Thermodynamics

THERMODYNAMICS Therm - Heat (non-organised) Dynamics - Motion (Caused by force) Action of force on moving body is ‘h’ as work. Conversion of non-organised form of energy into organised form of energy. Main aim of thermodynamics is to convert heat into work. Thermodynamics is branch of science deal with the heat and work interaction and its effect on properties of system.

Importance of thermodynamics: 1. There is a property which distinguish this subject from other sciences that is temperature and everything have certain temperature. So, everything comes under consideration of thermodynamics. 2. The subject thermodynamics is based on certain laws like law of conservation of mass, etc. Laws are the statements which are universally true but can’t be proved mathematically. 3. Most importantly thermodynamics tells us “what thins are not possible”. Example: According to 2nd law of thermodynamics no heat engine can give 100% efficiency.

1

HOW TO STUDY: There are two ways to study thermodynamics 1) Microscopic view point (Statistical thermo) 2) Macroscopic view point. 3) Postulate thermodynamics (recent)

1) Microscopic view point: In microscopic view point, behaviour of each & every molecule is taken into consideration & to know the overall behaviour we will use same statistical means. That why it is ‘k’ as statistical thermodynamics. Only useful for low densities. This approach is used for ramified gas (low pressure gas) (vacuum).

2) Macroscopic view point: In macroscopic point of view, our attention is focused on certain quantity of matter without going to the event occurring at molecular level. It is also known as classical thermodynamics average behaviour at molecule taken into the consideration. In our course, we will follow classical thermodynamics only. In classical thermodynamics, we will try to establish relation between measurable and non-measurable properties Measurable properties → temperature, pressure, volume Non-measurable properties → enthalpy, entropy, internal energy, etc. 2

CONCEPT OF CONTINUUM: ρ(dm/dV)

dV

dm

Domain of Molecular effect

Domain of Continumm ρ(dm/dV)

ρ

Volume

V

In microscopic point. of view, we always concern with a small volume which is very large compared to molecular dimension. Concepts of continuum give the criterion to apply macroscopic point of view. Even the small volume contains very large no. of molecules. We have to consider that small volume in such a way that statistical a varying is meaningful and material can be considered as continuous.

Thermodynamic system, surrounding & boundary: Surrounding System

Universe

 System is defined as quantity of matter or region in a space where our attention is focused.  Everything external to the system is surrounding. 3

 The thing which separates system from surrounding is known as boundary Thermodynamic boundary

Real

Fixed

Imaginary

moving

Note: To get works from close system minimum one boundary should be moving. Thermodynamic boundary 1. Adiabatic (No heat transfer high thermal resistance.) (R th → ∞) 2. Diathermal Low thermal resistance (R th → 0)

TYPES OF SYSTEM: 1. Closed system: In closed system, there is no mass transfer across the boundary. It has fixed mass ‘k’ as control mass. It is also c/a control mass of system. Energy transfer may take place. Example: ⅰ) Rigid container 4

Gas m = Constant Q

Q

T

For closed system minimum one boundary should be open. Here, all boundaries are fixed. ∴W=0 ⅱ) Piston-cylinder arrangement without piston Q

m = Constant

W

Q

W≠0 (W≠0 one moving boundary)

(∴ One moving Boundary)

2. Open system: In open system, mass transfer may also take place along with energy-transfer. Example: Turbine, pump, nozzle, compressor, radiator, etc. Q

m1

blades W

m2

Journal open system

5

For open system analysis attention is focused on certain volume in space surrounded by a surface known as control surface. Matter as well as energy can cross the control volume.

3. Isolated system: In isolated system neither mass nor energy interaction take place. Example: Thermal flask which is perfectly insulated the best example for isolated system is our universe.

THERMODYNAMIC PROPERTIES: Properties are characteristic by which you can identify the system. Thermodynamic properties 1. Intensive (Independent of mass or size of system) Example: Temperature, Density, Specific heat, Pressure, etc., Velocity. 2. Extensive (Depends on mass or size of system) Example: Volume, Mass, Energy, Weight

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M

M

M

2

2

V

V

V

2

2

P

P

T

T

ρ

ρ

P T ρ Note:

 Ratios of two extensive properties are always intensive. ρ=

M V

(M → Intensive, V → Intensive, ρ → extensive)

 Specific properties are always intensive properties. Specific property = extensive property = ext. prop/mass (Per unit mass)  Properties are independent of past history. Properties are the point or state function.  Properties are exact differential. If z is function o0f x & y and can be represented as dz = Mdx + Ndy And z is exact differential if ∂M

(

∂N

)

∂y x = constant

= ( )y = constant ∂x

E.g. for ideal gas, function is z=∫ dz =

7

dT T dT T

V

− dp T

V

− dp T

1

M=

T V

N=

T

dx = dT x=T dy = dp y=P (

∂1 T

∂P

) =0 T

V

∂(− ) T

(

∂T

) = 0 For ideal gas PV = RT P V

=

T

R P

R

∂(− P)

(

∂T

) =0 P

Hence, it is a property.

STATE OF SYSTEM: State represents condition of system. When all properties of system have definite value than system is said to be in state. State of system is represented by a point in a diagram where properties are co-ordinate.

P1

1

P

Process 2

P2 V1

V2 V

8

GIB’S PHASE RULE: In a gib’s phase rule, Let there are total M number of properties of the system in which N are independent it means if we fix N number of independent properties the M - N properties will automatically have fixed. To find no. of independent properties required to identify the gib’s phase rule is used. According to which, P+F=C+2 Where, P = no. of phases. F = Degree of freedom /no of independent properties to identify the system C = no. of components P=1

Water H2 O

C=1 F =? P+F=C+2 F=2 Two independent properties like pressure& temperature. Example: 23°C @1atm pressure is complete description of system Steam Water P = 2 C = 1 F =? P+F=C+2 F=1 9

Either temperature or pressure is independent property. Steam Water Ice

P = 3 C = 1 F =? P+F=C+2 F = 0 → Triple point of water Note: Invariable chemical composition irrespective of phase is k as pure substance (e.g. water is pure substance.) Air is not pure substance in liquid phase as chemical composition changes when it is converted from air to liquid.  At triple point of water all properties are fixed,  Ttp = 0.01℃/273.15 K Ptp = 0.006 bar

PROCESS AND PATH: Change of state is known as process and series of equilibrium states through which a system passes during its process is known as path of process. 1 Equilibrium states

path 10

2

Process

Reversible

Irreversible

Reversible: A process is said to be reversible if it reverses then follow the same path without leaving any effect on the system as well as surrounding. 2 1

Q1

Q2

 For reversible process, it must follow same path and should not affect system& surrounding.

Irreversible Irreversibility mainly caused due to friction.  All spontaneous process is irreversible. For the processes to be reversible if friction is zero then process must be slow.

Quasi – Static process: Quasi means almost. Can't decide 1 exact path P W 2 V

So, divide block into different pieces, 11

P

Actual path V

Frictionless quasi static process is known as reversible process.  All reversible process is quasi static but all quasi-static need not be reversible.

THERMODYNAMIC CYCLE: A system is said to be undergone a cycle if initial & final state of system is same. 1

1

2

The minimum no. of process required to complete a cycle is 2. Through no cycle has 2 no of process in practical. For a cycle, change in property is zero. 3 step cycle --------.

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THERMODYNAMIC EQUILIBRIUM: A system is said to be in thermodynamic equilibrium if there is no any unbalances potential (driving force) within the system. To meet the condition of thermodynamic equilibrium three types of equilibriums are required. 1) Thermal equilibrium [equality of temperature/no heat transfer] 2) Mechanical equilibrium [equality of all those properties which may cause work transfer (or) no work transfer] 3) Chemical equilibrium [A system is said to be in chemical equilibrium where there is no any mass transfer or chemical reaction takes place within system.]

IDEAL GAS: Obeys ideal gas equation. At all pressure and temperature. PV = mRT P → Absolute pressure (k pa) V → Volume (m3 ) M → Mass (kg) R → Characteristic gas constant (kJ/kg k) T → Absolute temperature For air, R = 0.287 kJ/kg k. A gas is said to be ideal gas if there are no any intermolecular forces. Any gas can behave like ideal gas at very low pressure (or) very high temperature.

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Ideal gas

Perfect gas

Semi-perfect gas

(Cp & Cv Are constant)

(Cp & Cv Depends on temperature)

No. of moles = mass (m)/molecular weight M m1 = n M ∴ PV = nMRT

AVOGADRO’S HYPOTHESIS: According to Avogadro’s hypothesis all gases at same pressure, temperature and volume, no of moles are constant. MR = Constant. = Universal gas constant 𝑅̅ ̅T PV = nR ̅ = 8.314 kj/kmol. k R Conservation of mass

Ma

Mb

∑ m = ma + mb + mc + ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

Pa

Pb

V = Va = Vb = Vc = ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

Va

Vb

Ta

Tb

Ra

Rb

ha

hb

Zeroth law of thermodynamics T = Ta = Tb = Tc = ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ Daltons law of partial pressure P = Pa = Pb = Pc = ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ̅T PV = nR ̅ Ta Pa Va = na R ̅T Pa V = na R ̅T Pb V = nb R 14

̅T Pc V = nc R ̅T (Pa + Pb + Pc + ⋯ ⋯ )V = (na + nb + nc + ⋯ ⋯ )R ̅T PV = ∑ nR P = Pa + Pb + Pc + ⋯ ⋯ ⋯ ⋯ ⋯ ⋯ ∑ n = na + nb + nc + ⋯ ⋯

Mole fraction: (x) It is defined as ratio of number of moles of a gas to the total number of moles in a mixture. n

xa = ∑ a = n

n

xb = ∑ b = n

n

xc = ∑ c = n

na na + nb + nc + ⋯⋯ nb na + nb + nc + ⋯⋯ nc na + nb + nc + ⋯⋯

(xa + xb + xc + ⋯ ⋯ ) = 1 ∴ xa + xb + xc + ⋯ ⋯ = 1 ̅T Pa V = na R ̅ T ⋯ ⋯ for ith gas → (ⅰ) Pi V = ni R ̅T PV = ∑ n R eqn (ⅰ) eqn(ⅱ)

=

Pi P

Pi P

→ (ⅱ)

= xi

15

n

= ∑i

n

MIXTURE OF IDEAL GASES: Equivalent gas constant: M1

M2 → Me = Equivalent mass constant

P1

P2 → P = P1 + P2 + P3 + ⋯ ⋯

V1

V2

→ V1 = V2 = V3 = ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

T1

T2

→ T1 = T2 = T3 = ⋯ ⋯ ⋯ ⋯ ⋯ ⋯

R1

R2

→ R e = Equivalent gas constant

n1

n2

→ ∑ n = n1 + n2 + n3 + ⋯ ⋯

P1 V = m1 R1 T P2 V = m2 R 2 T P3 V = m3 R 3 T (P1 + P2 + P3 + ⋯ ⋯ )V = (m1 R1 + m2 R 2 + m3 R 3 + ⋯ ⋯ )T →(ⅰ) PV = ∑ m R e T →(ⅱ) From equate (ⅰ) & (ⅱ) ∑ m R e = (m1 R1 + m2 R 2 + m3 R 3 + ⋯ ⋯ ) Re =

m1 R1 + m2 R2 + m3 R3 + ⋯⋯ m1 + m2 + m3 + ⋯⋯

Equivalent gas constant m=nM m1 = n1 M1 m2 = n2 M2 m3 = n3 M3 (m1 + m2 + m3 + ⋯ ⋯ ) = (n1 M1 + n2 M2 + n3 M3 + ⋯ ⋯ ) → (ⅰ) 16

∑ m = ∑ n Me →(ⅱ) Me =

n1 M1 + n2 M2 + n3 M3 + ⋯⋯ n1 + n2 + n3 + ⋯⋯

Equivalent molecular weight Note: m

Mass fraction (M. F) = ∑ a

m

Mass fraction = mass of molecule/total mass Relation between mass fraction and mole fraction Pi V = mi R i T → (ⅰ) PV = ∑ m R e T → (ⅱ) (ⅰ) − (ⅱ) Pi P

mi

=∑

m

×

Ri Re

xi = M. F ×

Ri Re

̅ M. R = R (M. F)i = xi × (M. F)i = xi ×

̅ R Me ̅ R Mi

Mi Me

Problems with answers: 1. An ideal gas mixture consists of 2 kilo mole of N2 and 6 kilo mole CO2 the mass fraction of CO2 in mixture is. N2 = 2 kilo mole CO2 = 6 kilo mole (M. F)CO2 = 17

mCO2 mN2 + mCO2

mCO2 = nCO2 MCO2 = 6 × 44 = 264 Kg mN2 = nN2 MN2 = 2 × 28 = 56 Kg (M. F) = 264/(264 + 56) M.F = 0.825 2. An ideal gas mixture, whose apparent molar mass 36 kj/k mol consist of N2 and 3 other if the mole fraction of nitrogen is 0.3 then what mass fraction of is N2 . M = 36 KJ/K mole No of gases = 4 xN2 = 0.3 xN2 = M. F ×

Ri Re

Me

= M. F × (

0.3 = (M. F)N2 =

Mi

)

36 28

(M. F)N2 = 0.23 3. An ideal gas mixture consist of 2 kilo mole of N2 and 4 kilo mole of CO2 the apparent gas constant of mixture is, nN2 = 2 nCO2 = 4 Re = =

M1 R1 n1 +M2 R2 n2 M1 n1 +M2 n2

(2+4) 8.314 [(2×28) + (4×44)]

R e = 0.215

18

=

(6×8.314)

KJ Kg

56 +176

K

4. The rigid tank is divided into 2 compartments by a partition. One compartment contains 3 k mole of N2 at 600 K pa & other compartment contains 7 K mole of CO2 at 200 K pa. Now the position is removed& two gases from homogeneous mixture & at 300 K pa. The partial pressure of N2 in mixture is ----nN2 = 3 K mole pN2 = 600 K pa nCO2 = 7 K mole pCO2 = 200 K pa xi =

3×28 3×28+7×28

x N2 = pi 300

3 3+7

= 0.3

= 0.3

pi = 90 Kpa 5. An 80 litre rigid tank contains an ideal gas mixture of 5gm of N2 and 5 gm of CO2 at specified pressure& temp. If N2 were separated from the mixture & stored at mixture temp. And pressure its volume would be -------80L

5gm N2 + 5gm CO2 T

P

5gm N2

T

nN2 = nCO2 = 19

5 28 5 44

P

PV = n RT p = pN2 + pCO2 pi p

=

5 28

5 5 + 28 44

= 0.611

pN2 = 0.611 p Now, PN2 VN2 = mN2 R N2 TN2 PV = mN2 R N2 TN2 VN2 V

=

P P N2

VN2 = 80 × 0.611 VN2 = 48.80 litre

ZEROTH LAW OF THERMODYNAMICS & CONCEPT OF TEMPERATURE: Temperature: (Degree of hotness and coldness) Macroscopic approach: The temperature is property by virtue of difference of that property heat flows. Microscopic approach: Temperature is defined as average kinetic energy of the molecules of a system. Internal energy: Sum of kinetic energy of the molecule.

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Zeroth law: When two bodies are in thermal equilibrium with third body then there are in equilibrium with each other. A A

B B

equilibrium equilibrium

B C

A is in equilibrium with C

C Reference Body

This is basic law used for temperature measurement.

Principle of thermometry: In thermometry principle, a property which vary with temperature. Is found first and with help of that property unknown temp. Can be measured. The property which helps to find unknown temp is known as thermometric property. Types of thermometer

Thermometric property

Hg glass tube

Length

thermometer Constant volume gas

Pressure

thermometer Constant pressure gas

Volume

thermometer Thermistor

Electric resistance

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thermistor R

R2 i G

R2

R4

Thermocouple (based on see back effect) emf T1

T2

See back effect: If two metals from junction such that both are at different temperature, then EMF is generated. What is Peltier effect? If two different metals form

Measurement of temperature: 1) Method used before 1954: In this method, two reference points are used a) Ice point of water (0°c) b) Steam point of water (100°c) T=a+bP At ice point Tice = a + b Pice 0 = a + b PI …….. (ⅰ) At steam point Tsteam = a + b Psteam 22

100 = a + b PS ……. (ⅰⅰ) From (ⅰ) & (ⅱ) b= T=

100 PS − PI

;a=

−100PI −(PI − PS )

T=

−100P

PI − PS

+ +

PS − PI

100PI 100 PS −PI 100 PS − PI

×P

×P

2) Method used after 1954: In this method, single reference point is used that is triple point of water (0.1°c or 273.15 K) T=aP At triple point, 273.15 = a Ptp a= T=

273.15 Ptp 273.15 Ptp

×P

Note: All temperature scales are arbitrary Up to 130°C → Refrigeration. Below 130°C → Cryogenic

Problems with solution: 1. The reading TA & TB of thermometer A & B agree at ice point and steam point. And related by equation, TA = l + mTB + nTB 2 When A reads 51°C then B reads 50°C. Determine the reading of A When B reads 25°C. TA = TB = 0 at ice point 23

L=0 TA = TB = 100℃ TA = l + mTB + nTB 2 100 = m × 100 + n × 1002 1 = m + 100n ……… (ⅰ) 51 = m × 50 + n × 502 51 = m × 50 + n × 2500n ……. (ⅱ) 1 = m + 100n 50 = m × 50 + n × 5000n −51 = m × 50 ± n × 2500n −1 = (2500)n n = −4 × 10−4 m = 1.04 TA = 25.75℃ TB = 25℃ 2. In a new temperature scale, say °P the boiling & freezing point of water at atmosphere are 100°P and 300°P respective co-relate this scale with centigrade scale and find the value of zero-degree P on centigrade scale. in °P

in °c

TB = 100°P

TB = 100℃

TF = 300°P

TF = 300℃

T=a+bP At ice point 0 = a + b (300) At steam point 100 = a + b (100) 24

100 = b (200) b = -0.5 a = 150 ∴ T = 150 - 0.5P At P = 0 °P

T = 150 °C

ENERGY INTERACTION: For closed system and its surrounding, energy interaction takes place in 2 ways. 1) Work interaction 2) Heat interaction

Work interaction: In mechanics, action of force on moving body is known as work. W = F × displacement.

m W Battery

Battery

Electrical work

Electrical work

25

m

m

Battery

Battery

No works

Mechanical works

In thermodynamics, the work is said to be done by system if sole effect external to the system can be reduced to the rising of weight. The weight may not be actually raised but the net effect external to the system can be reduced in form of lifting weight, work transfer is a boundary phenomenon, it is recognised only when it crosses the boundary of system. At boundary if work is done by temperature difference then it is heat transfer otherwise work transfer.

Heat Transfer

Electrical Work Transfer

Convection of work transfer:  Work done by the system is +ve  Work done on the system is –ve

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W

W

Gas

Gas

Expansion Process +ve work

Compression Process -ve work

Expansion work is always +ve whereas compression work is always –ve.

CLOSED SYSTEM WORK/NON-FLOW WORK: (Displacement work/p. dv work) P2V2

P1V1 Gas

1

P1 P

System

P

A

2

P2

V1 dV

dx

V2 V

Closed system reversible process P = F/A → F = PA dw = F × dx = PA × dx dw = P dv 2

Total work w = ∫1 P dv (Closed system reversible process) dA = P dv

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2

Total area A = ∫1 P dv (Closed system reversible work)  Area under the curve when projected on volume axis gives closed system reversible work on P-V diagram. 1 b a

P

2

V

Through the ends points are same for process A and B but the work transfer is not same because area below the curve is not same. Therefore, work transfer depends on path followed by system. Work transfer is path function it is not a property. It is inexact differential.

CLOSED SYSTEM WORK FOR VARIOUS PROCESSES: 1) Constant volume process:

Gas Q Q Rigid container with all boundaries fixed

V = Constant Ideal gas equation 28

P1 P2

T1

=

T2

dv = 0 W = ∫ P dv = 0 Heat supplied 1

2 P

Heat rejected 1

2 V

2) Constant pressure processes:

P 1

2

V1

V2

P = Constant Ideal gas equation V1 V2

=

T1

V∝T

T2

dp = 0 2

2

W = ∫1 P dv = P ∫1 dv = p[v]12 W = P (V2 − V1 )

29

pWt = 2 bar

Patm

W

W pg = 2 bar

Gas

pg = pWt + patm Q

Q Heat supplied 1

P

2 Heat rejected

2

1 V

3) Isothermal process (Constant temperature process): Q Q

T = Constant Ideal gas equation V1 V2

=

T1

PV = mRT

T2

PV = C P1 V1 = P2 V2 C

P= 2

V

2 C

V

W = ∫1 P dv = ∫1 ( ) dv = C ln [ 2 ] s V

W = P1 V1 ln

V2 V1

V1

= P2 V2 ln

W = Pinitial Vinitial ln

V1

Vfinal Vinitial

Piston - independent 30

V2

Pressure - dependent Isothermal expansion is possible when heat is added to system where as isothermal compression is possible if heat is rejected from the system. Heat supplied

P

Heat rejected

V

Note: Isothermal curve on p-v diagram is rectangular hyperbola.

4) Adiabatic process: (No heat transfer) Ideal gas equation for adiabatic process, PV γ = C γ = Adiabatic index γ=

cp cv

2

W = ∫1 P dv P1 V1 γ = P2 V2 γ 2 C

W = ∫1

Vγ

w = c[ w=c[

dv

−Vγ+1

2

]

−γ+1 1

V2 γ+1 − V1 γ+1 −γ+1

31

]

w=

P2 V2 .V2 γ+1 − P1 V1 .V1 γ+1 −γ + 1

w= w=

P2 V2 − P1 V1 −γ+1 P1 V1 − P2 V2 γ−1

1 2

Expansion

P Gas

2 Compression 1 V

Note:  Different form of ideal gas equation for adiabatic process. PV γ = C P1 V1 γ = P2 V2 γ PV V γ−1 = C PV = mRT TV γ−1 = C T1 V1 γ−1 = T2 V2 γ−1 T2 T1

=

P1 P2

V1 γ−1 V2 γ−1

=

V2 γ V1 γ 1

V1 V2 T2 T1

=

P γ ( 2) P1 P2

=( ) P1

32

γ−1 γ

5) Polytrophic process: Ideal gas equation, PV n = C n → Polytrophic index. 1
&

P1 V1 − P2 V2

wpolytropic =

n −1

Note: All process in p-v diagram can be represented by PV k = C where K is dependent on type of process for constant pressure process, K=0 Isothermal process, K=1 Adiabatic, k = γ Polytrophic, k = n Constant volume = k = ∞  Representation of various processes on p-v diagram. K=0 K=1 K=n

P

K=ꝏ For Expansion Increasing K=r

K=ꝏ

K=n K=1

K=0 For Compression

33

K = r Increasing V

SLOPE OF VARIOUS LINES ON P-V DIAGRAM: 1. Constant pressure line:

P

V

θ=0 Slope = m = 0 Slope = tanθ = tan0 = 0

2. Constant volume process: P

V

θ = 90° Slope = tan 90° m=∞

3. Isothermal process: P

V

Slope =

dy dx

m=− 34

= p v

dp dv

dp dv

=

d

c

( )

dv v

pdv + vdp = 0 dp dv

=−

p v

slope = −

p v

4. Adiabatic line: PV γ = C Pγ V γ−1 dv + V γ dp = 0 Dp dv

p

= γ (− ) v

p

Slope of adiabatic curve = γ (− ) v

 Slope of adiabatic curve on p-v diagram = γ (slope of isothermal curve on p − v) Note:  Net work done in cycle is equal to the area of closed region or area of cycle.  All clockwise cycle on p-v diagram are powers producing cycle. (example: - Carnot cycle, etc.) All anticlockwise cycles on p-v diagram are power consuming cycle. (example: - refrigeration cycle) W = ∫ P dv (Closed system, reversible process & energy should cross the boundary.)

35

Problems and Solutions: 1. An imaginary engine receive heat and perform work on slowly moving piston in such way that cycle of operation of 1 kg of working fluid can be represented as a circle of 10cm diameter on p-v diagram. In which 1cm = 300k pa & 1cm = 0.1m3 /kg find network during a cycle. Network = Area of circle π

= × (d2 ) × 300 × 0.1 4

Wnet = 2356.1KJ/Kg 2. An engine cylinder has piston area of 0.12m2 & contains gas at pressure of 1.5M Pa the gas expands according to a process which is represented by a straight line on p-v diagram. the pressure is 0.15MPa calculate work done by gas on piston, if stroke is 0.3m. A = 0.12m2

P1 =1.5M Pa

P2 = 0.15M Pa

L = 0.3m

1.5 Mpa = 1500 kPa P1 0.15 Mpa = 150 kPa P2

Clearance Volume

V1

V2

0.3 m Stroke length

V2 − V1 = 0.3 × 0.12 = 0.036m3 Wnet = area under the curve 1 Wnet = (150 × 0.3 × 0.12) − [0.3 × 0.12 × (1500 − 150)] 2 36

Wnet = 29.7 KJ 3. A gas expands from pressure P1 to P2 (P2 =

P1 10

). If pressure

expansion is isothermal the volume at end of expansion is V2 = 0.55m3 . If process of expansion is adiabatic then volume at expansion is? 1

P1

pv1.4 = c

T=C

P2

2

21

V2 0.55

V1

P1 V1 = P2 V2 P1 P2

= 10

10 × V1 = 0.55 V1 = 0.055 Now, for adiabatic process, P1 V1 γ = P2 V2 γ P1 P2

=(

10 = (

V2 γ V1

)

V2

γ

)

0.055

γ = 1.4 1 γ

10 = (

V2 0.055

)

V2 = 0.286m3

37

4. The gas space above water above water in storage tank contains N2 AT 25°C & 100K Pa the total volume is 1m3 & there is 500kg water at 25°C and additional 500kg is now forced into tank assuming constant temperature throughout find final pressure on N2 & work done on N2 in the process. 250C 100 kPa N2

water

P1 = x

250C

P1 = 100 kPa

V2 4m3

Volumewater = V1 = V2 = 0.5m3

500

V1

= 0.5m3

1000

V1 = 3.5 m3

P1N2 = 100K Pa P2N2 =? Final volume, v11 = 1m3

v21 = 3m3

P1N2 V1N2 = P2N2 V2N2 (100 × 3.5 = P2N2 × 3) P2N2 = 116.66KPa Wnet = P1 V1 ln

V2 V1 3

= 100 × 0.5 ln ( ) = −53.95 KJ 3.5

Work done on N2 = 53.95 KJ. 38

5. A piston cylinder device contains 0.05m3 of gas initially at 200k pa at this state a linear spring which have spring constant of 15K N/m is just touching the piston is transmitted to gas causing the piston rise and compress the spring until volume inside a cylinder is double. If cross section are of piston is 0.25m2 find final pressure inside the cylinder & work done by the gas. P= Stroke length = P=

F A

k×x

=

A

0.05 0.25

150 × 0.2 0.25

×

V2 − V1 A

= 0.2

= 120

P2 = Patm + Pp + Ps = 200 + 120 P2 = 320 K Pa  If there were no spring pressure will remain constant P1

200 kPa

P2

V1

V2

Pat

Ppisto

m

n

200 kPa

39

Patm

Ppiston

Patm Ppiston Psurrounding

100 kPa

Work done by gas = W@constant pressure + Wspring = Wexpansion + Wspring 1

= P(V2 − V1 ) + kx 2 2

1

= 200(1 − 0.05) + × 150 × (0.2)2 2

Wgas = 13 KJ Short trick to find area 2 320 200

1

0.05

0.1

Wgas = A (below the curve) = 0.05 × 200 + 0.05 × 120 Wgas = 13 KJ

40

6. An insulated vertical tank contains 0.1kg of organ gas with the help of piston as shown in fig. the mass of piston is 5 kg & initially at rest on the bottom of cylinder the cylinder is connected to nitrogen cylinder the valve the valve is opened and nitrogen slowly enter to the cylinder. During this process piston is lifted to a height of 10cm by nitrogen. The initial pressure & temperature of organ is 1 bar and 300k and final temperature of organ is 320k calculates the work done by organ and by𝑁2 . R = 0.208 KJ/Kg K γ = 1.67 for organ Ar = 0.1 kg 1 bar 300k

10 cm

N2 100 bar

morg = 0.1 Kg

For organ

mp = 5 Kg

PV = m RT

P1N2 = 100 bar

1 × V = 0.1 × 0.208 × 300

x = 10cm

V1Ar = 6.24 T2

P1Ar = 1 bar

T1

T1Ar = 300K

=( )

γ−1 γ

V1

V2 = 5.34m3

T2Ar = 320K Work done by organ WAr =

V2

P1 V1 − P2 V2 γ−1

41

= =

mR (T1 − T2 ) γ−1 0.1×0.208(300 − 320) 1.67 − 1

WAr = −0.627 KJ Now, work done by N2 , WN2 = War + Wpiston = 0.627 + ( = 0.627 +

mgh 1000

)

5×9.81×0.1 1000

WN2 = 0.6249 KJ 7. A sealed elevated storage tank of capacity 50m3 initially containing air at 1 bar & 25°C is to be pumped with water from a lake also @ 25°C g = 9.7 m/s2 the pump is operated until the 3

tank is full and during this operation the temp of air & water do 4

not change the average elevation of water in tank is 35m above the surface of lake. Take ρwater = 1000kg/m3 & assume air as ideal gas. P

P2

P1

V1

V2 V

1 bar = 1.01 × 102 k pa Volume of tank = 50 m3 Pair = 1 bar 42

Tair = 25℃ = 298K Twater = 25℃ g = 9.7 m/s2 ρwater = 1000 kg/m3 3

volume = Vwater = × 50 = 37.5 4

mwater = 1000 × 37.5 = 37500 kg h = 35m P.E = mgh P2 P1

P2 = 100 ×

=

V1 V2

50 12.5

= 400kpa

Work done = P1 V1 ln

V2 V1 12.5

= 100 × 50 ln (

50

)

= −6931.47 KJ Wpump = 6931.47 +

9.7 × 35 × 37500 1000

Wpump = 6931.47 + 12731.25 Wpump = 19662.7KJ Wpump = Wair compressure + Wp.e

HEAT Heat is transient form of energy which transfers due to negative temperature gradient. (High temp to low temp). Heat has its meaning only if it is crossing the system. Q∝m 43

Q ∝ ∆T Q ∝ m∆T Q = mc∆T C=

Q m∆T

(Specific heat) Put m = 1 ∆T = 1

∴C=Q

Specific heat is amount of heat required to increase the temperature of 1-unit mass of substance through 1°c temperature.

SPECIFIC HEAT OF GASES:

m

m

Ti Tf

Ti Tf QV

QP

dv = 0

dv ≠ 0

w=0

w≠0

Only intermolecular energy

Intermolecular energy + external energy

Qp > Qv mcp ∆T > mCv ∆T Cp > C v γ=

Cp

γ>1

Cv 44

Specific heat at constant pressure is greater than specific heat at constant volume because Cp includes internal energy as well as external work whereas Cv includes only internal energy. For monatomic gases (Ar, He etc.) γ = 1.67 For diatomic gas, γ = 1.4 For polyatomic gases, γ = 1.33

 Specific heat of solids and liquids: In case of solids & liquids with the applications of pressure, change in volume is negligibly small (incompressible). Therefore, for solid& liquid, Cp = Cv = C Note: γ=

Cp Cv

=1+

2 n

Where, n = Degree of freedom of module. For monatomic molecule, n = 3 For diatomic molecule, n = 5 For triatomic molecule, n = 6

FIRST LAW OF THERMODYNAMICS: First law of thermodynamics is based on law of conservation of energy heat & works are different form of same entity called energy which is conserved. 45

Statement of closed system undergoing cycle: First law of thermodynamic for closed system undergoing cycle: “For a closed system undergoing a cycle heat transfer is equal to the network transfer”. ∑ Q = ∑ w (Valid for closed system undergoing cycle)

Result from first law of thermodynamic: 1) Heat is a path function: c

P b a

V

For cycle 1a2b1 (dQ)1a2 + (dQ)2b1 = (dW)1a2 + (dW)2b1 … … . . (ⅰ) For cycle 1a2c1 (dQ)1a2 + (dQ)2c1 = (dW)1a2 + (dW)2c1 … … … (ⅱ) Equation (ⅰ) - (ⅱ) (dQ)2b1 − (dQ)2c1 = (dW)2b1 − (dW)2c1 … … . . (A) Work is a path function, (dW)2b1 ≠ (dW)2c1 (dW)2b1 − (dW)2c1 ≠ 0 (dQ)2b1 − (dQ)2c1 ≠ 0 (dQ)2b1 ≠ (dQ)2c1

46

Conclusion: Through end points are same for path b & c but heat transfer is not same. Therefore, heat transfer is a path function. 2) From equation (A): (dQ)2b1 − (dQ)2c1 = (dW)2b1 − (dW)2c1 (dQ − dW)2b1 = (dQ − dW)2c1 = dE The quantity (dQ − dW) is same for path b & c hence, it does not depend on the so it must be a property, this property is known as energy. dQ − dW = dE (First law of thermodynamics for closed system undergoing a process) 3) Energy of system can be divided as 1. Macroscopic 1

a. Kinetic energy ( mv 2 ) 2

b. Potential energy (mgh) 2. Microscopic a. Translational motion of molecule b. Rotational motion of molecule c. Vibration motion of molecule [Internal energy [energy associated with molecules] E = K. E + P. E + I. E dE = d(K. E) + d(P. E) + d(I. E) Neglected, change in K.E & P.E neglected. dE = dU 47

dQ = dU + dW (First law of thermodynamics for closed system undergoing a process neglecting change in K.E & P.E of system.) 4) Energy of isolated system is constant dQ = dE + dW For isolates system, dQ = 0 dW = 0 dE = 0 E=c Energy for universe is constant. 5) Perpetual motion of 1st kind: (PMM-1) Perpetual motion = Continuous motion i.e. (cycle) The word perpetual means continuous. If a system is producing continuous work, then it must be operating on cycle PMM-1 is a machine which give continuous work without any energy input. Q=0

PMM – 1

W

For cycle, [cycle is clockwise (work)] ∴ (Power producing) ∑Q = ∑w Q=0 48

W=0 So, according to 1st law PMM-1 is not possible. If such a device is developed, then it will violate 1st law of thermodynamics. For ideal gas, dU = mcv dT ; U = f(T) only dH = mcp dT (Joules law)

ENTHALPY: (H) In thermodynamics the term U + PV appears frequently so for our convince this term is taken as enthalpy. It is extensive property % specific enthalpy is given as, H = U + PV Specific enthalpy h = U + PV (h → Intensive) In closed system, Internal energy is very important. But in open system, Enthalpy plays vital role. PV = Flow work U = Internal energy

Heat transfer in various non-flow/closed systems: 1) Constant volume process: V=C dQ = dU + dW For constant volume process dW = 0 dQ = dU 49

For ideal gas, dU = mcv dT ⇒ dQ = dU = mcv dT 2) Constant pressure process: P = Constant dw = PdV = d(PV) dQ = dU + dW = dU + d(PV) dQ = d(U + PV) dQ = d(H) For ideal gas ∴ dH = mcp dT dQ = dH = mcp dT 3) Isothermal process: PV = C & T=C dQ = dU + dW For ideal gas U = f(T) If T = Constant, then U = Constant ∴ dU = 0 dQ = dW dW = P1 V1 ln

V2 V1

∴ dQ = dW = P1 V1 ln

50

V2 V1

4) Adiabatic process: dQ = 0 dQ = dU + dW dU + dW = 0 dW = −dU Note: Show that Cp − Cv = R for ideal gas H = (U + PV) dH = d(U + PV) = dU + d(PV) dH = mCp dT dU = mCv dT mCp dT = mCv dT + d(mRT) Cp − Cv = R Valid for ideal gas Cp Cv

= γ & Cv =

R γ −1

, Cp =

R γ−1

5) Polytrophic process: dQ = dU + dW = dU +

P1 V1 − P2 V2 n−1

For ideal gas, dQ = mcv dT +

P1 V1 − P2 V2

dQ = mcv (T2 − T1 ) + =

mcv mR

n−1 P1 V1 − P2 V2 n−1

( P2 V2 − P1 V1 ) +

51

P1 V1 − P2 V2 n−1

dQ = [

P2 V2 − P1 V1 γ−1

]−[

= P2 V2 − P1 V1 [ =

n−γ γ−1

dQ =

− 1[

P1 V1 − P2 V2

1 γ−1

n−1

−

1 n−1

P2 V2 − P1 V1 n−1

]

]

]

γ−n P1 V1 − P2 V2 [ ] γ−1 n−1

dQ =

γ−n γ−1

Wpoly

SPECIFIC HEAT FOR POLYTROPHIC PROCESS: dQ = =

γ − n P1 V1 − P2 V2 [ ] γ−1 n−1 γ − n mR (T1 − T2 ) γ−1

=m (

[

n−1

n−γ n−1

n−γ n−1

]

cv dT

cv → Cpoly )

Convection of heat transfer heat supplied to system +ve heat rejected for system -ve Cpoly =

n−γ n−1

cγ

γ>n>1 Cpoly is always negative Cpoly = −ve

52

In polytrophic process,

system

Q

If we supply the heat, then also the temperature of system decreases. This is because in such a polytrophic process work transfer is more than heat transfer and excess amount of work comes from the internal energy of the system as there is decrease in internal energy. Temperature of system decreases. Note: Show that PV γ = Constant for ideal gas undergoing adiabatic process. For a process, dQ = dU + dW dQ = dU + PdV For adiabatic process, dQ = 0 For ideal gas, dU = mcv dT 0 = mcv dT + PdV … … … … . . (ⅰ) mcv dT = −PdV (H = U + PV) [dU + d(PV)] dH = dU + PdV + VdP dH = dU + d(PV) dH = dU + PdV + VdP 53

For ideal gas, mcp dT = 0 + VdP … … … . (ⅱ) Equation (ⅱ)/Equation (ⅰ) →

cp cv

=

VdP −PdV

dv

dP

v

v

γ( ) = − dP v

dv

+ γ( ) = 0 v

Integrating, ∫

dP P

dv

+ ∫γ( ) = 0 v

lnP + γ lnV = ln C PV γ = C (Ideal gas undergoing adiabatic process)

SPECIAL CASE OF WORK TRANSFER:

Electric works by The system

B

54

1.

Electric works on The system B

w = −ve; V = Constant dV = 0 PdV = 0 Cannot apply W = PdV as PdV is displacement work And in this system work is electrical.

Gas

Vacuum

System → gas + vacuum W=0 (If partition is removed); V = constant PdV = 0

55

2.

Gas

Vacuum

dV ≠ 0 (Change is volume non-zero) PdV = 0 w=0 Since, there is nothing to resist hence, work done is zero. This condition is called as free expansion. Free expansion is defined as the expansion against vacuum or unrestricted expansion. Free expansion work is zero because as the gas is expanding against the vacuum there is no resistance offered from surrounding and hence work is zero.

Gas

Vacuum

dQ = dU + dW dU = 0 U = Constant; Ui = Uf For ideal gas, U = f(T) Ti = Tf H = U + PV 56

For ideal gas, H = f(T) + mRT H = f(T) Hi = Hf

Problems with Solutions: 1. A well-sealed room contain 60kg of air at 200° k pa & 25℃ , now solar energy enters the room at an average rate of 0.8 KJ/sec. while a 120Watt fan is turned on to circulate the air in room. If heat transferred through the wall is zero, the air temperature in room after 30min.

- 0.12 kJ/sec for air Cv = 0.718 KJ/Kgk Cp = 1.005 KJ/Kgk R = 0.287 KJ/Kgk ; γ = 1.4 m = 60kg For air Cv = 0.718 T = 25℃

P = 200Kpa

Q = 0.8KJ/sec P = W = 120 Watt = 0.12 KJ/sec t = 30min 57

dQ = dW + dU 30 × 60 × 0.8 = dU − 0.12 × 30 × 60 dU = 1656 mCv dT = 1656/[60 × 0.718(T2 − 25)] T2 = 63.44℃ 2. A two-kilo watt base board electric resistance heater in a vacant room is turned on & kept on for 15min. the mass of air in room at 75kg and room is perfectly sealed find the temp rise of air at end of 15min. W = 2 kW

P = 2Kw = 2KJ/sec m = 75kg dQ = 0 dW = dU mCv dT = 2 × 15 × 60 75 × 0.718 × (dT) = 2 × 15 × 60 dT = 33.42℃ 3. Room contains 60kg of air at 100 k pa & 15℃ the room has 250watt refrigerator 120W TV, a 1 KW electric resistance & 150W fan. During cold winter day it is observed that the refrigerator & TV &fan & electric resistance heater is running continuously but air temp in room remains constant. The rate of heat loss from the room in KJ/hr. 58

dQ = dW + dU dQ = mCv dT − (0.250 + 0.120 + 1 + 0.05) × 60 × 60 dT = 0 dQ = 0 − 5112 dQ = −5112KJ/hr Heat loss dQ = 5112KJ/hr 4. Helium at 20 atm and 40℃ is contained in small steel cylinder having a volume of 15cm3 = 15 × (10−2 )3 m3 . The cylinder is placed in large container having a volume of 1500cm3 . The large cylinder is perfectly evacuated and insulated by an appropriate means helium discharged to fill the container calculate final pressure of helium. 15cm3 20 atm

Work = 0 (free expansion) Tinitial = Tf Tf = 40℃ dT = 0 dU = 0 Now, Tinitial = Tf 59

1500cm3

Pi Vi = Pf Vf 20 × 15 = Pf × 1500 Pf = 0.2 atm 5. A gas undergoes a thermodynamics cycle consisting of following processes (ⅰ) Process 1-2 → Constant pressure P = 1.4 bar ; V1 = 0.028m3 ;

W1−2 = 10.5KJ

(ⅱ) Process 2-3 → Compression with PV = C U3 = U2 (ⅲ) Process 3-1→ Constant volume U1 − U3 = −26.4 KJ (i)

Find heat transfer & change in internal energy for 1-2

(ii) Find heat transfer in 2-3 (iii) Find heat transfer in 3-1 P1

1.4 = P2

3

1

2

V2

V1

(ⅰ) W1−2 = 10.5KJ 10.5 = P(V2 − V1 ) 10.5 = 1.4 × 100 × (V2 − 0.028) V2 = 0.103m3 dQ = dW + dU dU = U2 − U1 60

= U3 − U1

(U2 = U3 )

dU = 26.4KJ dQ = 10.5 + 26.4 dQ = 36.9KJ (ⅱ) dQ = dW + dU dU = 0

(U2 = U3 ) dQ = dw = P2 V2 ln

V2 V1

= 1.4 × 102 × 0.103 × ln (

0.103

)

0.028

dQ = −18.78KJ (ⅲ) 3-1 dQ = dW + dU = (U1 − U3 ) + 0 dQ = −26.4KJ 6. A rigid insulated tank of 3m3 volume have two compartments. 1st compartment of 1m3 contains ideal gas at 0.1MPa & 300K. while the 2nd compartment contains same gas 1MPa & 100K. If the partition is removed, then determine the final temp& pressure of gas. 1m3 0.1 MPa 300K

2m3 1 MPa 1000K

Pf Tf

V1 = 1m3 ; V2 = 2m3 ; Vf = 3m3 dQ = 0 Q loss = Q gain 61

P1 V1 = M1 R1 T1 & P2 V2 = M2 R 2 T2 Same gas P1 V1 = n1 R1 T1 0.1 × 103 × 1 = n1 × R × 300 ̅ n1 = (3 × 10−1 )/R n1 = 0.360 P2 V2 = n2 R 2 T2 2 × 103 × 1 = n2 × R × 1000 ̅ n2 = (2 × 10−3 )/R n2 = 2.40 Q loss = Q gain m2 (1000 − Tf ) = m1 (Tf − 300) Mn2 (1000 − Tf ) = M n1 (Tf − 300) n2 (1000 − Tf ) = n1 (Tf − 300) 2.4(1000 − Tf ) = 0.36(Tf − 300) 6.66(1000 − Tf ) = Tf − 300 Tf = 900K Pf Vf = ∑ nR f Tf Pf × 3 = (2.4 × 0.36) × 0.8314 × 900 Pf = 700 KPa Pf = 0.7 MPa

62

7. A gas of mass 1.5kg undergoes a quasi-static expansion which follows the relationship P = a + bV where a & b are constant the initial & final pressure are 1000kg & 200kpa respectively & corresponding volumes are 0.2m3 & 1.2m3 . The specific internal energy is given as μ = 1.5 ρV − 85KJ/kg where 𝜌 in k pa & ϑ = m3 /kg. Calculate the net heat transfer & maximum internal energy of gas during expansion. m = 1.5kg P = a + bV Pi = 1000KPa 1

(1000)Pi

2

(200) Pi

0.2

1.2

Pf = 200KPa Vi = 0.2m3 Vf = 1.2m3 μ = 1.5; ρV − 85 KJ/kg P = a + bV (Linear relation) 1000 = a + b(0.2) 200 = a ± b(1.2) 800 = b(−1) b = −800 63

a = 1000 + 800(0.2) a = 1160 0.2

ui = 1.5 × (1000) × ( ) − 85 = 115 1.5

1.2

uf = 1.5 × (200) × ( ) − 85 = 155 1.5

dU = md(Ui ) = 40 × 1.5 = 60 dQ = dU + dW = 60 + (1 × 200) +

1 × 1 × 800 2

dQ = 660 KJ m × U = 1.5PV × m − 85m U = 1.5PV − 85 × 1.5 U1 = 1.5P1 V1 − 8.5 × 1.5 U2 = 1.5P2 V2 − 8.5 × 1.5 dU = 60 KJ dQ = 60 + 600 = 660 KJ U = 1.5PV − 85 × 1.5 dU dV

=

=

d dV d

dV

[ 1.5(a + bV) × V − 85 × 1.5]

[ 1.5(aV + bV 2 ) − 85 × 1.5]

= 1.5a + 1.5 × 2 × b × V − 85 × 1.5 = 1.5 × 1160 + 1.5 × 2 × (−800) × V − 127.5 = −1600 × V + 1612.5 V = 0.725m3 Umax = 1.5 × (1160 − 800 × (0.725)) × 0.725 − 85 × 1.5 Umax = 503 KJ 64

OPEN SYSTEM: When there is mass transfer along with energy transfer across system boundary then system is called as open system.

Steady flow: A flow is said to be steady flow, if properties do not change w. r. to time at given location.

Uniform flow: A flow is said to be uniform if properties do not change w.r.to location (space) at a given time. Note: In closed system we talked about fixed mass but in open system we will talk about mass flow rate. In open system analysis, a fixed volume is considered known as control volume. In control volume mass can enter or leave energy can enter or leave but volume of system remains constant.

MASS BALANCE: (Law of conservation of mass) If there is no accumulation of mass (control volume). Then, rate of mass entering is equal to rate of mass leaving. In case of steady flow, there is no accumulation of mass. m1 = m2 = m → Steady flow ρ= ρAL time

= 65

m vol m time

m = ρAL For steady flow m1 = m2 ρ1 A1 V1 = ρ2 A2 V2 → continuty equation In case of unsteady flow, m1 ≠ m2

OPEN SYSTEM WORK 1) External work 2) Flow work Open system work

External work

Flow work

External Work: The work transfer across the control volume other than due to normal fluid force Example: shaft work

Flow work: It is the work involved in causing the fluid element either enter to C.v or leave the C.v. Flow work is analogous to “displacement work”.

66

Work: 1 dm,d v P

dv = ϑdm 2

dWflow = PdV = Pvdm dWflow dm

=Pv

flow × work mass

=Pv

Total flow work = P v M = PV At energy, flow work = −P1 V1 At exit, flow work = P2 V2

ENERGY BALANCE: 1) Steady flow: In steady flow there is no accumulation of mass & energy within the system (Control volume). Mass entering m1 = mass leaving m2 Energy entering E1 = energy leaving E2

67

Z1

P1 V1 C1 U1 M1

WCV

2

Q

Z2

P2 V2 C2 U2 M2

m1 = m2 = m E1 = E2 = E 1

E1 = m1 C1 2 + m1 gZ1 + U1 + Q 2

1

E2 = m2 C2 2 + m2 gZ2 + U2 + W 2

We have W = WCv − P1 V1 + P2 V2 For steady flow, E1 = E2 1 2

1

m1 C1 2 + m1 gZ1 + U1 + Q = m2 C2 2 + m2 gZ2 + U2 + W 2

m1 = m2 = m ; h1 +

C1 2 2

W = WCv − P1 V1 + P2 V2

+ gZ1 + q = h2 +

C2 2 2

+ gZ2 + WCv

1st law of thermodynamics for open system steady flow.

Examples of Steady flow process: 1) Nozzle: Nozzle is a device which is used to increase the velocity of fluid in expanse of pressure energy.

68

1

2

Q = 0; Z1 = Z2 WCv = 0 h1 +

C1 2 2

+ gZ1 + 0 = h2 + h1 +

C1 2 2

→

J Kg

C1 2 2

C2 2 2

= h2 +

; C1 →

m s

h1 +

2000

C2 2

= h2 +

2000

C2 >>>> C1 C1 Can be neglected, When C1 value is net given h1 = h2 +

C2 2 2000

C2 = √2000(h1 − h2 ) 2) Turbine: It is used to produce work W = +ve

Z1

Z2

Q=0

Z1 = Z2

69

C2 2 2

; h → KJ/Kg

Hence, C1 2

+ gZ2 + 0

Neglecting kinetic energy changes, h1 +

C1 2 2

+ gZ1 + q = h2 +

C2 2 2

+ gZ2 + wT

C1 − C2 Very small So, it can be neglected WT = h1 − h2 (Steady flow, fully insulated and change in K.E & P.E is neglected) 3) Compressor:

1

2

It consumes work, to compress the gas. Q = 0; Z1 = Z1 WC = −ve WC = h2 − h1 4) Throttling process: Example: a) Flow through very small opened valve. b) Flow through very small opening. 2

1

Q = 0; Z1 = Z1 WC = 0 No change in kinetic energy. h1 = h2 Throttling is an Iso-Enthalpy process. It is used in refrigerator. 70

OPEN & CLOSED SYSTEMS OF REVERSIBLE WORK: Open system reversible work Wopen = ∫ −vdP

Closed system reversible work 1 P dP

2

V

V

WClose = PdV dA = vdP Total area =∫ VdP  Area under the curve when projected on pressure axis gives open system reversible work.

OPEN SYSTEM WORK FOR VARIOUS PROCESSES: 1) Constant volume process: V=C Wopen = v(P1 − P2 )  Flow does not take place from high pressure to flow pressure it takes place from high energy to low energy.

71

2) Constant pressure process: P=C dP = 0 Wopen = 0

3) Isothermal pressure: Pv = Constant v=

C P C

Wopen = ∫ − dP P

Wopen = −C[ln P]12 Wopen = C ln

P1 P2

Wopen = P1 V1 ln P1 V1 = P2 V2 ⇒

P1 P2

Wopen = P1 V1 ln

P1 P2

=

V2 V1

V2 V1

= Closed system work In isothermal process, Wopen = Wclosed

72

Problems and solutions: 1. At nozzle inlet enthalpy of fluid passing 3000 KJ/Kg & velocity is 60m/sec at discharge end the enthalpy is 2762KJ/Kg. the nozzle is horizontal & perfectly insulated A1 = 0.1m2 ; V1 = 0.187m3 /Kg; V2 = 0.498m2 /Kg a. Find the velocity at nozzle exit. b. Find mass flow rate. c. Exit area of nozzle. for nozzle h1 +

V1 2 2000

3000 +

= h2 +

602 2000

V2 2 2000

= 2762 +

V2 2 2000

V2 = 692.53 m/s ρ1 V1 A1 = A2 ρ2 V2 = m 1

=

VA ϑ1 1 1

=

1 0.187

× 0.1 × 60

m = 32.08 V1 A1 ϑ1 0.1×60 0.187

=

=

A2 V2 ϑ2

A2 ×692.53 0.498

A2 = 0.023m3

73

2. In test of water jacketed compressor, the shaft work required is 90KJ/Kg during compression increase in enthalpy at air is 30KJ/Kg & increase in enthalpy of circulating is 40KJ/Kg. the change in velocity is negligible the atm of heat lost from compressor by air. dQ = dU + dW W = h2 − h1 Water

1

2

Water

For air, h1 + Q = h2 + W Q = (h2 − h1 ) + W = 30 − 90 = −60 Heat last by air = +60 Heat loss by air = 60 = Q water + Q atm Q atm = 20KJ

74

3. A turbine operates under steady flow condition receiving steam at following state pressure is 1.2MPa. Temperature 188°C. Enthalpy 2785 KJ/Kg velocity 33.33m/s and elevation 3m. The steam leaves the turbine at following state. Pressure 20KPa enthalpy 2512 KJ/Kg velocity 100m/s elevation 0 heat is lost to surrounding at the rate of 0.29 KJ/sec. If rate of steam flow through the turbine is 0.12 Kg/sec what is power output of turbine in KW. P1 = 1.2MPa T1 = 188℃ V1 = 33.33m/sec h1 = 2785 KJ/Kg ; Z1 = 3m h2 = 2512 KJ/Kg ; V2 = 100m/s Z2 = 0 Q lost(surrounding) = 0.29 KJ/sec m = 0.42Kg/sec; P =? m[h1 +

V1 2 2

+ gZ1 ] + q = m[h2 +

0.42 [2785 +

(33.33)2 2000

= 0.42[2512 +

+

(100)2 2000

9.81 1000

+

V2 2 2

× 3] + 0.29

9.81 1000

× 0] + WT

WT = 112.5 KJ/sec

75

+ gZ2 ] + WT

STEADY FLOW ENERGY EQUATION FOR MULTI STREAM FLOW: m1 m3 W m2 m4 Q

Steady flow equation (energy balance equation) ṁ 1 [h1 + = ṁ 3 [h3 +

C1 2 2 C3 2 2

+ gZ1 ] + ṁ 2 [h2 + + gZ3 ] + ṁ 4 [h4 +

C2 2 2 C4 2 2

+ gZ2 ] + Q + gZ4 ] + WCv

Mass balance equation: ṁ 1 + ṁ 2 = ṁ 3 + ṁ 4

Problems with solutions: 1. The steam supply to the engine comprise two stream which mix before enetering engine one steam is at rate of 0.01Kg/sec with enthalpy 2952KJ/Kg and velocity 20m/s. the other steam is supplied at 0.1Kg/sec with enthalpy 2569KJ/Kg and velocity 120m/s at the exit from the engine fluid leaves at 2 stream one of water at rate of 0.001Kg/sec with enthalpy 420KJ/Kg and other of a stream the fluid velocity at the exit is neglegibly small the engine develoip shaft power of 25KW heat transfer is negligible evaluate the enthalpy of second exit stream & mass flow rate. 76

ṁ 1 = 0.01Kg/sec; h1 = 2952KJ/Kg; C1 = 20m/s ṁ 2 = 0.1Kg/sec; h2 = 2569KJ/Kg; C2 = 120m/s ṁ 3 = 0.001Kg/sec; h3 = 420KJ/Kg; WCv = 25KW; Q = 0 Mass balance equation ṁ 1 + ṁ 2 = ṁ 3 + ṁ 4 ṁ 4 = 0.109Kg/sec Effect of Z is negligible Hence, gZ1 = gZ2 = gZ3 = gZ4 = 0 ṁ 1 [h1 +

C1 2 2

+ gZ1 ] + ṁ 2 [h2 +

= ṁ 3 [h3 + ṁ 1 [h1 +

C1 2 2000

C3 2

C3 2 2000

(20)2 2000

+ gZ2 ] + Q̇

2

+ gZ3 ] + ṁ 4 [h4 +

+ 0] + ṁ 2 [h2 +

= ṁ 3 [h3 + 0.01 [2952 +

2

C2 2

C2 2 2000

= 0.001 [420 +

(0)2

2

+ gZ4 ] + WCv

+ 0] + Q̇

+ 0] + ṁ 4 [h4 +

] + 0.1 [2569 +

C4 2

C4 2 2000

(120)2 2000

+ 0] + WCv

]+0

] + 0.109[h4 ] + 25

2000

h4 = 2401 KJ/Kg 2. The stream of air & gasoline vapour in ratio of 14:1 by mass enter in a gasoline engine at temperature 30°C & leave as combustion product and temperature of 190°C. the engine has specific fuel consumption 0.3 Kg/Kwhr. The net heat transfer rate from the fuel air stream to the jacketed cooling water & to the surrounding is 35Kw. The shaft power delivered by the engine is 26Kw 77

compute the increase in specific enthalpy of fuel air stream assuming the changes in K.E & P.E is negligibley small. ma

=

mg

14 1

SFC = 0.3 Kg/Kwhr Q = 35 Kw W = 26Kw ma = m1 & mg = m2 Air m Gasoline

ṁ(h1 ) + Q = ṁ(h2 ) + WCv ṁ(h2 − h1 ) = −35 − 26 h2 − h1 = −

61 m

Specific fuel consumption is defined for output power (SFC = To produce specific power how much fuel consumption is required.) SFC = 0.3Kg/Kwhr =

0.3×26

Kg/sec

3600

ṁ g = m2 = 2.16 × 10−3 Kg/sec mair = m1 = 14 × 0.0021 = 0.03Kg/sec ṁ = ṁ g + ṁ a ṁ = 0.321Kg/sec ∴ ∆h =

−61 m

= −1877KJ

78

UNSTEADY FLOW: Many flow problems such as filling up & evacuated cylinder are not steady in unsteady state the rate at which the mass of fluid within control volume is accumulated is equal to the net rate of mass flow across the control volume.

Mi

Me

I = Condition of fluid at inlet E = Condition of fluid at outlet1 1 = Initial condition of system 2 = Final condition of system (C.V) Mass balance, Rate of accumulation of mass within system = Rate of mass entering - Rate of mass leaving mi = Mass entering me = Mass leaving

Mass balance: The Rate of accumulation of mass within system = Rate of mass entering - Rate of mass leaving. dm

(

)

dt Cv

= Mass accumulation within control volume 79

dm

(

)

= (

dt Cv dm

(

)

dt Cv

dmi dt

dme

)−(

dt

)

= m i − me

This is mass balance for unsteady state flow.

Energy balance: Ei = Energy of fluid at inlet Ei = mi hi + mi

Ci 2 2

+ mi gzi + Q

Ee = Energy of fluid at outlet Ei = me he + me

Ce 2 2

+ me gze + WCv

A/c to energy balance equation for unsteady state flow: The rate of energy accumulated within the control volume is equal to difference between rate of energy entering to rate of energy leaving. dE

( )

dt Cv

dE

( )

dt Cv

=

d dt

[(mi hi + mi

Ci 2 2

dEi

=( =

dt

d dt

)−(

dEe dt

)

[Ei − Ee ]

+ mi gzi + Q) − (me he + me

Ce 2 2

+ me gze +

WCv )] Now neglecting K.E & P.E changes Energy = K.E + P.E + I.E dU

( )

dt Cv

dU

( )

dt Cv

=

= d dt

d dt

E = I.E [mi hi + Q̇ − me he − WCv ]

(mi hi ) + Q̇ −

d dt

(me he ) − WĊ v

hi & he Are constant with respect to time. 80

dU

( )

dt Cv

dU

( )

dt Cv

= hi

d dt

(mi ) + Q̇ − he

d dt

(me ) − WĊ v

= ṁ i hi + Q̇ − ṁ e he − WĊ v (Energy balance)

CHARGING OF TANK: An insulated storage tank that is initially evacuated is connected to a supply pipe line carrying a fluid which specific internal energy is Ui & specific enthalpy hi the valve is open & fluid flow in the tank through supply pipe line & reaches the pressure same as pressure of supply pipe line show that final specific internal energy of final temperature of fluid inside the tank if fluid flowing is ideal gas and supply line temp is Ti . Ui, hi, Ti, Pi

Mi = 0 Me = 0 Q=0

W=0

Mass balance, dM

(

)

dt Cv

= ṁ i − ṁ e

M1 − M2 = ṁ i − ṁ e M2 = ṁ i …….. (ⅰ) 81

Energy balance, (

dM

)

dt Cv

= ṁ i hi + Q − ṁ e he − WĊ v U1 − U2 = ṁ i hi U1 = 0 U2 = m2 hi U2 m2

= hi

U2 = hi …...... (ⅱ) For ideal gas du = Cv dT dh = Cp dT u2 − u1 = Cv (T2 − T1 ) T1 → 0 u1 → 0 u2 = Cv T2 hi = Cp Ti (he = 0 Te = 0) From (ⅱ) For ideal gas, Cv T2 = Cp Ti Cp

T2 = ( ) Ti Cv

∴ T2 = γTi

T2 > Ti

 There no problem with 1st law of thermodynamics it is quantitative law. Process is possible or not is not the criterion of 1st law. 82

SECOND LAW OF THERMODYNAMICS:  According to 1st law of thermodynamics, energy is conserved during process or cycle and energy balance hold good but it does not give any information regarding process is possible or not.  It is the 2nd law of thermodynamics, which provide the criterion for possibility of any process and give the direction for particular process through the concept of entropy. It is also known as directional law.  1st law is quantitative law of energy and 2nd law is quantitative law of energy. Note: All natural spontaneous process is occurring only in one direction (high potential to low potential) Thermal Reservoir

Source T1

Sink T2 Q2 (Absorb)

Q1 (Supply)

Source: Source thermal reservoir which can supply any amount of energy without undergoing any temp change Example: Sun

83

Sink: Sink is thermal reservoir which can absorb any amount of energy without undergoing any temperature change. Example: Atmosphere 2nd law given by joule mathematical representation – Clausius

2nd law of thermodynamic for cycle: Joules experiments demonstrated that work can be completely converted into heat but complete conversion of law grade energy (heat) to higher grade energy (work) is not possible for a cycle.

KELVIN – PLANK STATEMENT OF SECOND LAW (PMM-Ⅱ): It is not possible to construct a device which continuously delivers the work while exchanging heat a single reservoir.

Q PMM-II

W

PMM-Ⅱ has 100% efficiency (i.e. not possible)

84

CONCEPT OF HEAT ENGINE: If more than one reservoir is used, then it is possible to convert low grade energy to high grade energy. Source T1 Power Producing Clockwise W

Q1 HE Q2 T2 Sink

1st law applied to system ∑ Q = ∑ QW Q1 − Q 2 = W η= η=1−

Q2 Q1

O/P I/P

=

W Q1

=

Q1 − Q2 Q1

Valid for reversible & irreversible both. (As no assumption mode.)

CLAUSES STATEMENT OF 2ND LAW: It is not possible to construct a device which operates on a cycle and transfer heat from low temperature to high temperature body without any energy input. Source T1 Q1 HE

W

Q2 T2 Sink

85

Heat at low temperature → low grade energy (low significance) Heat at high temperature → high grade energy (high significance)

CONCEPT OF REFRIGERATOR: Refrigerator is a device which creates & maintains lower temperature than surrounding as lower temperature is to maintain continuously the system must operate on cycle. ∑ Q = ∑ QW

1st law,

Q 2 − Q1 = −W Source T1 Q1 R

Power Consuming anticlockwise cycle W

Q2 T2 Sink

Coefficient of performance (COP) = Desired effect /input =

Q2 W

COP =

(Valid for reversible & irreversible) Q1 Q1 − Q2

86

……… (1)

HEAT PUMP: Heat pump is device which maintains higher temperature than surrounding. Source T1 Q1 HP

W

Q2 T2 Sink

(COP)HP =

desired effect

(COP)HP =

input Q1 Q1 − Q2

(COP)HP − (COP)R = 1 (COP)HP = 1 + (COP)R (Valid for same temperature limit) Note: Clauses & Kelvin-plank statements are parallel statements of 2nd law of thermodynamic if we violate any one of them automatically another is violated.

87

VIOLATION OF CLAUSES: Source T1 Q1

W = Q1 – Q2

at (T1) Q1

Q1

HE

Q1

HE

Q2

Q3

Q2

T2 Sink Violation of Clausius law

Q3 T2

Violation of Kelvin plank law

Violation of Kelvin-plank statement: T1 Q1 W

T1 Q1

Q1 W

Re

W

Q3 = Q1 + Q2 Re

Q3

W

Q2

T2

T2 Violation of Clausius law

Violation of Kelvin plank law

T1 Q3 = Q1 + Q2

Q1 HE

W

W

R Q2

T2 Violation of Clausius statement

88

CARNOT CYCLE: Qin

1

2

4

3 Q Rejected

1-2 → Isothermal expansion (heat add) 2-3 → Adiabatic expansion (Q = 0) 3-4 → Isothermal compression. 4-1 → Adiabatic compression. Carnot cycle consist of two isothermal & two adiabatic process isothermal process is very slow whereas, adiabatic vs. very fast. Hence, the combination of these two processes in cycle is not possible. Carnot cycle is theoretical cycle which is used for compression of other actual cycles.

CARNOT THEOREM: For various cycles operated between same temperature limit non-has efficiency greater than reversible cycle. Let us consider two engines E1 & E2 E1 is a reversible engine & E2 is irreversible engine in same temp difference. T1 Q1

Q1 W

E1

E2 Q3

Q2 T2

89

W

E1 → Rev engine E2 → Irre engine Let us assume, ηirr > ηrev W2 Q1

>

W1

(W2 > W1 )

Q2

Now reversible the direction in reversible engine, T1 Q1

Q1

E1 Q2

W1 W2

Q1

E2

Q1

E1

Q3 Wnet = W2 – W1

W1 W2

E2

Q2 Q3 W2 – W1 T2

T2

Violation of Kelvin-plank so our assumption is wrong, ∴ ηrev > ηirr A cycle is said to be reversible cycle if all the processes of cycle are reversible if single cycle is irreversible, then cycle is irreversible cycle.

CLAUSIS THEOREM: (Clausis inequality) According to this theorem, the cyclic integral of

dQ T

is less than or

equal to zero. ∮ ⇒∮ ∮

dQ

dQ T

T

dQ T

≤0

= 0 ….. Reversible cycle

< 0 …... Irreversible cycle

90

If ∮

dQ T

> 0 ….. Cycle is not possible ∮

dQ

≤0

T

This is 2nd law of thermodynamics for cycle.

REVERSIBLE HEAT ENGINE: T1 Q1 HE

W

T2

Closed system undergoing cycle. Apply 1st law ∑Q = ∑W Q1 - Q 2 = W Apply 2nd law For reversible heat engine ∮ Q1 T1

−Q2

+(

T2

dQ T

=0

)=0

Q1 T1

-

Q1 T1 Q1 Q2

(T is in K)

Q2 T2

Q2

= =

=0 T2

T1 T2

91

ɳ=1-

Q2 Q1

It is valid for reversible as well as irreversible For reversible, Q1 Q2

T1

=

T2

ɳ=1-

T2 T1

It is only valid for reversible.

REVERSIBLE HEAT PUMP: T1 System Q1 HP

W

T2

Apply 1st law ∑Q = ∑W Q 2 - Q1 = -W ∴ W = Q1 - Q 2 Apply 2nd law ∮

dQ T

= 0 …… Reversible. Q2 T2

-

Q1 Q2

Q1 T1

=

92

=0

T1 T2

Cop =

Q1 Q1 − Q2

=

Cop =

1 Q 1 − Q2 1

T1 T1 − T2

It is only for reversible process Similarly, for reversible refrigerator: Cop =

T2 T1 − T2

It is valid for reversible only.

COUPLED ENGINE: (Engine in series) T1 Q1 E1

W1

Q T Q2 E2

W2

T2

W1 = Q1 – Q ɳ1 = 1 -

T T1

W2 = Q – Q 2 ɳ2 = 1 -

93

T2 T

=

1 T

1 − T2 1

a. When work from both the engine is same W1 = W2 Q1 – Q = Q – Q 2 Q1 Q T1 T

–1=1-

Q2

–1=1-

T2

T=

Q T

T1 T1 + T2

b. Efficiency of both engines are same ɳ1 = ɳ2 1-

T T1

=1-

T2 T

T = √T1 × T2 Note: 1.

T2 T1

=

T2 T

×

T T1

(1 − ɳ) = (1 - ɳ2 ) (1 - ɳ1 ) 1 − ɳ = 1 - ɳ1 - ɳ2 + ɳ1 ɳ2 ɳnet = ɳ1 + ɳ2 - ɳ1 ɳ2 ɳ > ɳ1 ɳ > ɳ2 2. Two constant property line can never intersect each other. Two adiabatic line can never intersect each other.

94

adiabatic isothermal P Qin

V

⇒ Only one reservoir producing work. Hence violation of kelvin planks statement

Problems with Solutions: 1. A heat engine working on Carnot cycle receive heat at 40KW from a source of 1200K & reject it to the sink at 300K. Find heat rejected. T1 Q2 W

HE T2

40 = 1 T2 T1

⇒

=

40 1200

T2 T1

Q2 Q1

=

Q2 300

Q 2 = 10KW 2. An inventor claims that heat engine has following specification. Power developed = 50KW. Fuel burned for hour = 3kg, heating value of fuel = 75000kJ/kg, temperature limits = 627 & 270 C. Cost = 30 Rs/kg, value of power = 5 Rs/Kwh

95

1st Method The performance of engine, W = 50KW Q1 = (3 × 75000) ÷ 3600 = 62.5KW ɳmax = 1 ɳactual =

T2 T1

output input

= 66.6%

=

Wout Qin

=

50 62.5

ɳactual = 80% ɳactual > ɳmax ….. not possible 2nd Method 900 K Qih = 62.5 HE

50

12.5 300 K

∮

dQ T

=

62.5 90

-

12.5 300

= 0.02 (+ve) Hence violation of 2nd law ∴ Not possible ɳmax = ɳactual → economical ɳmax > ɳactual → possible ɳmax < ɳactual → not possible 96

3. If efficiency of Carnot engine is given as 0.75. if cycle direction is reversed then value of cop for Carnot refrigerator. Q1

0.75 =

Q1 − Q2

0.75 =

1 Q

1 − Q2 1

Q2

1− ⇒

Q1

Q2 Q1

(Cop)ref =

=

=

4 3

T2 T1

T2 T1 − T2

=

1 T

1 − T2 1

(Cop)ref = 0.33 4. In given figure E is heat engine with efficiency 0.4, and R is refrigerator given that Q 2 + Q 4 = 3Q1 . Then what is cop of refrigerator. Q1 Q1 E

R Q2

Q3

-Win = - Q 3 + Q 4 -0.4Q1 = - Q 3 + 2.4Q1 Q 3 = 2Q1 0.4 =

Q1 Q1 − Q2

-0.4 Q 2 = Q1 – 0.4 Q1 Q 2 = 0.6Q1 W = Q1 - Q 2 97

W = 0.4 Q1 ɳr =

Q3 Q4 − Q3

Q 2 + Q 4 = 3Q1 0.6 Q1 + Q 4 = 3Q1 Q 4 = 2.4 Q1 2Q1

(Cop)r =

2.4Q1 − 2Q1

=5

5. Using the engine of 30% thermal efficiency to drive the refrigerator having Cop = 5. What is the heat input into the engine for each mega joule removed from cold body by refrigerator? T1 Q1 W

R Q2 T2

T1 Q3 W E Q4 T2

ɳ = 30% = 0.03 Cop = 5 5=

Q2 Q1 − Q2

5Q1 - 5Q 2 = Q 2 98

5Q1 = 6Q 2 Q1 = 1.2Q 2 0.3 =

Q3 − Q4 Q3

0.3Q 3 = Q 3 − Q 4 Q 4 = 0.7Q 3 Q1 = 1.2 MJ Q1 − Q 2 = Q 4 − Q 3 1.2 MJ – 1 = Q 3 (0.3) Q 3 = 0.66MJ 6. A heat engine is used to drive a heat pump. The heat transfers from heat engine and from heat pump are used to heat the water, circulating through the radiator of building the efficiency of heat engine is 27% & COP heat pump 4. Evaluate the ratio of heat transfer to circulating water to heat transfer to the heat engine. T1 Q1

W = Q1 - Q2

E Q2 T2

T3 Q3 W = Q1 - Q2

P Q4 T4

ɳ=

99

Q1 − Q2 Q1

0.27 = Q2 Q1

Q1 − Q2 Q1

= 0.73

Win = Q1 − Q 2 = 0.27 Q1 COP =

Q3 Q3 − Q4

Q 3 = 4 (Q 3 − Q 4 ) 3Q 3 = Q 4 Q 4 = 3Q 3 (Q1 − Q 2 ) + Q 4 = Q 3 0.27 Q1 + 4Q 3 = Q 3 0.27 Q1 = - 3Q 3 Q2 + Q3 Q1

=

(0.73 + 1.08)Q1 Q1

= 1.81

7. A house is to be maintained at temperature of 200 C by means of heat pump pumping heat from atmosphere. Heat losses through walls of house are estimated as 0.65KW per unit of temperature difference between the inside of house & atmosphere. If atmospheric temperature is - 100 C. What is minimum power required to dry the pump. It is proposed to use the same pump to cool the house in summer for same room temperature, same heat loss rate and same power input to the pump what is the maximum permissible atmospheric temperature.

100

T1 = 200C Q1 W = Q1 - Q2 Q2 T2 = 100C

T1 = 200 C T2 = −100 C Q1 = 0.65 × (T1 - T2 ) = 0.65 × (293 - 263) Q1 = 19.5 kW T1

=

T2

Q1 Q2

Q 2 = 19.5 ×

263 293

Q 2 = 17.5 Wpower = 19.5 – 17.5 = 2 kW T1 = ? Q1 W = Q1 - Q2 Q2 T2 = 200C

Q1 − Q 2 = 2kW Q1 = 2 + Q 2 Q1 = 2 + 19.5 Q1 = 21.5 101

T1

=

T2 x

=

200

Q1 Q2 21.5 19.5

x = 323K x = 500 C ∴ T1 = 500 C 8. The refrigerator is maintained at 20 C. every time the door opened warm material is placed inside introducing an average of 420KJ but making only a small change in temperature of refrigerator. The door is opened 20 times in day & refrigerator operates at 15% ideal COP. The cost of work is 32 paisa/kWhr. What is monthly bill for refrigerator. The atmospheric temperature 300 C. T1 = 313 Q1 Win

R Q2 T2 = 279

Q1 = 420 × 20 × 30 = 8400 (COP)R =

15 100

×

273 + 2 28

(COP)R = Win =

= 1.47

Q1 Win

420 × 20 × 30 1.47

Win = 171428.57 KJ/sec 102

KJ × hr

Win = 171428.57

3600 × sec

= 47.61 KW/hr Cost = 47.61 × 0.32 = ₹15.80

ENTROPY: ∮

dQ T

≤ 0 → 2nd law for cycle.

Case 1: Reversible cycle, 1

c

P

b a

2 V

For reversible cycle, |a2 b| dQ

dQ

( ) T

+( ) T

1a2

2b1

= 0 …… (1)

For reversible |a2 c|, dQ

dQ

( ) T

+( ) T

1a2

2c1

= 0 …… (2)

Equation (1) & (2) dQ

( ) T

dQ

2b1

-( )

dQ

( ) T

dQ

The quantity ( ) T

reversible

T

2c1

=0

dQ

2b1

= ( ) T

2c1

is same for both the path b & c. it only

depends on end points. So, it must be a property.

103

1

c

P

b

2 V

This property is known as entropy. Entropy (C ′ ) = Entropy Entropy is a property it does not depend upon the path. Entropy is same if the end points are same. Entropy depends only on end points. Hence, as long as end points are same the entropy remain same for reversible as well as irreversible path. Between two end points entropy is only defined for reversible path & same for irreversible path. In order to find out entropy change of irreversible path it must be replaced by reversible path between same end points.

ENTROPY CHANGE OF SYSTEM FOR REVERSIBLE PROCESS: Case 1: When heat is added to the system. ds =

dQ T

ds = + ve s2 − s1 = +ve The entropy of system increases when heat is added to a system. 104

dQ

ds = ( ) T

reversible

This is 2nd law of thermodynamics for process. Irreversible cycle 1

c

P

2 V

For reversible cycle |2b|, dQ

dQ

( ) T

1a2

+( ) T

dQ

=0

dQ

( ) T

2b1

1a2

= -( ) T

2b1

For irreversible cycle |a2 c ′ |, dQ

dQ

( ) T

1a2

+( ) T

dQ

<0

dQ

-( ) T

2c′ 1

2b1

< ( ) T

2c′ 1

dQ

ds > ( ) T

2c′ 1

dQ

ds > ( ) T

dQ

( ) T

irreversible dQ

reversible

>( ) T

irreversible

ds = Entropy

Case 2: When heat is rejected, ds =

dQ

105

T

ds = -ve s2 − s1 = -ve Entropy of a system decreases when heat is rejected.  Entropy of universe always increases

Case 3: Reversible adiabatic process dQ = 0 ∴ ds = 0 s2 − s1 = 0 s2 = s1 In this process entropy is same hence it is called isentropic process. For reversible process entropy of system is constant. Note: System entropy can increase or decrease or become constant depending upon type of heat transfer.

ENTROPY CHANGE FOR IRREVERSIBLE PROCESS: dQ

ds > ( ) T

irreversible

dQ

ds = ( ) T

irreversible 1

P

+ (δs)generation

d c

2 V

106

For reversible process, (δs)generation = 0 The end points are same for irreversible path c ′ & d′ , The entropy change will be same but dQ

( ) T

So, ((δs)generation )

2c′ 1

dQ

2c′ 1

≠( ) T

2b′ 1

≠ ((δs)generation )

2b′ 1

So, Entropy is a property so it does not depend on the path followed by process, but entropy generation is not a property. It depends on path followed by the process. (δs)generation is measure of irreversibility of a system. In case of reversible process, (δs)generation = 0 and if δs for 2c ′ 1 is greater than 2d′ 1 then c ′ will be more irreversible. (δs)2c′ 1 > (δs)2d′ 1 For irreversible path, dQ

ds = ( ) T

irreversible

+ (δs)generation

For non- adiabatic & heat rejection dQ

( ) T

irreversible

=-a

At same time (δs)generation = a ∴ ds = -a + a = 0 s2 − s1 = 0 107

s2 = s1 (Isentropic) Reversible adiabatic process is always isentropic but isentropic process need not to be reversible adiabatic. A Non-adiabatic irreversible process also can be isentropic.

ADIABATIC PROCESS: 1. Reversible dQ

ds = ( ) T

reversible

ds = 0 s2 = s1 (Isentropic)

2. Irreversible dQ

ds = ( ) + (δs)generation T

ds = + ve Note: Adiabatic process need not to be isentropic always. If adiabatic process is reversible then only it is isentropic & if it is reversible then it must be non-isentropic.  During adiabatic process, system entropy never decreases. Physical meaning of entropy: Entropy is the measure of molecular disturbance. Greater the disturbance greater the change in entropy.

Entropy principle: Entropy of isolated system never decreases. dQ

ds ≥ ( ) T

108

For isolated system, dQ = 0 ∴ ds ≥ 0 Universe is best example of isolated system. Hence, entropy of universe always increases. (ds)universe ≥ 0 (ds)system + (ds)surrounding ≥ 0 Entropy of system can increase or decrease or become constant. Entropy surrounding can increase, decrease or become constant. But entropy of universe never decreases. Note: a/c to 1st law, entropy of universe is constant and a/c to 2nd law, entropy of universe never decreases.

TEMPERATURE – ENTROPY DIAGRAM: (T-S DIAGRAM) T

S

Area of strip dA = T dS Total area below curve, 2

A = ∫1 T ds dQ

ds = ( ) T

reversible

dQ = T ds → Reversible 109

2

Total heat transfer = ∫1 T ds Area under the curve when projected on entropy axis on T-S diagram gives reversible heat transfer.

REPRESENTATION OF CARNOT CYCLE IN T-S DIAGRAM: 1

1

Qin

P

2

T1 2 T2

4

4

3

QR

S

V

Q in = T1 ds ɳ=1-

3

Q R = T2 ds QR Qin

=1-

ɳc = 1 -

T2 ds T1 ds

T2 T1

 Area under T-S cycle gives net heat transfer a/c to first law ∑ Q = ∑ W. Hence, area under cycle gives network transfer.  Area under (P-V) = Area under (T-S) Note: Second law is directional law with the help of entropy principle we can determine the direction for particular process. Isolated Q A

B TB

TA 110

(∆S)Isolated ≥ 0 (∆S)A + (∆S)A ≥ 0 −Q TA

Q[

+

Q TB

TA − TB TA TB

≥0 ]≥0

(ds)Isolated = 0

TA = TB

(ds)Isolated > 0

TA > TB

(ds)Isolated < 0

TA < TB

The process is possible or only that direction of process is possible in which either entropy of universe is constant or increasing.

COMBINED 1ST & 2ND LAW OF THERMODYNAMICS: dQ = dU + dW → 1st law for all process dW = P dV → Reversible process valid dQ = dU + P dV → Valid for reversible process ds =

dQ T

→ Valid for reversible process

dQ = T ds T dS = dU + P dV Valid for all process 1st T- dS equation This equation is valid for reversible as well irreversible process because it connects various properties. T ds = du + P dv Specific form (divided by mass) H = U + PV 111

dH = dU + P dV + V dP

(dQ = dU + P dV)

dH = dQ – V dP ⇒ dQ = dH – V dP T ds = dH + P dv 2nd T- ds equation for all process. Slope of various lines on T-S diagram PVr = C V = C T P=C

T=C S

1. Slope of isothermal process = Tan 0 = 0 2. Slope of adiabatic line = Tan 90 = ∞ 3. Slope of constant volume line: v=C ∴ dv = 0 We know, T ds = dU + P dV (dU = Cv dT)

T ds = dU

T ds = Cv dT dS dT dT dS

=

T Cv

=

Cv T

→ Slope of constant volume line in T-S diagram

112

4. Slope of constant pressure line: P=C We have, T ds = dh – V dP dP = 0 ∴ T ds = Cp dT ∴

dT dS

=

T Cp

→ Slope of constant pressure line in T – S diagram Cp > Cv

∴ Slope of constant volume line on T – S diagram > Slope of constant pressure line on T – S diagram

OPEN SYSTEM REVERSIBLE WORK: 1st law for steady flow open system: h1 +

C21 2

+ gz1 + q = h2 +

C22 2

+ gz2 + Wopen

Neglecting K.E & P.E change h1 + q = h2 + Wopen Now, q = (h2 - h1 ) + Wopen dq = dh + Wopen For reversible process, T ds = dh + Wopen ……. (1) Now, we have, T ds = dh – V dP ……... (2) From (1) & (2) 113

dWopen = - V dP Wopen = - ∫ V dP Assumption made 1. Steady flow 2. Open system 3. Neglect K.E & P.E 4. Reversible system

ENTROPY CHANGE FOR IDEAL GAS: Case 1: Temperature & volume We know, T ds = du + P dv For ideal gas T ds = Cv dT + P dv ds =

Cv T

P

dT +

T

dv

Pv = mRT ⇒ ds = Cv

P T

=

dT

R V

+R

T

dv V

Integrating S2 - S1 = Cv ln

T2 T1

+ R ln

V2 V1

Change of entropy in terms of temperature & volume

114

Case 2: Temperature & pressure T ds = dh – v dP For ideal gas T ds = Cp dT - v dP ds =

Cp

dT -

T

v

dP

T

For ideal gas Pv = mRT v

mR

T

P

∴ = ds =

Cp

dT -

T

S2 - S1 = CP ln

T2 T1

R

dP

P

+ R ln

P2 P1

Change in entropy in terms of T & P

Case 3: pressure & volume S2 - S1 = Cv ln = Cv ln

T2 T1 T2 T1

= Cv [ln

+ R ln

V1

+ (CP − Cv ) ln

T2 T1

= Cv [ln [

V2

− ln

T2

×

T1

V2 V1

V1 V2

T1

=

T2 V1

]] + CP ln

T1 V2

P2 V2

=

T2 P2 P1

P

V2

P1

V1

S2 - S1 = Cv [ln [ 2 ]] + CP ln 115

V1

] + CP ln

For ideal gas P1 V1

V2 V2 V1 V2 V1

Change in entropy in terms of P & v

ENTROPY CHANGE FOR RESERVOIR: Reservoir: 1. Source Source T1 Q1

ds =

dQ T

=

−Q1 T1

2. Sink Sink T2 Q2 dQ

ds =

dT

=

Q2 T2

ENTROPY CHANGE FOR FINITE TEMPERATURE BODY (SOLID & LIQUID): m c

Ti

Q

Tf

ds =

dQ

116

T

ds =

mc dT T

2

T

dT

i

T

∫1 ds = ∫T f mc S2 - S1 = mc ln

Tf Ti

CHANGE OF ENTROPY DUE TO MIXING OF TWO FLUIDS: m1

m2

m1 + m2

c1

c2

Tf

T1

T2

1st law ⇒ heat lost by hot fluid = Heat gain by cold fluid m1 c1 (T1 - Tf ) = m2 c2 (Tf - T2 ) Tf =

m1 c1 T1 + m2 c2 T2 m1 c1 + m2 c2

∆S = ∆S1 + ∆S2 = m1 c1 ln

Tf T1

+ m2 c2 ln

If both fluids are same, m1 = m2 = m c1 = c2 = c Tf =

T1 + T2

∆S = mc [ln

2 Tf T1

117

+ ln

Tf T2

]

Tf T2

∆S = mc [ln

T2f T1 × T2

]

Problems with Solutions: 1. 1600kJ of energy is transferred from a heat reservoir at 800K to another heat reservoir 400K the amount of entropy generated during process will be 800 K

a

1600kJ

b

400 K

∆S = (∆S)a + (∆S)b =

−1600 800

+

1600 400

=-2+4 ∆S = 2 kJ/K 2. A system of 100kg mass undergoes a process in which its specific entropy increases from 0.3 kJ/Kg K to 0.4 kJ/kg K. At same time entropy of surrounding decreases from 80kJ/K to 75 kJ/K. The process is (∆S)system = 0.1 × 100 = 10kJ/K (∆S)surrounding = -5 kJ/K (∆S)universe = 5 kJ/K ∴ (∆S)universe > 0 → Irreversible process

118

3. The heat added to a closed system during a reversible process is given by Q = αT + βT 2 where α & β are constant. The change of entropy system as its temperature changes from T1 to T2 is equal to ds =

dQ T

Q = αT + βT 2 dQ = α dT + 2βT dT dQ T 𝟐 dQ

∫𝟏

T

=

α dT + 2βT dT T 𝟐α

= ∫𝟏

T

2

dT+ ∫1 2 β dT P

S2 - S1 = α ln [ 2 ] + 2β (T2 - T1 ) P1

4. Heat loss through plane wall at 600 Watt. If inner & outer surface temperature of wall are 200 C & 50 C respectively. The rate of entropy generated within a wall is 50C

200C 600 W

(∆S)wall =

−600 293

+

600 278

∆S = 0.110 W/k 5. A steam is condensed at constant temperature of 300 C as it flows through the condenser of power plant by rejecting heat at rate of 55 M Watt. The rate of entropy changes of steam as it flows through a condenser 119

∆S =

−55 273+30

= -0.18 M Watt/K

6. Liquid water enters in an adiabatic piping system at 150 C at the rate of 8kg per second. If the water temperature rises by 0.20 C during the flow due to friction the rate of entropy generated in pipe is (∆S)universe = (∆S)system + (∆S)surrounding (∆S)surrounding = 0 → Adiabatic pipe (no heat loss) ∆S = mc ln

Tf Ti

= 8 × 4.18 × 103 × ln

0.2+273 15+273

∆S = 23 w/K 7. A reversible heat engine receives 6kJ of heat from thermal reservoir at temperature 800K & 8kJ of heat from another reservoir at 600K. if it rejects the heat to third reservoir at temperature 100K. then thermal efficiency of engine is T1 = 800 K Q1 = 6kJ

Q2 = 8kJ T2 = 600 K

HE

T3 = 100 K

1st law ∑W = ∑Q W = Q1 + Q 2 – Q W=6+8–Q 2nd law, for cycle 120

W

∮ 6 800

+

dQ T 8

600

=0 -

Q 100

=0

Q = 2.08kJ ∴ W = 11.92 ≈ 12 Now, ɳ =

W Q1 + Q2

=

12 14

= 0.85

ɳ = 85% 8. A resistor 30Ω is maintained at constant temperature of 270 C while current of 10A is allowed to flow for 1 sec. determine entropy change of resistor & universe. If the resistor initially at 270 C is now insulted & same current is passed for same time. Determine entropy change of entropy & resistor. Specific heat reservoir is 0.9kJ/kg K and mass of reservoir is 10gm. i = 10 A 30 Ω

R = 30 Ω I = 10A Time = 1sec C = 0.9kj/kg K m = 10g Case 1: Not insulated, (∆S)R = mc ln

Tf Ti

(∆S)R = 0 → (heat generated dissipated to surrounding) 121

& (no heat is contained by resistor) (∆S)universe = (∆S)R + (∆S)surrounding =0+ =

I2 × R ×time 273+27

(10)2 × 30 ×1 273+27

= 10 J/k Case 2: 30 Ω i

Fully insulated (∆S)R = mc ln

Tf Ti

Q = I2 × R × time = mc dT (10)2 × 30 × 1 = 10 × 10−3 × 0.9 × (dT) × 103 dT = 333.33 Tf - Ti = 333.33 Ti = 300K ∴ Tf = 633.33K (∆S)R = mc ln

Tf Ti

= 10 × 10−3 × 0.9 × ln = 6.7 × 10−3 = 6.7 J/k

122

633.33 300

9. 10gm of water at 200 C is converted into ice at −100 C at constant atmospheric pressure assuming the specific heat of liquid water remains constant at 4.2 J/gm K and that of ice half of the value. And taking latent heat of fusion of ice at 00 C is 335 J/gm. Calculate total entropy change of system. 200 C water → −100 C ice m = 10g Cw = 4.2 J/gm K CI = 2.1 J/gm K Latent heat = 335 J/g sensible 00C 20 C Heat loss 0

Latent

-100C

200C 1

Heat loss

00C

2 3 -100C

(-ve sign is not taken at sensible heat because formula itself is compensating) (∆S)system = (∆S)1 + (∆S)2 + (∆S)3 = mCw ln

Tf Ti

-

Latent 273+0

= 10 × 4.2 × ln

273 293

(∆S)system = -16 .09 J/K

123

+ mCice ln -

335 ×10 273

Tf Ti

+ 10 × 2.1 × ln

263 273

10. Calculate the entropy of universe as a result of following processes. (a)

A copper block of 600gm mass with CP of 150 J/K at 1000 C is placed in a lake at 80 C.

(b)

The same block at 80 C is dropped from height of 100m into the lake.

(c)

Two such a blocks at 1000 C & 00 C are joined together.

(a) Copper block

m = 600 gm E = 150 J/k

Ti = 1000 C Tf = 80 C T

(∆S)universe = (mc ln f )

Ti system(block) T

= (mc ln f )

Ti block

= 150 ln

273+8 273+100

+

+

= (∆S)surrounding(lake)

Q T

Q T

Lake is reservoir. And block temperature is going to be equal to temperature of reservoir after placed in lake. Q = mc (Ti − Tf )w = 150 × (100 – 8) = 13800J (∆S)universe = 150 ln

281 373

+

13800 273+8

(∆S)universe = 6.63 J/K 124

(b)

100

(∆S)universe = (∆S)cu + (∆S)lake = =

mgh 273+8 0.6 ×9.81 ×100 273+8

(∆S)cu = 0 because when copper block is dropped P.E is generated given to lake & there is no temperature change in copper block. (c)

Tf =

T1 + T2 2

= 50 ∆S = (∆S)A + (∆S)B = mc ln

T2f T1 T2 (273+50)2

= 150 ln [(100+273)

(273+0)

]

= 3.64 J/K 11. (a)

1 kg of water at 273K is brought into contact with a reservoir at 373K when water has reached at 373K. Find entropy change of water heat reservoir & universe.

(b)

If water is heated from 273K to 373K by 1st bringing it in contact with reservoir at 323K & then with reservoir at 373K. what will be entropy change of universe.

125

(c)

Explain how water might be heated from 273 to 373K. with almost no change in entropy of universe.

(a)

(∆S)universe = mc ln

Tf Ti

= 1 × 4.18 × ln

373 273

= 1.30 kJ/K

(Q)surrounding = mcP ∆T = 1 × 4.18 × 100 = 41.8 kJ/K (∆S)universe = (∆S)water + (∆S)lake = 1.30 = 1.30 -

dQ T 418 373

(∆S)universe = 0.179 ≈ 0.18 (b)

T

T

Ti

Ti

(∆S)universe = [mc ln ( f)] 1 + [mc ln ( f)] 2 + (∆S)surrounding 323

373

Q1

273

273

T1

323

373

209

273

273

323

= mc [ln (

) + ln (

)] -

-

Q2 T1

Q1 = mc (373 – 323) = 209 Q 2 = mc (323 – 273) = 209 (∆S)universe = mc [ln (

) + ln (

)] -

-

209 373

= 0.09 kJ/K (c)

When infinite number of reservoir are used then the change the entropy leads to zero. From (a) & (b) we conclude that when number of reservoir increases the change in entropy tends to zero.

126

ENERGY: 1. Low grade 2. High grade According to 2nd law of thermodynamics, complete conversion of low grade energy into high grade is not possible. energy

Low grade energy

Unavailable energy

max. of low grade energy which can be converted

min. of low grade energy which must be rejected

The part of low grade energy which is available for conversion is known as energy. And part of low grade energy which must be rejected known as unavailable energy. (or) The maximum amount of work that can be obtained from a system is known as energy. If system is performing on a cycle, then energy is known as available energy and for process energy is known as availability.

Available energy: The maximum amount of work that can be obtained from a cycle is known as available energy.

127

DEAD STATE: When system is in equilibrium with its surrounding then it is said to be in dead state. At dead state, pressure & temperature of system is equal to the atmospheric pressure & temperature. P = P0 = 1bar T = T0 = 300K T1 Q1 HE

W Q2

T2

To get maximum work from system, the system must be operating on cycle (Carnot cycle) Q1

1

2

T1

T2

Q2 4

3 ∆S

For given source temperature T1 and for given heat input Q1 and work output is maximum when T2 (rejection temperature of heat) is maximum. The lowest possible temperature of heat rejection is atmospheric temperature. 128

ɳ = [1 −

T2 T1

W

]=

W = Q1 [1 −

Q1

T2 T1

]

T2 = T0 ∴ Wmax = AE = Q1 [1 −

T0 T1

]

T0 = Atmospheric temperature AE = Available energy Unavailable energy (UAE)= T0 ∆S Unavailable energy is minimum amount of energy which must be rejected & the area below atmospheric temperature always represents unavailable energy in T – S diagram. For maximum available energy 1. Cycle must be Carnot 2. T2 = T0 AE = Q1 [1 − = Q1 - Q1

T0 T1

]

T0 T1

AE = Q1 - T0 ∆S

(∆S =

Q1 T1

)

AE = Q1 – UAE ∴ AE +UAE = Q1 Q input = AE + UAE UAE ⇒ Minimum heat which must be rejected. Loss of available energy (Increase in unavailable energy) When heat is transferred in finite temperature difference. 129

Whenever heat is rejected at atmospheric temperature then the energy is unavailable energy. Note: Qin

T1 T2

Qin

T0

Loss in AE or Increase in UAE

∆S ∆SS

Q in = T1 ∆S Q in = T2 ∆S ′ Increase in UAE = T0 (∆S ′ - ∆S) = T0 ( Loss of AE = Q in T0 (

Qin T2

−

T1 − T2 T1 T2

Qin T1

)

)

Heat at high temperature has greater significance 1st law quantitative law whereas 2nd law is quantitative law.

AVAILABILITY: The maximum work that can be obtained in a process. When system comes equilibrium with surrounding. T2 = T0 P2 = P0

130

Availability in Closed system:

System

Maximum work from closed system (∆S)universe ≥ 0 (∆S)system + (∆S)surrounding ≥ 0 For reversible process (∆S)system + (∆S)surrounding = 0 (∆S)system = - (∆S)surrounding ……. (1) (dQ)system = - (dQ)surrounding ……. (2) (dQ)surrounding = T0 (∆S)surrounding (dQ)system = (dU)system + (dW)system (dW)system = (dQ)system - (dU)system (W)system = - (dQ)surrounding - (U2 - U1 ) (W)system = - [T0 (∆S)surrounding − (U2 − U1 )] Wmaximum = (U1 − U2 ) - [T0 (−∆S)system ] Wmaximum = (U1 − U2 ) - [T0 (S1 − S)] Maximum work from a close system undergoing a Process

131

SURROUNDING WORK: P0

Gas

Wsurrounding = P0 (V2 - V1 ) Wuseful = Wmaximum − Wsurrounding = [(U1 − U2 ) − [T0 (S1 − S)]] - [P0 (V2 − V1 )] Wuseful = (U1 - T0 S1 + P0 V1 ) - (U2 - T0 S2 + P0 V2 ) Wuseful = ∅1 - ∅2 ∅1 = U1 - T0 S1 + P0 V1 ∅2 = (U2 - T0 S2 + P0 V2 ) ∅ = U - T0 S + PV Where ∅ is known as availability function for closed system.

MAXIMUM WORK FOR OPEN SYSTEM: 1st law for open system steady flow h1 +

c21 2

+ gz1 + q = h2 +

c22 2

+ gz2 + Wc

Neglect K.E & P.E change h1 + q = h2 + W W = (h1 - h2 ) + q Wmaximum = (h1 - h2 ) - T0 (S1 − S2 ) Maximum work from open system Wmaximum = (h1 - T0 S1 ) - (h2 - T0 S2 ) Wmaximum = φ1 - φ2 132

Where, φ = Availability function for open system φ = h - T0 S Irreversibility:

Total energy

UAE AE

use UAE full irr work

The actual work is always less than idealised reversible work and the difference of this two is irreversible. I = Wreversible - Wactual

GOUY – STODOLA THEOREM: It states that the rate of loss of available energy (rate of increase of unavailable energy) is proportional to rate increase of entropy of universe. I ∝ (∆S)universe I = T0 (∆S)universe

Second law efficiency: ɳ11 =

ɳactual ɳreversible

For power producing devices, ɳ11 =

Wactual Wreversible

Power consuming devices, ɳ11 =

Wreversible Wactual

133

Problems with Solutions: 1. 500kJ of heat is removed from a constant temperature heat reservoir maintained at 835K. the heat is received by a system at constant temperature of 720K. The temperature of surrounding is 280K. Draw the T-S diagram and calculate net loss of available energy as a result of this irreversible heat transfer. 500 kJ

835 T

720

500 kJ

280

Loss of dry energy ∆S

S ∆S’ T1 − T2

Loss of available energy = Q in T0 (

T1 T2

= 500 × 280 (

)

835 − 720

)

(835)(720)

= 26.77kJ 2. A 1000kg steel block at 1200K is to be cooled to 400K. If it is desired to use steel as source of energy. Calculate the available & unavailable energy, ambient temperature is 300K specific heat capacity of steel is 0.5kJ/kg K. Unavailable energy + Available energy = Q Q = mc ∆T = 1000 × 0.5 × (1200 – 400) = 400MJ 134

1000 kJ 1200 K

Q

400 K HE

AE

(UAE) 300

UAE = T0 (∆S)surrounding = T0 [−(∆S)surrounding ] T

= T0 [−mc ln ( f)] Ti

= 300 [−1000 × 0.5 ln (

400 1200

)]

= 164791.84 kJ UAE = 164.79 MJ ∴ AE = 400 – 164.79 = 235.21MJ Q = AE + UAE 3. In particular, chemical plant 5000kg of vapour of substance produced at 10000 C while 2nd unit is in same plant produced 5000kg of same substance at 6000 C. This two products are mixed together in insulated container & mixture is used as a source for heat engine till the temperature is reduced to 400 C. a recently recruited suggested that if two products are used separately as source for heat engine additional work can be obtained. The ambient temperature can be obtained & specific heat of product is 1kJ kg/K 135

m = 1000 Tf = 800 K

Q

400C HE

AE

(UAE) 250C

m1 = 5000kg m2 = 5000kg m = 1000kg T1 = 10000 C T2 = 6000 C Tf1 =

T1 + T2 2

= 8000 C

Tf2 = 400 C C = 1kJ Q = mc dT = 5000 × 1 × 760 = 3800MJ UAE = T0 (∆S)surrounding = T0 [−(∆S)surrounding ] T

= T0 [−mc ln ( f)] Ti

= 298 [−10000 × 1 ln (

273 + 40

)]

273 + 800

= 3671.3 MJ AE = 128.7MJ 136

m = 10000 Tf = 800 K

Q

5000 10000C

400C

Q1

HE

AE

400C

+

HE

(UAE)

5000 600

Q2

400C HE

AE1

(UAE1)

250C

(UAE2)

250C

250C

Q1 = mc dT = 5000 × 1 × 960 = 4800MJ UAE1 = T0 (∆S)surrounding = T0 [−(∆S)surrounding ] T

= T0 [−mc ln ( f)] Ti

= 298 [−5000 × 1 ln (

273 +10

)]

273 +1000

= 2090.36MJ AE1 = 2709.64 Q 2 = mc dT = 5000 × 1 × 560 = 2800MJ UAE2 = T0 (∆S)surrounding = T0 [−(∆S)surrounding ] T

= T0 [−mc ln ( f)] Ti

313

= 298 [−5000 × 1 ln (

873

)] = 1528.34MJ

AE2 = 1271 Additional work = AE1 + AE2 - AE = 52.7MJ 137

AE2

POWER PLANT ENGINEERING POWER ENGINEERING 1. Steam power plant (Rankine cycle) 2. Gas turbine plant (Brayton cycle) 3. IC Engine a. Otto cycle b. Diesel cycle c. Dual cycle 4. Hydro power plant

THERMAL CYCLE 1. Power cycle It is a thermodynamic cycle where power is produced in expense of heat. a. Vapour power cycle Working fluid changes its phase during the cycle. Example: [Rankine] b. Gas power cycle Working fluid always in gaseous phase Example: Brayton cycle, Otto, Diesel cycle

2. Refrigeration cycle Lower temperature is created and maintenance in expense of mechanical energy input.

138

GAS TURBINE POWER PLANT: Air standard cycle: 1. Air is used as working substance throughout the cycle & treated as ideal gas. 2. CP & CV do not change with respect to temperature. 3. Closed system is considered which undergoes a cyclic process, therefore there is no any intake or exhaust. 4. The combustion process replaced by equivalent amount of heat addition from external source. Thus there, is no any change in chemical equilibrium of working fluid. 5. The exhaust process is replaced by equivalent amount of heat rejection process, 6. Compression & expansion process in cycle is considered as reversible adiabatic.

Carnot cycle (Ideal air standard cycle): Source

Diathermal cover

T1 Qin Adiabatic cover QR Sink T2

139

1

1

2

Qin

P

2 4

T 4

3

QR

3

V

IDC

S

ODC

Process

du

1-2 Isothermal

0

dw P1 V1 ln

expansion

dQ V2 V1

P1 V1 ln

V2 V1

(heat add) 2-3 Adiabatic

Cv (T3 - T2 )

-Cv (T3 - T2 )

0

expansion 3-4 Isothermal

0

P3 V3 ln

compression

V4 V3

P3 V3 ln

(heat rejected) 4-1 Adiabatic

Cv (T1 - T4 )

-Cv (T1 - T4 )

0

∑ du = 0

∑W

∑Q

compression

∴ T3 = T4 , T2 = T1 ∑ du = Cv (T3 - T2 + T1 - T4 ) ∑ du = 0 → For a cycle ∑ du must be zero. 1st law, ∑ Q = ∑ W → Valid 140

V4 V3

QR

ɳ=1-

Qin V

=1-

P3 V3 ln V3 4 V

P1 V1 ln V2 1 V

=1-

RT3 ln V3 4 V RT1 ln 2

……. (1)

V1

4-1→ V4 γ−1

T1

=( )

T4

V1

1

⇒

V4 V1

=

T γ−1 ( 1) T4

=

T γ−1 ( 2) T3

2–3→ T2 T3

V3 γ−1

=( ) V2

T4 T1 V1 γ−1

( ) V4

1

V3

⇒ =

V2 T3 T2

V2 γ−1

=( ) V3

V3 V4

=

⇒

V1 V4

=

V2 V3

V2 V1

ɳ=1-

T3 T1

CLASSIFICATIONS OF GAS TURBINE POWER PLANT: Based on Brayton or joule cycle 1. Open system cycle 2. Closed system cycle

141

1. Open system Brayton cycle: 2

CC High P2, T2

Low P1, T1

Constant pressure

W

C

T

1

1

Qin

1

Qin

P

2

3

2

T

QR

4

4

3

QR

1

V

S

Air is sucked in compressor and combusted in combustion chamber using fuel and then expand in turbine after that gases are released to atmosphere. Working fluid is air only. All processes are reversible. Process

du

dw

dQ

1-2 Adiabatic

Cv (T2 - T1 )

wC = h2 - h1

0

Cv (T3 - T2 )

0

Q in = h3 - h2

Cv (T4 - T3 )

w T = h3 - h4

0

Cv (T1 - T4 )

0

Q R = h4 - h1

compressor 2-3 constant pressure heat addition 3-4 Adiabatic expansion 4-1 Constant pressure heat rejection 142

∑ du = 0 ∴ It is property ∑ du = Cv [T2 - T1 + T3 - T2 + T4 - T3 - T4 + T1 ] = 0 As du is property ∑ w = h2 - h1 + h3 - h4 ∑ Q = h3 - h2 - h4 + h1 ∴ ∑w = ∑Q It is satisfying 1st law ɳ=1=1=1ɳ=1-

QR Qin h4 − h1 h3 − h2 CP (T4 − T1 ) CP (T3 − T2 ) T T1 ( 4 − T1 ) T1 T

T2 ( T3 − T2 ) 2

Pressure ratio, γP =

P2 P1

=

P3

=

P4

After compression Before compression

Process, 1-2 T2 T1

P2

=( )

γ−1 γ

P1

γ−1 γ

= (γP )

Process, 3-4 T3 T4

P3

=( )

γ−1 γ

P4

T2 T1

=

T3 T4

143

⇒

γ−1 γ

= (γP ) T4 T1

=

T3 T2

ɳ=1-

T1 T2 1

⇒ɳ-

γ−1

(γP ) γ 1

ɳB = 1 (γP

γ−1 ) γ

When γP ↑ ɳ ↓ γ↑ ɳ↓ γ2

γ1 γ2 > γ1

η

γp

1

Limitation of open Brayton cycle, 1. In open Brayton cycle, air is used as working fluid so, gamma value is limited to 1.4 2. Only good quality fuel can be used because mixture directly go through turbine.

2. Closed gas turbine cycle (Closed Brayton cycle): 2

3

Heat exchanger

C

T

1

4 in

Cooling water

144

out

Advantages of closed Brayton cycle: 1. Low quality fuel can be used. (Solid fuel) 2. Any working fluid can be used which have higher gamma value.

ANALYSIS OF BRAYTON CYCLE: 3"

Tmax

2"

T

2' 2

Tmin

3'

3

4

1

4'

S

1

ɳB = 1 (γP

γ−1 ) γ

For any thermal system, Tmaximum is fixed value due to metallurgical condition of system and Tminimum is fixed due to surrounding condition. Carnot cycle is a reversible cycle which have maximum mean temperature of heat addition & minimum mean temperature of heat rejection (Tminimum ) that’s why Carnot cycle have maximum efficiency. For any other cycle it is a limiting efficiency. Even for Brayton cycle limiting efficiency is Carnot efficiency which is at γP = (γP )maximum 1

ɳB = 1 -

γ−1

(γP ) γ γ−1 γ

T2 = T1 (γP ) ɳB = ɳ C 145

1

1-

(γP )maximum

γ−1 γ

(γP )maximum = (

=1-

Tminimum Tmaximum

Tminimum Tmaximum

γ−1 γ

)

At maximum efficiency condition, network done is zero. When ɳ = Maximum Wnet = 0 Work will be produced but compressed work is increased. Among all the power cycles gas power cycle has minimum network rather than getting maximum efficiency. Brayton cycle is designed for maximum work. Maximum efficiency has no importance. In air crafts, open Brayton cycle is used. In Brayton cycle, T1 → Tminimum T3 → Tmaximum γ−1 γ

T2 = Tminimum (γP ) T4 =

Tmaximum γ−1

(γP ) γ

Wnet = WT - WC = (h3 - h4 ) - (h2 - h1 ) = CP [T3 - T4 - T2 + T1 ] = CP [Tmax −

Tmaximum (γP

γ−1 ) γ

γ−1 γ

− Tminimum (γP )

For (γP ) optimum value, 146

+ Tminimum ]

dWnet

=0

dγP γ−1

[0 − Tmax (

γ

γ−1

) (γP

−Tmax (

−( )−1 ) γ

γ−1 γ

) (γP ) Tmax Tmin

γ−1

− Tmin (

−2γ + 1 ) γ

(

Tmax Tmin

) (γP γ−1

= −Tmin (

−1

= (γP )

γ

γ−1

γ

( )−1 ) γ −1

) (γP )

−2γ + 1 ) γ

(γ)−(

= (γP )

(

2(γ − 1) ) γ γ

(γP )optimum =

T 2(γ − 1) ( max ) Tmin

1

= (γmax )2

(γP )optimum = √γmax

Wmax Wnet

1 γp

(γp)optimum

(γp)max

T2 for maximum work condition, γ−1 γ

T2 = T1 (γP )

γP = (γP )optimum = √γmax ∴ T2 = T1 (γP )

147

γ−1 2γ

+ C] = 0

(γ)

γ T3 2(γ−1)

T2 = T1 (( ) T1

γ−1 γ

)

T2 = √T1 T3 For maximum work, T3

T4 =

γ−1

[(γP )optimum ] γ

T4 = √T1 T3 For minimum work conditions, T2 = T4 = √T1 T3 W = CP [T3 - T4 - T2 + T1 ] For maximum work, T2 = T4 = √T1 T3 Wmax = CP [T3 - √T1 T3 - √T1 T3 + T1 ] = CP (√T3 − √T1 ) What is efficiency at Wmax 1

ɳ=1(γP

γ−1 ) γ

1

=1-

γ−1

(γPmax ) 2γ

ɳoptimum = 1 - √ In a gas cycle, Main aim is to find all temperatures In vapour cycle, Main aim is to calculate heat energy point.

148

T1 T3

2

GAS TURBINE CYCLE WITH MACHINE EFFICIENCY: 3 T

2

2'

3'

4 1 1'

4'

S

ɳC =

Wrev Wactual

=

CP [T2 − T1 ] CP [T′2 − T1 ]

T2′ =? T3 → given T3

T4 =

γ−1

(γP ) γ

ɳT = ɳT =

Wactual Wrev

CP [T3 − T′4 ] CP [T3 − T4 ]

(The change in actual curve (1′ - 2′ ) due to Internal irreversibility) T4′ =? WT = CP [T3 − T4′ ] WC = CP [T2′ − T1 ] Q in = CP [T3 − T2′ ] Q R = CP [T4′ − T1 ] ɳ=

WT − WC Qin

In pipe flow, pressure gradient is negative i.e., pressure decrease. 149

ACTUAL BRAYTON CYCLE:

T

2

3 3'

2'

4 1 1'

4'

S

Point 4 is not achieved because the pressure at 4 will be equal to 1 & no pressure difference leads to no flow hence, some pressure difference is kept. Pressure is kept minimum than P4 .

FREE SHAFT TURBINE: 2

3

CC

C

HPT 4

LPT 5

Frequency =

PN 60

f = 50Hz P = 12 N = 250rpm

150

3 2

T

WHPT WLPT 4 1 S

 WHPT = WC CP [T3 − T4 ] = CP [T2 − T1 ]  WLPT = Wgen CP [T4 − T5 ] = Wgen γP =

P2 P1

=

P3 P5

=

P3 P4

×

P4 P5

γP = (γP )HPT × (γP )LPT In free shaft turbine, the compressor is operated by high pressure turbine & generator is operated by low pressure turbine the exhaust of high pressure turbine goes directly to low pressure turbine which operates the generator.

ANALYSIS OF COMBUSTION CHAMBER:  Open combustion chamber: mf CV ɳc Air T2

(ma + mf)

CC T3

Cpa

Cpe

Energy balance equation, ma CPa T2 + mf CV ɳC = (ma + mf ) CPe T3 151

Dividing by mf ma mf

(CV = Calorific value)

CPa T2 + CV ɳC =

(ma + mf ) mf ma

(

mf

CPe T3

= Air fuel ratio)

(AFR) CPa T2 + CV ɳC = (AFR + 1) CPe T3

 Closed combustion chamber: CC mf CV ɳc ma Cpa

ma Cpa

T2

T3

ma CPa T2 + mf CV ɳC = ma CPe T3 (AFR) CPa T2 + CV ɳC = (AFR) CPe T3

REFRIGERATION IN GAS TURBINE: 6 CC 4

3

5

2 C

T

1

It is method of heat recovery. In regeneration system heat of exhaust gases is used to heat compressed air as a result sir enters in combustion chamber is at higher temperature & amount of heat supplied in combustion chamber is decreased. So, thermal efficiency

152

of system increases but there is no change in turbine & compressor work. 4 After using regenerator

Qin 3 2

5

6

Heat regenerated

Before using regenerator 1 QR

After

before

Heat recovered from exhaust gases

Heat supplied to compressed air in regenerator

In generation. The mean temperature of heat addition increase & mean temperature of heat rejection decreases so, the thermal efficiency of system increases. ɳB = 1 -

(Tm )R (Tm )C

Effectiveness of regenerator: 4 5 2

6

3

(∆T)act

(∆T)max

1

Effectiveness (e) =

153

(∆T)act (∆T)max

5 Maximum 3 temperature at point 3 for 100% 3' efficiency

6

Counter flow heat exchanger

e=

2

T3 − T2 T5 − T2

For e = 1 1=

T3 − T2 T5 − T2

⇒ T3 = T5 Heat received from exhaust gases = Heat supplied to compressed gas ma CPe (T3 − T2 ) = ma CPa (T3 − T2 ) CPe = CPa ∴ T5 − T6 = T3 − T2 & e = 1 then T3 = T5 Then, T6 = T2 For ideal regenerator, there are two conditions. 1. T3 = T5 2. T6 = T2 Regenerator are used in power plants only.

Efficiency of regenerator: Q in = h4 − h3 Q R = h6 − h1 ɳR = 1 -

QR Qin

=1-

154

h6 − h1 h4 − h3

Now, for ideal regenerator, ɳR = 1 -

h2 − h1 h4 − h5 T

=1-

T1 [(T2 ) − 1] 1

T

T4 [1 − T5 ]

…… (1)

4

T2 T1 T4 T5

γ−1 γ

= (γP )

γ−1 γ

= (γP )

γ−1

Tmin ((γP ) γ −1)

ɳIR = 1 Tmax (1−(γP

ɳIR = 1 -

Tmin Tmax

γ−1 ) γ ) γ−1 γ

(γP )

Efficiency & γP

ηSB

Simple Brayton cycle

η

P 1, 2

ηIR γp = 1 (γp)optimum

3, 4

V

Diagram ⇒ When γP = 1 &

P2 P1

=1&

P3 P2

=1

At intersection point, ɳSB = ɳIR 1

1(γP

γ−1 ) γ

=1-

155

Tmin Tmax

γ−1 γ

(γP )

Tmax Tmin

= (γP )

γ−1 ) γ

2(

γ

γP =

T 2(γ − 1) ( max ) Tmin

= (γP )optimum

At γP = γP(optimum) There is no regeneration possible whether you use regenerator or not. At low γP , Efficiency of the ideal regenerative cycle is more compared to higher value of γP . 3

T

2 4 1 S

γP < (γP )optimum ⇒ ɳIR > ɳSB γP = (γP )optimum ⇒ ɳIR > ɳSB γP > (γP )optimum ⇒ ɳIR < ɳSB

REHEATING IN GAS TURBINE CYCLE: CC

3

Reheater

5

4 2 C

HPT

LPT

1 6

156

3

Q2

Q1 2

T

4

Q1

5 P 2

3

Q1 5

6 4

1 6 Extra work due to reheat

1

V

S

In reheating, the expansion is done in two stages, with intermediate reheat. With only reheating efficiency of cycle decreases. In reheat cycle, scope of regeneration increases. So, reheat cycle with regeneration efficiency must be increased.

INTERCOOLING IN GAS TURBINE: Inter cooler

3

4

C.C.

5

2

W

LP

HP

C1

T

C2

1 6

5 4 T

4 6

2 3

3

5 2

P 6

1

1 S

V

157

5 (∆T)S

4'

(∆T)IC

4

T

2 3

6

1 S

Intercooling is used to decrease the compression work. In case of intercooling mean temperature of heat addition decreases whereas mean temperature of heat rejection is same. So, the efficiency of cycle is decreases. In intercooling, the scope regeneration increases so, if we use intercooling with regeneration efficiency will increase, intercooling is used at high pressure ratio.

BRAYTON CYCLE WITH MANY STAGE OF INTERCOOLING, REHEATING & PERFECT REGENERATION: P2 P1

T

S

For one intercooler, reheater & generator Qin

Th Q1 P = C

P=C Q2

T T

QR

L

S

158

For perfect regenerator, Q1 = Q 2 ɳ=1-

QR Qin

=1-

ɳ=1-

TmR Tma

TL Th

Brayton cycle with large no. of intercooler & reheater approximated to Erikson cycle. In Erikson cycle, two process are isothermal & two are constant pressure. And mean temperature of heat addition is equal to Carnot & maximum & mean temperature of heat reject is also equal to Carnot & minimum so, the efficiency of Erikson cycle is maximum & equal to Carnot. A regenerator is a type of heat exchanger which never be 100% so, the efficiency of Erikson cycle is always less than Carnot. Note: ɳCarnot = ɳsterling = ɳErikson

STERLING CYCLE: Qin 4

3 V=C

V=C 2 QR

1

In sterling cycle, mean temperature of heat addition is maximum and equal to Carnot & mean temperature of heat rejection & equal to Carnot. So, efficiency of sterling cycle & Carnot cycle is same. ɳsterling = 1 159

TL Th

IC ENGINE CYCLES ENGINE: Any form of energy is converted into mechanical energy. Heat engine {Heat → Work} 1. I.C Engine 2. E.C Engine Example: Steam engine Closed gas engine

I.C ENGINE Chemical energy of fuel

Combustion

Heat

Expansion

TDC L = stroke length BDC L = 2r D → Bore L → Stroke π

Vs = D2 L 4

160

Mechanical energy

D VC TDC L

VS BDC

r

r

r

CLASSIFICATIONS OF I.C ENGINE 1. 4-Stroke a. SI (Petrol) b. CI (Diesel) 2. 2-Stroke a. SI (Petrol) (Not used due to pollution) b. CI (Diesel)

Volume efficiency: ɳv = = =

ma mswept ma ρa Vs mass of air that is actually injected Mass associated with swept volume

Ratio of actual mass of air to the mass associated with swept volume. 161

VALVE TIMING DIAGRAM FOR 4-STROKE ENGINE: EVC

spark

compression

IVO suction

expansion

expansion

For Ideal

IVC EVO IVO TDC

EVC For actual

BDC IVC

EVO

The angle of IVo , EVo , EVs , IVs are difference for different engine.

TWO STROKE ENGINE: TDC

BDC

162

EPC IPC

IPO

EPO

Volume efficiency is greater than 1.

Super charger: It is used to increase ɳv

SC

Turbocharger: Compressed air

Exhaust gases

C

T

Air

AFR =

ma mf

i. Stoichiometric mixture Minimum amount of air required to burn fuel completely. Generally, it is (15:1). 163

ii. For rich mixture (3 to 14:1) It is used for starting, cold starting, Acceleration & Power. iii. Lean mixture (Ratio more than 15) Rich Lean

Strich

If we use more lean mixture, more power will be lost in exhaust.

AIR STANDARD OTTO CYCLE: (SI Engine) (Petrol) 3 Qin

Expansion

Heat add 2 Compression

P

Heat rejection

Exhaust 0

1 Power consuming anticlock wise

Suction V

IDC

ODC 3

T

Power producing clock wise 4 QR

Qin 2 V=C 4 V=C

QR

1 S 164

ɳ=1-

QR Qin

Q in = mCv (T3 - T2 ) Q R = mCv (T4 - T1 ) ɳo = 1 -

T4 − T1 T3 − T2 T

ɳo = 1 -

1 T3 T2 (T 2

− 1)

Volume before compression

Compression ratio =

Volume after compression

r= r=

T1 (T4 − 1)

Vs + Vc Vc V1 V2

=

V4 V3

Isentropic equation 1-2, T2 T1

V1 γ − 1

=( ) V2

= (r)γ − 1 ……. (1)

Constant volume heat addition, 2-3 ……. (2) T3 T4

V4 γ − 1

=( ) V3

= (r)γ − 1 γ2

ɳ0

γ1

γ

γ=1

From (1) & (2) ⇒

T4 T1

=

165

T3 T2

ɳo = 1 -

T1 T2 1

= 1 - (r) γ − 1 r↑ ɳo ↑ γ ↑ ɳo ↑

Problems with Solutions: 1. For an engine operating on air-standard Otto cycle clearance volume is 10% of swept volume & specific heat ratio of air is 1.4. Find air standard efficiency. r=

Vs + Vc Vc

=

Vs + Vc Vc

=

1 + 0.1 0.1

1

=

1.1 0.1

= 11

1

∴ ɳo = 1 - (r)γ − 1 = 1 - (11)1.4 − 1 ɳo = 0.616 ɳo = 61.6% For SI Engine, 9 ≤ r ≤ 12 → To avoid knocking r is always less than 12. For CI Engine, 16 ≤ r ≤ 24 2. An engine working on air standard Otto cycle has cylinder diameter 10cm. & stroke length 15cm. the ratio of specific heat for air is 1.4. Clearance volume is 196.3 CC & the heat supplied per kg per cycle is 1800kJ/kg. then work output per cycle per kg of air is D = 10cm L = 1.5cm 166

γ = 1.4 Vc = 196.3 CC Q in = 1800kJ/kg ɳ=1-

QR Qin

π

Vs = × 102 × 15 = 1178.09 4

r=

1178.09 + 196.3 196.3

r = 7.001 1

ɳ = 1 - (r)γ − 1 1

= 1 - (7)0.4 = 0.54 0.54 =

Wout Qin

=

Wout 1800

Wout = 973.5kJ 3. In an air standard Otto cycle, the maximum and minimum temperature 14000 C and 150 C the heat supplied is 1800kJ. Calculate compression ratio cycle efficiency. Also calculate maximum to minimum pressure ratio in cycle. 3

400 800kJ T

2 4

1 S P3 P1

=

167

P3 P4

×

P4 P1

T1 = 15 + 273 T3 = 1400 + 273 Q in = 800kJ T2 = T1 (r)γ − 1 Q S = Cv (T3 - T2 ) T3 = 1400 800 = 0.178 (273 + 1400 - T2 ) T2 = 559.7K 1

r=

T γ−1 ( 2) T1 559.7

=(

1 1.4 − 1

)

15+273

= 5.4

1

ɳ = 1 - (5.4)0.4 = 48.46% P3 P1

=

P3 P2

×

P2 P1

1-2 ⇒ T2

P2

γ−1 γ

=( )

T1

P1

P2 P1

= 10.23

2-3 ⇒ V=C P3 P2 P3 P1

=

T3 T2

=

1400 + 273 559.7

= 2.98

= 10.23 × 2.98 = 30.57

168

DIESEL CYCLE: Qin 3

2

4

P

QR 1 V 3 T

P=C 2 4 V=C 1 S

Q in = mCp (T3 - T2 ) Q R = mCv (T4 - T1 ) r=

Volume before compression Volume after compression

r=

V1 V2

For heat addition, constant volume is better than constant pressure. Firstly, systems of cycle were made and then to analyse cycles were made ɳ=1ɳ=1=1-

QR Qin

Cv (T4 − T1 ) Cp (T3 − T2 ) 1 (T4 − T1 ) γ (T3 − T2 )

169

T

ɳD = 1 -

4 1 T1 (T1 − 1)

γ T2 (T3 − 1) T 2

ɳD = 1 -

γ

1 γ(r)

γ−1 [

rc =

rc − 1 rc − 1

]

V3 V2

4. A diesel engine has compression ratio of 17 and cut off take place at 10% of stroke assuming γ = 1.4 the air standard efficiency is rc ⇒ V3 - V2 = 0.1 (V1 - V2 ) r = 17 r = 1.4 V3 - V2 = 0.1 (V1 - V2 ) V3 V2

V1

– 1 = 0.1 (

V2

− 1)

rc – 1 = 0.1 (r – 1) = 0.1 (17 – 1) rc = 2.5 ɳD = 1 -

1 1.4 (17)1.4 − 1

×

[2.61.4 − 1] [2.6 − 1]

ɳD = 0.59 5. In CI engine, the inlet air pressure is 1bar and pressure at end of isentropic compression is 32.42 bar, the expansion ratio is 8, assuming γ = 1.4. Find the efficiency in percentage. P1 = 1bar P2 = 32.42 bar Expansion ratio = 170

V4 V3

=8

rc = rP = V3 V2

=

V3

×

V4

P2 P1 V1 V2

V3 V2

= 32.42 11.99

=

8

= 1.5

1

r=

V1 V2

=

P γ ( 2) P1

ɳD = 1 -

1

= (32.42)1.4 = 11.99

1 1.4 (11.99)1.4 − 1

×

[1.51.4 − 1] [1.5 − 1]

= 0.595 ɳD = 59.5%

COMPARISON OF OTTO & DIESEL CYCLE: Case 1: For same compression ratio & same heat input

T

2' 2 1

3 3'

V=C

4

P=C

1'

4'

V=C

S

T3 > T3′ as temperature at constant volume is greater than temperature constant process. 3 P 3'

2 2'

4' 4

1 1' V 171

T2 = T1 (r)γ−1 ɳ=1-

QR Qin

(Q R )D > (Q R )o ∴ ɳD > ɳ o

Case 2: For same compression ratio & same heat rejection. 3 3

3'

T

2

4

3'

P

4'

2' 4

2'

4'

2

1' 1

1'

1 S

V

(Q in )D > (Q in )o ɳD > ɳ o ɳ=1-

QR Qin

Case 3: For same maximum temperature & same heat rejection. 3

3

3'

3'

P=C

T

2'

2

P V=C

2'

4

4 4'

2

4' 1' 1'

1

1 S

V

(Q in )D > (Q in )o ɳD > ɳ o 172

If compression ratio is same then ɳo > ɳD otherwise ɳD > ɳo , In general, ɳD > ɳo → (Because of high compression ratio)

MEAN EFFECTIVE PRESSURE: Specific fuel consumption: (S.F.C) =

Mf Power output

S.F.C.

N

Variation of S.F.N with respect to rpm. For Otto cycle, 3 P

4

2

Pm 1

V2

V

V1

V2

V1

Wnet = Pm (V2 - V1 ) It is seen that the pressure inside the cylinder is continuously varying, during a cycle. Mean effective pressure is an imaginary & constant pressure defined in such a way that Wnet / Cycle = Pm (V1 - V2 ) V1 - V2 = Swept volume = VS 173

Wnet / Cycle = Pm VS = Pm A × L Power =

Wnet Cycle

Power =

×

Cycle Sec

Pm A L × n 60

ɳ = N → 2 Stroke N

ɳ=

2

→ 4 Stroke

Problems with Solutions: 1. A 4-Stroke diesel engine, has displacement volume of VS = 0.0259 m3 . The engine has output of 950KW at 2200 rpm. Find mean effect pressure in M pa. N

Power = 950 =

Pm A L × 2 60

Pm (0.0259) L ×

2200 2

60

Pm = 2000 K pa Pm = 2 M pa 2. In an Otto cycle, compression ratio is 10. The condition at the beginning of compression process is 100 K pa & 270 C. Heat added at constant volume is 1500kJ/kg. While 700kJ/kg of heat is rejected during other constant volume process in the cycle. Gas constant R is 0.287 kJ/kg K. Find Pm in K pa. r = 10 P1 = 100 K pa T1 = 300K 174

Q in = 1500 Q R = 700 kJ/kg R = 0.287 ∴ CP - C V = R Power = Wnet = 1500 – 700 = 800 kJ/kg P1 V1 = m R T1 100 V1 = m × 0.287 × 300 V1 = 0.861 m3 /kg V2 V1

=

1 10

⇒ V2 = 0.0861 m3 /kg

∴ 800 = Pm × (V1 - V2 ) 800 = Pm × (0.861 - 0.0861) Pm = 1032 K pa 3. SI engine working on Otto cycle, has compression ratio 5.5. The work output per cycle (Area of P-V diagram) is equal to 23.625 × 105 Vc J. Where VC is clearance volume in m3 . Find the mean effective pressure Pm . W = 23.625 × 102 VC kJ r=

V1 V2

= 10 =

Vs + Vc Vc

= Vs = 4.5 Vc

23.625 × 102 VC = Pm × (V1 - V2 ) 23.625 × 102 VC = Pm × 4.5 Vc Pm = 0.525 M pa

175

PROPERTIES OF PURE SUBSTANCE: Pure substance: Homogeneous & invariable chemical composition irrespective of phase. Heat 1. Sensible heat (Heat required to increase the temperature of body) Q = mc ∆T 2. Latent heat (Heat required to change the phase) Q = mL

Phase transformation of pure substance (water): Pressure ↑ Boiling point ↑ Latent heat ↓Pressure ↓ Melting point ↑ Critical point

P2 = C P1 = C

L L+V -

+

P = Patm V

T 1000C

Saturated Vapor line

Saturation liquid line

250C

P2 > P1 > P h Patm

P = W/A

W W PW Water

LH = 2257 kJ/kg 1000C Water 1000C Steam LH = 335 kJ/kg 00C Water

0

0 C ice

176

 Critical point is a point where saturated liquid line meets saturated vapour line.  With the increase of pressure boiling point increases and latent heat of vaporization decreases.  At critical point, latent heat of vaporisation is zero.  Slope of saturated liquid line is positive, slope of saturated vapour line is negative & at critical point slope of vapour dome is zero. P2 P P1

TSat1

TSat2

T

Pressure at critical point Pcritical = 221.2 bar Tcritical = 374.50 C Saturation curve P1

L P2

T P1 > P2

P

P2

2

3

P1

1

∆VL

S

V

∆VV V

In case of liquid, change in volume is negligible as it is treated as incompressible (∆VL ). Which change in volume for gases is high as it is compressible 177

(∆VV ). Water, antimony & Bismuth T↑ V↓ P = Patm P1 P1 < P2 T

Triple point line

P

0

Minimum 4 C volume 00C temp -350C

1

V

 At 40 C water has minimum volume and we know that density is reciprocal of specific volume so, density is maximum at 40 C. Fusion curve Vaporizatio L n curve

S

L

G

+ + Triple point

Ptripple + S

G

Sublimation curve

Normal substance

P L + Triple point of water

0.006 bar S

+ V 273.16 K T

178

Note: Only three substances (Water, antimony and bismuth) have negative slope of fusion curve whereas all other substance have +ve slope of fusion curve. If any substance heated below its triple point pressure, then solid become vapour directly & process is called sublimation.

VARIOUS REGIONS IN TEMPERATUREENTROPY DIAGRAM: 1. Subcooled region: Tact < Tsat Degree of sub cooling = (Tsat - Tact ) T Tsat

T1

Sub cooled region

1

S

2. Superheated region: Tact Tsat

Tact > Tsat

Tact > Tsat Degree of superheating = (Tact - Tsat ) 179

3. Wet region:

Tact = Tsat Steam Water

MV ML

For wet region, Tsat = Tact It is a region where liquid & gas both exist in equilibrium.

Dryness fraction (x): It is defined as the ratio of mass of vapour to the total mass of mixture. Mass of vapour

x=

Total mass

x=

mV mV + mL

Note: Dryness fraction of saturated liquid line = 0 Dryness fraction of saturated vapour line = 1

SPECIFIC VOLUME OF WEIGHT STEAM: V = VL + VV mv = mL vL + mV vV v= v= mL m

mL vL + mV vV

mL m

m

vL +

=1–x

mV m mV m

vV =x

∴ v = (1 – x) vL + x vV 180

v = vL + x (vV - vL ) v = vf + x (vg - vf ) This equation is valid only in wet region. Similarly, S = Sf + x (Sg - Sf ) u = uf + x (ug - uf )

ENTHALPY AT VARIOUS POINTS: Case 1: When point is on saturated vapour curve.

T

1 ∆h = latent heat ∆h hf

hg

h

∆h = Latent heat h1 = hg = hf + ∆h LH = hg - hf

Case 2: When point is on wet region. 1 T

h

181

∆h = latent heat h1 = hf + x (hg - hf ) h1 = hf + x L.H

Case 3: When point is on superheated region. 1

T Tsup

Tsat ∆h

In this region gas act as ideal gas

∆h = CPV (Tsup – Tsat) hf

hg

h1

h

h1 = hf + ∆h h1 = hg + CPv (Tsup - Tsat )

Case 4: When point is on subcooled region. T Tsat 1 T1

h1

hf h

h1 = hf - ∆h h1 = hf - Cw (Tsat - T1 ) Cw = C of water = 1.48 kJ/kg 182

ENTROPY OF VARIOUS POINTS: Case 1: When point is on saturated vapour curve. 1 T

∆S Sf

Sg

S

S1 = Sg = Sf + ∆S S1 = Sg = Sf + L.H Tsat

L.H Tsat

= Sg - Sf

Case 2: When point is on wet region.

1 T

Sf S

Sg

S1 = Sf + x (Sg - Sf ) S1 = Sf + x [

183

L.H Tsat

]

Case 3: When point is on superheated region. Tsup T Tsat

∆S Sf

Sg

S

S1 = Sg + ∆S For ideal gas ∆S = CP ln

T1 Tsat

– R ln

∴ S1 = Sg + CP ln

Pf Pi

Tsup Tsat

Case 4: When point is on subcooled region.

T Tsat

1

T1 ∆S Sf S

Sg

S1 = Sf - ∆S S1 = Sf – C ln

Tsat T1

(-ve sign is already considered) S1 = Sf − C ln 184

Tsat T1

MOLLIER’S CHART: Mollier chart (h-S diagram)

1. Complete Mollier chart SVL

x=1 x = 0.9 x = 0.8 x = 0.7 x = 0.6 x = 0.5 x = 0.4 x = 0.2 x=0

h SL L

S

We generally use steam having, x = 0.8

2. Industrial Mollier chart

h

x=1 x = 0.9

x = 0.6

x = 0.7 S

P1 P2 T1 = C T2 = C x=1

T1 > T2

h

S

T ds = dh – V dP dh dS

P=C

=T

Slope of constant pressure line in h-S diagram 185

Note:  During phase change, constant temperature lines and constant pressure lines are identical.  The slope of constant pressure line is constant within the vapour dome & increasing in super-heated region on h-S diagram.  Constant pressure lines are diverging in superheated region.

Reference point for steam table: For steam table, reference point is triple point of water. Internal energy and entropy of saturated water at triple point is arbitrarily taken as zero. (Uf )tp = 0 (Sf )tp = 0 (hf )tp = (Uf )tp + Ptp Vtp

DETERMINATION OF DRYNESS FRACTION (X): 1. Throttling calorimeter: T1

1

2 T2

Steam main (Wet steam) P1

2

1

P2

186

P1

P2 Superheat region

h

1

2 P2 < P1

Wet region

T2 < T1 S

Wet steam is throttled and it became superheated steam after throttling. So, by measuring pressure and temperature the final enthalpy can be find out and in throttling process, final enthalpy & initial enthalpy is same. So, with help of h2 we can find the dryness fraction of h2 . h1 = hf + x (hg - hf ) h2 = hsup @ P2 & T2 h1 = h2 hf + x (hg - hf ) = hsup @ P2 T2 This method is only used for high quality steam. i.e., x > 0.9

2. Separating & throttling calorimeter: 1 Steam main

2

3

T2

m2 P2

m1 m2

187

P1

P2

h

3

2 1

S

m = m1 + m2 h2 = hf + x2 (hg - hf ) h3 = hsup @ P2 & T2 h2 = h3 x2 =? x1 = = x2 =

Mass of vapour at 1 Total mass (m1 + m2 ) Mass of vapour at 2 (m1 + m2 )

Mass of vapour at 2 m2

……. (1)

……. (2)

From (1) & (2) x1 =

x 1 × m1 (m1 + m2 )

3. Electrical calorimeter: T2 1

2

Steam main P2

P1 Q

m

188

h1 = hf + x (hg - hf ) …… (a) P1 mw

m=

time

h2 = hsup @ P2 & T2 mh1 + Q = mh2 h1 =? x =? h1 = hf + x (hg - hf )

THERMODYNAMIC RELATIONS Principle of exact differential: Let z is function of x & y z = f(x, y) ∂z

∂z

dz = ( )

∂x y=constant

dx + ( )

∂y x=constant

dz = M dx + N dy ∂z

∂z

∂x y

∂y x

M=( ) N=( ) ∂M

∂

∂y x

∂y

(

) =[

∂z

∂M

(

(( ) )] ∂x y

) =(

∂y x ∂N

)

∂x ∂y y,x

( ) =( ∂x y

∂2 z

∂2 z

)

∂x ∂y x,y

If z is exact differential then, ∂N

∂M

∂x y

∂y x

( ) = (

)

These properties are exact differential. 189

x

dy

Maxwell’s relation: 1. T ds = du + P dv du = T ds – P dv Compare with, dz = M dx + N dy M=T

-P = N

ds = dx

dz = du

dv = dy

We know that u is the property. So it is satisfying condition of exact differential. So,

∂T

∂P

∂v S

∂S V

( ) = ( )

This is 1st Maxwell equation. 2. We know, T ds = dH – v dP dH = T ds + v dP Compare with dz = M dx + N dy ∴M=T

S=x

N=v

ds = dx

dy = dP

we know H is property, so,

y=P

∂T

∂P

∂v S

∂S V

( ) = ( )

This is 2nd Maxwell equation.

190

3. From Gibb’s function: G = H – TS dG = dH – T dS – S dT dG = v Dp – S dT

(dH – T ds = v dP)

M = v; dx = dP; N = -S dy = dT So, we know that G is a property, ∂v

∂s

∂T P

∂P T

( ) = ( )

This is 3rd Maxwell equation. 4. Helmholtz function: F = U – TS dF = dU – T ds – S dP = - P dV – S dP M = -P; dx = dv; dz = dF N = -S; dy = dP We know F is a property, ∂P

∂s

∂T V

∂v T

( ) = ( )

This is 4th Maxwell equation. Note: S

P

T

(

191

−dS dP

∂V

) = ( ) T

∂T P

P

S

T

∂v

∂P

∂S P

∂S S

( ) = ( )

Partial derivative relation (Cyclic relation): Let x, y, z having a functional relation in such a way that f(x, y, z) = 0. Among the variable x, y, z any one of two others. x = ∅ (y, z) y = ∅ (z, x) z = ∅ (x, y) x = ∅ (y, z) ∂x

∂x

∂y z

∂z y

dx = ( ) dy + ( ) dz ……… (1) z = ∅ (x, y) ∂z

∂z

∂x y

∂y x

dz = ( ) dx + ( ) dy ……… (2) Put (2) in (1) ∂x

∂x

∂z

∂z

∂y z

∂z y

∂x y

∂y x

dx = ( ) dy + ( ) [( ) dx + ( ) dy ] ∂x

∂x

∂z

∂y z

∂z y ∂y x

∂x

∂z

dx = [( ) + ( ) ( ) ]dy + ( ) ( ) dx ∂z y ∂x y

∂x

∂x

∂z

∂y z

∂z y ∂y x

dx = [( ) + ( ) ( ) ]dy + dx ∂x

∂z

∂x

( ) ( ) =-( ) ∂z y ∂y x

192

∂y z

∂y

∂x

∂z

∂x z

∂z y ∂y x

( ) ( ) ( ) = -1 → Cyclic reaction Among the thermodynamic variables P, V and T this relation can be ∂P

∂V

∂T

∂V T

∂T P

∂P V

∂P

∂V

∂T

∂V T

∂T P

∂P V

written as ( ) × ( ) × ( ) = -1 ( ) × ( ) × ( ) = -1 This is valid for all type of equations & processes. For ideal gas, P=

nRT V nRT

V=

P

T= −nRT V2

×

nR P

×

PV nR

V nR

=

−nRT PV

⇒1

THERMODYNAMIC RELATIONS T – ds equation: 1. S = f(T, V)

T V=C

S ∂S

∂S

∂T V

∂V T

ds = ( ) dT + ( ) dV ∂S

T ds = Cv dT + T ( ) Dv ∂V T

193

∂P

By Maxwell relation ( )

∂T V ∂T

T

∂S V

Cv

Slope of V = C line = ( ) = ∂S

∴ Cv = T ( )

∂T V

∂P

T ds = Cv dT + T ( ) dV → 1st T dS equation …. (1) ∂T V

2. S = f(T, P) T V=P

S dS

dS

dT P

dP T

ds = ( ) dT + ( ) dP dS

dS

dT P

dP T

T ds = T ( ) dT + T ( ) dP dS

dV

dP T

dT P

( ) =( ) ∂T

Slope of ( )

∂S P

P = C line =

T CP

∂S

CP = T ( )

∂T P ∂S

T ds = CP dT + T ( ) dP ∂P T

∂V

By Maxwell relation = ( )

∂T P

∂V

T ds = CP dT – T ( ) dP → 2nd T dS equation …. (2) ∂T P

194

Equating (1) & (2) ∂P

∂V

∂T V

∂T P

Cv dT + T ( ) dV = CP dT – T ( ) dP ∂P

∂V

∂T V

∂T P

∂P

∂V

∂T V

∂T P

(-Cv + CP ) = T [( ) dV + ( ) dP] (CP - Cv ) dT = T [( ) dV + ( ) dP] dT =

T (CP − Cv )

∂P

∂V

∂T V

∂T P

[( ) dV + ( ) dP]

T = f(P, V) ∂T

∂T

∂P V

∂V P

dT = ( ) dP + ( ) dV ……. (3) T (CP − Cv )

∂P

∂T

∂T V

∂V P

( ) =( )

[(CP − Cv ) = R is valid for ideal gas only] ∂P

∂V

CP − Cv = T ( ) ( )

∂T V ∂T P

From cyclic relation, ∂V

( )

∂T P=Constant

∂V

∂P

∂P T

∂T V

= [− ( ) × ( ) ] 2

∂P

CP − Cv = -T ( )

∂T V 2

∂P

CP − Cv = -T ( )

∂T V

∂V

( )

∂P T

∂V

( )

∂P T

This relation is valid for all. …… (4) CP − Cv = +ve CP > Cv γ=

CP Cv

>1

Hence, CP is always greater than Cv For ideal gas, 195

PV = m RT mR 2 −mRT

CP − Cv = -T(

) (

V

m2 R2

=T(

V2

)

P2 mRT

)(

P2

)

CP − Cv = mR When ⇒ m = 1 Defined for unit mass (Specific properties) ∴ CP − Cv = R Volume expansivity: It shows the variation of volume with respect to temperature under isobaric condition. 1

∂V

V

∂T P

α= ( ) Isothermal compressibility:

It shows the variation of volume w.r.t pressure under isothermal condition. 1 ∂V

β=- ( )

V ∂P T

Isentropic compressibility: It shows the variation of volume w.r.t pressure under adiabatic condition. 1 ∂V

KS = - ( )

V ∂P S

From (4) ∂P

CP − Cv = -T ( )

∂T V

196

2

∂V

( )

∂P T

Cyclic relation, ∂P

∂V

∂T

∂V T

∂T P

∂P V

( ) ( ) ( ) = -1 ∂P

∂P

∂V

∂T V

∂V T ∂T P

( ) =-( ) ( ) ∂P

2 ∂V

∂V

CP − Cv = -T {( ) ( ) } ( ) ∂V T ∂T P

∂P T

∂V 2

∂P

CP − Cv = -T {( ) ( ) } ∂V T ∂T P

From co-relations, ∂V

( ) = αV ∂T P ∂P

−1

∂V P

αV

( ) = CP − Cv = -T {

−1 βV

CP − Cv = T( CP − Cv =

(αV)2 }

α2 V β

)

T α2 V β

Energy equation: We know, T ds = dU + P dV dU = T ds – P dV ∂P

= [Cv dT + T ( ) dV] – P dV ∂T V

∂P

∴ dU = Cv dT + [T ( ) − P] dV → For all gaseous. ∂T V

For ideal gas, dU = Cv dT

197

PV = mRT ⇒ P = mR

dU = Cv dT + [T (

V

mRT V

) − P] dV

∴ dU = Cv dT We know, T ds = dH – V dP dH = T dS + V dP ∂V

= [CP dT − T ( ) dP] + V dP ∂T P

∂V

dH = CP dT − [T ( ) − V] dP ∂T P

For ideal gas, dH = CP dT

JOULE – THOMSON (OR) JOULE – KELVIN EFFECT: T1

T2

P2

P1

Steady flow, h1 +

C21 2

+ gz1 + Q = h2 +

C22 2

+ gz2 + WCv

h1 = h2 (Q = 0 and WCv = 0)

198

Heating region Cooling region + T 1' 2'

2 -

1

h6 h5 h4 h3 h2 h1

Inversion curve P

Inversion curve: This curve decides whether the fluid will cool or heat after throttling. Example: Expansion process 1-2 → Temperature increases 1′ -2′ → Temperature decreases

Slope of isenthalpic curve: ∂T

μ=( )

∂P h

μ = Joule-kelvin coefficient μ = +ve → Cooling μ = -ve → Heating μ = 0 → Same temperature For maximum cooling effect, throttling must start from inversion point.

199

Joule-Thomson coefficient [(Slope of isenthalpic curve) on the T-P diagram is positive in cooling region. and its negative in heating region]

CLAUSIUS - CLAYPERON EQUATION: S = f(T, V) ∂S

∂S

∂T V

∂V T

dS = ( ) dT + ( ) dV During phase change, T = C; P = C; V↑ ∂S

ds = ( ) dV ∂V T

By Maxwell equation, ∂S

∂P

∂V T

∂T V

( ) =( ) ∂P

ds = ( ) dV ∂T V

∂P

dS

∂T V

dV

( ) = During phase change, dP

( )

dT Saturated

=

Sg − Sf Vg − Vf

dP

Vg − Vf = ( )

dT Saturated

ds = Sg − Sf =

(Sg − Sf )

dQ T h g − hf TSaturated

dT

Vg − Vf = ( )

[

hg − hf

dP Saturated TSaturated

200

]

VAPOUR POWER CYCLE: (Steam Power Plant) Qin

3 B

[Adiabatic] W T

2 4 P WP

Condenser 1

QR

IDEAL VAPOUR POWER CYCLE: (Carnot Vapour Power Cycle) Qin Th

2

3

T TL

4

1 QR S

1-2 → Adiabatic Compression 2-3 → Constant temperature heat addition 3-4 → Adiabatic Expansion 4-1 → Constant temperature heat rejection ɳC =

Th − TL Th

In this cycle efficiency does not depend of the working fluid. The efficiency of Carnot cycle does not depend on type of working fluid. For gas power cycle, it is very difficult to achieve constant temperature heat addition process and constant temperature heat

201

rejection process. But, it is quite easy to achieve for two phase system so, process 2-3 and 4-1 is easy to achieve in Carnot vapour cycle. In adiabatic expansion (3-4) the turbine has to handle low quality steam which is major cause of corrosion and wear. Which leads to cavity formation. It is not practical to design compressor which handle two phases of fluid simultaneously. In 2-3 process, temperature beyond 3 can’t be increased and will move to superheated region & temperature will increase but 2-3 is constant temperature process.

RANKINE CYCLE: 3 3 Qin

Qin

2

2

T

T 4

1

4

1

QR

QR

S

S

1-2 → Adiabatic compression in pump 2-3 → Constant pressure heat addition process (Boiler) 3-4 → Adiabatic expansion (Turbine) 4-1 → Constant pressure heat rejection (Condenser) If you don’t have time take pump work WP = 5

202

Qin 3

2

Qin

Ph

3 Ph

2 P

h 1

PL

4 PL

1

4

QR

QR V

S

Analysis of simple Rankine cycle: Whole cycle is closed system. Process 1-2: (Reversible adiabatic compression in pump) 1st law (Steady flow energy equation): Open system → Compressor h1 +

C21 2

+ gz1 + q = h2 +

C22 2

+ gz2 + Wcv

Neglect change in K.E & P.E WP = h2 - h1 h1 = hf @ L.P (Condenser pressure)  If we neglect pump work, h1 = h2  If we consider pump work, WP = ∫ v dP = v1 [P2 - P1 ] vf @ L.P Process 2-3: Constant pressure heat adds pressure (Boiler) Q in = h3 - h2  If point 3 is in saturated vapour line, h3 = hg @ H.P (Boiler pressure) 203

 If point 3 is in superheated region. h3 = hsup @ H.P & Tsup & h3 = hg + CP (Tsup - Tsat ) Process 3-4: Reversible adiabatic expansion (Turbine) By 1st law SFEF WT = h3 - h4  If point 4 is on saturated vapour line. h4 = hg @ L.P (C.P)  If point 4 is in wet zone, S8 = S4 S8 sup @ H.P & Tsup S4 = Sf + x (Sg - Sf ) @ L.P (C.P) x =? h4 = hf + x (hg - hf ) @ L.P (C.P) C.P ⇒ Compression precision Process 4-1: Constant pressure heat rejection (Condenser) Q R = h4 - h1 ɳR = ɳR =

WT − WP Qinp

(h3 − h4 ) −(h2 − h1 ) (h3 − h2 )

ɳCarnot > ɳRankine (Mean temperature heat addition is less than Carnot) 204

Mean temperature heat rejection is equal to Carnot. ɳRankine is more in all cycle.

Specific steam consumption: It is mass of steam per unit power SSC =

ms Poutput

SSC of Rankine is less than SSC of Carnot hence, Rankine is used in all steam power plant for same power, Carnot required more steam than Rankine. SSC decides the size of steam power plant.

Method to improve efficiency of Rankine cycle: 1. Increasing boiler pressure: BP2 > BP1 Tmean (heat addition) ↑ η ↑ quality of steam ↓ BP2 3' 3 2' T

2

1

4'

4

S

BP2 > BP1 Tmean (Heat addition) ↑ ɳ↑ Quality of steam ↓ 205

BP1

2. Decreasing condenser pressure: 3

T

PC1

2

PC2 4

1

X=1 X = 0.5

S

PC1 > PC2 Tmean (heat rejection) ↑ η ↑ quality of steam ↓ In condenser there will be leakage problem due to air entrance. If air entrance condenser it will go to pump & pump will stop working. Pump can’t deal with gaseous. If we decrease the condenser pressure, then efficiency of Rankine cycle increases b/c mean temperature of heat rejection decrease while mean temperature of heat addition is almost same/constant. 3. Superheating: 3' 3 T

2 4' 4

1 S

Tmean (heat addition) ↑ η ↑ 206

quality of steam ↑ 4. Reheating in steam turbine: Reheater

Q

5

B 4

3

2 T

3

5

4 1

W HPT

2 4'

LPT

6 6 P

S

Condenser 1

Poutput ↑ Qintput ↑ ɳ=

Poutput ↑ Pintput ↑

3

2 P

4

Constant pressure 5 6

1 V

It is not possible to conclude whether efficiency of heat cycle increases or decreases. If mean temperature of heat addition increases due to reheating, then efficiency decreases. If mean temperature of heat addition decreases due to reheating the efficiency decreases.

207

REFRIGERATION: (IN RANKINE CYCLE) (Feed water heater) (1-y) kg 6

B 5 4 HP T P2

LPT (1-y) kg

y kg

7

6 3

FWH

1 kg water

2 P1 1 Condenser (1-y) kg water

5 4

Rankine 2'

3' 1 kg 6

3 2

P

7

1 1'

4' Refrigeration

V

WT = (h5 - h6 ) + (1 – y) (h6 - hf ) Work of turbine decreases. ɳ=

Wnet ↓ Qinp ↓

(WP ≈ C, Wnet ↓, Q inp ↓)

Tm (heat add)↑ ɳ↑ Quality of steam = Constant 208

WORK & HEAT Indicator diagram mechanism: ad

ld

Mean effective pressure is defined as ρm =

ad K ld

Where, ad = Area of indicator diagram ld = Length of indicator diagram K = Spring constant

REFRIGERATION: Refrigeration is the process of creating & maintaining lower temperature than surrounding. The lower temperature is to be maintained continuously so system must operate on cycle. According to clausius 2nd law statement, it is not possible to transfer heat from low temperature body to high temperature body

209

without any energy input so, for all refrigeration system work input is required and cycle is power absorbing cycle. (Anticlockwise P-V diagram)

Refrigeration effect: It is the amount of heat which is to be extracted from lower temperature in order to maintain lower temperature known as refrigeration effect. T1

Winp

R Q2 = Refrigeration effect T2

(COP)R =

Desired output Input

(COP)R = (COP)R =

RE Winp

×

m m

=

=

RE Winp

=

Q2 Q1 − Q2

Q2 Q1 − Q2

Refrigeration capacity Power input

(COP)R =

=

RC Pinp

RC Pinp

Unit of refrigeration (Ton of refrigeration): 1 ton of refrigeration is an amount of heat which is to be extracted from 1 ton of water at 00 C in order to convert it into 00 C of ice in 24 hours. 1 ton = 3.5KW 1 TR = 3.5KW Latent heat of water =

335 24 × 60 × 60

210

= 3.5kJ/S

IDEAL REFRIGERATION CYCLE: (Reverse Carnot cycle) 3

Q1

Th

2

T TL

4

1

Q2 S

(COP)R =

Tl Th − Tl

VAPOUR COMPRESSION REFRIGERATION CYCLE: QR 3 Ph

Condenser 2 Ph 1

4

PL C

PL Evaporator Qin

Ph 3

QR

2 PL

T 1 4'

4 Q in

3----4 highly irreversible process & enthalpy increases 211

3

2 QR

Ph

P

4 Qin

PL

1

h

Analysis of vapour compression cycle: Process 1-2: Reversible adiabatic (isentropic) compression in compressor. Apply steady flow energy equation By 1st law, for open system, h1 +

C21 2

+ gz1 + Q = h2 +

C22 2

+ gz2 + Wcv

Neglect K.E & P.E Q=0 Wcv is negative Hence, Wcv = -Wcv Wc = h2 - h1 h1 = hg @PL (or) evaporator pressure. Process 2-3: Constant pressure heat rejection in condenser. Q R = h2 - h3 h2 → hsup @PH & Tsup h3 = hf @PH h4 = h3

212

Process 3-4: Isenthalpic expansion in throttling valve W3−4 = 0 h3 = h4 Process 4-1: Constant pressure heat addition ∴ Q in = R.E = h1 - h4 (COP) =

R.E

=

Winp

R.E Wcompressor

(COP) =

=

h1 − h4 h2 − h1

h1 − h3 h2 − h1

Why isentropic expansion not used in 3-4? 1. At point 3 fluid is liquid which are incompressible hence change is volume is less. 2. Hence, work is not more. 3. If turbine is used the cost of refrigerator will increase & will not be economical. 4. Difference in entropy of 4 & 4′ is less.

Problems with Solutions: 1. In a 5KW cooling capacity refrigeration system operating on V-C cycle. Refrigerant enters evaporator with enthalpy 75kJ/kg, and leave with enthalpy of 183kJ/kg. The enthalpy of refrigerant after compression is 210kJ/kg. Calculate the COP, power input to the compressor in KW, rate of heat transfers at condenser. h4 = 75 kJ/kg h1 = 183 kJ/kg 213

h2 = 210 kJ/kg (COP)R =

h1 − h4 h2 − h 1

Now, (COP)R = Q inp =

5 4

183 − 75

=

210 − 183 RC

Qinp

=

=4

5 Qinp

= 1.25KW

Q 2−3 = h2 − h3 = 135 kJ/kg Pinp = ṁ (h2 − h1 ) 1.25 = ṁ (210 − 183) ṁ = 0.048 kg/S Now, Q R = ṁ (h3 − h2 ) = 0.048 × 135 Q R = 6.48KW

EFFECT OF VARIOUS PARAMETERS ON THE PERFORMANCE OF V-C CYCLE: 1. Decrease in evaporator pressure: 3

2

Ph

P 1

PL

4 4'

PL’ h

PL′ < PL RE ↓ Winp ↑ (COP)R ↓ 214

2. Increase in condenser pressure: 3'

2'

3

Ph Ph

2 P

4'

4

PL

1

h

RE ↓ Winp ↑ (COP)R ↓

3. Superheating: 3' 3

2

2'

P

4'

1

4

1'

h

RE ↑ Win ↑ (COP) =

RE ↑ Win ↑

In case of superheating COP may increase or decrease depending upon refrigerant. In case of ammonia (NH3 ), superheating results decrease in COP. Whereas in case of R12 , COP increases due to result of superheating.

215

4. Sub cooling: 3' 3

2

P

4'

1

4 h

RE ↑ Win = Constant (COP) = ↑

DESIGNATION OF REFRIGERANT: Case 1: Saturated hydro-carbon. If the chemical formula of refrigerant is Cm Hn FP Clq then it is designated as R – (m – 1) (n + 1) P Where, N + p + q = 2m + 2 Example: CCl2 F2 m = 1, n = 0, p = 2, q = 2 R-012 ⇒ R-12 Reverse R-011 (m-1) – 0 ⇒ m = 1 n+1=1⇒n=0 p=1 We know, n + p + q = 2m + 2 216

0 + 1 + q = 2 (1) + 2 q=3 CFCl3 ⇒ R-123 m–1=1⇒m=2 n+1=2⇒n=1 p=3 n + p + q = 2m + 2 1 + 3 + q = 2(2) + 2 q=2 C2 HF3 q 2

Case 2: Unsaturated hydro-carbon If chemical formula of unsaturated refrigerant is Cm Hn FP Clq then refrigerant is designated as, R - 1(m – 1) (n + 1) p Where, n + p + q = 2m Example: C2 H4 m=2 n=4 p=0 q=0 ∴ R-1150 217

GAS REFRIGERATION CYCLE: This system is developed by belcoolman. In gas refrigeration cycle, air is used as working fluid. If works on reverse Brayton cycle. It is generally used for refrigeration in aircraft application due to its low weight per tonne of refrigeration. QR Heat Exchanger

3

2 1 We 4 PL

Heat Exchanger

C

WC

Qin

Condenser & Evaporator are heat exchangers when there is phase change. Here air is used hence, here only heat exchanger. In gas refrigeration cycle if isenthalpic throttle valve is used then h1 = h2 But for ideal gas as air is used as fluid dh = CP dT 0 = CP dT

CP ≠ 0

dT = 0 …… This is not

218

3

QR

QR

3

2

2 T

P 4

Qin

1

4

S

Qin V

Process 1-2: Reversible adiabatic compression, WC = h2 - h1

Process 2-3: Constant pressure heat rejection, Q R = h2 - h3

Process 3-4: Reversible adiabatic expansion, We = h3 - h4

Process 4-1: Constant pressure heat addition, Q in = h1 - h4 (COP) = (COP) = (h

Qin Wnet

h 1 − h4 2 − h1 ) (h3 − h4 )

Wnet = WC - We = (h2 − h1 ) - (h3 − h4 ) = (h2 - h3 ) - h1 + h4 = (h2 - h3 ) – (h1 - h4 ) 219

1

T1 − T4 2 − T1 ) − (T3 − T4 )

(COP) = (T

T1 − T4 2 − T3 ) − (T1 − T4 )

(COP) = (T

Divide (T1 − T4 ) 1

(COP) = (COP) =

(T − T ) ( (T2 − T3 )) −1 4 1

1 T T3 (T2 − 1) 3 ( ) −1 T1 T4 ( − 1) T4

1→ 2 ⇒ Pressure ratio, ꝏ

Cop

1

rp

P2

rP = T2 T1

P3

=

P1

P4

γ−1 γ

= (rP )

3→4 ⇒

T3 T4

⇒ T2 T3

=

γ−1 γ

= (rP ) T2 T1

T1

=

T3 T4

…….. (1)

T4

1

COP = T3 T4

−1

1

∴ (rP

γ−1 ) γ

220

−1

VAPOUR ABSORPTION REFRIGERATION SYSTEM: QR 3

Condenser 2

General Vapour Refrigeration System

1

4 C

PL Evaporator Qin QR Condenser

G NH3(g)

NH3(Ɩ)

NH3 + H2O (Ɩ)

H 2O P

WP

NH3(Ɩ + g) A Evaporator NH3(g)

H2O

NH3 + H2O (Ɩ)

Qe

In vapour absorption refrigeration system compressor is replaced by an absorber, pump & generator & pressure reducing valve.

221

Solar refrigeration system works on vapour absorption cycle. The most popular vapour absorption system is NH3 - H2 O system where NH3 is refrigerant & H2 O is absorber. This system is generally used where waste heat is available. In vapour absorption cycle heat is rejected in condenser as well as absorber. TG QG Qg + Qe

VA

T0

Qe Te

(∆S)universe ≥ 0 Qa Qa

Qa [

1 Ta

Qe Qa

+

Qe Qe

−

≤[

1 T0

-

Qa + Qe T0

] ≥ Qe [

T0 − Ta

COP ≤ [

T0 Ta

][

Ta − T0 Ta

≥0 1 T0

T0 Te Te − T0

][

1

−

Te

]

]

Te T0 − Te

]

PSYCHOMETRY: Air conditioning: It is a simultaneous control of temperature humidity & velocity of air.

Psychometry: It is branch of science deal with study of moist air.

222

MOIST AIR In atmospheric thermodynamics, air that is a mixture of dry air and any amount of water vapour. Generally, air with a high relative humidity. 1. Dry air (ma ) 2. Water vapour (mv )

PROPERTIES OF MOIST AIR: 1. Specific humidity: (𝛚) It is the ratio of mass of vapour to the mass of dry air. ω=

mv ma

Ma Mv

ƩM = Ma + Mv

Va Vv

Va = Vv = V

Pa Pv

Pt = Pa + Pv

Ta Tv

Ta = Tv = T

Pa V = ma R a T …… (1) Pv V = mv R v T …… (2) (1) (2)

⇒

Pa Pv

=

ma mv

×

Ra Rv ̅

̅⇒R=R MR = R

M

ω= ω=

mv ma mv ma

223

= =

Ra Rv ̅ R Ma ̅ R Mv

× ×

Pv Pa Pv Pa

ω=

mv ma

ω= ω=

mv ma

=

Pv Pa

×

Mv Ma

18 Pv 29 Pa

= 0.622

Pv Pa

Pt = Pa + Pv Pa = Pt - Pv w=

mv ma

= 0.622

Pv Pt − Pv

∴ w = f(Pv ) Specific humidity is function of partial pressure of vapour.

2. Relative humidity: (∅) It is defined as the mass of vapour to the mass of vapour under saturation condition. ∅=

mv mvs

Mvs

Mv

Ma

Ma PvV = mvRvT (1) (2)

1

⇒

PvsV = mvsRvT

Pv Pvs

=

mv mvs

2

=∅

Dry bulb temperature: It is a temperature measured by ordinary thermometer. Wet bulb temperature: It is a temperature measured by thermometer. When its bulb is covered with wet cloth. 224

Wet bulb temperature of air under saturation condition.

Pv < Pvs Unsaturated air

Dew point temperature: It is a temperature at which water vapour starts condensing. (or) It is a temperature at which 1st droplet of water is formed. Note: Pvs

Pv

DBT DBT

WBT

WBT DBT DBT Saturated Unsaturated

For unsaturated air DBT > WBT > DPT. For saturated air, DBT = WBT = DBT WBT is saturation temperature corresponding to Pvs whereas DPT is saturation temperature corresponding to Pv .

225

Problems with Solutions: 1. For a moist air, total pressure of 101K pa contain 10gm of vapour per kg of dry air at 300 C. find out the relative humidity. ∅= Pv Pvs

Pv Pvs

mv

=

mvs

=∅

Pvs = Tsat @300 C ω= ⇒

mv ma

0.01

⇒ (1 +

1

Pv

= 0.622

Pt − Pv

= 0.622 ×

0.01 0.622

Pv 101 − Pv

(101 − Pv ) = Pv

0.01 0.622

) Pv =

0.01 0.622

× 101

⇒ Pv = 1.6 K pa ∅=

Pv Pvs

1.6

=

4.2069

= 36%

Degree of saturation: It is a ratio of specific humidity to specific humidity at saturated condition. μ=

w ws

ω = 0.622 ωs = 0.622 μ=

Pv P t − Pv Pvs Pt − Pvs

Pv Pt − Pv Pvs Pt − Pvs

226

Pv

μ=

[

Pvs

μ = ∅[

Pt − Pvs Pt − Pv

Pt − Pvs Pt − Pv

]

]

Enthalpy of moist air: H = Ha + Hv Ha = Enthalpy of dry air Hv = Enthalpy of moist air mh = ma ha + mv hv h= h=

ma ha + mv hv m

ma m

ha +

mv m

hv

m ≈ ma h = ha + w hv Specific enthalpy of dry air (𝒉𝒂 ): Specific enthalpy of saturated dry air at 00 C is arbitrarily taken as 0. t

h

ha = 0

00C

For ideal gas, dh = CP dT (h - ha ) = CP (t - t 0 ) At t 0 = 00 C, ha = 0 CP = 1005 kJ/kg K ha = 1005 t 0 C 227

h = CP t

Specific enthalpy of water vapour: The specific enthalpy of saturated water at 00 C arbitrarily taken as 0. t T 00C

LH

∆h

hv = 0

hv

hv = 0 + LH + ∆h = LH + CP (t – 0) hv = LH + CPv t In general, LH = 2500 CPv = 1.88 hv = 2500 + 1.88t

Problems with solutions: 1. Degree of saturation of air at 300 C, 100K pa is 24%. The saturation pressure of water vapour at 300 C is 4K pa then find the relative humidity and specific humidity. μ = 0.24 μ=

Pv Pvs

[

Pt − Pvs Pt − Pv

Pt = 100K pa Pvs = 4 K pa

228

]

0.24 =

Pv 4

[

100 − 4 100 − Pv

]

0.24 × 4 (100 - Pv ) = Pv (96) 100 × 0.24 × 4 = Pv (96 + 0.24 × 4) Pv = 0.99K pa ∅=

Pv Pvs

=

0.99 4

ω = 0.622

= 0.2475 Pv Pt − Pv

ω = 6.219 × 10−3

kg of vapour kg of air

2. Dew point temperature of air at atmospheric pressure (101.3 K pa) is 180 C & DBT = 300 C. The saturation pressure of vapour at 180 C and 300 C are 2.06K pa & 4.24 K pa respectively. Find enthalpy of moist air. 300C T 180C

LH

∆h

hv = 0

DPT = 180 C DBT = 300 C Pvs = 2.06K pa P = 4.24 K pa h = ha + w hv ha = CPa × t = 1.005 × 30 229

hv

ha = 30.15 hv = LH + CPv (30 – 18) hv = 2522.56 ω = 0.622

Pv Pt − Pv

= 0.622 ×

2.06 101.3 − 2.06

ω = 0.0129 DPT = Tsat @PV Pv = Psat @DPT Pv = 2.06K pa ∴ h = ha + ω hv = 301.5 + 0.0129 × 2522.5 h = 62.7 kJ/kg 3. Atmospheric air at dry bulb temperature 400 C and wet bulb temperature of 200 C undergoes a process in the wet bulb temperature remain constant. Wet bulb depression at exit is 0.25 times of that of inlet then find DBT at exit. Wet bulb depression = DBT – WBT DBT = 400 C, (WBT)inlet = 200 C = (WBT)exit (WBT)exit = 0.25 (WBT)inlet (DBT − WBT)exit = 0.25 (DBT − WBT)inlet = 0.25 (40 – 20) =5 (DBT)exit = 5 + 20 (DBT)exit = 250 C 230

4. Moist air with dry bulb temperature of 400 C has relative humidity of 50% atmospheric pressure is 1.01bar. Saturation pressure of vapour at 400 C is 7.38 K pa & saturation pressure of vapour at 1500 C is 4.578 bar. Find the specific humidity. Moist air as stated above compressed to 5.05 bar and its corresponding dry bulb temperature is 1500 C then find relative humidity of compressed air.

P = 1.01 bar 400C DBT = 400 C, ∅ = 0.50, P = 1.01 bar Pvs @400 C = 7.38 K pa Pvs @1500 C = 4.758 bar ω =? i. ⇒ ω = 0.622

Pv Pt − Pv

∅=

Pv Pvs

0.50 = ∅ =

Pv 7.36

⇒ Pv = 3.69 ω1 = 0.622

3.69 101 − 3.69

ω1 = 0.0235 kg of vapour/kg of air

231

ii. ⇒ ω1 = ω2 Since pressure & temperature both are in increased hence, mass of system remains same. P = 5.05 bar 1500C

ω2 = 0.622

Pv Pt − Pv

0.0235 = 0.622 ×

Pv 505 − Pv

Pv = 19.53 K pa ∅=

Pv Pvs

=

19.53 4.758 × 102

∅ = 0.04106 ∅ = 4.1% 5. Moist air at pressure of 100 K pa. it compressed to 500 K pa & then cooled at constant pressure to 350 C in cooler the air at entry to the cooler is unsaturated & become just saturated at the exit of cooler. During the entire process there is no condensation take place. The saturation pressure of vapour at 350 C is 5.628 K pa then find the partial pressure of vapour of moist air at the entry of compressor. Pt1 = 100 K pa Pt1 = 500 K pa T2 = 350 C

232

Compressor

Cooler

Pt = 100 kPa

Pt = 500 kPa

Pt = 500 kPa ϕ=1 T = 300C

As there is no condensation so, the specific humidity at all the points are same. Pvs @350 C = 5.628 K pa ∅=1=

Pv Pvs

ω3 = 0.622

⇒ Pvs = Pv = 5.628 Pv

Pt − Pv

= 0.622

5.628 100 −5.628

ω3 = 7.08 × 10−3 ω3 = w1 = 0.622

Pv Pt − Pv

7.08 × 10−3 = 0.622 × 7.08 × 10−3 = 0.622 ×

Pv 100 − Pv

Pv Pt − Pv

⇒ Pv = 1.125K pa

DEVELOPMENT OF PSYCHOMETRIC CHART:

ω = 0.622 Pv/Pt - Pv Pv

-200C

-10

0

10

20

40 30

50

T (DBT)

We can measure ω = 0.622 233

Pv P t − Pv

Pt = Patm = 101 K pa Constant relative humidity lines. ϕ=1 Pv

0.8 = ϕ 0 .6 = ϕ 0.4 = ϕ

ω

T (DBT)

P2 > P1 Tsat 2 > Tsat 1 P2 T

P1

Tsat2

Tsat1

S

Different lines on psychometric chart: 1. Constant specific humidity lines

ω

T (DBT)

234

2. Constant dry bulb temperature line

ω

T (DBT)

En

th a

lp y

sc al e

3. Constant enthalpy line

ω

T (DBT)

h = ha + ω (hv ) h = CP + w (2500 + 188t) At different t & w. find h & then keeping h constant vary w & t 4. Constant wet bulb temperature line: Though there is small deviation between constant enthalpy & constant wet bulb temperature line but for all calculation purpose this deviation is neglected. 5. Dew point temperature lines: Dew point temperature is saturation temperature corresponding to Pv . As long as partial pressure of vapour is constant its due Pt , temperature is also constant. If Pv is constant then ω is also constant. hence, constant due point temperature line follows constant ω line. 235

ω

Pv

Constant DPT Lines T (DBT)

Note:

DPT

WBT DBT

For unsaturated air, DPT < WBT < DBT From gib’s phase rule, P+F=C+2 No. of phase = P = 1 → Gases are present in 1st phase No. of component = 2 (Dry air, Water vapour) 1+F=2+2 F=3 Minimum three properties are required to identify point. 1st property is fixed in the chart. i.e., Pt = 101.3 K pa Hence, we require only two point to identify point.

236

ϕ=

1

0 ϕ=

.8

ϕ=

ω

0.6

T (DBT)

For moist air, degree of freedom is 3 but on psychometric chart only two properties are required to fix the state b/c the psychometric chart is drawn for a fixed total pressure [3rd parameter is already fixed] i.e., Pt = 101.3K pa

BASIC PSYCHOMETRIC PROCESS: 1. Sensible heating: Tc t1 mv1 ma1

t2 mv2 = mv1 ma2 = ma1

ω2 = ω1 ∅2 < ∅1 ω2 = ω1

ϕ1

ϕ2 2

1

t1

237

t2

ω

Bypass factor: In this sensible heating process, the maximum temperature of air which is possible is coil temperature but some of the air will bypass without any contact to heater/heating coil, so the final temperature t 2 is less than coil temperature. h2 h1

t1

t2

Bypass factor =

tc TC − t2 TC − t1

Coil efficiency: It is a ratio of actual temperature rise to the maximum possible temperature rise. By SFE, h1 +

C21 2

+ gz1 + Q = h2 +

C22 2

+ gz2 + W

Q = h2 - h1 Work is zero as air is a system. ɳC =

t2 − t1 TC − t1

⇒ BPF + ɳC = 1

2. Sensible cooling: It is process of decreasing temperature without any change in specific humidity.

238

Cooling Coil t1 mv1 ma1

t2 mv2 ma2

Cooling Coil (Refrigerant flow through coil) t2 < t1 ω=

mv ma

mv1 = mv2 → (TC = abv DPT) m a1 = m a2

ω

1

2

t2

t1

T

Dew point temperature is temperature Below. Which condensation will start. For sensible cooling, the coil temperature must be greater than dew point temperature.

3. Humidification: It is the process of increasing specific humidity at constant dry bulb temperature.

1 ω

2

239

4. Dehumidification: It is a process of decreasing specific humidity at constant dry bulb temperature.

1 ω

2

Note: It is not possible to achieve pure humidification and pure dehumidification because this processes are always associated with temperature change.

5. Heating and humidification:

2 ω

1

6. Cooling and humidification:

2 ω

1

Below process is adiabatic process, 240

t1 mv1 ma1

t2 mv2 ma2 Cooling water

t 2 < t1 m a1 = m a2 mv1 > mv2 Apply SFE, h1 + Q = h2 + W → K.E & P.E neglected. ∴ h1 = h2 Conclusion: Cooling & humidification follows constant enthalpy line.

7. Cooling & humidification:

2 ω 1

DBT

t2

241

t1

Cooling coil t1 mv1 ma1

t2 mv2 ma2

t 2 < t1 m a1 = m a2 mv1 > mv2 In summer air conditioning, cooling and dehumidification process is used. To achieve cooling & dehumidification coli temperature must be less than dew point temperature.

8. Heating & dehumidification:

Silica or alumina gel

ω

1

2 T

Silica gel, alumina gel can absorb moisture. This process is also known as chemical humidification or adiabatic dehumidification. Certain chemicals like silica gel or alumina gel are used for absorbing moisture. When moisture is to be absorbed then latent heat of condensation released and hence its dry bulb temperature increases. 242

9. Sensible heat factor: Sensible heat factor is a ratio of sensible heat to the total heat.

L.

S. H

h1

H

h2

In ge re n a ch atu No per tem

2 ω 1

T

SHF =

SH SH+LH

GAS & AIR COMPRESSOR Compressor is a device which convert mechanical energy into pressure energy. Open system reversible work W = ∫ −v dP

Compression without any clearance volume: PVn = C PV = C PVr = C 1 211 2

2

P2

P2

1

P1

V1 243

P1

1-2 → Adiabatic compression 1-2′ → Poltropic compression 1-2′′ → Isothermal compression P2 2 P1 2' T 2" 1

S

The process may be adiabatic or isothermal or any general polytrophic process depending upon application. Case 1: Adiabatic process dW = V dP For adiabatic process, PV γ = C γ

γ

P1 V1 = P2 V2 = PV γ 1

V=

P γ ( 1) P

V1 1

Wopen = ∫ −V dP = 1 γ

= -V1 (P1 ) 1 γ

P

= -V1 P1 [

P γ ∫ − ( P1 ) 1

1 γ ∫ ( P) 1 2

1−

γ 1

1−γ 244

]

dP

V1 dP

= Wopen =

−γ

1 γ

γ−1 γ

γ−1

V1 P1 [P2

−γ

1 γ

γ−1

Wopen =

− (P1 )

γ−1 γ

V1 P1 (P1 ) γ γ−1

γ−1 γ

]

γ−1 γ

P2

[( ) P1

P2

V1 P1 [( )

γ−1 γ

P1

− 1]

− 1]

Case 2: Polytrophic process Wopen =

−n

P2

n−1 n

V1 P1 [( )

n−1

P1

− 1]

Case 3: Isothermal process PV = C V=

C P

Wopen = ∫ −V dP = -C ∫ Wopen = - V1 P1 ln

1 P

dP

P2 P1

COMPRESSORS 1. +ve Displacement type: In this type, pressure is increased by decreasing the volume with the help of positive displacement of piston. Example: Reciprocating compressor It is best suited for very high pressure up to 1000 bar & low discharge. 245

(Q < 9 m3 /s) Example: In refrigerator mining work, etc., 2. Dynamic type:  In dynamic type, pressure is increased by dynamic action of gas.  The K.E imparted to the gas by rotation of impeller & this K.E change into pressure energy partially on impeller & rest is diffuser. (a)

Centrifugal compressor Best suited for medium discharge & medium pressure 10 bar < P < 100 bar 9 m3 /s < Q < 300 m3 /s

(b)

Axial compressor Best suited for low pressure & very high discharge.

246

SINGLE STAGE SINGLE ACTING RECIPROCATING COMPRESSOR WITH CLEARANCE VOLUME:

Expansion

3 TDC BDC

Delivery Process 2 PVr = C P n=C V

Compression

1

4 Suction Process VC

V

TDC

VS

BDC

On downward motion of piston from point 3 the pressure in the cylinder fall and then atmospheric air enter to the cylinder through the suction valve. When piston goes up the suction & delivery both valves are closed & air is compressed till the delivery wall open due to difference in pressure in cylinder & delivery maintained & when piston goes down on next downward stroke the air trapped in clearance volume expands & the pressure fall to suction pressure then inlet valve opens & same cycle repeated. Winput = WC - We For polytrophic process. W=

n n−1

P2

n−1 n

V1 P1 [( ) P1

247

− 1]

−n

VP n−1 4 4

P3

n−1 n

[( ) P4

− 1]

P2 = P3 , P1 = P4 W=

n

P2

n−1

Win =

[( )

n

n−1 n

P1

P2

n−1

[( )

− 1] (V1 P1 - V4 P4 )

n−1 n

P1

− 1] (P1 V1 - P4 V4 )

P1 = P4 = P Win =

n n−1

P2

[( )

n−1 n

P1

− 1] P (V1 - V4 )

Va = V1 - V4 Vs = V1 - V3 Win =

n n−1

P2

[( )

n−1 n

P1

− 1] P (V1 - V4 )

Volumetric efficiency: It is ratio of actual volume of gas taken into the cylinder during suction stroke to the piston displacement volume (Swept volume).

248