111302 aero engineering thermodynamics by mr.suresh chandra khandai

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111302 Aero Engineering Thermodynamics by Mr.Suresh Chandra Khandai. Unit - I. Thermodynamic Systems, States and Processes. Objectives are to: define thermodynamics systems and states of systems explain how processes affect such systems - PowerPoint PPT Presentation

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111302 Aero Engineering Thermodynamics

byMr.Suresh Chandra Khandai

Unit - I

Thermodynamic Systems, States and Processes

Objectives are to:• define thermodynamics systems and states of systems• explain how processes affect such systems• apply the above thermodynamic terms and ideas to the laws of

thermodynamics

“Classical” means Equipartition Principle applies: each molecule has average energy ½ kT per in thermal equilibrium.

Internal Energy of a Classical ideal gasInternal Energy of a Classical ideal gas

At room temperature, for most gases:

monatomic gas (He, Ne, Ar, …) 3 translational modes (x, y, z)

kTEK2

3

diatomic molecules (N2, O2, CO, …) 3 translational modes (x, y, z) + 2 rotational modes (wx, wy)

kTEK2

5

pVkTNU2

3

2

3

Internal Energy of a Gas

pVkTNU2

3

2

3

A pressurized gas bottle (V = 0.05 m3), contains helium gas (an ideal monatomic gas) at a pressure p = 1×107 Pa and temperature T = 300 K. What is the internal thermal energy of this gas?

J105.705.0105.1 537 mPa

Changing the Internal Energy

U is a “state” function --- depends uniquely on the state of the system in terms of p, V, T etc.

(e.g. For a classical ideal gas, U = )

WORK done by the system on the environment

Thermal reservoir

HEAT is the transfer of thermal energy into the system from the surroundings

There are two ways to change the internal energy of a system:

Work and Heat are process energies, not state functions.

Wby = -Won

Q

Work Done by An Expanding Gas

The expands slowly enough tomaintain thermodynamic equilibrium.

PAdyFdydW

Increase in volume, dV

PdVdW +dV Positive Work (Work isdone by the gas)

-dV Negative Work (Work isdone on the gas)

A Historical Convention

Energy leaves the systemand goes to the environment.

Energy enters the systemfrom the environment.

+dV Positive Work (Work isdone by the gas)

-dV Negative Work (Work isdone on the gas)

Total Work Done

PdVdW

f

i

V

V

PdVW

To evaluate the integral, we must knowhow the pressure depends (functionally)on the volume.

Pressure as a Function of Volume

f

i

V

V

PdVW

Work is the area underthe curve of a PV-diagram.

Work depends on the pathtaken in “PV space.”

The precise path serves to describe the kind of process that took place.

Different Thermodynamic Paths

The work done depends on the initial and finalstates and the path taken between these states.

Work done by a Gas

Note that the amount of work needed to take the system from one state to another is not unique! It depends on the path taken.

We generally assume quasi-static processes (slow enough that p and T are well defined at all times):

This is just the area under the p-V curve

f

i

V

Vby dVpW

V

p p

V

p

V

dWby = F dx = pA dx = p (A dx)= p dV

Consider a piston with cross-sectional area A filled with gas. For a small displacement dx, the work done by the gas is:

dx

When a gas expands, it does work on its environment

What is Heat?

Q is not a “state” function --- the heat depends on the process, not just on the initial and final states of the system

Sign of Q : Q > 0 system gains thermal energyQ < 0 system loses thermal energy

Up to mid-1800’s heat was considered a substance -- a “caloric fluid” that could be stored in an object and transferred between objects. After 1850, kinetic theory.

A more recent and still common misconception is that heat is the quantity of thermal energy in an object.

The term Heat (Q) is properly used to describe energy in transit, thermal energy transferred into or out of a system from a thermal reservoir …

(like cash transfers into and out of your bank account)

Q U

An Extraordinary Fact

The work done depends on the initial and finalstates and the path taken between these states.

BUT, the quantity Q - W does not dependon the path taken; it depends only on the initial and final states.

Only Q - W has this property. Q, W, Q + W,Q - 2W, etc. do not.

So we give Q - W a name: the internal energy.

-- Heat and work are forms of energy transfer and energy is conserved.

The First Law of Thermodynamics

(FLT)

U = Q + Won

work doneon the system

change intotal internal energy

heat added

to system

or

U = Q - Wby

State Function Process Functions

1st Law of Thermodynamics

• statement of energy conservation for a thermodynamic system• internal energy U is a state variable• W, Q process dependent

system done work : positive

system addedheat : positive

by

to

W

Q

WQU

The First Law of Thermodynamics

bydWdQdE int

What this means: The internal energy of a systemtends to increase if energy is added via heat (Q)and decrease via work (W) done by the system.

ondWdQdE int

. . . and increase via work (W) done on the system.

onby dWdW

Isoprocesses

• apply 1st law of thermodynamics to closed system of an ideal gas

• isoprocess is one in which one of the thermodynamic (state) variables are kept constant

• use pV diagram to visualise process

Isobaric Process• process in which pressure is kept constant

Isochoric Process• process in which volume is kept constant

Isothermal Process• process in which temperature is held constant

Isochoric (constant volume)

Thermodynamic processes of an ideal gas( FLT: U = Q - Wby )

V

p

1

2

pVTNkU

0pdVWby

FLT: UQ Q

Temperature changes

Isobaric (constant pressure)

V

p

1 2

VpTNkU FLT: VpWUQ by 1

VppdVWby

Q

p

Temperature and volume change

Isothermal (constant temperature)

2

1

V

V 1

2by V

VnNkTdVpW

0U

FLT: byWQ

p

V

1

2

( FLT: U = Q - Wby )

V

1p

Q

Thermal Reservoir

T

Volume and pressure change

The First Law Of Thermodynamics

§2-1.The central point of first law

§2-2. Internal energy and total energy

§2-3.The equation of the first law

§2-4.The first law for closed system

§2-5.The first law for open system

§2-6.Application of the energy equation

§2-1.The central point of first law

1.Expression In a cyclic process, the algebraic sum of the

work transfers is proportional to the algebraic sum of the heat transfers.

Energy can be neither created nor destroyed; it can only change forms.

The first law of thermodynamics is simply a statement of energy principle.

§2-1.The central point of first law

2.Central point The energy conservation law is used to

conservation between work and heat.

Perpetual motion machines of the first kind.(PMM1)

Heat: see chapter 1; Work: see chapter 1;

§2-2.Internal Energy

1.Definition:

Internal energy is all kinds of micro-energy in system.

2. Internal energy is property

It include:

a) Kinetic energy of molecule (translational kinetic, vibration, rotational energy)

b) Potential energy

c) Chemical energy

d) Nuclear energy

§2-2.Internal Energy

3.The symbol u: specific internal energy, unit –J/kg, kJ/kg ; U: total internal energy, unit – J, kJ;4.Total energy of system E=Ek+Ep+U Ek=mcf

2/2 Ep=mgz ΔE=ΔEk+ΔEp+ΔU per unit mass: e=ek+ep+u Δe=Δek+Δep+Δu

§2-3. The equation of the first law

1. The equation

( inlet energy of system) – (outlet energy of system) = (the change of the total energy of the system)

Ein-Eout=ΔEsystem

§2-4.The first law in closed system1. The equation

Ein-Eout=ΔEsystem

WQ

§2-4.The first law in closed system

Q-W=ΔEsystem=ΔU

Q=ΔU+W

Per unit mass:

q= Δu+w

dq=du+dw

If the process is reversible, then:

dq=du+pdv

This is the first equation of the first law.

Here q, w, Δu is algebraic.

§2-4.The first law in closed system

The only way of the heat change to mechanical energy is expansion of working fluid.

§2-5. The first law in open system

1. Stead flow

For stead flow, the following conditions are fulfilled:

① The matter of system is flowing steadily, so that the flow rate across any section of the flow has the same value;

② The state of the matter at any point remains constant;

③ Q, W flow remains constant;

§2-5. The first law in open system2. Flow work

Wflow=pfΔs=pV

wflow=pvp

V

§2-5. The first law in open system

3. 技术功 “ Wt” are expansion work and the

change of flow work for open system.

4. 轴功 “ Ws” is “ Wt” and the change of kinetic

and potential energy of fluid for open system.

§2-5. The first law in open system

5. Enthalpy

for flow fluid energy:

+mcf2/2+mgzU+pV

H =U+pV unit: J, kJ

For Per unit mass:

h=u+pv unit: J/kg, kJ/kg

§2-5. The first law in open system6. Energy equation for steady flow open system

U1+p1V1H1, mcf1

2/2, mgz1

U2+p2V2H2 , mcf22/2, mgz2

QW

§2-5. The first law in open system

12

111 2

1mgzcmHQE fin

22

222 2

1mgzcmHWE fsout

0 systemE

0)2

1()

2

1( 2

2221

2111 mgzmcHWmgzcmHQ fsf

§2-5. The first law in open system

sf WzmgcmHQ 2

2

1

0)2

1()

2

1( 2

2221

211 gzchwgzchq fsf

Per unit mass:

sf wzgchq 2

2

1

§2-5. The first law in open system

vdphq

If neglect kinetic energy and potential energy , then:

twhq

If the process is reversible, then:

This is the second equation of the first law.

W

§2-5. The first law in open system7. Energy equation for the open system

Inlet flows Out flows

Q

1

2

… …

i

Open system

1

2

……

j

§2-5. The first law in open systemEnergy equation for the open system

...2

...2

...

)2

1()

2

1( systemjjfjj

n

i

iifi

n

iis EmzgchmzgchWQ

§2-6. Application of The Energy Equation1. Enginea). Turbines energy equation:

Ein-Eout=Esystem=0

Wi=H2-H1

wi=h2-h1

U1+p1V1H1, mcf1

2/2, mgz1

U2+p2V2H2

mcf22/2, mgz2

Q WiQ≈0

=0

=0

§2-6. Application of The Energy Equation

1. Engine

b). Cylinder engine energy equation:

Wt=H2-H1+Q=(U+pV) 2-(U+pV) 1 +Q

Q Wt

H1

H2

Ek1, Ep1≈0

Ek1, Ep1≈0

§2-6. Application of The Energy Equation

2. Compressors

Energy equation:

Wc=- Wt =H2-H1

Wc

H1

H2

Ek1, Ep1≈0

Ek1, Ep1≈0

Q≈0

§2-6. Application of The Energy Equation

3. Mixing chambers

Energy equation:

m1h1 + m2h2 -m3h3=0

Cold water: m1h1

hot water: m2h2

Mixing water: m3h3

§2-6. Application of The Energy Equation

4. Heat exchangers

Energy equation:

m1h1

m 2 h 2

m3h3

m4h4

m5h5

m6h6

(m1h1 + m2h2 + m3h3)-(m4h4 + m5h5 + m6h6)= 0

§2-6. Application of The Energy Equation

5. Throttling valves

Energy equation:

h1 -h2 =0

h1

h2

Unit - II

Air Cycles

OTTO CYCLE

OTTO CYCLE

Efficiency is given by

Efficiency increases with increase in compression ratio and specific heat ratio (γ) and is independent of load, amount of heat added and initial conditions.

1

11

r

r

1 0

2 0.242

3 0.356

4 0.426

5 0.475

6 0.512

7 0.541

8 0.565

9 0.585

10 0.602

16 0.67

20 0.698

50 0.791

CR ↑from 2 to 4, efficiency ↑ is 76%

CR from 4 to 8 efficiency is 32.6

CR from 8 to 16 efficiency 18.6

OTTO CYCLEMean Effective Pressure

It is that constant pressure which, if exerted on the piston for the whole outward stroke, would yield work equal to the work of the cycle. It is given by

21

32

21

VV

Q

VV

Wmep

OTTO CYCLEMean Effective Pressure

We have:

Eq. of state:

To give:

rV

V

VVVV

11

1

1

1

2121

1

101 p

T

m

RMV

r

TMRmp

Q

mep1

1

10

132

OTTO CYCLEMean Effective Pressure

The quantity Q2-3/M is heat added/unit mass equal to Q’, so

r

TRmp

Q

mep1

1

10

1

OTTO CYCLEMean Effective Pressure

Non-dimensionalizing mep with p1 we get

Since:

1011

1

1

TR

mQ

rp

mep

10 vcm

R

OTTO CYCLEMean Effective Pressure

We get

Mep/p1 is a function of heat added, initial temperature, compression ratio and properties of air, namely, cv and γ

11

1

1

11

rTc

Q

p

mep

v

Choice of Q’

We have

For an actual engine:

F=fuel-air ratio, Mf/Ma

Ma=Mass of air,

Qc=fuel calorific value

M

QQ 32

cyclekJinQFM

QMQ

ca

cf

/

32

Choice of Q’

We now get:

Thus:

M

QFMQ ca

rV

VVAnd

V

VV

M

MNow a

11

1

21

1

21

rFQQ c

11

Choice of Q’

For isooctane, FQc at stoichiometric conditions is equal to 2975 kJ/kg, thus

Q’ = 2975(r – 1)/r

At an ambient temperature, T1 of 300K and cv for air is assumed to be 0.718 kJ/kgK, we get a value of Q’/cvT1 = 13.8(r – 1)/r.

Under fuel rich conditions, φ = 1.2, Q’/ cvT1 = 16.6(r – 1)/r.

Under fuel lean conditions, φ = 0.8, Q’/ cvT1 = 11.1(r – 1)/r

OTTO CYCLEMean Effective Pressure

Another parameter, which is of importance, is the quantity mep/p3. This can be obtained from the following expression:

1

11

11

13

rTcQrp

mep

p

mep

v

Diesel CycleThermal Efficiency of cycle is given by

rc is the cut-ff ratio, V3/V2

We can write rc in terms of Q’:

1

111

1c

c

r

r

r

11

1

rTc

Qr

pc

We can write the mep formula for the diesel cycle like that for the Otto cycle in

terms of the η, Q’, γ, cv and T1:

11

1

1

11

rTc

Q

p

mep

v

Diesel CycleWe can write the mep in terms of γ, r and

rc:

The expression for mep/p3 is:

11

11

1

r

rrrr

p

mep cc

rp

mep

p

mep 1

13

DUAL CYCLE

Dual Cycle

The Efficiency is given by

We can use the same expression as before to obtain the mep.

To obtain the mep in terms of the cut-off and pressure ratios we have the following expression

11

111

1cpp

cp

rrr

rr

r

Dual Cycle

For the dual cycle, the expression for mep/p3

is as follows:

11

111

1

r

rrrrrrrr

p

mep cppcp

Dual Cycle

For the dual cycle, the expression for mep/p3

is as follows:

11

111

1

r

rrrrrrrr

p

mep cppcp

3

1

13 p

p

p

mep

p

mep

Dual Cycle

We can write an expression for rp the pressure ratio in terms of the peak pressure which is a known quantity:

We can obtain an expression for rc in terms of Q’ and rp and other known quantities as follows:

rp

prp

1

1

3

Dual Cycle

We can also obtain an expression for rp in terms of Q’ and rc and other known quantities as follows:

111

11

pvc rrTc

Qr

c

vp r

rTcQ

r1

111

Unit – IV & V

Refrigeration &

Air Conditioning

Objectives

• Basic operation of refrigeration and AC systems• Principle components of refrigeration and AC

systems• Thermodynamic principles of refrigeration cycle • Safety considerations

Uses of Systems

• Cooling of food stores and cargo• Cooling of electronic spaces and

equipment– CIC (computers and consoles)– Radio (communications gear)– Radars– ESGN/RLGN– Sonar

• Cooling of magazines• Air conditioning for crew comfort

Definition Review

• Specific heat (cp): Amount of heat required to raise the temperature of 1 lb of substance 1°F (BTU/lb) – how much for water?

• Sensible heat vs Latent heat

• LHV/LHF

• Second Law of Thermodynamics: must expend energy to get process to work

Refrigeration Cycle

• Refrigeration - Cooling of an object and maintenance of its temp below that of surroundings

• Working substance must alternate b/t colder and hotter regions

• Most common: vapor compression– Reverse of power cycle– Heat absorbed in low temp region and

released in high temp region

Generic Refrigeration Cycle

Thermodynamic Cycle

TypicalRefrigeration

Cycle

Components

• Refrigerant • Evaporator/Chiller • Compressor• Condenser• Receiver• Thermostatic

expansion valve (TXV)

Refrigerant• Desirable properties:

– High latent heat of vaporization - max cooling– Non-toxicity (no health hazard)– Desirable saturation temp (for operating pressure)– Chemical stability (non-flammable/non-explosive)– Ease of leak detection– Low cost– Readily available

• Commonly use FREON (R-12, R-114, etc.)

Evaporator/Chiller

• Located in space to be refrigerated

• Cooling coil acts as an indirect heat exchanger

• Absorbs heat from surroundings and vaporizes– Latent Heat of Vaporization– Sensible Heat of surroundings

• Slightly superheated (10°F) - ensures no liquid carryover into compressor

Compressor

• Superheated Vapor:– Enters as low press, low temp vapor– Exits as high press, high temp vapor

• Temp: creates differential (T) promotes heat transfer

• Press: Tsat allows for condensation at warmer temps

• Increase in energy provides the driving force to circulate refrigerant through the system

Condenser

• Refrigerant rejects latent heat to cooling medium

• Latent heat of condensation (LHC)

• Indirect heat exchanger: seawater absorbs the heat and discharges it overboard

Receiver

• Temporary storage space & surge volume for the sub-cooled refrigerant

• Serves as a vapor seal to prevent vapor from entering the expansion valve

Expansion Device

• Thermostatic Expansion Valve (TXV)

• Liquid Freon enters the expansion valve at high pressure and leaves as a low pressure wet vapor (vapor forms as refrigerant enters saturation region)

• Controls:– Pressure reduction– Amount of refrigerant entering evaporator

controls capacity

Air Conditioning

• Purpose: maintain the atmosphere of an enclosed space at a required temp, humidity and purity

• Refrigeration system is at heart of AC system

• Heaters in ventilation system

• Types Used:• Self-contained• Refrigerant circulating• Chill water circulating

AC System Types

• Self-Contained System– Add-on to ships that originally did not have AC

plants– Not located in ventilation system (window unit)

• Refrigerant circulating system– Hot air passed over refrigerant cooling coils

directly

• Chilled water circulating system– Refrigerant cools chill water– Hot air passes over chill water cooling coils

Basic AC System

Safety Precautions

• Phosgene gas hazard– Lethal – Created when refrigerant is exposed to high

temperatures

• Handling procedures– Wear goggles and gloves to avoid eye irritation and

frostbite

• Asphyxiation hazard in non-ventilated spaces (bilges since heavier than air)

• Handling of compressed gas bottles

THANK U

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