Thermodynamics 1 - Energy Analysis of Closed Systems

Chapter 4 Thermodynamics 1 1

Thermodynamics I

Introduction 1. Basic Concepts of Thermodynamics 2. Energy, Energy Transfer, and General Energy Analysis 3. Properties of Pure Substances 4. Energy Analysis of Closed Systems 5. Energy and Mass Analysis of Control Volumes 6. The Second Law of Thermodynamics 7. Entropy 8. Steam Power Cycle Applications Examples


Overview

4-1 Specific Heats 4-2 Moving Boundary Work 4-3 Internal Energy, Enthalpy and Specific Heats

of Ideal Gases 4-4 Internal Energy, Enthalpy and Specific Heats

of Solids and Liquids 4-5 Energy Balance for Closed System


Specific Heat Capacity and Heat Capacity

Specific heat is the energy required to raise the temperature of a unit mass of a substance by one degree in a specified way.


Specific Heat Capacity V=const: ( , )P=const: ( , )differentiated:

V T

u u v Th h p T

udu d du TT

vv

∂ ∂

==

∂ = + ∂

P T

hdh dTT

dPP

h∂ ∂ = + ∂

∂

V VV V

P PP P

u Uc CT Th Hc CT T

∂ ∂ = = ∂ ∂ ∂ ∂ = = ∂ ∂

or

or


Specific Heat for Ideal Gases

, ,

( )

( )

( ) ( )v av p a

V P

V P

V

v

P

u u T

h u Pvh u RT h h T

Pv RT

du dhdu dT dh dTdT dT

du c dT d

u u c T T h h c T T

h c dT

u u u c dT h h h c dT

=

= + = + ⇒ ==

= =

− ≈

⇒ = ⇒ =

⇒ ∆ = − = ⇒ ∆ = − =

− − ≈ −∫ ∫

From Joule's experiment:

therefore:

differentiated:

2 2

2 1 2

2 1 2 1 2 1 2

1 1

1

1

, ,( ) ( )v av p av

v p

U U C T T H H C T Tu c T h c T

− ≈ − − ≈ −

≈ ≈ 2 1 2 1 2 1 2 1


Specific Heat Relations for Ideal Gases

P V

P V

P

V

V P

h u RTdh du RdTc dT c dT RdT

c c R

ckc

kc R c Rk k

= += += +

⇒ = +

=

⇒ = =− −

Specific Heat Ratio

and11 1

• k=1.667 for monatomic gases • k=1.4 for diatomic gases


(kJ/kg)

For small temperature intervals, the specific heats may be assumed to vary linearly with temperature.

Internal energy and enthalpy change when specific heat is taken constant at an average value

The relation ∆ u = cv ∆T is valid for any kind of process, constant-volume or not.


Internal Energy, Enthalpy and Specific Heats of Solids and Liquids

( )

( ) ( )

P V

V

av

av

c c c

du c dt c T dT

u u u c T dT c T T

dh du vdP P du vdPdvh u v P c T v P

= =

= =

∆ = − = ≅ −

= + + = +∆ = ∆ + ∆ = ∆ + ∆

∫

Internal Energy Changes:

Enthalpy Changes:

2

2 1 2 11

0

• Constant spec. volume -> incompressible


Internal Energy, Enthalpy and Specific Heats of Solids and Liquids

@ , @

@ , @ @

::

( )

(

av

av

av

P T f T

P T f T f T

h u v P c T v P

h u c T

P const h u c TT const h v P

h h v P P

h hh h v P

∆ = ∆ + ∆ ≅ ∆ + ∆

∆ = ∆ ≅ ∆

= ∆ = ∆ ≅ ∆= ∆ = ∆

− = −

≅

≅ + −

For Solids:

For Liquids: (e.g. heaters)(e.g. pumps)

Enthalpy for compressed liquid:

2 1 2 1

)satP


Moving Boundary Work Moving boundary work (P dV work): The expansion and compression work in a piston-cylinder device.

The work associated

with a moving

boundary is called

boundary work.

A gas does a differential amount of work δWb as it forces the piston to move by a differential amount ds.

Quasi-equilibrium process: A process during which the system remains nearly in equilibrium at all times.

Wb is positive → for expansion Wb is negative → for compression


Moving Boundary Work

The area under the process curve on a P-V diagram represents the boundary work.

The boundary work done

during a process

depends on the path

followed as well as the end states.

The net work done during a

cycle is the difference

between the work done by the

system and the work done on the

system.


Isochoric/Isometric Process • Constant Volume Process:

– Ideal Gas

P

V

1

2

bW PdV

PV mRTPV mRT

P T Por constP T T

= =

==

⇒ = =

∫2

1

1 1

2 2

1 1

2 2

0


Isochoric/Isometric Process • With the moving boundary work

W=0 for the isochoric process to work heat needs to be extracted:

• Only the internal energy of the system changes.

P

V

1

2

( )

TvT

Tv T

Q U U

m c dT

mc T T

= − =

= =

= −

∫ 2

1

2

1

12 2 1

2 1


Isobaric Process • Constant Pressure Process:

– Ideal Gas

V P

V

1 2 ( )

( )

bW PdV P dV

P V VmR T T

PV mRTPV mRT

V T Vor constV T T

= = =

= −= −

==

⇒ = =

∫ ∫2 2

1 1

2 1

2 1

1 1

2 2

1 1

2 2

Law of Gay-Lussac


Isobaric Process • The isobaric process needs a heat

transfer that compensates for the volume change while keeping the pressure constant:

• The heat transfer is equal to the enthalpy change of the system.

V P

V

1 2

( )

T

pT

Tp T

Q H H

m c dT

mc T T

= − =

= =

= −

∫2

1

2

1

12 2 1

2 1


Isothermal Process • Constant Temperature Process:

P

V

1

2

ln ln

b

mRTPV

mRTW PdV dVV

V PmRT PVV P

PV mRTPV mRT

PV PV or PV const

=

⇒ = = =

= =

==

⇒ = =

∫ ∫2 2

1 1

2 1

1 2

1 1

2 2

1 1 2 2Law of Boyle-Mariotte


Isothermal Process • Since the temperature of the system is

constant, the internal energy of the system does not change. This implies that the heat transfer associated with an isothermal process must be:

• The heat transfer is equivalent to the

moving boundary work.

P

V

1

2

ln ln

bQ W

V PmRT PVV P

= =

= =

12

2 1

1 2


Polytropic Processes

• Constant PVn Process: – Ideal gas

,

ln ln ln ,

( ) , for ideal gas

n

b n

b

PV constconstW PdV dVV

PV PV nn

V V VPV PV PV nV V V

mR T TW nn

=

⇒ = = =

−= ≠

− = = = =

−⇒ = ≠

−

∫ ∫2 2

1 1

2 2 1 1

2 2 21 1 2 2

1 1 1

2 1

11

1

11


Polytropic Processes

• Special polytropic processes: – Constant pressure: n=0 – Constant volume: n= – Isothermal & ideal gas: n=1 – Adiabatic & ideal gas: n=k=Cp/Cv

• The last process is know as isentropic. Cp and CV are specific

heats of the substance at constant pressure and constant volume.


Isentropic Process

• Since the isentropic process is adiabatic no heat transfer will occur:

• For isentropic processes the following relationships are valid:

P

V

1

2

kkk

Q

T V PT V P

−−

=

= =

12

11

1 2 1

2 1 2

0


Isentropic Process • The work for isentropic processes is

given by:

P

V

1

2 b

kk

k

kk

W PdV

dVPVV

PV Vk V

PV T PV Pk T k P

−

−

= =

= =

= − = −

= − = − − −

∫

∫

2

1

21 1 1

11 1 1

2

1

1 1 2 1 1 2

1 1

11

1 11 1


Isentropic Process • The work for isentropic processes is also

given by:

• The moving boundary work is completely converted into internal energy and vice versa.

P

V

1

2

( )

( )

( )

b

v

PV TWk T

mR T Tk

P T PTkmc T T

= − = −

= − =−

= − =−

= −

1 1 2

1

2 1

2 2 1 1

2 1

11

11

1


Energy Balance for Closed System

>Heat Transfer< >Work Transfer< >Mass flow<

A closed system involves only heat transfer and work.


Energy balance for a constant-pressure expansion or compression process

HWU b ∆=+∆

For a constant-pressure expansion or compression process:

An example of constant-pressure process

General analysis for a closed system undergoing a quasi-equilibrium constant-pressure process. Q is to the system and W is from the system.


A piston cylinder contains 25g of saturated water vapour at a constant pressure of 300kPa. An electric heater (0.2A and 120V) heats the water for 5 mins. There is a heat loss of 3.7kJ. Determine the final temperature of the steam.

Closed System - Example


Examples

• A 0.1 m3 rigid tank contains steam initially at 500 kPa and 200ºC. The steam is now allowed to cool until its temperature drops to 50 ºC. Determine the amount of heat transfer during this process and the final pressure in the tank.

• A piston-cylinder device initially contains 0.5 m3 of nitrogen gas at 400 kPa and 27ºC. An electric heater with the device is turned on and is allowed to pass a current of 2A for 5 min from a 120V source. Nitrogen expands at constant pressure, and a heat loss of 2800 J occurs during the process. Determine the final temperature of the nitrogen.

Documents

Thermodynamics 1 - Energy Analysis of Closed Systems