mechanical engineering · 2017-10-15 · mechanical engineering – gate exam universal gas...

Mechanical Engineering – GATE Exam

Thermodynamics

Symbol/Formula Parameter

M Molar mass (M/)

m Mass (M)

M

mn

Number of moles ()

E Energy or general extensive property

m

Ee

Specific molar energy (energy per unit mass) or general extensive property per unit mass

eMn

Ee

Specific energy (energy per unit mole) or general extensive property per unit mole

P Pressure (ML-1T-2)

V Volume (L3);

Specific volume or volume per unit mass, v (L3M-1) and the volume

per unit mole v (L3

-1)

T Temperature (Θ)

Density (ML-3); = 1/v.

x Quality

U Thermodynamic internal energy (ML2T-2);

Internal energy per unit mass, u (L2T-2), and the internal energy per

unit mole, u (ML2T-2

-1)

H = U + PV Thermodynamic enthalpy (ML2T-2);

Enthalpy per unit mass, h = u + Pv (dimensions: L2T-2) and the

internal energy per unit mole h (ML2T-2

-1)

S Entropy (ML2T-2Θ-1);

Entropy per unit mass, s(L2T-2Θ-1) and the internal energy per unit

mole s (ML2T-2Θ-1

-1)

W Work (ML2T-2)

Q Heat transfer (ML2T-2)

uW : The useful work rate or mechanical power (ML2T-3)

m : The mass flow rate (MT-1)


2

2V

: The kinetic energy per unit mass (L2T-2)

gz: The potential energy per unit mass (L2T-2)

Etot: The total energy = m(u +

2

2V

+ gz) (ML2T-2)

Q : The heat transfer rate (ML2T-3)

dEcvdt

: The rate of change of energy for the control volume.(ml2t-3)

M Molar mass (M/)

m Mass (M)

M

mn

Number of moles ()


m

Ee


eMn

Ee



V Volume (L3);

Specific volume or volume per unit mass, v (L3M-1) and the volume

per unit mole v (L3

-1)

T Temperature (Θ)


x Quality

U Thermodynamic internal energy (ML2T-2);

Internal energy per unit mass, u (L2T-2), and the internal energy per

unit mole, u (ML2T-2

-1)

H = U + PV Thermodynamic enthalpy (ML2T-2); we also have the enthalpy per unit mass, h = u + Pv (dimensions: L2T-2) and the internal energy

per unit mole h (ML2T-2

-1)

S Entropy (ML2T-2Θ-1);

Entropy per unit mass, s(L2T-2Θ-1) and the internal energy per unit

mole s (ML2T-2Θ-1

-1)


W Work (ML2T-2)




2

2V




2

2V

+ gz) (ML2T-2)


dEcvdt

: The rate of change of energy for the control volume.(ml2t-3)

M Molar mass (M/)

m Mass (M)

M

mn

Number of moles ()


m

Ee


eMn

Ee



V Volume (L3); we also have the specific volume or volume per unit

mass, v (L3M-1) and the volume per unit mole v (L3

-1)

T Temperature (Θ)


x Quality

U Thermodynamic internal energy (ML2T-2); we also have the internal energy per unit mass, u (L2T-2), and the internal energy per unit

mole, u (ML2T-2

-1)

H = U + PV Thermodynamic enthalpy (ML2T-2); we also have the enthalpy per unit mass, h = u + Pv (dimensions: L2T-2) and the internal energy

per unit mole h (ML2T-2

-1)


S Entropy (ML2T-2Θ-1); we also have the entropy per unit mass, s(L2T-

2Θ-1) and the internal energy per unit mole s (ML2T-2Θ-1

-1)

W Work (ML2T-2)




2

2V




2

2V

+ gz) (ML2T-2)


dEcvdt

: The rate of change of energy for the control volume. (ml2t-3)

Unit conversion factors

For metric units

Basic:

o 1 N = 1 kg·m/s2;

o 1 J = 1 N·m;

o 1 W = 1 J/s;

o 1 Pa = 1 N/m2.

Others:

o 1 kPa·m3 = 1 kJ;

o T(K) = T(oC) + 273.15;

o 1 L (liter) = 0.001 m3;

o 1 m2/s

2 = 1 J/kg.

Prefixes (and abbreviations):

o nano(n) – 10-9

;

o micro() – 10-6

;

o milli(m) – 10-3

;

o kilo(k) – 103;

o mega(M) – 106;

o giga(G) – 109.

o A metric ton (European word: tonne) is 1000 kg.

For engineering units


Energy:

o 1 Btu = 5.40395 psia·ft3 = 778.169 ft·lbf = (1 kWh)/3412.14 = (1 hp·h )/2544.5 =

25,037 lbm·ft2/s

2.

Pressure:

o 1 psia = 1 lbf/in2 = 144 psfa = 144 lbf/ft

2.

Others:

o T(R) = T(oF) + 459.67;

o 1 lbf = 32.174 lbm·ft/s2;

o 1 ton of refrigeration = 200 Btu/min.

Concepts & Definitions

Formula Units

Pressure FP

A

Pa

Units 21 1 /Pa N m 51 10 0.1bar Pa Mpa

1 101325atm Pa

Specific Volume Vv

m

3 /m kg

Density m

V

1

v

3/kg m

Static Pressure Variation P gh , Pa

Absolute Temperature ( ) ( ) 273.15T K T C

Properties of a Pure Substance

Formula Units

Quality vapor

tot

mx

m (vapour mass fraction)

1liquid

tot

mx

m (Liquid mass fraction)

Specific Volume f fgv v xv

3 /m kg

Average Specific Volume (1 ) f gv x v xv (only two phase mixture) 3 /m kg

Ideal –gas law cP P cT T

1Z

Equations Pv RT PV mRT nRT


Universal Gas Constant 8.3145R /kJ kmol K

Gas Constant RR

M M = molecular mass

/kJ kg K

Compressibility Factor Z Pv ZRT

Reduced Properties r

c

PP

P , r

c

TT

T

Work & Heat

Formula Units

Displacement Work 2 2

1 1W Fdx PdV

J

Integration 2

2 11

( )W PdV P V V J

Specific Work Ww

m (work per unit mass)

/J kg

Power (rate of work) W FV PV T W

Velocity V r /rad s

Torque T Fr Nm

Polytropic Process ( 1)n 1 1 2 2

n n nPV Const PV PV nPv C

Polytropic Exponent 2

1

1

2

ln

ln

PP

nV

V

n=1 1 1 2 2PV Const PV PV

Polytropic Process Work 1 2 2 2 1 1

1( ) 1

1W PV PV n

n

J

n=1 2

1 2 2 21

lnV

W PVV

J

Adiabatic Process 0Q

Conduction Heat Transfer dTQ kA

dx , k =conductivity

W

Convection Heat Transfer Q hA T , h =convection coefficient W

Radiation Heat Transfer 4 4( )s ambQ A T T W

Terminology: Q1 = heat Q2 = heat transferred during the process between state 1 and state 2


Q = rate of heat transfer W = work

1W2 = work done during the change from state 1 to state 2 W = rate of work = Power. 1 W=1 J/s

The First Law of Thermodynamics

Formula Units

Total Energy ( ) ( )E U KE PE dE dU d KE d PE J

Energy 2 1 1 2 1 2dE Q W E E Q W J

Kinetic Energy 20.5KE mV J

Potential Energy 2 1 2 1( )PE mgZ PE PE mg Z Z J

Internal Energy liq vap liq f vap gU U U mu m u m u

Specific Internal Energy of Saturated Steam (two-phase mass average)

(1 ) f g

f fg

u x u xu

u u xu

/kJ kg

Total Energy 2 2

2 12 1 2 1 1 2 1 2

( )( )

2

m V VU U mg Z Z Q W

J

Specific Energy 20.5e u V gZ

Enthalpy H U PV

Specific Enthalpy h u Pv /kJ kg

For Ideal Gasses ( )Pv RT and u f T

Enthalpy h u Pv u RT

R Constant ( ) ( )u f t h f T

Specific Enthalpy for Saturation State (two-phase mass average)

(1 ) f g

f fg

h x h xh

h h xh

/kJ kg

Specific Heat at Constant Volume

1 1v

v v v

Q U uC

m T m T T

( ) ( )e i v e iu u C T T

Specific Heat at Constant Pressure

1 1p

p p p

Q H hC

m T m T T

( ) ( )e i p e ih h C T T

Solids & Liquids Incompressible, so v=constant

c pC C C (Tables A.3 & A.4)

2 1 2 1( )u u C T T

2 1 2 1 2 1( )h h u u v P P


Ideal Gas h u Pv u RT

2 1 2 1( )vu u C T T

2 1 2 1( )ph h C T T

Energy Rate ( )E Q W rate in out

2 1 1 2 1 2 ( )E E Q W change in out

First-Law Analysis for Control Volume

Formula Units

Volume Flow Rate V VdA AV (using average velocity)

Mass Flow Rate Vm VdA AV A

v (using average values)

/kg s

Power pW mC T

vW mC T

Vmv

W

Flow Work Rate flowW PV mPv

Flow Direction From higher P to lower P unless significant KE or PE

Total Enthalpy 212toth h V gZ

Instantaneous Process

Continuity Equation

. .C V i em m m

Energy Equation

. . . . . .C V C V C V i tot i e tot eE Q W m h m h First Law

2 21 1( )2 2i i i e e e

dEQ m h V gZ m h V gZ W

dt

Steady State Process A steady-state has no storage effects, with all properties constant with time

No Storage . . . .0, 0C V C Vm E

Continuity Equation

i em m (in = out)

Energy Equation

. . . .C V i tot i C V e tot eQ m h W m h (in = out) First Law

2 21 1( )2 2i i i e e eQ m h V gZ W m h V gZ

Specific Heat Transfer

. .C VQq

m

/kJ kg

Specific Work . .C VW

wm

/kJ kg

SS Single Flow Eq.

tot i tot eq h w h (in = out)

Transient Process Change in mass (storage) such as filling or emptying of a container.


Continuity Equation

2 1 i em m m m

Energy Equation

2 1 . . .C V C V i tot i e tot eE E Q W m h m h

2 2

2 1 2 2 2 2 1 1 1 11 1

2 2E E m u V gZ m u V gZ

2 2

. 2 2 2 2 1 1 2 1 . .. .

1 12 2C V i tot i e tot e C V

C V

Q m h m h m u V gZ m u V gZ W

The Second Law of Thermodynamics

Formula Units

All ,W Q can also be rates ,W Q

Heat Engine HE H LW Q Q

Thermal efficiency

1HE LHE

H H

W Q

Q Q

Carnot Cycle 1 1L L

Thermal

H H

Q T

Q T

Real Heat Engine

1HE LHE Carnot HE

H H

W T

Q T

Heat Pump HP H LW Q Q

Coefficient of Performance

H HHP

HP H L

Q Q

W Q Q

Carnot Cycle H H

HP

H L H L

Q T

Q Q T T

Real Heat Pump

H HHP Carnot HP

HP H L

Q T

W T T

Refrigerator REF H LW Q Q

Coefficient of Performance

L LREF

REF H L

Q Q

W Q Q

Carnot Cycle L L

H L H L

Q T

Q Q T T

Real Refrigerator

L LREF Carnot REF

REF H L

Q T

W T T

Absolute Temperature L L

H H

T Q

T Q


Entropy Formula Units

Inequality of Clausis 0

Q

T

Entropy

rev

QdS

T

/kJ kgK

Change of Entropy 2

2 1

1 rev

QS S

T

/kJ kgK

Specific Entropy (1 ) f gs x s xs

f fgs s xs

/kJ kgK

Entropy Change

Carnot Cycle Isothermal Heat Transfer:

2

1 22 1

1

1

H H

QS S Q

T T

Reversible Adiabatic (Isentropic Process):

rev

QdS

T

Reversible Isothermal Process:

4

3 44 3

3 rev L

QQS S

T T

Reversible Adiabatic (Isentropic Process): Entropy decrease in process 3-4 = the entropy increase in process 1-2.

Reversible Heat-Transfer Process

2 2

1 22 1

1 1

1 1 fg

fg

rev

hqQs s s Q

m T mT T T

Gibbs Equations Tds du Pdv

Tds dh vdP

Entropy Generation gen

QdS S

T

irr genW PdV T S

2 2

2 1 1 2

1 1

gen

QS S dS S

T

Entropy Balance Equation Entropy in out gen

Principle of the Increase of Entropy

. . 0net c m surr gendS dS dS S

Entropy Change

Solids & Liquids 2

2 1

1

lnT

s s cT

Reversible Process: 0gends

Adiabatic Process: 0dq


Ideal Gas Constant Volume:

2

22 1 0

11

lndT v

s s Cv RvT

Constant Pressure: 2

22 1 0

11

lndT P

s s Cp RPT

Constant Specific Heat: 2 22 1 0

1 1

ln lnT v

s s Cv RT v

2 22 1 0

1 1

ln lnT P

s s Cp RT P

Standard Entropy

0

00

Tp

T

T

Cs dT

T

/kJ kgK

Change in Standard Entropy 0 0 2

2 1 2 11

lnT T

Ps s s s R

P

/kJ kgK

Ideal Gas Undergoing an Isentropic Process

2 22 1 0

1 1

0 ln lnT P

s s Cp RT P

0

2 2

1 1

RCpT P

T P

but 0 0

0 0

1p v

p p

C CR k

C C k

,

0

0

p

v

Ck

C = ratio of specific heats

1

2 1 2 1

1 2 1 2

,

k k

T v P v

T v P v

Special case of polytropic process where k = n: kPv const

Reversible Polytropic Process for Ideal Gas

1 1 2 2

n n nPV const PV PV 1 1

2 1 2 2 1

1 2 1 1 2

,

nn nnP V T P V

P V T P V

Work 2 2

2 2 1 1 2 11 2

1 1

( )

1 1n

PV PV mR T TdVW PdV const

V n n

Values for n Isobaric process: 0,n P const

Isothermal Process: 1,n T const

Isentropic Process: ,n k s const

Isochronic Process: ,n v const


Second-Law Analysis for Control Volume Formula Units

2nd Law Expressed as a Change of Entropy

. .c mgen

dS QS

dt T

Entropy Balance Equation rateof change in out generation

. . . .C V C Vi i e e gen

dS Qm s m s S

dt T

where . . . . ...C V c v A A B BS sdV m s m s m s

and . . ...gen gen gen A gen BS s dV S S

Steady State Process . . 0C VdS

dt

. .

. .

C Ve e i i gen

C V

Qm s m s S

T

Continuity equation i em m m . .

. .

( ) C Ve i gen

C V

Qm s s S

T

Adiabatic process e i gen is s s s

Transient Process . .

. .

C Vi i e e genC V

Qdms m s m s S

dt T

. .2 2 1 1 1 2. .

0

t

C Vi i e e genC V

Qm s m s m s m s dt S

T

Reversible Steady State Process

If Process Reversible & Adiabatic

e is s e

e i

i

h h vdP

2 2

2 2

( )2

( )2

i ei e i e

e

i ei e

i

V Vw h h g Z Z

V VvdP g Z Z

If Process is Reversible and Isothermal . .

. .

. .

1 C Ve i C V

C V

Qm s s Q

T T

or . .C Ve i

QT s s q

m

e

e i e i

i

T s s h h vdP


Incompressible Fluid

2 2

02

e ie i e i

V Vv P P g Z Z

Bernoulli Eq.

Reversible Polytrophic Process for Ideal Gas

e

n n

i

w vdP and Pv const C

1

1 1

e e

ni i

e e i i e i

dPw vdP C

P

n nRPv Pv T T

n n

Isothermal Process (n=1) ln

e e

ei i

ii i

PdPw vdP C Pv

P P

Principle of the Increase of Entropy

. . 0net C V surrgen

dS dS dSS

dt dt dt

Efficiency

Turbine a i e

s i es

w h h

w h h

Turbine work is out

Compressor (Pump)

s i es

a i e

w h h

w h h

Compressor work is in

Cooled Compressor Tw

w

Nozzle 2

2

12

12

e

es

V

V Kinetic energy is out

Note:

°F = (°C x 9/5) + 32

°C = (°F - 32) x (5/9)

°K = °C + 273

Q = mC∆T thermal energy = mass x specific heat x change in T

Q = mHf thermal energy = mass x heat of fusion

Q = mHv thermal energy = mass x heat of vaporization

∆L = αLi∆T change in length = coefficient of expansion x initial length x change in T

∆V = βVi∆T change in volume = coefficient of expansion x initial volume x change in T

∆U = Q – W internal energy = heat energy - work


Plausible Physical Situations

Name State Variables P V T ΔEth Q Ws W

Insulated sleeve or rapid process

Add weight to or push down on piston

Adiabatic compression

PVγ = Const; TVγ-1 = Const

Up Down Up

nCvΔT > 0

0 -nCvΔT < 0 nCvΔT > 0

Insulated sleeve or rapid process

Remove weight from or pull up on piston

Adiabatic expansion

PVγ = Const; TVγ-1 = Const

Down Up Down

nCvΔT < 0

0 -nCvΔT > 0 nCvΔT < 0

Heat gas Locked piston or rigid container

Isochoric V fixed; P α T

Up Fixed Up

nCvΔT > 0

nCvΔT > 0

0 0

Cool gas Locked piston or rigid container

Isochoric V fixed; P α T

Down Fixed Down

nCvΔT < 0

nCvΔT < 0

0 0

Heat gas Piston free to move, load unchanged

Isobaric expansion

P fixed; V α T

Fixed Up Up

nCvΔT > 0

nCpΔT > 0

PΔV > 0 -PΔV < 0

Cool gas Piston free to move, load unchanged

Isobaric compression

P fixed; V α T

Fixed Down Down

nCvΔT < 0

nCpΔT < 0

PΔV < 0 -PΔV > 0

Immerse gas in large bath

Add weight to piston

Isothermal compression

T fixed at temperature of bath, PV = Const

Up Down Fixed

nCvΔT = 0

nRT*ln(Vf/Vi) < 0

nRT*ln(Vf/Vi) < 0

-nRT *ln(Vf/Vi) > 0

Immerse gas in large bath

Remove weight from piston

Isothermal expansion

T fixed at temperature of bath, PV = Const

Down Up Fixed

nCvΔT = 0

nRT*ln(Vf/Vi) > 0

nRT*ln(Vf/Vi) > 0

-nRT* ln(Vf/Vi) < 0

Unknown Unknown No Name PV/T = Const

? ? ?

nCvΔT ΔEth + Ws

∫PdV = ± area under curve in PV diagram

-∫PdV = ± area under curve in PV diagram

mechanical engineering · 2017-10-15 · mechanical engineering – gate exam universal gas...

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