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Notes 15 Chapter 5 Components of Thermodynamic CyclesPowerPoint Presentation

Notes 15 Chapter 5 Components of Thermodynamic Cycles

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Notes 15Chapter 5Components of Thermodynamic Cycles

Components of Thermodynamic Cycles

- What can be the components of thermodynamic cycles?
- Turbines, valves, compressors, pumps, heat exchangers (evaporators, condensers), mixers,

Review of 1st Law

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

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

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 Diffusers

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

V2, P2

V1, P1

General shapes of nozzles and diffusers

nozzle

diffuser

Subsonic flow

nozzle

diffuser

Supersonic flow

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

If we have steady state, steady flow, then:

And

We continue with conservation of energy conservation of mass

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: conservation of mass

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 conservation of massProblem 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 conservation of mass

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 conservation of mass

INLET

T1 = 300C

P1 = 100 kPa

V1 = 250 m/s

= 7 kg/s

OUTLET

P2 = 167 kPa

V2 = 35 m/s

Diffuser

Sample Problem: Assumptions conservation of mass

- SSSF (Steady state, steady flow) - no time dependent terms
- adiabatic
- no work
- potential energy is zero
- air is ideal gas

How do we get specific volumes? conservation of mass

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 conservation of mass

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 conservation of mass

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. conservation of mass

and

Throttling Devices (Valves) conservation of mass

Typical assumptions for throttling devices conservation of mass

- 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: conservation of mass

Apply assumptions from previous page:

0

0

0

0

We obtain:

or

Discussion Question conservation of mass

Does the fluid temperature

increase,

decrease, or

remain constant

as it goes through an adiabatic valve?

h const. conservation of mass

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 implications: conservation of mass

If fluid is an ideal gas:

Cp is always a positive number, thus:

TEAMPLAY conservation of mass

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?

TEAMPLAY conservation of mass

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 conservation of mass

- 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 conservation of mass

We’ll assume steady state,

Sometimes neglected

Almost always neglected

Compressors, conservation of masspumps, 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 conservation of mass

- 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

Sample Problem conservation of mass

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.

Assumptions conservation of mass

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

Continuing with the solution.. conservation of mass

Get h1 and h2 from air tables

Follow through with solution

Heat Exchangers and mixing devices conservation of mass

- 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 conservation of mass

- Automotive - radiator
- Refrigeration - evaporators/condensers
- Power production - boilers/condensers
- Power electronics - heat sinks
- Chemical/petroleum industry- mixing processes

Heat Exchangers conservation of mass

Condenser/evaporator for heat pump conservation of mass

Something a little closer to home.. conservation of mass

Heat Exchangers conservation of mass

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

Question…... conservation of mass

- At steady flow, what is the relationship between

Conservation of Mass

What assumptions to make? conservation of mass

- 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... conservation of mass

If we have steady flow, then:

And

Fluid A

Fluid B

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

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

Implications:

No heat transfer from the control volume.

Fluid A

Fluid B

Heat Exchangers conservation of mass

- 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 conservation of masslooks pretty complicated:

We know from conservation of mass:

Conservation of energy for the heat exchanger conservation of mass

Apply what we know about the mass flow relationships:

Heat Exchangers conservation of mass

- 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 Exchangers conservation of mass

0

0

0, (sometimes negligible)

0, (usually negligible)

0, (sometimes negligible)

0, (usually negligible)

Heat Exchangers conservation of mass

- And we are left with

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

QUESTION….. conservation of mass

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

Heat Exchangers conservation of mass

The energy equation for one side:

Or dividing through by the mass flow:

Example Problem conservation of mass

Refrigerant 134a with a mass flow rate of 5 kg/min enters a heat exchanger at 1.2 MPa and 50C and leaves at 1.2 MPa and 44C. 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 diagram conservation of mass

AIR OUTLET

T3=42C

P3 = 101 kPa

R-134a INLET

T1=50C

P1 = 1.2 MPa

R-134a OUTLET

T2=44C

P2 = 1.2 MPa

AIR INLET

T3=34C

P3 = 101 kPa

State assumptions conservation of mass

- 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 conservation of mass

0

0

0

Apply assumptions

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

From R-134a tables conservation of mass

h1 = 275.52 kJ/kg

h2 = 112.22 kJ/kg

Plugging back into energy equation:

On to part (b) of the problem conservation of mass

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’t conservation of mass

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!!!!! conservation of mass

We can write:

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

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