UNIT – III

PART – A 1. State Boyle’s law. Boyle’s law states. “The volume of a given mass of gas varies inversely as its absolute pressure, when the temperature remains constant.

1v

p

2. State Charle’s law. Charle’s law states, “The volume of a given mass of a gas varies directly as its absolute temperature, when the pressure remains constant.

V T 3. State Joule’s Law. Joule’s law states, “The internal energy a given quantity of gas depends only on the temperature”. 4. State Regnault’s law Regnault’s law states that CP and Cv of a gas always remains constant. 5. State Avogadro’s law. Avogadro’s law states. “Equal volumes of different perfect gases at the same temperature and pressure, contain equal number of molecules”. 6. State Daltons law of partial pressure. Dalton’s law of partial pressure states. “The total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by individual gases if each one of them occupied separately in the total volume of the mixture at mixture temperature”. P = P1 + P2 + P3 + …… Pk 7. Distinguish between ideal and real gas. An ideal gas is one which strictly follows the gas laws under all conditions of temperature and pressure. In actual practice, there is no real gas which strictly follows the gas laws over the entire range of temperature and pressure. However hydrogen, oxygen, nitrogen and air behave as an ideal gas under certain temperature and pressure limits.

8. Define Joule – Thomson Co – efficient. The Joule – Thomson co – efficient is defined as the change in temperature with change in pressure keeping the enthalpy remains constant. It is denoted by the

h

T

P

9. Define co-efficient of volume expansion and isothermal compressibility. Co-efficient of Volume expansion: The co – efficient of volume expansions is defined as the change in volume with

change in temperature per unit volume keeping the pressure constant. It is denoted by

p

1 v

v T

Isothermal compressibility: It is defined as the change in volume with change in pressure per unit volume keeping the temperature constant. It is denoted by K

1 vK

v P

10. What is compressibility factor? We know that, the perfect gas equation is pv = RT. But for real gas, a correction factor has to be introduced in the perfect gas equation to take into account the deviation of real gas from the perfect gas equation. The factor is known as compressibility factor (z) and is defined by

PvZ

RT

11. State the assumptions made in kinetic theory gases.

1. There is no intermolecular force between particulars. 2. The volume of the molecules is negligible comparison with the gas

PART – B 1. A Mixture of 2 kg Oxygen (M=32kg/kgmol) and 2kfg argon (m=40 Kg/kmd) is present in an insulated piston in cylinder arrangement at 100kpa, 300K. The piston now compresses the mixture to half of its initial volume. Find the final pressure, temperature and piston work, assume c, for oxygen and argon for oxygen and argon as 0.6618kj/kgk and 0.3122KJ/kgK respectively System : Closed Process : Adiabatic compression Working fluid : Mixture of oxygen and argon.

1. Mo2 = 32kg/kg mol 2. Mar = 40kg/kg mol 3. T1 = 300t\k

Known 4. P1 = 100kpa 5. V2 = ½ V1 6. Cv02= 0.6618kj/kgk 7. Cv02= 0.3122kJ/kgK

Diagram: 2KgO2

2kg Ar at 100 kpa, 300K To find: 1. p2; 2. T2 ; 3.1W2

Analysis : a. p1 = 11

1

MRTV

p

Where m=mo2+mAr

=2+2 =4kg

22

2 2

2 8.314 2 8.314 =

2+2 32 2 2 40

kj =0.234

kgk

ARAR

AR AR

mo mR Ro R

mo m mo m

x x

1

3

2 2

2

2 2 0.23 300

100

=2.806m

( ) ( )

( )r AR

v

AR

x xV

mo Cvo m CvC

mo m

Diagram:

2 0.6618 2 0.3122

2 2

0.487

0.487 0.234

0.721

p v

kJ

kgK

C C R

0.721

0.487

1.480

p

v

C

C

It is given that

2kg O2

+

2kg Ar

at 100kPa, 300K

11

2

VV

Initial State

Final State

12

2

VV

=1.4033 Since the process is adiabatic, assuming the process to be reversible

1

2 1

1 2

1

2 1

1 2

0.48

2

2

2

=300(2)

=418.K

p

T V

T V

T V

T V

mRT

V

2 2 1 11 2

4 0.234 418.4

1.403

279.15

1

279.15 1.403 100 2.806

1.48 1

231.3

kpa

p V pVW

kJ

2. Consider an ideal gas at 300k and 0.86m3/kg. As a result of some disturbance that state

of the gas changes to 302K and 0.87m3.kg. Estimate the change in the pressure of the gas as

a result of these disturbances.

Let p= f(T.v)

p pdp =

TT dv

v

For an ideal gas pv=RT

2

2

2

v

T

RTp

v

p RdT

T v

p Rtdvand

v v

RdT Rtdvdp

v v

From the given data dT=2K dv=0.01m3/kg Substituting these values we get

0.287 2 0.287 301 0.01

0.865 0.865

=-0.461kpa

dp

3. Derive the equation.

2

2

P

T p

C vT

p T

and prove that Cp of an ideal gas is a function of T only. From I law of thermodynamics. Dq=dh-vdp For an ideal gas undergoing constant pressure process Dq=Cp dt Or Tds = Cp dt

v

v

sC T

T

Differentiating w.r.t. v at constant T we get

=

v

vT

v

TT

v

C sT

v v v

or

C sT

v T v

p

T T

From Maxwell relation

2

2

v

T v

C pT

v T

For an ideal gas Pv=RT

0

0

v

v

T

RTp

v

p R

T v

p R

T T T v

Cor

v

Cv is a function of temperature alone. 4. Show that the relation p(v-b)=RT satisfies cyclic relation. where b and R are constant.

cyclic relation in terms of pa v and T can be expressed as

1T p v

p v T

v T p

where,

2( )T T

p RT RT

v v v b v b

1

1

1 =

R/v-b

v-b =

R

v

v

v

T

pp

T

p RT

T v b

2

R v =

P

-( )

R =

p(v-b)

p p

T p v

u RT RTb Since b

T T p p

p v T RT R v b

v T p v b p R

=-1

5. Determine hfg for water at 1000C from the following data. T(0C) p(kpa)

95 84.55 100 101.35 105 120.82

Consider the clausius – clpeyron eqn

1

2 1 2

4

1 1

120.82 1 1

8.31484.55 368 378

18.016

0.357 (1.558 10 )

2291 /

fg

fg

fg

fg

hp

p R T T

hIn

h

h kJ kg

T1=95+273 =368K T2=105+273 =378K p1=84.55kpa

p2=120.82kpa 7. Determine the Joule Thomson coefficient for a van der walls gas given by the

equation.2

( )a

p v b RTv

Prove that for large volume (or low pressure), the inversion temperature is equal to za/br.

joule-Thomson Coefficient

p

1 =

Cp

T

p

TT v

p

Differentiating

2

2 3

2 2 3

2 3

( ) . . . @constant,

2( )

2 2

2

p p

p

p

ap v b RTwr t toT P we get

v

a T a Tp v b R

v p v p

v a a abp R

T v v v

v a abp R

T v v

2 3

2

2 3

2

2 3

1

2

( )1

2

2 31

=2

ph

p

p

T RT

a abp C pv v

ap v b

v

a abC pv v

a abbp

v va abC

pv v

For large value of v at low-pressure.

ph

p

p

21

=C

1 2 =

C

1 2 =

C

abv

T vp p

ab

pv

ab

RT

When

inversion

inversion

0

2athen b=

RT

2or T

h

T

p

a

bR

8. Determine the change of internal energy, enthalpy and entropy for an isothermal process

when the gas obeys Vander Wall’s equation.

System : A van der Waal’s gas under going a process

To obtain : An expression for each of the following

1)u2-u1

2)h2-h1

3)s2-s1

We know that from Eqn. 9.25

T v

u pT p

v T

For Van der Waals gas

2

2

( )

v

ap v b RT

v

RT ap

v b v

p R

T v b

2

2

2

Therefore

=

T

v

T T

u RTp

v v b

ap p

v

p a

T v

adu dV

v

2 1

1 2

2 1

1 2

1 1( )

1 1( )

u u T aV V

u u aV V

We know that

U2=h2-p2v2

U1=h1-p1v1

There fore

H2 –h1 = (u2-u1)+(p2v2-p1v1)

2 2 1 1

1 2

1 1( )p v p v a

V V

Equation 9.34 gives

vCv

dT pds dv

T T

But from Equations 9.25

1

v p

p up

T v T

It has already been proved that for Vander Waals gas

2

2

1Therefore

T

v

u a

v v

p ap

T v T

Substituting the above expression in the expression for ds we get

s v

2

v

v

dT 1d =C

T

dT 1 =C

T

dT =C

T

ap dv

v T

RTdv

v b T

dvR

v b

Let us Consider an is other malo process

21

1

( )s T

V bs s RIn

V b

9. Prove that Cp-Cv=R for an ideal gas.

System : Ideal gas undergoing process

To prove : Cp-Cv=R

Proof : From Equation 9.37

For an ideal gas

pv=RT

p v

p v

v pC C T

T T

Therefore

p

p

p v

v R

T p

v R

T V

R RC C T

p V

RTR

PV

R

10. Calculate the specific volume of dry saturated steam at a pressure of 147 kPa at which

the values of temperature T and latent heat L are 110.790C and 2223.3KJ/kg respectively.

Further, saturation temperature of steam at pressure of 157 kPa is 112.740C . Neglect the

specific volume of water.

From Clausius – Clapeyron equation

3

( )

2223.3 112.74 110.79

273 110.79 157 147

1.13 /

fg

fg

fg

g

fg

jdp

dT T V

h dpOr V

T dT

h T

T P

Kg

11. Prove that constant pressure lines in the wet region of a mollier diagram are straight and

not parallel and that the slope of a constant pressure line in the superheat region increases

with temperature.

Let h=f(s.p)

Also dh=Tds+vdp

Comparing the coefficients of dh, we get

p s

p

h hdh ds dp

s p

hT

s

This relation give the slop of the isobars in the Mollier diagram as shown below.

h P=C

dh

ds

s

In the wet region for a given pressure temperature remains constant and hence it is a straight line.

Higher the pressure, higher will be the saturation temperature and higher will be the slope. there fore isobars slope upward more steely as the pressure increases.

As temperature increases beyond the saturation line, that is, in the superheated vapour region, these lines bend slightly upward in that region.

13. AVessel of volume 0.28 m3 contain 10 kg of air at 302K. Determine the pressure exerted by the air using

(1) Perfect gas equation (2) Vander Waals equation (3) Generalised compressibility chart.

[Tale Critical temperature of air is 132.8 K; Critical pressure of air is 37.7 bar] Given data: Volume, v = 0.28 m3 Mass, m = 10 kg Temperature T = 302.8K Critical Temperature (Tc) = 132.8K Critical Pressure (pc) = 37.7 bar

= 37.7 100 kN/m2 To find: Pressure (p) Solution: 1. Perfect gas equation:

2

2

pv = mRT

mRT p =

v

10 0.287 302 p = [ R for air is0.287 kJ/kgk]

0.28

= 3095.5 N/m

p = 3095.5 kPa

[ 1 N/m 1 pascal]

2. Vander Waals equation:

2

qp (v-b) = RT ....(A)

v

2 2 2 2

c

c

27R (T ) 27 (0.287) (132.8)a =

64p 64p

[Critical Pressure, pc = 37.7 bar = 37.7 100 kN/m2]

2[ 1 bar = 100 kN/m ] 2 227 (0.287) (132.8)

a64 (37.7) (100)

a = 0.162.

We know that

c

c

-3

RT 0.287 132.8b

8p 8 37.7 100

b = 1.26 10

Substituting a, b values in Vander Waals Equation

-3

2

0.162(A) p v-1.26 10 0.287 302

v

Where

V - Specific volume

Volume v 0.28 v=

mass m 10

v = 0.028 m3/kg Substituting Specific volume

-3

2

2

0.162p+ 0.028-1.26 10 0.287 302

(0.028)

p = 3034.7 kN/m

3. Generalized compressibility chart:

R

c

R

cR

c

3

R

T 302 T = 2.27K

T 132.8

T 2.27 K

vP 0.028 37.7 100v

RT 0.287 132.8

v 2.76 m /kg

Reduced temperature is 2.27 K and reduced specific volume is 2.76 m3/kg both are intersects at one point. Mark this point on compressibility chart. From chart, corresponding (z) value is 0.99. We know that,

2

pvCompressibility factor (z) =

RT

p 0.028 0.99 =

0.287 302

p = 3064.5 kN/m

Results :

1. Pressure p (By perfect gas equation) = 3095.5 kN/m2 2. Pressure p (By Vander Waals equation) = 3034.7 kN/m2 3. Pressure p (By compressibility chart) = 3064.5 kN/m2

15. Compute the specific volume of steam at 0.75 bar and 570 K using Vander Waals equation. Take critical temperature of steam is 647.3 K and Critical pressure is 220.9 bar.

Given data:

Pressure, p = 0.75 bar = 0.75 100kN/m2 = 0.75 100 kPa

2[ 1 bar = 100 kN/m 100kPa]

Temperature, T = 570 K Critical Temperature, Tc = 647.3 K

Critical Pressure, pc = 220.9 100 kN/m2 To find Specific volume (v) Solution : We know that Vander Waals equation

2

2 2

c

c

ap (v - b) = RT

v

27R (T )Where, a =

64p

Universal Gas Constant 8.314Where, R = kJ/kgK

Molecular weight of Steam 18

R 0.462 kJ/kgK

2 2

c

c

-3

27 (0.462) (647.3)a =

64 (220.9) 100

a = 1.70

RT 0.462 647.3We know, b =

8p 8 220.9 100

b = 1.69 10

Substituting a, b and pressure and temperature values in Vander Waals equation.

3

2

2

1.700.75 100 v 1.69 10 0.462 570

v

1.7075 (v - 1.69 10-3) = 263.34

v

By trial and error method, we get Specific volume v = 3.58m3/kg Result :

Specific volume v = 3.58 m3/kg. 16. An ideal gas mixture consisting of 3 kg of air and 7 kg of nitrogen at a temperature of

25C occupies a volume of 1m3. Determine the specific enthalpy, the specific internal energy and specific entropy of the mixture. Assume that air and nitrogen are ideal gases. Take R for air is 0.287 kJ/kgK and for nitrogen is 0.297 kJ/kgK. The other properties of air and nitrogen are given as under.

Name of the gas Properties

H, kJ/kg U, kJ/kg S, kJ/kg

Nitrogen 309.64 221.11 6.46

Air 298.52 212.90 2.35

Given data : For air Mass, ma = 3 kg

Temperature, T = 25C + 273 = 298 K Volume, v = 1m3 Gas constant, R = 0.287 kJ/kgK Enthalpy, h = 298.52 kJ/kgK Internal energy, U = 212.90 kJ/kg Entropy, S = 2.35 kJ/kgK For Nitrogen Mass, mn = 7 kg Temperature, T = 298 K Volume, v = 1m3 R = 0.297 kJ/kgK h = 309.64 kJ/kg U = 221.11 kJ/kg S = 6.46 kJ/kgK To find :

1. Specific volume (v) of the mixture 2. Pressure (p) of the mixture 3. Specific Enthalpy (h) of the mixture 4. Specific Internal energy (U) of the mixture 5. Specific Entropy (S) of the mixture

Solution:

a

a

3

a

n

n

3

n

volume v 1Specific Volume of air, v

mass m 3

v 0.333m /kg

volume 1 1Specific Volume of nitrogen, v

mass m 7

v 0.142m /kg

a a aa

a

2

a

m R TPressure of air, P pv = mRT

v

3 0.287 298 p

1

p 265.5 kN/m

Pressure of nitrogen,

n n n

n

2

n

m R T 7 0.297 298p

v 1

p 619.5 kN/m

For mixture

a nTotal mass, m = m m 3 7 10kg

m = 10 kg

Temperature, T = 298 K Volume, v = 1m3

m

3

a n

v 11. Specific Volume of mixture, v

m 10

= 0.1m /kg

2. Pressure of mixture, p = p p 265.5 619.5

= 876.07 kN

2

a a n n

/m

m h m h3. Specific Enthalpy of mixture, h =

m

3 298.52+ 7 309.64 =

10

= 306.34 kJ/kg

a a n n

a a n n

m U m U4. Internal energy of mixture, U =

m

3 212.90+7 221.11 =

10

= 218.64 kJ/kg

m S m s5. Specific Entropy of mixture, S =

m

3 2.35+7 6.46 =

10

= 5.227 kJ/kg.

Results:

1. Specific Volume of the mixture, vm = 0.1m3/kg 2. Pressure of the mixture, p = 876.07 kN/m2 3. Specific Enthalpy of the mixture, h = 306.34 J/kg 4. Specific Internal energy of the mixture, U = 218.64 kJ/kg 5. Specific Entropy of the mixture, S = 5.227 kJ/kg.

17. A perfect gas of 0.25 kg has a pressure of 298 Kpa, a temperature of 80C, and a volume of 0.08m3. The gas undergoes an irreversible adiabatic process to a final pressure of 350 kPa and find volume of 0.10 m3, work done on the gas is 25 kJ. Find cp, cv. Given data : m = 0.25 kg p1 = 298 Kpa

T1 = 80C + 273 = 353 K v1 = 0.08m3 p2 = 350 Kpa v2 = 0.10m3 W = -25 kJ [Work done on the gas in Negative valve] To find: Cp and Cv Solution: We know that, Perfect gas equation P1v1 = mRT1

1 1

1

p v 298 0.08 R= 0.270

mT 0.25 353

Characteristic gas constant, R = 0.270 kJ/kgK Similarly P2v2 = mRT2

2 2

2

2

p v 350 0.10 T

mR 0.25 0.270

T 518.5K

We know that

Heat transfer, Q = W + U

v 2 1 v 2 1

v

v

v

p v

p

Q W mC T T U= m C T T

Q = -25 + 0.25 C (518.5 353)

For adiabatic process, Q = 0

0 = -25 + 0.25 C (518.5 353)

C 0.604 kJ/kgK

We know that, R = C C

0.270 = C 0.604

p C 0.874 kJ/kgK

Results: Cv = 0.604 kJ/kgK Cp = 0.874 kj/kgK

h pP

T 1 vT v

P C T

From equation, we can determine the Joule-Thomson coefficient () in terms of measurable properties such as pressure (p), temperature (T), specific volume (v) and Cp. Let, Enthalpy is a function of pressure and temperature. i.e. h = f(p, T)

pT

h h dh= dp dT

p T

For Throttling process, Enthalpy remains contact H = C dh = 0

Substitute dh value in equation.

pT

pT

h h O = dp dT

p T

h h dp dT 0

p T

Divided by dT

h pT

h pT

pT h

p T

p

p T

h p h 0

p T T

h p h

p T T

h 1 h T =

p T p

h 1 h

p p

h 1 h C

p p

The property p

T

h C

p

is known as constant temperature coefficient.

18. Prove that internal energy of an ideal gas is a function of temperature alone. Solution: We know that, ideal gas Equation Pv = RT

RT

p ...(A)v

We know that, Internal energy Equation

v

pdU C dT T dv pdv

T

Divided by dv [From Equation (30)]

v

T T v

v

T v T

T

T

U T p C T p

v v T

U p T T p C 0

v T v

U RT T p [From Equation (A)]

v T v

U R T p

v v

= p - p

T

RT p =

v

U 0

v

If the temperature remains constant, there is no change in internal energy with volume. Hence internal energy is a function of temperature alone. 19. Prove that specific heat at constant volume (Cv) of a Vander Waals gas is a function of temperature alone? Solution: We know that, Vander Waals equation

2

2

2

v

2

2

v

ap (v - b)= RT

v

a RT p +

v (v b)

RT a p =

(v-b) v

p R

T v b

p 0

T

We know that,

2v

2

T

2v

2

T

C pT

v T

C p0 0

v T

Thus if the temperature remains constant, there is no change in specific heat with specific volume. Hence specific heat of constant volume is function of temperature alone.

20. Find the value of (h)T for a fluid that obeys the equation of state

2

RT aP

v v

Solution: We know that,

T

v

U p T p

v T

p dU = T p dv ........(A)

T

We know that

2

2

2

2

RT aP (Given)

v v

RT a P -

v v

RT a P

v v

RT a p dv dv

v v

2

1

2

1

T 2

v

2v

v

v

1 2

a RT (dU) dv [From equation (A) and P = ]

v v

U = dvv

-1 U = a

v

1 1 U = a .......(B)

v v

We know that,

2 2 1 1 2 2 1 1

1 2

2

12 2

1 2 2 2 1 1

2

1 2 1 2

1 2

a ah U P v Pv P v p v

v v

a a RT a RT ah v v

v v v v v v

RT a P = (Given)

v v

a a a a h =

v v v v

2a 2a =

v v

1 2

1 1 ( h)r = 2a .

v v