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THERMODYNAMICS

Lecture 3: Work and heat – FirstLaw of Thermodynamics

Thermodynamics

Thermodynamic Systems and P-V Diagrams

Ideal gas law: P V = n R TFor n fixed, P and V determine the “state” of the system

T = P V/ (n R)U = (3/2) n RT = (3/2) P V

Examples:which point has highest T ?

2which point has lowest U ?

3to change the system from 3 to 2,

energy must be added to system.

Work done in a thermal process is given by the area underneath the P,V graph.

V

P1 2

3

V1 V2

P1

P3

Thermodynamics

Work Done by a System (P=constant)

WW Δy

W = F s = P A Δ y = P ΔVW > 0 if ΔV > 0

expanding system does work on surroundings (W positive)W < 0 if ΔV < 0

contracting system : work is done on the system by surroundings (W negative)W = 0 if ΔV = 0

system with constant volume does no work

GasP,V1,T1

GasP,V2,T2

System: GasSurroundings:Piston, walls

Isobaric process

Thermodynamics

Classification of Thermal Processes

Isobaric : P = constant, W = P ΔVIsochoric : V = constant, W = 0 => ΔU = QIsothermal : T = constant, W = n R T ln(Vf/Vi) (ideal gas)Adiabatic : Q = 0 => W = - ΔU= -3/2 n R ΔT (ideal gas)

Thermodynamics

V

P 1 2

34

V

PW = PΔV (>0)

1 2

34

ΔV > 0 V

PW = PΔV = 0

1 2

34

ΔV = 0

V

PW = PΔV ( 0

Wtot = ??Isobaric Process: P=constantIsochoric Process: V=constant

Thermodynamics

V

P1

3

2

Now try this: What is the total work doneby system when going fromstate 1 to state 2 to state 3 andback to state 1 ?

V1 V2

P1

P3Area = (V2-V1)x(P1-P3)/2

Thermodynamics

First Law of ThermodynamicsExample

V

P1 2

V1 V2

P2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant pressureP=1000 Pa, where V1 =2m3 and V2 =3m3. FindT1, T2, ΔU, W, Q.

1. P V1 = n R T1 ⇒ T1 = P V1/(nR) = 120K2. P V2 = n RT2 ⇒ T2 = P V2/(nR) = 180K

3. ΔU = (3/2) n R ΔT = 1500 J or ΔU = (3/2) P ΔV = 1500 J

4. W = P ΔV = 1000 J > 0 Work done by the system (gas)

5. Q = ΔU + W = 1500 J + 1000 J = 2500 J > 0 heat gained by system (gas)

Thermodynamics

First Law of ThermodynamicsExample

2 moles of monatomic ideal gas is takenfrom state 1 to state 2 at constant volumeV=2m3, where T1=120K and T2 =180K. Find Q.

1. P V1 = n R T1 ⇒ P1 = n R T1/V = 1000 Pa

2. P V2 = n R T1 ⇒ P2 = n R T2/V = 1500 Pa

3. ΔU = (3/2) n R ΔT = 1500 J

4. W = P ΔV = 0 J

5. Q = ΔU + W = 0 + 1500 = 1500 J=> It requires less heat to raise T at const. volume than at const. pressure.

V

P

2

1

V

P2

P1

Thermodynamics

Concept Question

Shown in the picture below are the pressure versus volume graphsfor two thermal processes, in each case moving a system from state A to state B along the straight line shown. In which case is the work done by the system the biggest?1. Case 1 2. Case 2 3. Same

A

B4

2

3 9 V(m3)

Case 1

A

B4

2

3 9 V(m3)

P(bar)

Case 2

P(bar)

correct

Net Work = area under P-V curveArea the same in both cases!

Thermodynamics

work W

insulation

Q=0ΔE=W= ΔU

Internal energy increases as workwas done on the system. Afterreaching by liquid of equilibrium we notice that increase of internalthermal energy occured.

That increase took place due to transfer of additional kinetic energyto the liquid particles.

Internal thermal energy is denoted by U. Then we can write:

⋅⋅⋅++++= chempk EEEUE (2.13)

Thermodynamics

Jeśli rozważymy proces cykliczny w którym następuje mała zmiana układu, wtedy I zasadę termodynamiki możemy zapisaćjako;

dEdWdQ =+

Ponieważ energia wewnętrzna jest funkcją stanu, jej zmiana dla cyklu zamkniętego jest równa zero.

∫ ∫ =+ 0dWdQ (2.14)

Energia wewnętrzna jest funkcją stanu opisaną przez

∫ = 0dE .

Thermodynamics

2.4 Heat and specific heat

Heat is a form of energy of interaction between thermodynamicalsystem and surroundings. Heat cannot be calculated in a form workis calculated (force times displacement). Hence heat and work aredifferent types of energy.

Heat is intuitively linked to temperature of the system, as it isincreased with amount of heat supplied to the system. Formerlyit was assumed that heat is a substance which can be transferred from one body to another.

It should be now apparent that heat is not the substancecontained in the system, but is revealed during the interactionbetween the system and surroundings during transition from one state to another.

Thermodynamics

yy dT

QdC ⎟⎟⎠

⎞⎜⎜⎝

⎛=

'

(2.15)

Let’s consider the thermodynamical system where changes ofinternal energy are rendered solely by change of internal thermalenergy U. Let’s also assume that influence by the kind of work canonly take place by the change of system volume. For such system the First Law of Thermodynamics reads:

dUpdVQddUWdQd

=−

=+'

''

Remembering the sign convention for work, the heat added to thesystem can be writen as;

pdVdUQd +='

The remainder of previous theory is the concept of specific heat:

Thermodynamics

In case of constant volume process dV=0. Then :

dTcdudTcdQ

VV

VV

== Small letters relate to relevant quantities

per unit mass.

In case of constant volume process the specific heat reads:

VV T

uc ⎟⎠⎞

⎜⎝⎛∂∂

= (2.16)

If the isobaric process is considered then the transfer of heatper unit mass is:

ppp pdvuQd +='

(2.17)

Thermodynamics

A new property of the system can be introduced - enthalpy.

pvuh += (2.18)

At constant pressure the enthalpy differential per unit mass is:

ppp dvpdudh += (2.19)

Then the heat of such process reads;

pppp dTcdhdQ ==

oraz

pp T

hc ⎟⎠⎞

⎜⎝⎛∂∂

= (2.20)

Specific heat is expressed in units kJ/kg ·K .

Thermodynamics

2.5 Specific heat of ideal gasIf gas is an ideal gas then internal energy and enthalpy arefuncitions of temperature.

In such case we can write

dTcuuuT

T v∫=−=2

112Δ dTchhh

T

T p∫=−=2

112Δ(2.21) (2.22)

In case of constant values of specific heat:

)( 1212 TTcuu v −=− )( 1212 TTchh p −=−(2.23) (2.24)

Remembering that dh=cpdT and du=cvdT. Hence dTccdudh vp )( −=− .

We know also that

dTRdupvddudh +=+= )( . .

Thermodynamics

Czyli dTccdTR vp )( −= oraz,

vp ccR −= (2.25)

Thermodynamics

How To Change the Temperature of a System (Solids or Liquids):

Add or subtract heatQ = heat = energy that flows from warmer to cooler systems.

Q = c m ΔTQ = amount of heat that must be supplied or subtracted to raise

or lower the temperature of mass m by an amount ΔT. Units of Q: Joules or calories

• 1 cal = 4.186 J• 1 kcal = 1 Cal = 4186 J

c = specific heat capacity: Heat required to raise 1 kg by 1oC.

Q = c m ΔT : “Cause” = “inertia” x “effect” (just like F=ma)cause = Qeffect = ΔTinertia = c m (mass x specific heat capacity)

ΔT = Q/(cm) (just like a = F/m)

Thermodynamics

Examples of Specific Heat Capacity

Substance c in J/(kg C)Aluminum 900Copper 387Iron 452Lead 128Water (15 C) 4186Ice (-15 C) 2000

Suppose you have equal masses of aluminum and copper at the sameinitial temperature. You add 1000 J of heat to each of them. Which one ends up at the higher final temperature

a) aluminumb) copperc) the same

correct

Q= c m ΔT

Thermodynamics

Analysis of processes in open systems

We are now aware of analytical form of First Law of Thermodynamicsand we know how to calculate thermodynamic properties ofsubstances.

Now let’s consider thermodynamic problems in systems, wheresubstance can cross the boundaries of the system, hence opensystems. In order to consider such processes the procedure must be followed which enables description of the system and processestaking place there. The following issues ought to be considered:

1. How the thermodynamic system under study is described?

2. What substance is contained in the considered system and do we know properties of that substance ?

3. Is the system open or closed?

4. How to describe the undergoing thermodynamical process?

5. Is the considered process stationary or variable in time?

Thermodynamics

After answers to these questions is found then the mathematicalanalysis of the problem must be conducted, for example:1. Draw the schematic of the system and determine inlets and

outlets for mass and energy,2. Mathematical formulation of available information,3. Performing the energy balance,4. Writing down relations describing the process,5. Combination of all information and obtaining the result.

We ought to remember the assumed sign convention!!!.

We ought not forget the energy conservation principle:

Energy which is received by the system is equal to energy whichleaves the system plus energy accumulated in the system.

Thermodynamics

Streamof coldwater

Hot water

Q added heat

System boundary

Defined system does not conform with the previous definitionsof thermodynamical system, where some specified amount ofmass was present.Let’s see now how can we describe the flow of mass. That can be done by definition of its rate.

An example of the thermodynamicalopen system can be the water heater.

Thermodynamics

1 2

A

m•

Δs

If we denote velocity as v=Δs/dt then we get an average velocityof flowing substance, density of which is denoted by ρ, and thenthe mass flowrate per unit time reads

Am ρ=. v (4.2)

Thermodynamics

Analysis of open system – control volume

In order to be able to analyse thermodynamical processes in opensystems we introduce the definition of control volume. It is a part of space, limited by the boundary, where the flow of mass andenergy is observed.

σ⎟⎠⎞

⎜⎝⎛

dtdm

•

im

A

•

emσ controlvolume

mass balance

Thermodynamics

σ⎟⎠⎞

⎜⎝⎛

dtdE

•

im

B

•

em

σ objętośćkontrolna

energybalance

ei ee

Letter e denotes amount of energy per unit mass.

dtQd '

dtWd '

Thermodynamics

Mass flowing into the control volume is equal to mass flowing out plus the amount of increase of mass inside the control volume.

edtdm

i mm••

+⎟⎠⎞

⎜⎝⎛=

σ

(4.3)

Our system can have several inletes and outlets. Then (4.3) reads:

∑∑••

+⎟⎠⎞

⎜⎝⎛=

eiedt

dmi mm

σ (4.4)

In order to balance energy in the entire control volume let’sconsider some specified mass moving through the control volume.

That corresponds to the behaviour of a closed system where mass motion takes place through the control volume

Thermodynamics

Such system can be exposed to pressure of surroundings. Thetransfer of heat outside the system boundary may take place orvarious forces performing work can be present.

Internal energy of the closed system can change due to the motionfrom one place to another, as well as due to change of velocity.

Independently from observed phenomena we can apply the principleof energy conservation. The total mass flowrate to and from thecontrol volume can be considered as a series of elements dm , i.e. small closed thermodynamical systems. We can regard that themass flowrate through the system transports internal energythrough the boundaries of our system.

The energy conservation equation for such system can be formulatesin the following way:

Thermodynamics

Transport of internal energy to the control volume +heat added to the control volume, +work done on all elements during their motion through thesystemare equalto the increase of internal energy inside the control volume+transport of internal energy outside the control volume.

Analityczny zapis jest następujący:

edtdE

dtWd

dtQd

i EE••

+⎟⎠⎞

⎜⎝⎛=++

σ

''(4.4)

where Ei and Ee denote transport of internal energy per unit oftime at inlet and outlet from the system, respectively. Thesequantities can be determined as:

. .

Thermodynamics

ee

ii

eme

emi

E

E••

••

=

=(4.5)

Equation (4.4) can be used in analysis of open systems. However, The work term can be expressed in a different manner. For themass to flow through the system the force is required. That forceis given through the pressure in the system.

The element of massy with volume A · Δs in order to be transportedto or from the control volume has to withstand the action of forcep·A on the distance Δs , where independently from mass Δs=V/A.

Work required to pump mass to or from the control volume is :

VpAVApsFdsFW ⋅=⋅⋅=⋅=⋅= ∫ Δ (4.6)

Thermodynamics

1 2

m•

The resulting work done on the system during thedisplacement of mass from point 1 to point 2 is:

Δs

A

2211 VpVpWwyp −=

p1V1 denotes the work performed on the volume during its inputinto the control volume, whereas p2V2 at removing it from thecontrol volume, respectively. The difference is equal to theresulting work.

pV denotes the work of the rate of mass and that quantitiy mustbe considered separately from the work introduced from outside.

Thermodynamics

The energy balance equation takes the form:

)('')( eeeezewiiii vpemdtdE

dtWd

dtQdvpem ++⎟

⎠⎞

⎜⎝⎛=+++

••

σ(4.7)

Summarising, vi,e is a volume of a unit of mass. Wzew worksupplied to the control volume be the external forces.

Equation (4.7) represents a general energy balance for the opensystem.

In case when considered open system behaves as a stationarysystem, i.e. there is no change of any parameter in the controlvolume in the period of time, then

00 =⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

σσ dtdmand

dTdE ,

THERMODYNAMICSThermodynamic Systems and P-V DiagramsWork Done by a System (P=constant)Classification of Thermal ProcessesFirst Law of Thermodynamics�ExampleFirst Law of Thermodynamics�ExampleConcept QuestionHow To Change the Temperature of a System (Solids or Liquids):Examples of Specific Heat Capacity