal gas

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Sample problem:diagram and basic information

INLET

T1=300C

P1=100 kPa

V1=250 m/s

m = 7 kg/s

OUTLET

P2=167 kPa

V2=35 m/s

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Sample Problem: apply basic equations

Conservation of Mass

mmm 21

2

22

1

11

ν

AV

ν

AVm

Solve for A2

2

22

V

νmA

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How do we get specific volumes?

Remember ideal gas equation of state?

RTP or

1

11 P

RT

2

22 P

RTand

We know T1 and P1, so v1 is simple. We know P2, but what about T2?

NEED ENERGY EQUATION!!!!

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Sample problem - con’tEnergy

V1 and V2 are given. We need h2 to get T2 and v2. If we assumed constant specific heats, we could get T2 directly

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Sample problem - con’t

However, use variable specific heats...get h1 from air tables at T1 = 300+273 = 573 K.

2

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2

222

2 m

s

kg

kJ10

s

m

2

)35()250(

kg

kJ578.73h

kg

kJ578.73h1

From energy equation:

kg

kJ4.609h2 This corresponds to an exit

temperature of 602.2 K.

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Now we can get solution.

kg

m0352.1

P

RT 3

2

22

2

24

3

cmm

10sm

35

kgm

0352.1s

kg 7

and

22 cm 2070A

2

22

V

νmA

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TEAMPLAYTEAMPLAY

Work problem 5-65

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Throttling Devices (Valves)

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Short tube orifice for 2.5 ton air conditioner

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Throttles (throttling devices)

• A major purpose of a throttling device is to restrict flow or cause a pressure drop.

• A major category of throttling devices is valves.

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Typical assumptions for throttling devices

• Do no work, have no work done on them

• Potential energy changes are zero

• Kinetic energy changes are usually small

• Heat transfer is usually small

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Look at energy equation:

Apply assumptions from previous page:

0 0 00

We obtain:

or

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Look at implications:

If fluid is an ideal gas:

cp is always a positive number, thus:

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Discussion QuestionDoes the fluid temperature

increase, decrease, or

as an ideal gas goes through an adiabatic valve?

remain constant

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TEAMPLAY

Refrigerant 134a enters a valve as a saturated liquid at 200 psia and leaves at 50 psia. What is the quality of the refrigerant at the exit of the valve?

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Turbine

• A turbine is device in which work is produced by a gas passing over and through a set of blades fixed to a shaft which is free to rotate.

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Turbines

)zg(zVV

)h(hwq 12

21

22

12 2

Sometimes neglected

)( 12 hhwq

Almost always neglected

We’ll assume steady state,

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Turbines

• We will draw turbines like this:

inlet

outlet

w

maybe q

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Compressors, pumps, and fans

• Machines developed to make life easier, decrease world anxiety, and provide challenging problems for engineering students.

• Machines which do work on a fluid to raise its pressure, potential, or speed.

• Mathematical analysis proceeds the same as for turbines, although the signs may differ.

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Primary differences

• Compressor - used to raise the pressure of a compressible fluid

• Pump - used to raise pressure or potential of an incompressible fluid

• Fan - primary purpose is to move large amounts of gas, but usually has a small pressure increase

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Compressors, pumps, and fans

Axial flow Compressor

Side view End view

Centrifugal pump

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Sample Problem

Air initially at 15 psia and 60°F is compressed to 75 psia and 400°F. The power input to the air is 5 hp and a heat loss of 4 Btu/lb occurs during the process. Determine the mass flow in lbm/min.

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Draw Diagram

15 psia 60 F

W sh = 5 hp

75 psia 400 F q = 4 Btu/lb

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Assumptions

• Steady state steady flow (SSSF)• Neglect potential energy changes• Neglect kinetic energy changes• Air is ideal gas

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What do we know?

INLET

T1 = 60F

P1 = 15 psia

OUTLET

T2 = 400F

P2 = 75 psia

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Apply First Law:

022 2

22

21

21

1

gz

Vhmgz

VhmWQ sh

0 00 0

Simplify and rearrange:

12 hhmWqm sh

12 hhq

Wm sh

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Continuing with the solution..

Get h1 and h2 from air tables

m2

m1 lb

Btu206.46h

lb

Btu124.27h

Follow through with solution

Btulbft

778lbBtu

124.27206.46lbBtu

4

min60s

shplbft

5505hp

mf

mm

f

min

lb2.46m m

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Work problem 5-73E

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TEAMPLAYTEAMPLAY

• Use EES and vary the exit pressure from 5 psia to 0.5 psia in increments of 1.0 psia. Show the results as a table and a plot.

• Open EES and put in the basic equation

out12out W)h(hmQ

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TEAMPLAYTEAMPLAY

• You will have to use some new features of EES– 1. Under options always check and set

unit system, if necessary.– 2. Under options, find function info,

and select fluid properties.– 3. For steam, use Steam_NBS.

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TEAMPLAYTEAMPLAY

• Parametric studies• Under “Tables”, select “New Parametric

Table”• Click and drag the variables you want to

see to the right--P2, Qdot, and h2.

• See that P2 is not specified in the problem statement in the “Equations Window”.

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TEAMPLAYTEAMPLAY

• Enter P2 via “Alter Values” under “Tables”

• Click on the column headings to be able to enter units.

• You must solve the table before you can plot it.

• Under “Calculate” select “Solve Table.”

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TEAMPLAYTEAMPLAY

• Under “Plot” select “New Plot Window” and “X-Y Plot”.