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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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,
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

general shapes of nozzles and diffusers
General shapes of nozzles and diffusers



Subsonic flow



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:


we continue with conservation of energy
We continue with conservation of energy

We can simplify by dividing by mass flow:




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


T1 = 300C

P1 = 100 kPa

V1 = 250 m/s

= 7 kg/s


P2 = 167 kPa

V2 = 35 m/s


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?



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


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.

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:





We obtain:


discussion question
Discussion Question

Does the fluid temperature


decrease, or

remain constant

as it goes through an adiabatic valve?

look at implications

h const.


P const.

h const.


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:


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?


Liquid + Vapor


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 = ?

  • 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.

We’ll assume steady state,

Sometimes neglected

Almost always neglected

  • We will draw turbines like this:



maybe q


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


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



Wshaft = 5 hp

75 psia


Q = 4 Btu/lb

  • 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?


T1 = 60F

P1 = 15 psia


T2 = 400F

P2 = 75 psia

apply first law
Apply First Law:





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 exchangers1
Heat Exchangers
  • Now, we must deal with multiple inlets and outlets:
  • 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:


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.


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, (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.


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



P3 = 101 kPa

R-134a INLET


P1 = 1.2 MPa



P2 = 1.2 MPa



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




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:





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