ThermodynamicsMacroscopic and Microscopic Point of view:

The study of thermodynamics begins with the study of restricted region in space or a finite portion of a matter called the system. Once the system is chosen the next step is to describe in terms of quantities that will be helpful for discussing the behavior of the system or its interaction with the surroundings or both.There are 2 general views adopted for description of the system.

Macroscopic Point of view:

Let us take a system, the contents of a cylinder of an automobile engine. The chemical analysis mixture of petrol and air before ignition and after the mixture is ignited there would be combustion describable in terms of certain chemical composition. The system occupies a certain volume and exists at certain pressure and temperature, All these quantities can be measured in the lab. The contents of the cylinder now being describable by means of and quantities namely composition, volume, pressure and temperature. These quantities refers to the gross characteristics of the system is a specification of a few fundamental measurable quantities.

Characteristics:

They involve no assumption regarding the structure of an atom.They are few in numberThey are suggested more or less by the sense perception They can be measured directly Eg: Pressure by pressure gauge.

Microscopic Point of view:

If a gas consists of ‘n’ no. of molecules at having same mass of each moving with a velocity independent of other. The position of any molecule is specified by 3 cartesian coordinates X,Y,Z. The velocity of 3 components be VX, VY,VZ. Therefore to describe the position of velocity of molecule 6 no.s are

required.This equivalent to a imaginary 6- ‘D’ space called phase space whose co-ordinates are XYZ+ VX VY VZ.

Characteristics:

Assumptions are made regarding the structure of matter that exist in molecular form.

Many quantities must be specified.

The quantities Specified are not suggested by our sense perception.

These quantities cannot be measured.

This type of description is used for statistical thermodynamics. The study from macroscopic point of view is

subject to classical thermodynamics.

Macroscopic v/s Microscopic:

Although the point of view seems different view and incomparative there is a relation between them where

both the points of view are applied to the same system at the end of the relation lies in the fact that few

measurable properties which deals with the macroscopic description are partly the average over the period of

a large no. of microscopic characters. Ex: Pressure is average rate of change of momentum due to all

molecular collision/ unit sec.

Thermodynamic system is defined as a quantity of matter of fixed mass upon which attention is focused on

thermodynamic study.

Any prescription and identifiable collection of matter across the boundary of which there is a transfer of mass and energy is called System.

Everything external to the system is called Surrounding.The system and surrounding is separated by means ofa system Boundary.It may be real or imaginary, movable or fixed body. The combined system and surrounding is commonly known as Universe.

System is divided into three types – 1. Closed System : If mass within the system remains constant and if there is only transfer of energy

across the boundary of the system, then it is called a closed system.

Ex: consider a system consisting of a gas as shown, if the heat is added to the gas it will expand and work will be done by the gas on the piston. Thus heat and work will cross the boundary of the system and volume of system will change due to expansion of gas but no mass crossesthe boundary of the system, such a system is called Closed system.2. Open System : If both mass and energy are allowed to cross the boundary of the system, then it is

called an Open System.

Ex: An air compressor in which air enters atlow pressure and leaves at high pressure, there is a transfer of energy across the boundary of a system. E.g. Gas turbine and air compressor.

3. Isolated System : In this system neither mass nor energy crosses the boundary of the system. Here there is no interaction between the system and the surrounding.

Ex: Thermo flask filled with hot coffee.

Boundary:The hypothetical envelope enclosing a system.

State:

When all the properties of a system have definite values the system is said to exist in a definite state properties are co ordinates to represent state of a system.Any operation in which one or more of the properties of a system changes is called change of state.

Process:

When a system changes from one equilibrium state to another equilibrium state a process is said to have taken place.

Cycle:

If the initial and final state of a system are identical, the system is said to have undergone a cycle i.e. the system in a given state goes through a number of processes and finally returns to its initial state. System is said to have undergone a cycle.Ex: H2O in a stem power plant.

Properties:

Any observable characteristics such as pressure, volume and temperature by which the physical condition of a system can be described is called property of a system. Thermodynamic properties can be divided into two groups –

1. Intensive property: If the value of a property is independent of mass of the system it is called the Intensive property. Ex. Velocity, weight, viscosity, temperature, pressure.

2. Extensive property: If the value of a property is directly proportional to the mass of the system it is called Extensive property. Ex. volume, area, density, energies of all kind, electric charge, magnetization.

The ratio of extensive property to the mass is called specific value of that property and it is an intensive property. Ex. Volume/ mass = specific volume

Equilibrium:

A system is said to exist in a state of thermodynamic equilibrium when there is no change in any macroscopic property of the system, if the system is isolated from the surrounding. The system is said to be in thermodynamic equilibrium if the condition for the following types of equilibrium are satisfied.

1. Mechanical equilibrium : If there are no unbalanced forces in the interior of the system and between the system and the surrounding, the system is said to be in state of mechanical equilibrium.

2. Chemical equilibrium : If there is no chemical reaction or transfer of matter from one part of the system to another, the system is said to exist in a state of chemical equilibrium.

3. Thermal equilibrium : For a system which is in chemical and mechanical equally if there is no change in the property of the system, the system is said to be in thermal equilibrium. Ex. co ordinates of system do not change.

When the condition of any one of the three types of equilibrium is not satisfied the system is said to be in Non equilibrium state.

Quasi static process:

When a process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all the times then it is called quasi static or quasi equilibrium or semi static process.

A quasi static process can be viewed as a slow process which allows the system to adjust itself internally so that properties in one part of the system do not change any faster than those at the other part.

Pure substance: A pure substance is one that has homogeneous & invariable chemical composition. It may exist in more than one phase but chemical composition is same in all phases. Every substance exists in any one of the 3 phases i.e. solid, liquid & gaseous state.Eg: Pure substance like H2O may exist in solid phase as ice, liquid phase as H2O & gaseous phase as steam. H2O, N

& He are all pure substance. It has uniform chemical composition. Classify the following as intensive & extensive properties: Temperature, pressure, mass, elevation, length, specific weight, specific volume, surface tension, enthalpy, volume, force Intensive properties Extensive propertiesTemperature, pressure, elevation, length, surface tension, mass, volume Specific weight enthalpy, entropy, force Classify the following into closed, open and isolated system Closed System:

1. Mixture of ice and water in conducting cylinder. 2. Pressure cooker 3. Storage battery producing electricity. Open System:1. Automobile engine 2. wind mill3. Reciprocating air compressor4. Steam turbine5. Water wheel6. Boiler

Isolated System:1. Thermos flask with hot coffee.

HEAT AND WORKWork: In mechanics “Work is said to be done when the point of application of force moves in the direction of force”.

Work is the form of interaction between a system and surrounding. In thermodynamics work is identified only when it crosses the boundary of the system “work is said to be done by a system if the sole effect on things external to the system can be deduced to the rising of a weight”. Here rising of weight is effect of force acting through a distance.

Let us consider battery and motor as a system as shown in fig.1. The motor is driving a fan the system is doing work upon the surrounding. When the fan is replaced by a pulley and weight as shown in fig.2. The weight may be raised by the pulley driven by motor. The sole effect of thing external to system is raising of weight.Work : When work is done by a system, it is taken to be positive and when work is done on the system, it is taken to be negative.Eg: A gas expanded against a piston is positive work. Piston compressing a gas is negative work. Positive work means energy leaves the system and negative work means energy is added to system.Heat: It’s a form of energy that is transferred across the boundary by virtue of temperature difference. The direction of heat transfer is taken from the high temperature system to low temperature system.

Heat flow into a system is taken to be positive and heat flow out of a system is taken to be negative. A process in which no heat crosses the boundary of the system is called adiabatic process (Q=0). A heat is a transient quantity which can be identified only as it crosses the boundary of the system.

Displacement work or pdv work or work done at the moving boundary of a simple compressible system in a quasi static process.

Consider a gas in a cylinder as a system having initially pressure and volume of P1 & V1. As it occupies a

new position the pressure and volume are P2 & V2 respectively. Consider the movement of piston through an infinitesimal distance ‘dl’ and if ‘A’ is area of c-s of piston. Then force acting on the piston F= Pressure×area=P×AThe amount of work done by the gas on the piston W.d.=Force ×dl ∂w= p×A×∂l ∂w= pdv Therefore dv=A×dl

When the piston moves from position 1 to 2 the total work done by the system will be ∂w 2 v2 v2

∫ δW = ∫ pδv W1-2

= ∫pdv

1 v1 v1

The magnitude of work is given by area under the path 1-2 as shown in pv-diagram. The integration pdv can

be performed only on a quasi static path.

1. A non-flow reversible process occurs for which p=3v2+1/v ; P→ bar, v→m

3 What will be the w.d.

when v changes from 0.5 to 1.5 m3 .

Sol: 1 bar = 1× 105 N/m2 ; V1=0.5; V2=1.5; p=3v

2+1/v

w= 1∫

2pdv

1.5 1.5

W= ∫ ( 3V2 + 1/V1) dV = [ V3 + logV] W= 4.348× 105N-m

0.5 0.5

2. A fluid system undergoes a non flow frictionless process from V1= 6m3

V2= 2 m3 . The process is given by P = 15/V + 2. Determine workdone ?

v2 2 2

W = v1∫ pdv = ∫ ( 15/V + 2) dV = [ 15logV + 2V] W=-24.479×105 N-m

6 6

-‘ ve sign indicates work is done on the system.

3. A non flow Quasistatic process, occurs for p = -3V + 16. What is the workdone when V changes from 2 to

6m3? v2 6 6

W = ∫ pdv = ∫ (-3V + 16) dV = [-3 V2 + 16 V] W = 16×105 N-m

v1 2 2

+ve indicates workdone is done by the system

4. When the valve of a evacuated bottle is opened atmospheric air rushes into it. If the atmospheric air is 95

KPa & 1.5m3 of air enters into the bottle calculate the work done by air.

Sol: Total work done = ∫ work done by bottle + ∫work done by air = 0 + ∫pδv.Wair = Pair Vair = -955×1.5 = -142.5 KJ

Work is done by the air on the system. Since free air boundary is contracting, the work is done on the bottle & it is –ve.

5. A gas in the cylinder and piston arrangement compressed the system. It expands from 1.5 -2 m3 by receiving

20×104 N-m of work from a pedal wheel. The pressure of gas remains constant at 6 bar. Determine the workdone.

v2

Sol: W.d. gas = ∫v1 pδv = 6 × 105 [2-1.5] = 30 × 104 N-m

Net W.d = W.dgas + W.dpedal

= 30 × 104 - 20× 104

= 10× 104 N-m

6. A piston and a cylinder machine contains fluid has a stirring device in the cylinder. The piston is frictionless and it is held down against the fluid to atmospheric pressure of 101.325 KPa. The stirring device is turned 10000 rev with an average torque against a fluid of 1.275 N-m. Meanwhile the pistonof 0.6m dia moves out of a distance of 0.8 m. Find the net work transfer for the system.

Sol: Dia of piston – 0.6m l = 0.8m P= 101.325 KPa W. d by the stirring device = 2πNT = 80.11 KJ or N-m

W.d. by the gas on the piston = P. A. L

= 101.325 × π/4 (0.6)2 × 0.8 = 22.92 KJ W.d. on the system by the stirring device = -80.11 KJ The workdone is positive by the pressure of air acting on the system Net W.d. = -80.11 + 22.92 = -57.19 KJ

Equality in temperature:

Consider 2 bodies one hot and other cold are placed in contact in an isolated system. After sometimes the hot body loses heat and becomes cold. The cold body becomes hot. This process continues till the exchange of heat attains a state of thermal equilibrium.. When two bodies have equal temperature then there is no change in any observable property when they are in thermal equilibrium.

Zeroth Law of Thermodynamics:

“If two systems are in thermal equilibrium with the third system, then the two are themselves in equilibrium with each other”

Ex: Let us apply the concept of equality of temperature to 3 system, 1, 2 & 3 . Let 1 be a container fitted with a pressure gauge.The system 2 & 3 are metallic spheres.Let system 1 be brought in contact with 2 if thereis no change in pressure of system 1. Then system 1& 2 are equal in temperature, similarly if the system 1 is brought in contact with 3 again there is no change in property of system1 then system 1 & 3 are in thermal equilibrium. The experiment shows that if system 2 is in contact with 3, if there is no change in property of system, then the system 2 & 3 are at equal temperatures. When two system are in thermal equilibrium, according to Zeroth law of thermodynamics.

Ice point: it is defined as a temperature of a mix of ice and H20 which is

inEquilibrium with saturated air at atmospheric pressure.

Steam point: It is defined as temperature of steam and H20 at atmospheric pressure.

Temperature scale: On a centigrade scale ice point is numbered as zero and steam point as 100 on the Fahrenheit scale.

The respective points are32oF & 212 oF.If solid ice liquid H20 and water vapour exist at equilibrium condition, the state of equilibrium is called

triple point of water.

Toc = ToK – 273 T oF =

ToR – 460

ToF = 1.8oc + 32 ToR = 1.8

ToK

International Temperature scale: The procedure for measuring temperature was adopted to provide a scale that could be used to calibrate scientific instruments. They specified certain fixed points at which instruments like thermometers, thermocouples, pyrometers, resistant thermometers etc... can be calibrated.

Sl.no Fixed point Temperatureoc     

1 Triple point of H2 -259.34

2Normal boiling point of 02 -182.962

3 Triple point of H20 0.014 Ice point 05 Steam point 1006 Sulphur point 444.67 Silver point 960.88 Gold point 1064.44

Problems:

1. The equation Rt= Ro(1+αT) is applicable to a resistant thermometers. Rt & Ro are the

values of resistance, at temperature toc and tok. The thermometer is calibrated by immersing

in boiling H20 273 ok + boiling sulphur 718ok. The instrument indicates the resistance values

at these two temperatures to be 14.7 Ω & 29.2 Ω respectively. Determine the temperature of fluid for which the corresponding thermometer resistance is 23.64 ΩΩSol: Rt1=14.7 Rt2=29.2 t1= 273 K t2= 718K

Rt= Ro(1+αt) 14.7=Ro(1+α (273))-----1 29.2= Ro(1+α(718))------2Equatingthe equations eq1/eq2 = 14.7/29.2=1+α273 1+α 718 α =5.6*10-3 Rt=Ro(1+αt)====>14.7=Ro(1+5.6*10-3*273) Ro=5.8

At 23.54 Ro=5.8 23.54= 5.8(1+5.6*10-3*T) T=543.7oK

Problem2. The temperature on a thermometric scale is defined in terms of a property ‘p’ by the relation t = a log p+b where a & b are constant. The temperature of ice point & steam

point are assigned 32 + 212. Experiment gives the values of ‘p’ 1.86 + 6.81 at ice point + steam point respectively. Evaluate‘t’ corresponding to p= 2.50.

Sol: t = a logep + b a =138.69

32= a loge

1.86 + b b = -54.06

212 = a loge

6.81+b

-180 = a (log 1.86 -log 6.81)

t = 138.69 loge 2.5 + (-54.06) t=73.02

3. A constant volume gas thermometer containing He gives the reading of gas pressure at 1000 + 1366 mm of Hg at ice point and steam point respectively. Assuming the linear relationship of the form t = a+b. Express the gas thermometer / Celsius temperature ‘t’ in terms of gas pressure ‘p’. What is the temperature recorded by the thermometer.

Sol: t = a + bp 0 = a + bp → 0 = a+b(1000) a = -1000b b = 0.2732 100= a + b(1366) = -1000b + 1360b = 366b a = -273.2 T = -273.2 + (0.2732) (1074) t = 20.2168

First-Law—of-Thermodynamics:

Joule’s Experiment: Take a quantity of H20 in a rigid vessel,which-was isolated

adiabatic, from its environment as shownin the vessel was fitted with a peddle wheel. The initial state ofthe system was determined by measuring the pressure-and temperature.As work was done on the system, by a peddle wheel which wasactuated by a falling weight ‘W’. Its temperature is increased. The system was then placed on a constant temperature bath to restore to its initial state. Thus the system had undergone one complete cycle. The net work input to the system was always α to the heat transferred out of the system. In other words, when a closed system undergoes any cyclic process then cyclic integral of work is α to cyclic integral of heat. Φ δW α φ δQ

Therefore φ δW = J φδQ where J → Joules constant

Statement of First Law of Thermodynamics:

The first law states that “ During any cycle a closed system undergoes, cyclic integral of heat is directly proportional to cyclic integral of work.

Φ δQ α φ δW Φ δW α J φ δQ where J→ proportionality constant→ Joule’ constant

Whenever a system executes a cyclic process the algebraic sum of heat transfer is equal to algebraic sum of work transfer. i.e. Q1 + Q2 + Q3 + Q4 + …………= W1 + W2 + W3 + W4 + ……….. ∑ δQ = ∑ δW

To show that internal energy is a property of a system consider a system, that undergoes a cycle changing its state from 1- 2 along path A &returns to its original state from state 2 to state 1 along two paths B or C.

From First law of thermodynamics, φ δQ = φ δW φ δQ - φ δW = 0 → 1

Now applying first law for cyclic process consisting of path A & B

Φ2(δQ-δW) + Φ1(δQ-δW) = 0 → 2 1 path A 2 path B

Consider another cycle in which the system changes its state from 1 to 2 along path A but returns from state 2 to 1 along path C

Φ2(δQ-δW) + Φ1(δQ-δW) = 0 → 3 1 path A 2 path C

Equating 2 & 3 Φ1(δQ-δW) = Φ1(δQ-δW) 2 path B 2 path C

The quantity (δQ-δW) is same for processes B & C but process B & C are two arbitrary process between 1 & 2 .Therefore The quantity (δQ-δW) does not depend on the path function between 1 & 2 but depends only on initial and final states. Hence it is a point function and represents a change in property of the system and is denoted by E. δQ-δW = δE δQ = δE+ δW

If this is integrated between 1 & 2

2 2 2 ∫

1 δQ = ∫

1 δE + ∫

1 δw Q

1-2= (E2-E1) + W

1-2

Q1-2 → Heat transferred between 1 & 2

E2→ Final value of energy

W

1-2→ Work transferred between 1 & 2

E1→ Initial value of energyThe energy includes internal energy, potential energy and kinetic energy.

ie.. δE = δU + δ(KE + δ(PE)

δE = δU + δ(½mv2) + δ(mgh) δ(KE) + δ(PE) = 0 When δE=δU δQ = δU+ δWSince δQ-δW is the property of the system. Internal energy is also a property of a system.If an Isolated system is considered the heat + work cannot cross the boundary of the system Q=W=0 → E2=E1ie.. the energy of an isolated system always remains constant.

Perpetual motion machine of 1 st kind [ PMM- I] A machine that creates its own energy or continuously produces work out of nothing is

called PMM- I. The machine violates first law of thermodynamics PMM- I is thusimpossible.

Problems1. In a cyclic process the heat transfers are 14.7KJ, -3.56KJ, & 31.5KJ, What is the net work done.

∑δQ = ∑W Q1+Q2+Q3+Q4 = W

14.7-25.2+31.54-3.56 = W

+ve === work is done by the system.

2. In a cyclic process the heat transfers are 10J,-24J,-3J & 31J determine the nett work done. ∑Q = ∑W 10-24-3+31 = W

W = 14 N-M Work is done by the system.

3. Closed system undergoes a process in which heat transfer from the system is 40 KJ & w.d on the system is 18 KJ. Calculate the change in internal energy. The system is brought back to its initial stage 20 KJ of work is done by the system heat transfer also takes place. Calculate magnitude & direction of heat transfer.

From law of thermodynamics

δQ = δE + δW Q1-δE + W1-40 =δE-18 δE = -40+18

δE = -22 KJ

i.e. The energy is increased by an amount of22 KJ

∑Q = ∑W Q1 + Q2 = W1 + W2

-40 + Q2 = -18 + 20

Q2 = 42 KJ

Heat is transferred into the system

4. A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 & 0.105 mpa to a final state of 0.15m3 & 0.105mpa. the pressure remaining constant during process . how much does the internal energy of the gas changes.When there is a heat transfer of 37.6 KJ of heat from the gas during the process.

δW = ∫v2

pdv = p(v2-v1) = 0.105(0.15-0.3) = 0.105(-0.15)

v1

w = -15.75x103 J w = -15.75 kj

δQ = δu + δw δu = E2 – E1

δu = -37.6-(-15.75) = -37.6+15.75

δu = -21.85 KJ

Steady flow energy Equation :.

Consider a steady flow system in which there is a mass m1 entering at section 1 & m2 leaving at section 2. There is no accumulation of mass or energy within the controlled volume & the properties at any location within the controlled volume are steady with time. Section 1 & 2 are at an elevation of Z1 & Z2 respectively w.r.t datum line of the controlled volume undergoes process there will be heat & work transfer applying 1st

law of thermodynamics Q-W=∆E W includes all types of external work and the flow work due to mass entering and leaveing the controlled volume

The following assumption are made for the analysis of the system:

1. the mass flow through the system remains constant m1=m2=m

2. There is no chemical composition of fluid

3. The state fluid at any point remains constant with the time 4. The rate of heat & work passing through the boundary of a system remains

5. The fluid or gas is uniform in composition

By the law of conservation of energy m1e1+m1p1v1+Q=m2e2+m2p2v2+w-----1

e = ∆u+∆KE+∆PE e = u+v2/2+Zg substituting value of ‘e’ in 1

m1(u1+v12/2+gZ1)+m1p1v1+Q = m2(u2+v2

2/2+Z2g)+m2p2v2+W

The enthalpy is given by h = u+pv

m1(h1+v12/2+gZ1)+Q = (h2+v2

2/2+gZ2)m2+W

Mass remains constant therefore m1=m2=m

Q-W = m(h2-h1) + v22 -v 1

2 + (Z2-Z1)g

2

Q-W = m∆(h+v22/2+gZ)

Where A1 & A2 are area of c.s in m2. m1 & m2 are mass flow rates in kg/sec

P1 & p2 are pressure9rates) in Pascal,

V1 & v2 are velocity in m/sec,

Z1 & Z2 are elevation above datum, Q = heat in joule, w = work in joule

Application of steady flow energy equation 1 . Work developing system Ex-: Engine, steam gas & H2O Turbines.2 . Work absorbing System Ex-: Pump, Compressor & refrigerator3 . Non work developing & absorbing system Ex-: Nozzles & Heat exchangers

Application of steady flow an equation for Throttling devices:

When fluid flow through a restricted passage like a partially opened valve or orifice or a porous plug there is an appreciable drop in pressure & flow is said to be throttled. Fig shows process of throttling by a partially opened valve fluid flowing in a insulated pipe from SFEE/

Q-W = (h2-h1 + v22 -v 1

2 + (Z2-Z1)g)m …….eq1

2

Q = 0 & W = 0 & potential energy is negligible. Than the equation becomes

0 = m((h2-h1) + ( v22 -v 1

2 ) )………..eq2

2

velocity inlet isequalto-outlet than h2=h1 therefore itbecomes V2=V1The enthalpy of fluid before & after throttling is equal.

Nozzle & diffuses

Nozzle is a device which increases the volume or KE of the fluid at the expense of the pressure drop where as diffuser increases the pressure of a fluid at the expense of the KE. The nozzle is insulated. Than the SFEE

Q-W = m((h2-h1) + v22 -v 1

2 + (Z2-Z1)g)

2 The W.d is zero, Q = 0, PE = Negligible

Than 0 = h2-h1 + v22 -v 1

2

2

v2 = √(2(h1-h2) + v12)

The inlet volume is negligible therefore v1 = 0

V2 = √(2(h1-h2))

Go to 15

8. 135KJ of work is done by each kg of fluid. The specific volume , pressure and velocity at the inlet are 0.37

m3/kg 600Kpa at 16 m/s. The inlet is 32m above the flow and the discharge is at float level. The discharge

conditions are 0.62m3/kg ,100Kpa and 270 m/s. Total heat loss between inlet and discharge is 9KJ/Kg of fluid. It is flowing through this apparatus. Thus the speci mter ener increase, or decrease and by how much?

Sol: V1 = 0.37 m3/kg P1 = 600Kpa V1 = 16m/s (V2= 0.62m3/kg) W= 135 KJ

V2 = 0.62 m3/kg P2 = 100Kpa V2= 270m/s Q= -9KJ

Q-W = m [(h2-h1) + V2 -V12 + g(Z2-Z1)]

2×1000 1000U2-U1 + -20.53KJ -ve since internal energy is increased

9. A steam turbine receives steam blow of 5000Kg/hr and delivers 550Kw. The heat loss from the basin and turbine is ~0 (i) Find change in enthalpy across the turbine if velocity at the entrance and difference in elevation at entrance+ exit are negligible. Take the velocity at the exit as 360 m/s. (ii) Find change in enthalpy across the turbine if velocity at entrance is 66 m/s and the inlet pipe is 3m above exhaust.

(i) Q=0 h2-h1=? W= 550Kw Z2-Z1= 0 m= 5000KG/hr = 5000 Kg/s = 5000 Kg/min

3600 60 1Kw = 60KJ/minV1 = 0 V2 = 360 m/s W= 550 KJ/min

60

Q-W = m [(h2-h1) + V2 -V12 + g(Z2-Z1)] h2-h1= -64.8

2×1000 1000

0- 550×60 = 5000 [(h2-h1) + (360) 2 -0 + g(0) ]

60 2×1000 1000

1Kw-hr = 3.6×106 N-m 1Kw-min = 3.6×10 6 N-m/min 60

Diameter ratio d1/d2 = √1000x10x4/√230.24x10-4x4 = 2.08 d1/d2 = 2.08

7 Air expands from 3bar to 1bar in a nozzle in velocity is 90 m/s & in temperature = 150oc,final velocity from SFEE k = 1.4 T1 = (150+273)from enthalpy equation ∆H = h1-h2 = cp(T1-T2)When pressure p1 & p2 the temperature are known

P1v1k = p2v2

k-----1 for a polytrophic process

P1/p2 = (v2/v1)K K—index constant for air

T1v1k-1 = T2V2

k-1----2 v2/v1 = (T1/T2)k-1 p1/p2 =(T1/T2)k/k-1

3 = (423/T2)1.4/0.4 T2 = 309k Cp of air 1.005 KJ/kg

∆H = 1.005(423.309) h1-h2 = ∆H = 114.5

v2 = √(2(h1-h2) + v12 = 487 m/s v2 = 487 m/s

8 Steam having specific enthalpy of 2930 KJ/kg flows through a turbine nozzle after expansion it leaves the nozzle with an enthalpy of 2255 kj/kg in adiabatic determine exit volume. If the inlet volume is 3600m/min.

v2 = √(2(h1-h2)103 +(3600/60))2 = √(2(2930-2255) 103 + 3600) = 1163.4 m/s

Reversible Process: The process said to be reversible if the system and all the elements of its surroundings can be restored to the respective initial stage after the process has occurred. Ex: Frictionless motion of solid, gradual extension of spring, slow frictionless adiabatic expansion of a gas, electrolysis of water.This process actually donot occur in nature. They are nearly idealization of actual process.

Irreversible Process: The change in the equilibrium of aa system and surrounding then the process that doesn’t come back to its initial state is called irreversible process.Ex: Combustion in an IC engine, diffusion of 2 unlike fluids into each other, electric current flows thru a resistor , flow of river water over a waterfall, plastic deformation of materials, magnetization of material exhibiting hystensis.

Turbine:-

Turbines & engines give power output whereas compressors & pump reg power for turbine which is well

insulated the flow velocities input are small & KE is neglected SFEE == Q-w=MA(h+v2/2+gz) Q = 0,

v2/2=0, gz=0

m(h2-h1)+w=0 w/m= h1-h2

In case of turbine h1>h2 i.e. the work is done by the fluid or gas at the expense of the enthalpy.

Compressor:- for a compressor work is done on the fluid & w is ‘-ve’ ... SFEE

w/m = h2-h1 the enthalpy of fluid m1 by an amt of work input.

Boiler:- is used to raise steam by supplying heat from an ext source the change in KE+PE is negligible. The work done is zero, than SFEE Q = (h2-h1)m.

Condenser:- is a device used for condensing the exhaust steam from turbine into water KE+PE & work done is zero therefore SFEE becomes Q = (h1-h2)m

1. A mix of gases expands at constant pressure from 1 mpa , 0.03m3 to 0.06 m3 with 84KJ +ve heat transfer. There is no work other than that is done on heat system, Find the energy for gaseous mixture.

V1 = 0.03m3 v2 = 0.06 m3 Q= +84KJ from 1st law of TD we have

Q-δw=∆E Q-W=∆E work done = ∫v2

pdv = 1x106(0.06-0.03)

V1 = 3 x 104 KJ

W.D = 30KJ ∆E = 54KJ

2. A turbine operating under steady flow condition receives 5000kg of steam per hour. Steam enters the turbine at a velocity of 3000m/min in elevation of 5m & specific enthalpy of 2787KJ/kg. it leaves the turbine at a velocity of 6000m/min at an elevation of 1m & specific enthalpy of 2259 KJ/kg what losses from the turbine to the surroundingamounts to 16,736KJ/hr. determine the power output of the turbine in kw.

m=5000kg/hr h1=2787KJ/kg h2=2259KJ/kg v1=3000m/min v2=6000 m/min z1=5m, z2=1m Q= -16,736KJ/hr.

from SFEE Q-W = m{(h2-h1) +v22- v1

1/2x1000 + g(Z2-Z1)}/1000

W=723.4849KW

-16736/60x60-w =5000/60x50{-528 + 27x103/60x2000 – 9.81(4)/1000}

-4.468-W = 1.39{-528 + 36x106 – 0.039} = 723.485 Kw

3. In a steady flow system 1kg of fluid per/sec enters the container at a pr. Of 1bar & leaves the system at a pr. of 10bar. The inlet & exit velocity are 40m/s & 20m/s respectively during the process 36000KJ of heat is transferred /hr to the control volume .The rise in enthalpy is 15KJ/kg. calculate the power dev. By the system neglect change in PE.

m=1kg/sec v1= 40 m/s v2= 20m/s Q = 36000x5/18 = 10,000 h2-h1 15 KJ

Ans W = 4.4 KW

4. A turbine operating under steady flow condition receives steam at 4500kg/hr. the steam enters the turbine at a velocity of 2500m/min at an elevation of 4m with specific enthalpy of 665 KJ/kg. it leaves the turbine at a velocity of 560 m/min at a elevation of 1m & specific enthalpy of 54KJ/min. The heat losses from the turbine to the surroundings amount to 4000 KJ/hr. determine the power o/p of machine in KW

m=4500 kg/hr v1=2500 m/min v2 = 560 m/min z1 = 4m Z2 = 1m, h1= 665 KJ/kg, h2=540KJ/kg, Q = 4000KJ/hr

Second Law of thermodynamics:Limitations of first law of thermodynamics:The 1st law states that” When a closed system undergoes any cyclic process, the cyclic integral of work= cyclic integral of heat ie energy from one transforms into another by an equivalent amount. The 1at law doesnot impose any restriction on direction of flow of heat and work. However complete conversion of heat into work is not possible as it is known from our experience. This kind of experimental evidence has led to the formation of second law of thermodynamics. A hot cup of coffee can by virtue of heat transfer to the surrounding but heat will not flow vice versa.1. Heat engine and heat power:

A heat engine is a thermodynamic cycle in which there is a net heat transfer to the system and net work transfer from the system without mass transfer across its boundary. The mass transfer across its boundary is called as heat engine. Ex: A steam power plant consists of a boiler, turbine, condenser and a pump in a steam engine but petrol and diesel engines are not heat engines since the mass flows in the form of air- fuel mixture and the exhaust gases crosses the boundary of the system.The H2o flows from pump to boiler in which an amount of heat Qh is transferred to H2o from a high

temperature reservoir like a furnace. The steam formed is expanded in the turbine delivering the net amount of work ‘W’. The steam then passes on the condenser where an amount of heat Ql is rejected from the steam.

The condensed H2o in a condenser is again pumped by pump to the boiler and the cycle repeats such a

device is called a heat engine.The substance to which and from which heat was transferred is called the working substance.From the 1st law of thermodynamics the netheat transfer = net workdone ie.. Qh- Ql =W

The efficiency of heat engine is defined as ratio of work output of cycle to heat input to the cycle. Thermal

efficiency is given by :

η = work done = W = Qh- Ql ηHE = 1- Ql

Heat supplied Qh Qh

Qh

Heat pump: If the cycle is performed in the opposite direction so that absorption of heat is at lower

temperature and reference of large amount of heat at ahigher temperature where net amount of workdone is

on the system. Such a device is called a heat pump or refrigerator of a reversed heat engine.

Example:-Vapour compressed refrigeration:

In this the working fluid NH4 or Freon 12is compressed in a compressor to

a higher temperature and pressure.It is passed into condenser where heat is rejected at constant pressure and

vapour becomes saturated liquid. This is expanded to a low pressure compressor in a throttle valve. The

temperature of liquid is reduced. This low temperature pressurized liquid is passed onto the evaporator where

it absorbs heat from a lower temperature body. Such a device is known as reversed heat engine. If the

purpose are utilized to deliver heat at high temperature aH.Then it is called as a heat pump whose purpose is

to abstract heat at lower temperature Ql then the device is called as a refrigerator.

The η of a refrigerator or heat pump is expressed by COP ( coefficient of performance).

COP of refrigerator: When the system works as a refrigerator then the main interest is refrigerating effect Ql+ energy input will be the work which is required for the plant.

Therefore COP = Heat absorbed at low temperature = Ql

Work supplied Qh- Ql

COPref = QlQh- Ql

COP of heat pump: When a system works as a heat pump is Qh the system will transfer to the surrounding

and energy input will be work.

Heat pump=heat delivered at high temperature =QH

Work supplied QH- QL

COPHp= QH QH- QL

Representation of heat engine and heat pump:

State of II nd law of thermodynamics: there are two statements and are as follows,

1. Kelvin Planck's statement &2. Clausius statement.

Kelvin Planck's statement: “it is impossible to construct a device which operates in cycle and produces no effect other then the rising of a weight and exchange of heat with a single reservoir“

It means that it is impossible to construct a HE which operates in a cycle and extracts a given amount of heat from the WTR and delivers an actual amount of work or “it is impossible to construct a W.E of 100% efficiency or its impossible to construct all heat into work.

Clausius statement : “ it is impossible to construct a device that operates in cycle and produces no effect other then transfer of heat from a colder body to hotter body”.

This means it is impossible to construct a refrigerator that operates without an input of work.

To show the violation of clausius statement implies violation of Kelvin Planck’s statement: Consider a HP and HE operates between the same 2 reservoirs i.e. HTR and LTR. The HP required no work to operate thus violating the clausius statement.

The HP extract an amount of heat from the LTR rejects the same amount of heat Q L to the HTR.

Now the H.E extracts an amount of heat Qh>QL from the HTR and rejects an amount of heat to QL to LTR.

The net amount of W.D W=Qh-Ql. Since there is no change in heat transfer in LTR the heat pump, the HTR and the HE together can be considered as a device which absorbs an amount of heat from the HTR and produces an equal amount of work W without produces any change in LTR which infact violates the Kelvin Planck’s statement. Hence the above statement is proved.

To show the violation of Kelvin Planck’s statement implies violation of clausius statement:

Consider a HE and HP working between HTR and LTR. The HE operates in cyclic process takes an amount of heat from HTR and converts it completely into work W violates K.P statement let this work= be utilized to drive a HP as shown. Let HP takes Ql amount of heat from LTR and pumps Qh+Ql amount of heat to HTR on analysis of a part of heat i.e. pumped to the high from LTR to HTR is delivered to HE while there remains a heat flow from LTR to HTR which infact violates clausius statement of second law of thermodynamics. Hence the above statement is proved.

Carnot cycle:

Process 1-2 isothermal expansion at constant temperature TH. Heat is transferred isothermally and reversibly

from HTR at temperature TH. to the working fluid (steam organ) the heat observed is QH.

Process 2-3 reservoir adiabatic expansion of a gas. The process is processed by thermodynamic insulated system carrying out expansion reversibly and the temperature of working fluid falls from temperature.

Process 3-4 reversible isotherm, compression of a gas at constant temperature TL with this prog system, is

brought in contact with LTR at temperature TL and heat is transfer, reversibly and isothermally from the

working fluid to LTR.Heat rejected during the process is QL.

Process 4-1 reverse adiabatic compression prog. The prog is performed by thermally regulating the system a carrying out compression prog, until working fluid is returned to its initial state. During this process the temperature of working fluid risen from lower temperature TL to higher temperature TH as shown in P-V

and T-S diagram. Since an amount of heat QH is transfer to the system to HTR and an amount of heat QL is rejected from

system to LTR. The net amount of heat transfer QW- QL will be numerically equal to the work done by the

system, I.e. W= QN- QLThermal efficiency of the Carnot engine is given by,Carnot= WD = QH - QL in terms of temperature Carnot= TH - TL QH QH TH

Carnot cycle for a heat pump:

The Carnot cycle is carried out in a reversed manner. The Carnot engine will work as a heat pump.Process 1-4 reverse adiabatic expansion process. It is carried out by thermally insulate of the system and carrying out the expansion process, until temperature of the working fluid form the TH to TL.

Process 4-3 reverse isothermal expansion. During the process, heat is transferred from LTR to working fluid.Process 3-2 reverse adiabatic compression process. During the process the system is thermally insulated and work is done on the system. The compression process is carried out until the temperature of working fluid, from TH to TL.

Process 2-1 reverse isothermal compression process. During this process, heat is transferred reversibly and isothermally, from the working fluid to HTR. The process is continued until initial state is reached,COP heat pump= QH = QH in terms of temperature COP = TH W QH - QL TH - TL It works as a refrigerator COPrefrig= QL = TL QH-QL Th-TL

Carnot cycle is not practicable due to the following reasons:1. Heat transfer at constant temperature is virtually impossible.

2. Reversible adiabatic expansion and compression without heat transfer requires a large displacement and the process must take place slowly. This is also not practicable. Therefore carnot cycle is only Ideal cycle or theoretical cycle.

1. A heat engine performs many cycles while doing work – 21.5×105 N-m and receiving a heat transfer of

90×105 N-m. Evaluate the efficiency of the engine and heat tranfer from working fluid.

W= 21.5×105 N-m ηHE = W = 23.8% W= QH-QL

QH=90×105 N-m QH QL= 68.5×105J

2. A heat engine working at a rate of 100Kw has an Efficiency of 20%. Evaluate the magnitude of heat transferred rate to and from the working fluid.

η=20% W= 100Kw 0.20= W/ QH QH= 500Kw

3. In a reversed heat engine the workdone on the engine is 75×103 N-m and heat transfer to the engine from

LTR is 220×103 J. Evaluate the heat transferred to the HTR and the COP of refrigerator and if it also works as a heat pump. Calculate its COP?

QH= W+ QL= (75+220)103 295×103J COPHE= QH = 295 = 3.93

W 75

COPrefrig= QL = 220×10 3 = 2.93

W 75×103

4. A heat engine is supplied with a heat of 1800 KJ\min and gives an output of 9.5 KW . Determine the thermal efficiency and the rate of heat rejection.

Q1=1800 KJ\sec W= 9.5 KJ\sec η = W= 31.6%

60 Q1W = Q1- Q2 Q2= Q1-W = 30 -9.5 = 20.5 KJ \sec

5. A cyclic heat engine operates between a source temperature of 800oc and a sink temperature of 30oc . What is the rate of heat rejection Kw + nett O/p of energy.

T1 = 800 + 273

T2 = 30 + 273

ηHE =W/ QH = QH-QL ηHE = T1- T2 = 71.76% QH T1

ηHE =W/ QH W= 0.7176 QH W = QH-QL QH = 3.54

QLQL= 1.393 Q2= 0.303 1 KW= QH-QL QL = 0.393 Kw QH= 1.393 kw

6. A domestic food freezer maintains a temperature of -15oc. The ambient temperature is 30oc. If heat leaks in the freezer at a continuous rate of 1.75 KJ/s. What is least power necessary to pump this heat continuously.

COPrefrig= T2 = -15 = -1/3 COPrefrig= 258/45 = 5.733

T1- T2 30 +15

COPrefrig= QL Qh = 2.05 KJ/s W= 0.305 Kw

Qh-QL

Perceptual motion machine II-Kind:- A machine, which extracts heat from an infinite and readily available source (thermal reservoir) and delivers equal amount of work container, to the surrounding such a device is PMM-II. The efficiency of such a machine would be 100%.

Carnot theorems:

Corollary 1: “ It is impossible to construct an engine that operates between two given reservoirs and is more efficient then a reversible engine operating between the same two reservoirs”.

Construct that there is an reversible engine operating between two given reservoirs that has a greater efficiency than a reversible engine operating between same two reservoirs. Let the heat TR to the IR, engine

the QH. The heat rejected is QL the work WI= QH-QL.Let the R, engine operates as a refrigerator or heat pump. The heat TR from LTR is QL. The heat TR to the

HTR is QH than the work required by the engine is WR = QH - QL.

nJ>nR => WI>WR=> QH -QL > QH - QL.

The engine can run as a HP or a refrigerator delivers an amount of positive work, then neet work done WN= WI - WRWN= QH - QL- QH + QL = QL - QL. WN= QL - QL.Now the HTR, IR and R, engine constitutes a PMM-II which operates in a cycle an extracts an amount of heat QL - QLand delivers an equal amount of work WNeet = QL - QL. Hence the theorem is

proved that it is impossible according to 2nd law of TD, therefore efficiency of IR, engine cannot be >

efficiency of R engine.

Corollary 2 :

“AD R engine have the same when operate between two given reservoirs”.Consider 2 R, engines R1+R2 working between 2 given reservoir. Let us assume nR1>nR2. Now R2 is

reserved to run as HP using some amount of work o/p of WR1 of the engine R1results in PMM-II from the

Carnot theorem. nR1>/nR2. Similarly assuming n R2>n R1 then reversing the engine R1 to work as a HP

and R2 as a HE. This also results in PMM-II therefore n R2>nR1.

Therefore, nR2>nR1.

Corollary 3: TD temperature scale or Abs temperature or Kelvin temperature scale: -

The zeroth law of TD provides a basis for temperature means, but the temperature depends on the thermometric property, of a particular substance and the mode of work of a thermometer.The efficiency of Carnot is independent of working fluid and depends only the temperature of 2 reservoirs. This provides a basis for develop of an absolute scale of temperature which is independent of nature of any thermometric substance.

Consider 3 reservoirs and 3 engines that operate on carnot cycle T1→ highest temperature , T3→ lowest temperature, T2→ intermediate. Engine marked R1-2

receives heat Q1 at temperature T1 and reject heat Q2 at T2. Engine R2-3receives heat Q2 at temperature T2 and rejects heat Q3 at T3. The engine R1-3receives heat Q1 at T1 and rejects Q3 at T3. When it is working betweent he same 2 reservoirs.Since η of carnot cycle, is a function of only the temperature. = 1- QL/QH =1- TL/ THη= Ψ (TL, TH)

If the engine R1-2 η12= 1- Q2/Q1 = 1- T2/T1 = Ψ (T2,T1)

η2-3 = 1- Q3/Q2 = 1- T3/T2 = Ψ (T3,T2)

η1-3 = 1- Q3/Q1 = 1- T3/T1 = Ψ (T3,T1)

where Ψ→ designate functional relations and is given by Ψ (T3,T1) = Ψ (T2,T1) * Ψ (T3,T2)→1The efficiency of engine R1-2 & R2-3 must be same together as the efficiency of the engine R1-3. Since they

operate between the same 2 reservoirs, at temperature T1, T3. This functional expression is expressed as Ψ (T1,T2) = f (T1)/f(T2)→2 Ψ (T2,T3) = f (T2)/f(T3)→3 Ψ (T1,T3) = f (T1)/f(T3)→4Substituting the values of the functions in the expression →1

f (T3)/f(T1) = f (T2)/f(T1) * f (T3)/f(T2) ie the temperature quantities is expressed in terms of heat quantity then Q3/Q1 = f (T3)/f(T1) OR QH/QL = TH/TL (general form)

Therefore η of carnot engine η= 1- TL/TH

Relationship between co& k

o :

If the heat engine is operating as on a carnot cycle the heat received at the temperature of steam point and heat rejected at a temperature of ice point is 100 ie. T steam point – T ice point = 100 →2

Tok = Toc + 273.16

Corollary 4: Clausius inequality theorem:

“ When a system executes a complete cyclic process than ∫ δQ/TAround the cycle less than 0 or in the limit equal to 0”Ie. Φ δq/T < 0

1. Proof for a reversible process: Φ δq/T = 0

Faraday’s Principle of increase of entropy:

“The entropy of an isolated system increases or the limit remains constant.”

Consider a system, undergoing a

cyclic process changing its state from 1 → 2along a reversible path R & returning to its original state 2 → 1 along two possible paths one irreversible & other reversible. Consider for the path R + R 2R 1R 2R 1I ∫ δQ/T = ∫δQ/T =0→1 For path R+I ∫δQ/T + ∫ δQ/T = 0→2 1R 2R 1R 2IFrom Clausius inequality theorem subtracting 2 from 1we get1R 1I 1I 1R∫ δQ/T = ∫δQ/T ∫δQ/T - ∫ δQ/T < 02R 2I 2I 2R 1R 1I 1The equating ∫ δQ/T = ds ∫δQ/T ≤ ∫ds ds ≥ ∫(δQ/T)I 2R 2I 2

If a system is isolated both heat to and from a system is o. Therefore δs ≥ 0

Combination of 1 st and 2 nd Law of thermodynamics:

Consider a system in a equilibrium state at temperature T suppose the system absorbs heat δQ & does work δW & its internal energy increases by δu then from 1st law of thermodynamics δq = δu + δW→1From 2nd law of thermodynamics the entropy relation ds = (δQ/T)Rev δQ= T ds→2

Substituting 2 in 1 Tds= δu+δw Tds = δu+pdv→3The enthalpy h = u + pv diff. dh = du +pdv + vdpdh= Tds+ vdp from 3

1. A carnot engine operates between T1 & T2o K. The output of engine is 0.6 times heat rejected. Given the temperature difference between source & sink is 200 k. Find the source & sink temperature. Find the thermal efficiency of the engine.

Sol: T1-T2 = 200o K W=0.6 Q2W= Q1-Q2 0.6 Q2 = Q1- Q2 Q1 = 1.6 Q2η= 1- Q2/Q1 = W/Q1 η = 37.55%

η= 1-T2/T1 37.55/100 = 1- 200/T1 T1= 533o K

T2= 333.3o K

2. A reversible engine works between 200oc & 60oc . Would the η be improved by raising the source

temperature to 300oc OR by lowering the sink temperature to 30oc.

Sol: ηHE = W/Q1 = (Q1-Q2)/Q1 = (T1-T2)/T1 = (260+273)-(60+273)

260+273

ηHE = 37.5%

T1 = 300oc T2 = 60oc ηHE = T1-T2 = 41.85%

T1

T1 = 260oc T2 = 30oc η = 43.15% T2 T2 T2 δQ/dT = 1KJ/K ∫ ds = ∫δQ/dT = ∫ 1 dT/T = log T2 – logT1 =1.87 KJ/K

T1 T1 T1

S2-S1 = 0.177 KJ/Kg K

Thermodynamic Process

Constant volume process: (Isochoric Process (CVP):

The movement of piston is prevented from moving so that the system volume remains unchanged. The

process is carried out by either by heating or by cooling . For a ConstantVolumeProcess P/T = C.

The workdone during the process : V2 δW= ∫ pdv = 0 W = 0 V1 1-2

The heat transferred from the 1st law of thermodynamics δQ = δu + δW δW = 0 Therefore δQ = δuThe change in internal energy δu = Cv dT u2-uT = Cv(T2-T1)

Change in enthalpy δh = Cp dt h2-h1 = Cp (T2-T1) 2 2 2Change in entropy ∫δs = ∫ δQ/T = ∫ Cv dT/T S2-S1 = Cv ln T2/T1 1 1 1

Constant pressure Process: (Isobaric process)

Consider a gas or fluid inside a cylinder as a system. The cylinder is closed by means of piston carrying weight. If heat is transferred the gases expand lifting the piston in upward direction. If process is carried out slowly pressure remains constant. The process is plotted in P-V diagram as shown.Ex: in boiler and condenser.

For CPP, PV/T = C P = C V/T = Constant V1/T1 = V2/T2 = C→1

2Workdone = W1-2 = ∫ pdv = P2(V2-V1)

1 2 2 2Heat transferred ∫δQ = ∫δu + ∫δW = u2- u1 + P(V2-V1) 1 1 1

= u2- u1 + PV2-PV1 = u2 + PV2 - u1 – PV1 = h2-h1 Q1-2 = h2-h1

Entropy δs = δQ/T we know δh = CpdT

Since heat transferred δQ = δH where δh = Cp dT S2-S1 = Cp ln T2/ T1 2 2∫δs =∫CpdT/T = Cp log T2/T1 1 1 S2-S1 = Cp ln V2/ V1

Constant temperature process: (Hyperbolic Process):

Many substance carry out such process when expanding in a fully resistant manner at constant temperature. For A CTP PV= CP1V1 = P2V2 = C

2 2 P1/P2 = V2/ V1 = C Workdone W1-2 = ∫ pdv = ∫ K/V dv From 1

PV = K P = K/V→1 1 1 W1-2 = K ln V2/ V1= P1V1 ln V2/ V1 = P2V2 ln V2/ V1

Heat transferred δQ = δu + δW δu = Cv dT (T is constant) = 0 u2 = u1Therefore δh = CpdT = 0 (T is const) h2= h1

Therefore δQ = δu + δW = δW (since δu=0) δQ = δW

Change in Entropy δs = δQ/T = PdV/T (since δQ = δW) 2 2We know PV = RT P/T = R/V ∫ δs = ∫ R/V dV ( R = Cp-Cv) 1 1 Universal Gas constant

S2-S1 = R ln V2/ V1 = (Cp-Cv) ln V2/ V1 S2-S1 = (Cp-Cv) ln V2/ V1

Isentropic process OR Reversible Adiabatic process:

An Adiabatic process is one in which no heat is transferred to or from the fluid or gas during the process. Such a process can be reversible or irreversible. A reversible adiabatic nonflow process is considered.

The general expression is PVγ = C→1 γ = K = 1.4 for air

2 2 2

Workdone W1-2 = ∫ pdv = ∫ C/Vγ dV = C ∫ V

-γdV = C [V2

1-γ – V1

1-γ]

1 1 1 1-γ

γ 1-γ γ 1-γ W1-2 = P2 V2 V2 - P1 V1 V1 = P2V2 – P1V1 (From 1) W = P2 V2 - P1 V1 1-γ 1-γ 1-γ

Heat transferred δQ = δu+δW For an Adiabaric process, δQ=0Therefore Q = δu+δW δu =-δW δu = Cv dT dh = Cp dT

Change in Entropy δs = δQ/T δQ = 0 .Therefore δs = 0 S2-S1=0 S2= S1

Polytropic Process PVn = C:

It is found that many process in practice approximate to a reversible law of the form

PVn = C. When n is a constant value. Both vapour and perfect gas obey this type of law in many non flow

processes. Such processes are internally reversible for a polytropic process.

1 1 1

Workdone = ∫ pdv = ∫ C/Vn dV = C ∫ V

-ndV = C [V2

1-n – V1

1-n]

2 2 2 1-n

n 1-n n 1-n W1-2 = P2 V2 V2 - P1 V1 V1 W1-2 = P2 V2 - P1 V1

1-n 1-n

Heat transferred δQ = u2 – u1 + R(T2 – T1) [ from PV = RT P1 V1 = R T1

1-n P1 V1 = R T2 ]

δu = Cv dT = Cv (T2– T1) u2 – u1 = Cv (T2– T1)

δu = Cp dT = Cp (T2– T1) h2 – h1 = Cp (T2– T1)

Change in Entropy δs = δQ/ T = [ Cv dT + R(T2 – T1) ] / T Since Cv dT= u2 – u1 1-n δs/δt = ds = Cv dT/T + pdV/T δu+δW = Cv dT/T + R/V dV T S2-S1 = Cv ln T2/ T1 + R ln V2/ V1

1. The pressure exerted by the air inside a vessel of volume 0.14 m3 was 106 N/m2. When temperature was

250oc due to cooling at constant volume the pressure falls to 3.5 * 105 N/m2. Determine the final temperature heat exchange & change in entropy if Cp value = 1.005 KJ/Kgk & Cv = 0.718 KJ/Kgk

Process

Constant pressure

Constant volume

Reversible adiabatic

Constant temperature Polytrophic

Index 'n' n = 0 n = ∞ n = r = k n = 1 n = n

PVT T2/T1 = V1/V2 P1V1 = P2V2 P1V1r = P2V2

rP1V1 = P2V2 P1V1

n = P2V2n

T2/T1 =

(V1/V2)r-1 T2/T1 = (V1/V2)n-1

T2/T1 =

(P2/P1)r-1/r T2/T1 = (P2/P1)n-1/n

Heat added Cp(T2-T1) Cv(T2-T1) δQ = 0

P1V1 lnV2/V1

δQ = CvdT+ R(T2-T1) 1-n

W.D ∫Pdv P(V2-V1) 0

P1V1-P2V2r-1

P1V1 lnV2/V1

P1V1-P2V2n-1

δs Cp ln T2/T1 Cv ln T2/T1 0 R ln P1/P2Cv lnT2/T1 + R ln V2/V1

V1 = 0.14m3 ; P1V1 = mRT1

m= P1V1/RT1T1 = (250+273) =523k ; R = 8314.3 = 287.6 J/kg k

m = 10 6 *0.14 = 0.931 ; 28.9 287.6*523 m = 0.931 kgP1V1/T1 = P2V2/T2 since (V1=V2) ; P1/P2 = T1/T2 T2 = 183.05k

Heat transferred δQ = mCv (T2-T1) ; δQ = -227kJ -ve sign indicates heat

Entropy s = Cv ln T2/T1 (m) s = -0.701 kJ/k transferred out of System

2) 1kg of air initially at a temperature of 165oc , pressure of 7bar is heated at constant pressure till volume is doubled. Determine work done, heat exchange, δs , Cp = 1.005KJ/kgk , Cv = 0.715KJ/kgk.

m = 1kg, T1 = 165+273 = 438K, V1 = V, V2 = 2V, P1 = 7bar

P1V1 = P2V2 V/438 = 2V/T2--- (1), T2 = 876K

T1 T2Heat transferred δQ = m.Cp (T2-T1) = 1*1.005(438) δQ = 440.49KJ Entropy = s = Cp ln T2/T1 (m) s = 0.697

W1-2 = P (V2-V1) V1 = 0.18m3 V2 = 0.36m3 from (1)

W1-2 = 7(0.18) = 126KJ (since 1bar = 1*105 N/m2)

3) Air at a pressure of 15bar & a temperature of 250oc expands according to law PV1.25 to a pressure of 1.5bar determined, w.d, δQ, and s. If it contains 0.9kg of air.

PVn = c PV1.25= C.

P1 = 15bar, P2 = 1.5bar, T1 = 523K, m = 0.9kg

n-1 1.25-1

n 1.25

T2/T1 = (P2/P1) T2/523 = (1.5/15)

T2 = 523(1.5/15)1/5 T2 = 329.9 K

W1-2 = P1V1 – P2V2 = RT1 – RT2 = R (T1-T2) * m

n-1 n-1 n-1 = 287.6(523-329.9)*0.9 W1-2 = 199.83KJ 0.25 δQ = m Cv (T2-T1) + W1-2 Q1-2 = 75.11KJ S2- S1 = mCv ln T2/T1 + mR ln V2/V1 S2- S1 = 0.178 KJ/KgK

S2- S1 = m (Cv ln T2/T1 + R ln (P1/P2)1/n)

4. 0.9 Kg of steam at a pressure of 15 bar and 250oc is expanded adiabatically & reversible tp a pressure of 1.5 bar γ= 1.4

γ-1

γW= P2V2- P1V1 T2/T1 = (P2/P1) γ -1 = mR(T2-T1) T2= 270.82 K -γ+1 W= 163.19 KJ

5. 1Kg of air initially at 1 bar and 156oc is compressed isothermally till the volume is reduced to 0.28 m3. Determine W, Q, δs, δu?

m = 1kg T = 429 K V2 = 0.28m3 m = P1V1/RT1

V1 = 1.234 m3 W = P1V1ln V2/V1= -183.03 KJ -ve’ sign indicates Wd on the system

δQ = δW = -183.03

δs = R ln V2/V1 = -426.57 KJ/K δu = 0

1000

6. A certain gas has Cp = 1.968 KJ/KgK & Cv = 1.507 KJ/KgK. Find its molecular weight & gas constant. A

Constant volume chamber of 0.3 m3 capacity contains 2 Kg of this gas at 5oc. Heat is transferred to the gas

until temperature is 100 oc. Find W, δQ, δs, δu, δh? _R = Cp- Cv = 0.461 MR= R = 8.314 M = 8.314 = 18.03 0.461Wd = ∫PdV = 0 ( CVP ) δQ = δu = Cv (T2 – T1) * m δQ = 286.33δh = mCp( T2- T1) = 373.92 KJ δh = 373.92S2- S1 = mCv ln T2/T1 δs = 0.88 KJ/KgK

7. 0.5 Kg of air compressed reversible & adiabatic from 80 KPa of 60oc to 400 Kpa & then expanded under constant pressure, to original volume . Sketch this process in PV diagram. Calculate δQ, δW, for the whole path.

P1V1 = mR T1 P1 = 80 Kpa P2= 400 Kpa

V1 = 0.598 m3 T1 = 333K m = 0.5

γ = 1.4 T2= 527.412 K

γ-1

γ

T2/T1 = (P2/P1)

W1-2= P1V1- P2V2

γ-1 W1-2 = mR (T2- T1) = -69.89 KJ W1-2= -69.89KJ

-γ+1 Q1-2= 0

γ γ γ γ γ

P1V1 = P2V2 P2 = P1(V1/ V2) V2 = P1V1 V2 = 0.819 m3

P2

CPP 2-3 W2-3 = ∫PdV = P (V3- V2) W2-3 = -163.6KJ V2/ T2 = V3/

T3 T3= 1668.7 K

Q2-3 = mCp(T3- T2) = 573.5 KJ

Total Wnett = W1-2 + W2-3 = 93.71 KJ Qnett = Q1-2 + Q2-3 = 573.5 KJ

Ideal Gases

A Thermodynamic system has a no of thermo static properties such as temperature, pressure, specific volume, electricalresistance, δu etc... Any equation that relates P, T and specific volume the substance, is called equation of state. (P, V, T)=0, if two of these parameters of property are known then the other one can be easily calculated by using equation of state. By Charles law P=R (T/V). PV=RT R= proportionality constant or Gas constant.

Ideal gas: the hypothetical gas which obeys the law PV=RT at all temperature and pressure is ideal gas. There is no ideal or perfect gas in nature. At a very slow pressureand hightemperature real property like N2, H2, O2, HC etc (remain)

behave nearly as the same way of ideal property. These property can change into liquid phase only if they are subjected to a great decrease in temperature and increase in pressure.

Vanderwaal’s Equation: the Vanderwaal’s equation by apply of laws of mechanics for in divide, molecules introduced two corrected terms in ideal gas equation and it is given

by (P+a/V2) (V-b)=RT. The coefficient ‘a’ was introduced for the existence if mutual attraction between

the molecules. The term O/V2= force of cohesion. The coefficient ‘b’ was introduced to account for the volume of molecules and is known as ‘co-volume’ real propertyhaveinter-molecular forces. The molecules which are close enough to a wall of container exerts an attractive force which prevents the partic, molecule to exert full pressure, on container wall as an ideal gas which is present in the container. This reducedcordinates In pressure, is internal pressure, vanderwaal suggested that internal pressure is

propositional to square of density and is =a/V2

i.e. pressure =kinetic pressure –internal pressure= (RT/V-a/V2),

P= (RT/V-a/V2).

a= constant accounts for reduction in total pressure, due to long range of attractive force.In real property the free volume is equal to total volume minus volumeof molecules themselves.Hence total volume =V-b, where V= accounts for reduction in volume due tofinite size of molecules of a real gas. This reduction in volume causes moremolecular collision and hence on increase in pressure, contents a and b canbe alter from the critical proper,

Pc= (a/27)*b2.

VC=3b Tc= (8a)/ (27Rb)

Compressibility chart:

Real gases are those that deviate form the relation PV=RT, so this relationhas to be modified in order to describe behavior of real gas. This is done bycompressibility factor Z=function of P and T and it is not constant for all the property. For an ideal gas Z=1. Since Z is a function of P and T.The plot is drawn Z v/s P with constant temperature. Lines on it as shown the values of Z for any gas can be determined, this is done by gasesin terms of reduced property.The reduced property is the ratio of exist of pressure, to critical pressure of a substance similarly for volume and temperature,Pr= (P/Pc) Vr= (V/Vc) Tr= (T/Tc).r = denotes reduced property,c = denotes property at critical state.

The relation among reduced property , Pr,Vr, Tr is known as law of corresponding states.

Ideal gas : it is a substance which has a equation of state PV=RT Where P= pressure in pascals,

V= specific volume in m3/kg, T= temparature in degree kelvin,

& R= universal gas constant in kj/kgok and it has different values for different property,multiple by ‘m’ we get PVm=mRT, but V=mv=total volume, therefore PV=mRT.--------------1If an ie=ideal gas undergoes a process from state 1 to 2 i,e. P1,V1,T1 and P2,V2,T2

with mass as constant, P1V1=mRT1 amd P2V2=mRT2,

therefore ( P1V1/ T1) = (P2V2/ T2)= mR=conatant.

One mole has a mass which is nearly equal to the molecular weight i,e mole of O2 =32kg,

one mole of N2=28kg if ‘n’ is the number of moles and ‘m’ is molecular weight of

gas then mass =m*n,substituing for mass then, PV=nMRT--------------2.

Avagadro’s law: it status that “Equal volumes of different gases at same temparature and pressure have same number of moles”. If they are two property occupaing the same volume and existing at same temparature and pressure then,

PAVA=nAMARATA and PBVB=nBMBRBTB,(PAVA/RATA)=(nAMA) and (PBVB)/( RBTB)=(nBMB) i,e nAMA= nBMB at constant P&T

Applying Avagadro’s law (PAVA/ nATA)=( RAMA) and (PBVB)/( nBTB)=( RBMB)

_ _I,e RAMA= RBMB=MR. Where MR=R and R=8314.3Nm/KJ mol k.

Mixture charecteristics: Total mass of the mixture consider a mixture of property a,b,c, etc., existing in equillibrium at presssure P, temparature T having the volume V. The total mass of the mixture = sum of the masses of the indivisual property ,mt = ma+mb+mc+md +----------

Mass function: it is defined as the ratio of mass of each component to the total mass of the mixture,ma/ mt=mfa, mb/ mT= mfb, and mc/ mT=mfc.

mfa+mfb+mfc+------------ =1

Mass fraction: it is the ratio of number of moles of each component to the total number of moles of the mixture.nt = na+nb+nc+--------

(na/ nt)=xfa (nb/ nt)=xfb (nc/ nt)=xfc (nd/ nt)=xfd

(xfa )+ (xfb )+ (xfc )+ (xfd )+------------- =1

Volume fraction: it is the ratio of volume of each component to the total volume of the mixture, i.e,va/vt =vfa vb/vT = vfb Vc/VT = vfc vfa+vfb+vfc+...... =1

Dalton’s law: total pressure of the mixture of a gas is equal to the sum of the pressure of the each component if it alone occupaies the volume of mixture of that temparature, i.e,pt = pa+pb+pc+-------- at constant temparature.

Gibb’s Dalton’s law: It states that “Entropy,enthalpy and internal energy of a mixture of gases is = Sum of Entropy,enthalpy and internal energy of individual component if it alone occupies the volume of mixture at that temparature”Ut =Ua+Ub+Uc H = Ha+Hb+Hc S = Sa+Sb+Sc at constant temparature

Relations involving pressure, volume & temparatures: For a mixture of gases ABC, _PV = mRT = nMRT = nRT _ _ _ _PVT = nT RT – (1), PVa = naRT – (2), PVb = nbRT – (3), PVc = nc RT – (4)Dividing equation (2),(3)&(4) by (1) Va/VT = na/nT, Vb/VT = nb/nT, Vc/VT = nc/nTVa/VT = xfa, Vb/VT = xfb, Vc/VT = xfc since: na/nT = xfa etc.We know that xfa + xfb + xfc + .............. = 1 i.e, Va/V + Vb/V + Vc/V = 1 There fore Va + Vb + Vc = V at constant temparature.

Amagat’s Ludel’s law : It states that the total volume of a mixture of gases is equal to the sum of the volume of individual components at the same pressure and temparature” For a mixture of gases Abc _

PV = nRT -- (5) _ _ _PaV = naRT PbV = nbRT PcV = ncRT –(6),(7) & (8)Dividing equation (6),(7)&(8) by (5), Pa/P = na/nT = x/a, Pb/P = nb/nT = x/b, Pc/P = nc/nT = x/c We know that xfa+xfb+xfc+...........= 1 = Pa/P + Pb/P + Pc/P = 1 There fore Pa + Pb + Pc = P at constant temparature.For a mixrure of gases PV = mTRT mT total mass, R Gas constant.For a mixrure of gases Abc PaV = maRaT, PbV = mbRbT, PcV = mcRcT P = Pa+Pb+Pc mTRT/V = maRaT/V+ mbRbT/V + mcRcT/V mTR = maRa+ mbRb+ mcRc R = 1/mT[maRa+mbRb+mcRc]—(A)

Molecular weight of mixture : From the relation Me = R for gases a,b,c _ _ Ra = R/Ma, Rb = R/Mb, _Rc = R/Mc substitute in (a) _ _ _R = 1/mT [maRa+mbR/Mb+mcR/Mc =1/mT [maRa/ Ma +mbR/Mb+mcR/Mc] _ _ _ _ _mTR/M = ma R/Ma+mbR/Mb +mcR/Mc = R[ma/ Ma + mb/Mb +mc/Mc]

1/M = [ma/mT Ma +mb/mT Mb+mc/mT Mc]

1/M = [mfa/Ma + mfb/Mb +Mfc/Mc]

M= 1____________ mfa/Ma + mfb/Mb +Mfc/Mc

3. Specific Weight of a mixture: mass basis:

1. Specific weight at constant volume: Consider the internal energy of a gas mixture Utotal = Ua + Ub . U = mu mTUT = maua + mbub + mcucUT = maua/mT + mbub/mT + mcuc/mT

UT = mfaua + mfbub + mfcuc→1 dUT = mfadua+ mfbdub+ mfcducdu= CvdT → CvdT = mfa Cva dT + mfb Cvb dT+ mfc Cvc dT

If it is on mole basis Cv dT= xfa Cva dT + xfb Cvb dT+ xfc Cvc dT

2. Specific heat at constant pressure on mass basis: HT= Ha+Hb+Hc H = mh mThT= maha+ mbhb+ mchc

Similiarly dh= mfadha+ mfbdhb+ mfcdhc We know that dh= CpdT

Therefore CpTdT= mfa Cpa dT + mfb Cpb dT+ mfc Cpc dT

1. Determine a. Pressure in the vessel. b. Mole fraction of each constituent. C. Gas constant of the mixture from the given data.

mCo2= 2Kg mN2= 1.5Kg T =300K V= 0.2m3 mT= 2+1.5 Kg =3.5Kg

Mass fraction: mf Co2= mCo2 = 2/3.5 = 0.571 mf N2 = 1.5/3.5 = 0.4285

_ mT _MR=R PV=MRT PCo2V = mCo2 RCo2 T MCo2 RCo2 = R

RCo2 = 8314.3/44 RCo2 = 188.96 RN2= 8314.3/28 = 296.94 N-m/KJK mol

PCo2= mCo2 RCo2 T = 2 ×188.96×300 = 566.88 KPa PCo2= 566.8 KPa

V 0.2 PN2= mN2 RN2 T = 1.5×296.94×300 = 668.11 Kpa PN2=668.11 KPa

V 0.2PT= PCo2 +PN2 = 566.88 KPa + 668.11 Kpa PT=1234.99 Kpa

Mole Fraction: m=nM nCo2= mCo2/ MCo2= 0.2/44 = 0.045 nCo2=0.045

n N2 =mN2/ MN2= 1.5/28 = 0.05357 nN2=0.05357 nT =0.099

xf Co2= nCo2 = 0.045/0.099 = 0.4589 xf N2= nN2 = 0.05357/ 0.099 = 0.5409

nT nT

Gas Constant: PV = mTRT 103×1234.98×0.2 = 3.5×R×300 R = 235.23 N-m/KJ molK

2. A vessel having a volume of 0.115 m3 is filled with a mix of 1.5 Kg Co2 & 1Kg 0f N2

At 25oc. Determine Mole fraction, mass fraction, gas constant.

mCo2= 1.5Kg mN2= 1Kg V= 0.115m3 T =273+25= 298 mT=1.5+1=2.5Kg

Mass fraction: mf Co2= 1.5/2.5 =0.6 mf N2 1/2.5=0.4

MR=R R Co2=83143/44 = 189.96 N-m/KJ molK

_MN2 RN2= R RN2= 8314.3/28 = 296.94 N-m/KJ molK

P Co2V= mCo2 RCo2T Therefore PCo2= 1.5×189.96×298 PCo2=738.36 KPa

0.115P N2V= mN2 RN2 T Therefore PN2= 1×296.94×298 PN2= 769.46 KPa

0.115 PT=1507.82KPa m=nM PT=1507.82KPa

nN2 =mN2/M = 1/28 =0.03571 nCo2= mCo2/M = 1.5/44 = 0.03409

nT= nN2/ nCo2 =0.0698 xf N2 = 0.03571/0.0698 xf N2 = 0.51159

xf Co2 = 0.03409/ 0.0698 xf Co2 = 0.4883