me2121-me1121e slides (chapter 2 2014) [compatibility mode]

Mechanical Engineering,NUS

• Definitions and Basic Concepts– What is a thermodynamic system?

ME2121/ TM1121- Thermodynamics

Chapter 2 – 1/ 19

A close system – is one where no mass leavesor enters the boundary.

An open system – is one where mass enters and leavesthe boundary.

Matter

Eg. Gases, liquids, solids

System boundary

Work transfer,

dW Heat transfer

dQ

Energy content of matter, dE = dU


• System characterizes by properties (P,T, V, U, H, S etc.),• Property measures the state of matter(s), it describes

two states to form a process,• property can be defined by other established properties,

e.g., enthalpy (H) is given by (in extensive form)


Chapter 2 – 2/ 19

work

dQ

dWGas, dU

Measured by P, T, V

U = U (P,T) or U = U (P,V)

and H =H (P,T) = H (P,V)= U + PV

PvuhmVp

mU

mH

Piston-cylinder


• Quantity of matters– Use SI units (kg, m, mol, s, J, radians, etc)– Mol. Is a basic measure of matter, eg. kmol of

Carbon (C12) has a mass of 12 kg. – For gases at STP, 1 kmol has 6.023 x 1026

gaseous molecules– Conversion between gravimetric (mass) analysis

to volumetric (molar) analysis is via the molecular mass (M),

• Eg., ni = mi / Mi


Chapter 2 – 3/ 19


• Cycle and property– A cycle requires two or more processes to complete– The summation of a thermodynamic variable for a

cycle leads to


Chapter 2 – 4/ 19

0Y = P

x= v

Path A

Path B

1(P1, v1)

2(P2, v2)

Equation of state (EoS), eg., PV = mRT

Or PV = nRo T


• Mathematical tests of a property– A property satisfies continuum requirement and from maths,

– and if continuous, then it satisfy the following identity (exactness requirement):

– Letting z= T, x = P, y= v, and writing for the total change in dT,– the function is T = T(P,v)


Chapter 2 – 5/ 19

dvvTdP

PTdT

Pv

xinChangedy

ldifferntiaPartial

xyz

yinchangedx

aldifferentiPartial

yxz

aldifferentiTotal

dz

yxxy yz

xxz

y

Order of differentiation is immaterial.


• Invoking exactness,

• For ideal gas, the equation of state is Pv=RT, differentiating twice

and

• The other requirement of a property is independency.


Chapter 2 – 6/ 19

RPvT 1

.

2

RvP

T 1.

2

Rv

TP

andRP

vT 1

vP

TPv

T..

22


• Test for independency (Tutorial#1, Q1) – P = P(T,v); and T=T(P,v)

and

– To show that the mathematical requirement for independency (by inter-substituting)


Chapter 2 – 7/ 19

dvvPdT

TPdP

Tv

1

TPv Pv

vT

TP

dvvTdP

PTdT

Pv

Re-arranging

Writing for total differentials

PV= (R)T , Equation of state

Step 1

Step 2

Step 3

P and V can take any value whenthe bracketed terms are zero.

Worked example 2.1

TTV

V P

11

PP

VV T

11

Consider an ideal gas in a system where the gas has its expansivity ( ) and compressibilty ( ) expressed as

The general expression for the variables can be given by

V=V(P,T)

From mathematics, the total differential of V is given by

dPPVdT

TVdV

V

T

V

P

Using the definitions and the experimental relation of the gas, the above can be re-written as

0

PdP

TdT

VdVor

PdPV

TdTV

VdPVdTdV

And taking log, we have, ln V- ln T + ln P =ln (constant),

)(tan

tan

mRistconstheand

tconsT

PV

That is, the equation of state (EoS)

Replace &

By experiments


• Zero Law of Thermodynamics

– Thermometric property of sensor,


Chapter 2 – 10/ 19

Thermometer AEquilibrium with each other

Master thermometer

Thermometer B

referenceorXX

1

2

1

2

Thermodynamic equilibrium

Thermodynamic equilibrium

propertyicthermometrXetemperaturempiricalPP

hh

XX

,1

2

1

2

1

2

1

2


Types of temperature measuring devices• a liquid-in-bulb device, called thermometer – (X=

expansion of liquid represented by height of column)• resistance change of an inert wire, called resistance-

temperature device or RTD, - (X=change in R (ohms))• the electro-motive force generated across a pair of wires

of different materials, called the thermocouples, (X= change in e.m.f.(micro-volts).

• the constant-volume bulb with pressure changes (mercury in tube), (X= change in gas pressures (Pa)).


Chapter 2 – 11/ 19


– Constant-volume bulb thermometer ( 3 – triple point of water, s – steam point)


Chapter 2 – 12/ 19

N2

O2

Air

H21.3650

1.3660

1.3670

1.3680

250 500 750 1000

P3 (Torr)

1.3661

Temperature to be determined

Ps/P3

P3 = is the pressureat the triple point, 1 torr = 1/1000 mm hg

Gas molecules inside the bulb

How do you calibrate?

Triple point of water – where all 3 –phase of water can co-exist.

• Reading of mercury height is an indicationof the Pressure (P) of gas in the bulb,• Immerse the bulb into a referencetemperature (i.e., the triple point of water),• Then immerse the bulb into an environmentof steam (boiling water) at 1 atm.,• Evacuate some gas out of the bulb andrepeat the processes,• Plot the ratio, Ps /P3 to P3 using theexperimental data


• How are the numerical numbers of an absolute temperature scale assigned?– Require two pieces of information,– (a) The number of divisions between freezing point and ice point,

– (b) The thermometric property ratio of a thermometer (that follows the thermodynamics law), eg., constant volume bulb thermometer.

– Solving,


Chapter 2 – 13/ 19

,""tan180100 Atconsaororis

3661.1iP

P

is

K

PP

i

si 15.273

3661.0100

1

100


• Other absolute temperature scales


Chapter 2 – 14/ 19

NBP of H2O

NFP of H2O

NBP of O2

Absolute Zero

373.15 100 671.67 212

K oC R oF

273.15 0 491.67 32

90 -183 162 -297

0 -273.15 0-459.67

Tutorial #1, Q4

• The thermometric property of a constant-volume bulb thermometer is

where subscript “s” refers to steam point, and “3” refers to the triple point of water (0.01 C).

• Needed two information to define a temperature scale: (i) the number of divisions between ice point and boiling of water, (ii) the ratio of Psteam to Pice = 1.3661 (see lecture notes also).

33 PPss

.deg57.18657.13650deg57.136,

)exp(3661.1

,50

si

i

s

i

s

is

andhaveweequationstwothesesolving

facterimentalanPP

and


• Example 2.2• The volume of mercury contained in a bulb type

thermometer is Vo (m3) at ice point. If the capillaryof thermometer has a cross sectional area Ao (m2)and the linear and volume expansion coefficientsof glass and mercury are

, respectively,

determine the change in height (m) of mercurycolumn for a small change in temperature, (K)?


Chapter 2 – 16/ 19

mG and


• Graphical interpretationFor a ΔT, increase in mercury volume

– increase in stem area


Chapter 2 – 17/ 19

h2

A2h2

V0=Vb1+A1h1

Vb2

t=0oCt= 0+Δt (C)

h1

V2=V,b2 + A2h2

)1(1112 tVtVVV mb

tofeffect

m

)21())(21(

)(

1

221

2112

tAttr

trrA

G

small

GG

tofeffect

G

K-1

)31(

)31(34

)(34

1

31

3112

tV

tr

trrV

Gb

G

tofeffect

Gb

Increase on bulb volume

stem

bulb


– re-arranging, we obtained


Chapter 2 – 18/ 19

ttA

Vh

G

Gm

o

o

213

Should h1 ≠ 0 , then initial volume of mercury in thermometer is V0 = Vb1 + A1h1.

(Vb1 +A1 h1)(1+βm Δt) = (Vb1 )(1+3αG Δt)+h A1 (1+2αG Δt)

You can re-arrange the expression to find the new height, h?.

Mercury volume New bulb volume New stem volume


• Home exercise (the case when h1 is not naught. May worth your while to derive!)

• and Tutorial #1- (4 challenging questions. Q1 & Q4 are past year exam questions)


Chapter 2 – 19/ 19

.,0

21

)2(3

1

1

11

1

112

solutionprevioustorevertequationthenhifand

tVhA

AtVhh

G

mGGm

Home assignment #1 (Chapter 2)

An experiment is conducted for four types of gases where the product,pressure and volume, (PV), of each gas is monitored at a reproducibletemperature of water, namely the triple point of water. These experimentsare repeated consecutively with the gas pressure in the apparatus isreduced, P→0, as shown in the figure below.

Write down the empirical temperature scale for the gases at any temperature, θ.

If the numerical number for the triple point of water is adopted internationally as θ* = T* = 273.16 K, then, write down the absolutely temperature scale T(K).

From the figure, demonstrate that the gas constant is 82.05 cm3.atm/mol.K or 8.314 J/mol.K. (Note: 1 atm = 1.01325 bar, 1 m3 = 106 cm3). Explain, briefly, why this is a universal value?

Example 2.1 (b) of Notes

v

T

P

dTTvdv

P

dpPvdv

T

dv

dPPRTdT

PR

dPPvdT

Tvdv

T

TP

2

Actual 3-D surface of specific volume

Here, R= Ro/M = 0.287 kJ/kg.K,Tmean= 318 K, Pmean= 112 kPa,using total differential,

kgm

xx

xxxxdv

/248.01455.01025.0

)1020(10112

31810287.0)40(1011210287.0

3

323

3

3

3

From the ideal gas equation, dv can also be estimated, i.e.,

./2499.07010.09510.0122

298287.0

102338287.0

3

1

1

2

2

12

kgm

x

xP

RTP

RTdv

vv

me2121-me1121e slides (chapter 2 2014) [compatibility mode]

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