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Notes 15 Chapter 5 Components of Thermodynamic Cycles

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

- Start simple
- nozzles
- diffusers
- valves

- Includes systems with power in/out
- turbines
- compressors/pumps

- Finish with multiple inlet/outlet devices
- heat exchangers
- mixers

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

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

V2, P2

V1, P1

This configuration is for sub-sonic flow.

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

V2, P2

V1, P1

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

Nozzles

nozzle

diffuser

Subsonic flow

nozzle

diffuser

Supersonic flow

- Steady state, steady flow
- No work - device produces or uses shaft work
- Heat loss/gain is often neglected
- Potential energy change is usually small

If we have steady state, steady flow, then:

And

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)

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

- State assumptions for the device
- Write basic form of conservation of mass and energy equations
- Apply assumptions to get to previous equation

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 300Â°C. Determine the required outlet area in cm2 if the mass flow rate is 7 kg/s and the final pressure is 167 kPa.

INLET

T1 = 300ï‚°C

P1 = 100 kPa

V1 = 250 m/s

= 7 kg/s

OUTLET

P2 = 167 kPa

V2 = 35 m/s

Diffuser

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

Conservation of Mass

Solve for A2

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!!!!

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

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

However, Iâ€™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.

and

- No work
- Potential energy changes are zero
- Kinetic energy changes are usually small
- Heat transfer is usually small
- Two port device

Apply assumptions from previous page:

0

0

0

0

We obtain:

or

Does the fluid temperature

increase,

decrease, or

remain constant

as it goes through an adiabatic valve?

h const.

T

P const.

h const.

s

If the fluid is liquid/vapor in equilibrium:

- During throattling process:
- The pressure drops,
- The temperature drops,
- Enthalpy is constant

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

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 = ?

- 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.

Weâ€™ll assume steady state,

Sometimes neglected

Almost always neglected

- We will draw turbines like this:

inlet

w

maybe q

outlet

- 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.

- 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 fans

Side view End view of pump of pump

Compressor

Air initially at 15 psia and 60Â°F is compressed to 75 psia and 400Â°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.

15 psia

60oF

Compressor

Wshaft = 5 hp

75 psia

400oF

Q = 4 Btu/lb

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

INLET

T1 = 60ï‚°F

P1 = 15 psia

OUTLET

T2 = 400ï‚°F

P2 = 75 psia

0

0

0

0

Simplify and rearrange:

Get h1 and h2 from air tables

Follow through with solution

- 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

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

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

- At steady flow, what is the relationship between

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?

If we have steady flow, then:

And

Fluid A

Fluid B

Iâ€™ve drawn the control volume around the whole heat exchanger.

Implications:

No heat transfer from the control volume.

Fluid A

Fluid B

- 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.

We know from conservation of mass:

Apply what we know about the mass flow relationships:

- 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.

0

0

0, (sometimes negligible)

0, (usually negligible)

0, (sometimes negligible)

0, (usually negligible)

- 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?

The energy equation for one side:

Or dividing through by the mass flow:

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

a) the heat transfer from the refrigerant

b) the mass flow rate of the air

AIR OUTLET

T3=42ï‚°C

P3 = 101 kPa

R-134a INLET

T1=50ï‚°C

P1 = 1.2 MPa

R-134a OUTLET

T2=44ï‚°C

P2 = 1.2 MPa

AIR INLET

T3=34ï‚°C

P3 = 101 kPa

- Steady state, steady flow
- No work
- Air is ideal gas
- Kinetic energy change is zero
- Potential energy change is zero

0

0

0

Apply assumptions

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

h1 = 275.52 kJ/kg

h2 = 112.22 kJ/kg

Plugging back into energy equation:

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

Start by writing the energy equation for the air side:

0

0

0

Simplify

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?

We can write:

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