Physics 231 lecture 31 engines and fridges
This presentation is the property of its rightful owner.
Sponsored Links
1 / 21

PHYSICS 231 Lecture 31: Engines and fridges PowerPoint PPT Presentation


  • 46 Views
  • Uploaded on
  • Presentation posted in: General

PHYSICS 231 Lecture 31: Engines and fridges. Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom. Metabolism. U=Q+W. Work done (negative). Heat transfer: Negative body temperature < room temperature. Change in internal energy: Must be increased: Food!. t. t.

Download Presentation

PHYSICS 231 Lecture 31: Engines and fridges

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Physics 231 lecture 31 engines and fridges

PHYSICS 231Lecture 31: Engines and fridges

Remco Zegers

Question hours: Thursday 12:00-13:00 & 17:15-18:15

Helproom

PHY 231


Physics 231 lecture 31 engines and fridges

Metabolism

U=Q+W

Work done (negative)

Heat transfer: Negative

body temperature < room temperature

Change in internal energy: Must be increased: Food!

PHY 231


Physics 231 lecture 31 engines and fridges

t

t

t

Metabolic rate

U = Q + W

Metabolic rate: rate in which food and oxygen are

transformed into internal energy (to balance losses due

to heat loss and work done).

|W/t|

Body’s efficiency:

|U/t|

PHY 231


Physics 231 lecture 31 engines and fridges

Body’s efficiency

U/t~oxygen use rate

can be measured

W/t can be measured

PHY 231


Physics 231 lecture 31 engines and fridges

Types of processes

A: Isovolumetric V=0

B: Adiabatic Q=0

C: Isothermal T=0

D: Isobaric P=0

PV/T=constant

First law of thermo-

Dynamics:

U=Q+W

PHY 231


Isovolumetric processes line a

Isovolumetric processes (line A)

  • V=0

  • W=0 (area under the curve is zero)

  • U=Q (Use U=W+Q, with W=0)

  • In case of ideal gas:

  • U=3/2nRT

  • if P then T (PV/T=constant)

    • so U=negative Q=negative

    • (Heat is extracted from the gas)

  • if P then T (PV/T=constant)

    • so U=positive Q=positive

  • (Heat is added to the gas)

p

v

PHY 231


Adiabatic process line b

Adiabatic process (line B)

p

  • Q=0

  • No heat is added/extracted from the

  • system.

  • U=W (Use U=W+Q, with Q=0)

  • In case of ideal gas:

  • U=3/2nRT

  • if T

    • U=negative W=negative

    • (The gas has done work)

  • if T

    • U=positive W=positive

  • (Work is done on the gas)

v

PHY 231


Isothermal processes

isothermal processes

p

  • T=0

  • The temperature is not changed

  • Q=-W (Use U=W+Q, with U=0)

  • if V

    • W=positive Q=negative

    • (Work is done on the gas

    • and energy extracted)

  • if V

    • W=negative Q=positive

  • (Work is done by the gas

  • and energy added)

v

PHY 231


Isobaric process

isobaric process

p

  • P=0

  • Use U=W+Q

  • In case of ideal gas:

  • W=-PV & U=3/2nRT

  • if V then T (PV/T=constant)

    • W: positive (work done on gas)

    • U: negative

    • Q: negative (heat extracted)

  • if V then T (PV/T=constant)

    • W: negative (work done by gas)

  • U: positive

    • Q: positive (heat added)

v

PHY 231


Cyclic processes

Cyclic processes

The system returns to its

original state. Therefore,

the internal energy must

be the same after completion

of the cycle (U=0)

PHY 231


Cyclic process step by step 1

W=-Area under P-V diagram

=-[(50-10)*10-3]*[(1.0-0.0)*105]

-½[(50-10)*10-3]*[(5.0-1.0)]*105=

=4000+8000

W=-12000 J

Cyclic Process, step by step 1

Process A-B.

Negative work is done on the gas:

(the gas is doing positive work).

U=3/2nRT=3/2(PBVB-PAVA)=

= 1.5*[(1E+5)(50E-03)-(5E+5)(10E-03)]=0

The internal energy has not changed

U=Q+W so Q=U-W=12000 J: Heat that was added to the

system was used to do the work!

PHY 231


Cyclic process step by step 2

W=-Area under P-V diagram

=-[(10-50)*10-3*(1.0-0.0)*105]=

W=4000 J

Work was done on the gas

Cyclic process, step by step 2

Process B-C

U=3/2nRT=3/2(PcVc-PbVb)=

=1.5[(1E+5)(10E-3)-(1E+5)(50E-3)]=-6000 J

The internal energy has decreased by 6000 J

U=Q+W so Q=U-W=-6000-4000 J=-10000 J

10000 J of energy has been transferred out of the system.

PHY 231


Cyclic process step by step 3

Cyclic process, step by step 3

Process C-A

W=-Area under P-V diagram

W=0 J

No work was done on/by the gas.

U=3/2nRT=3/2(PcVc-PbVb)=

=1.5[(5E+5)(10E-3)-(1E+5)(10E-3)]=+6000 J

The internal energy has increased by 6000 J

U=Q+W so Q=U-W=6000-0 J=6000 J

6000 J of energy has been transferred into the system.

PHY 231


Summary of the process

-AREA

Summary of the process

A-B

B-C

C-A

PHY 231


What did we do

What if the process was done in

the reverse way?

Net work was performed on the

gas and heat extracted from the gas.

We have built a heat pump!

(A fridge)

What did we do?

The gas performed net work (8000 J)

while heat was supplied (8000 J):

We have built an engine!

PHY 231


More general engine

More general engine

turns water to steam

heat reservoir Th

W=|Qh|-|Qc|

efficiency: W/|Qh|

e=1-|Qh|/|Qc|

Qh

work is done

the steam moves the piston

work

engine

W

Qc

The efficiency is determined

by how much of the heat you

supply to the engine is turned

into work instead of being lost

as waste.

the steam is condensed

coldreservoirTc

PHY 231


Physics 231 lecture 31 engines and fridges

Reverse direction: the fridge

heat is expelled to outside

heat reservoir Th

Qh

work is done

a piston compresses the coolant

work

engine

W

Qc

the fridge is cooled

coldreservoirTc

PHY 231


The 2 nd law of thermodynamics

The 2nd law of thermodynamics

1st law: U=Q+W

In a cyclic process (U=0) Q=W: we cannot do more work

than the amount of energy (heat) that we put inside

2nd law: It is impossible to construct an engine that,

operating in a cycle produces no other effect than the

absorption of energy from a reservoir and the performance

of an equal amount of work: we cannot get 100% efficiency

What is the most efficient engine we can make

given a heat and a cold reservoir?

PHY 231


Carnot engine

AB isothermal expansion

BC adiabatic expansion

W-, T-

W-, T-

W+, Q-

Tcold

DA adiabatic compression

CD isothermal compression

Carnot engine

W-, Q+

Thot

PHY 231


Carnot cycle

Carnot cycle

inverse Carnot cycle

Work done by engine: Weng

Weng=Qhot-Qcold

efficiency: 1-Tcold/Thot

A heat engine or a fridge!

By doing work we can

transport heat

PHY 231


Next lecture

Next lecture

Entropy and examples

PHY 231


  • Login