Notes 15 chapter 5 components of thermodynamic cycles
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Notes 15 Chapter 5 Components of Thermodynamic Cycles. Components of Thermodynamic Cycles. What can be the components of thermodynamic cycles? Turbines, valves, compressors, pumps, heat exchangers (evaporators, condensers), mixers,. Open Systems - Control Volume. Where Exactly Are They Used?.

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Notes 15 Chapter 5 Components of Thermodynamic Cycles

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Notes 15 chapter 5 components of thermodynamic cycles

Notes 15Chapter 5Components of Thermodynamic Cycles


Components of thermodynamic cycles

Components of Thermodynamic Cycles

  • What can be the components of thermodynamic cycles?

  • Turbines, valves, compressors, pumps, heat exchangers (evaporators, condensers), mixers,


Open systems control volume

Open Systems - Control Volume


Where exactly are they used

Where Exactly Are They Used?


Vapor cycle power plants

Vapor-cycle Power Plants


Inside the household refrigerator

Inside The Household Refrigerator


Review of 1 st law

Review of 1st Law


Applications to some steady state systems

Applications to some steady state systems

  • Start simple

    • nozzles

    • diffusers

    • valves

  • Includes systems with power in/out

    • turbines

    • compressors/pumps

  • Finish with multiple inlet/outlet devices

    • heat exchangers

    • mixers


You will need to use all that you ve learned to date

You will need to use all that you’ve learned to date

  • Conservation of mass

  • Conservation of energy

  • Property relationships

  • Ideal gas equation of state

  • Property tables

  • Systematic analysis approach


Nozzles and diffusers

Nozzles and Diffusers

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

V2, P2

V1, P1

This configuration is for sub-sonic flow.


Nozzles and diffusers1

Nozzles and Diffusers

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

V2, P2

V1, P1


Notes 15 chapter 5 components of thermodynamic cycles

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

Nozzles


General shapes of nozzles and diffusers

General shapes of nozzles and diffusers

nozzle

diffuser

Subsonic flow

nozzle

diffuser

Supersonic flow


Common assumptions we ll make for nozzles and diffusers

Common assumptions we’ll make for nozzles and diffusers

  • Steady state, steady flow

  • No work - device produces or uses shaft work

  • Heat loss/gain is often neglected

  • Potential energy change is usually small


We start our analysis of diffusers and nozzles with the conservation of mass

We start our analysis of diffusers and nozzles with the conservation of mass

If we have steady state, steady flow, then:

And


We continue with conservation of energy

We continue with conservation of energy

We can simplify by dividing by mass flow:

0

0

0

q = 0 (adiabatic)

w = 0 (these are not work producing devices; neither is work done on them)


We can rearrange to get a much simpler expression

We can rearrange to get a much simpler expression:

With a nozzle or diffuser, we are converting flow and internal energy, represented by Dh =D(u+Pv) into kinetic energy, or vice-versa.


System approach to problem solving

System Approach to Problem Solving

  • State assumptions for the device

  • Write basic form of conservation of mass and energy equations

  • Apply assumptions to get to previous equation


Sample problem

Sample Problem

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


Sample problem diagram and basic information

Sample problem: diagram and basic information

INLET

T1 = 300C

P1 = 100 kPa

V1 = 250 m/s

= 7 kg/s

OUTLET

P2 = 167 kPa

V2 = 35 m/s

Diffuser


Sample problem assumptions

Sample Problem: Assumptions

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

  • adiabatic

  • no work

  • potential energy is zero

  • air is ideal gas


Sample problem apply basic equations

Sample Problem: apply basic equations

Conservation of Mass

Solve for A2


How do we get specific volumes

How do we get specific volumes?

Remember ideal gas equation of state?

or

and

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

NEED ENERGY EQUATION!!!!


Sample problem energy eqn

Sample Problem:Energy Eqn

V1 and V2 are given. We need h2 to get T2 and v2.

If we assumed constant specific heats, we could get T2 directly


Sample problem con t

Sample problem - con’t

However, I’m going to use variable specific heats.. we get h1 from air tables at T1 = 300 + 273 = 573K.

From energy equation:

This corresponds to an exit temperature of 602.2 K.


Now we can get solution

Now we can get solution.

and


Throttling devices valves

Throttling Devices (Valves)


Typical assumptions for throttling devices

Typical assumptions for throttling devices

  • No work

  • Potential energy changes are zero

  • Kinetic energy changes are usually small

  • Heat transfer is usually small

  • Two port device


Look at energy equation

Look at energy equation:

Apply assumptions from previous page:

0

0

0

0

We obtain:

or


Discussion question

Discussion Question

Does the fluid temperature

increase,

decrease, or

remain constant

as it goes through an adiabatic valve?


Look at implications

h const.

T

P const.

h const.

s

Look at implications:

If the fluid is liquid/vapor in equilibrium:

  • During throattling process:

  • The pressure drops,

  • The temperature drops,

  • Enthalpy is constant


Look at implications1

Look at implications:

If fluid is an ideal gas:

Cp is always a positive number, thus:


Teamplay

TEAMPLAY

Refrigerant 12 enters a valve as a saturated liquid at 0.9607 Mpa and leaves at 0.1826 MPa. What is the quality and the temperature of the refrigerant at the exit of the valve?


Teamplay1

TEAMPLAY

Liquid + Vapor

R12

Liquid R12

State (2)

Liquid+vapor, x=?

Psat = 0.1826 MPa

Tsat = ?

Hliq = ?

State (1)

Liquid saturated, x=0

Psat = 0.9607 MPa

Tsat = ?

Hliq = ?


Turbine

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.


Turbines

TURBINES

We’ll assume steady state,

Sometimes neglected

Almost always neglected


Turbines1

Turbines

  • We will draw turbines like this:

inlet

w

maybe q

outlet


Compressors pumps and fans

Compressors, pumps, and fans

  • Machines developed by engineers to make life easier, decrease world anxiety, and provide exciting engineering problems from the industrial revolution for 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.


Primary differences

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


Compressors pumps and fans1

Compressors, pumps, and fans

Side view End view of pump of pump

Compressor


Sample problem1

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/lbm occurs during the process. Determine the mass flow in lbm/min.


Draw diagram

Draw Diagram

15 psia

60oF

Compressor

Wshaft = 5 hp

75 psia

400oF

Q = 4 Btu/lb


Assumptions

Assumptions

  • Steady state steady flow (SSSF)

  • Neglect potential energy changes

  • Neglect kinetic energy changes

  • Air is an ideal gas


What do we know

What do we know?

INLET

T1 = 60F

P1 = 15 psia

OUTLET

T2 = 400F

P2 = 75 psia


Apply first law

Apply First Law:

0

0

0

0

Simplify and rearrange:


Continuing with the solution

Continuing with the solution..

Get h1 and h2 from air tables

Follow through with solution


Heat exchangers and mixing devices

Heat Exchangers and mixing devices

  • Heat exchangers are devices which transfer heat between different fluids

  • Mixing devices (also called open heat exchangers) combine two or more fluids to achieve a desired output, such as fluid temperature or quality


Heat exchangers are used in a variety of industries

Heat exchangers are used in a variety of industries

  • Automotive - radiator

  • Refrigeration - evaporators/condensers

  • Power production - boilers/condensers

  • Power electronics - heat sinks

  • Chemical/petroleum industry- mixing processes


Heat exchangers

Heat Exchangers


Condenser evaporator for heat pump

Condenser/evaporator for heat pump


Something a little closer to home

Something a little closer to home..


Heat exchangers1

Heat Exchangers

  • Now, we must deal with multiple inlets and outlets:


Question

Question…...

  • At steady flow, what is the relationship between

Conservation of Mass


What assumptions to make

What assumptions to make?

  • See any devices producing/using shaft work?

  • What about potential energy effects?

  • What about kinetic energy changes?

  • Can we neglect heat transfer?


Apply conservation of mass on both streams

Apply conservation of mass on both streams...

If we have steady flow, then:

And

Fluid A

Fluid B


Conservation of energy can be a little more complicated

Conservation of energy can be a little more complicated...

I’ve drawn the control volume around the whole heat exchanger.

Implications:

No heat transfer from the control volume.

Fluid A

Fluid B


Heat exchangers2

Heat Exchangers

  • Now if we want the energy lost or gained by either fluid we must let that fluid be the control volume, indicated by the red.


Conservation of energy looks pretty complicated

Conservation of energylooks pretty complicated:

We know from conservation of mass:


Conservation of energy for the heat exchanger

Conservation of energy for the heat exchanger

Apply what we know about the mass flow relationships:


Heat exchangers3

Heat Exchangers

  • Generally, there is no heat transfer from or to the overall heat exchanger, except for that leaving or entering through the inlets and exits.

  • So,

  • And, the device does no work,

  • Also, potential and sometimes kinetic energy changes are negligible.


Heat exchangers4

Heat Exchangers

0

0

0, (sometimes negligible)

0, (usually negligible)

0, (sometimes negligible)

0, (usually negligible)


Heat exchangers5

Heat Exchangers

  • And we are left with

The energy change of fluid A is equal to the negative of the energy change in fluid B.


Question1

QUESTION…..

How would the energy equation differ if we drew the boundary of the control volume around each of the fluids?


Heat exchangers6

Heat Exchangers

The energy equation for one side:

Or dividing through by the mass flow:


Example problem

Example Problem

Refrigerant 134a with a mass flow rate of 5 kg/min enters a heat exchanger at 1.2 MPa and 50C and leaves at 1.2 MPa and 44C. Air enters the other side of the heat exchanger at 34 C and 101 kPa and leaves at 42 C and 101 kPa. Calculate:

a) the heat transfer from the refrigerant

b) the mass flow rate of the air


Draw diagram1

Draw diagram

AIR OUTLET

T3=42C

P3 = 101 kPa

R-134a INLET

T1=50C

P1 = 1.2 MPa

R-134a OUTLET

T2=44C

P2 = 1.2 MPa

AIR INLET

T3=34C

P3 = 101 kPa


State assumptions

State assumptions

  • Steady state, steady flow

  • No work

  • Air is ideal gas

  • Kinetic energy change is zero

  • Potential energy change is zero


Start analysis with r 134a

Start analysis with R-134a

0

0

0

Apply assumptions

We can get h1 and h2 from tables. The refrigerant mass flow is given.


From r 134a tables

From R-134a tables

h1 = 275.52 kJ/kg

h2 = 112.22 kJ/kg

Plugging back into energy equation:


On to part b of the problem

On to part (b) of the problem

We want to get the mass flow of the air...

Start by writing the energy equation for the air side:

0

0

0

Simplify


Sample problem con t1

Sample Problem, Con’t

If air is an ideal gas, then we can rewrite the enthalpy difference as:

Rearrange to solve for mass flow:

How do we get the heat transfer rate to/from the air?


Almost there

Almost there!!!!!

We can write:

Get specific heat from table: Cp = 1.006 kJ/kg.K


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