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1 Lec 13: Machines (except heat exchangers)

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2 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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3 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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4 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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5 Nozzles and Diffusers Nozzle--a device which accelerates a fluid as the pressure is decreased. V 1, p 1 V 2, p 2 This configuration is for subsonic flow.

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6 Nozzles and Diffusers Diffuser--a device which decelerates a fluid and increases the pressure. V 1, p 1 V 2, p 2

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

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8 General shapes of nozzles and diffusers

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9 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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10 TEAMPLAY 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 lb m have to fall in a standard earth gravity field to “match” this much energy? Example 5-12 on p. 175 has an enthalpy change h 1 - h 2 less than 20 Btu. What does your result mean physically for a nozzle or diffuser?

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11 We start our analysis of diffusers and nozzles with the conservation of mass If we have steady state, steady flow, then: And

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12 We continue with conservation of energy We can simplify by dividing by mass flow: Applying the definition that w=0 and using some other assumptions... 000

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13 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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14 Sample Problem

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15 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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16 Sample problem:diagram and basic information INLET T 1 =300C P 1 =100 kPa V 1 =250 m/s m = 7 kg/s OUTLET P 2 =167 kPa V 2 =35 m/s

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17 Sample Problem: apply basic equations Conservation of Mass Solve for A 2

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18 How do we get specific volumes? Remember ideal gas equation of state? or and We know T 1 and P 1, so v 1 is simple. We know P 2, but what about T 2 ? NEED ENERGY EQUATION!!!!

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19 Sample problem - con’t Energy V 1 and V 2 are given. We need h 2 to get T 2 and v 2. If we assumed constant specific heats, we could get T 2 directly

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20 Sample problem - con’t However, use variable specific heats...get h 1 from air tables at T 1 = 300+273 = 573 K. From energy equation: This corresponds to an exit temperature of 602.2 K.

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21 Now we can get solution. and

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22 TEAMPLAY Work problem 5-65

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23 Throttling Devices (Valves)

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24 Short tube orifice for 2.5 ton air conditioner

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25 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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26 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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27 Look at energy equation: Apply assumptions from previous page: 00 0 0 We obtain: or

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28 Look at implications: If fluid is an ideal gas: c p is always a positive number, thus:

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29 Discussion Question Does the fluid temperature increase, decrease, or as an ideal gas goes through an adiabatic valve? remain constant

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30 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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31 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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33 Turbines Sometimes neglected Almost always neglected We’ll assume steady state,

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34 Turbines We will draw turbines like this: inlet outlet w maybe q

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35 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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36 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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37 Compressors, pumps, and fans Axial flow Compressor Side viewEnd view Centrifugal pump

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40 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 lb m /min.

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41 Draw Diagram

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

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43 What do we know? INLET T 1 = 60F P 1 = 15 psia OUTLET T 2 = 400F P 2 = 75 psia

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44 Apply First Law: 00 0 0 Simplify and rearrange:

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45 Continuing with the solution.. Get h 1 and h 2 from air tables Follow through with solution

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46 TEAMPLAY Work problem 5-73E

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47 TEAMPLAY 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

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48 TEAMPLAY 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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49 TEAMPLAY Parametric studies Under “Tables”, select “New Parametric Table” Click and drag the variables you want to see to the right--P 2, Qdot, and h 2. See that P 2 is not specified in the problem statement in the “Equations Window”.

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50 TEAMPLAY Enter P 2 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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51 TEAMPLAY Under “Plot” select “New Plot Window” and “X-Y Plot”.

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