MOHAMMED ABASS ALI

Fundamentals of mechanical engineering

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What is thermodynamics

Thermodynamics is the branch of physics that studies

the effects of temperature and heat on physical systems

at the macroscopic scale. In addition, it also studies the

relationship that exists between heat, work and energy.

Thermal energy is found in many forms in today’s

society including power generation of electricity using

gas, coal or nuclear, heating water by gas or electric.

THE BASICS OF THERMODYNAMICS

Basic concepts

Properties are Features of a system which include

mass, volume, energy, pressure and temperature.

Thermodynamics also considers other quantities that

are not physical properties, such as mass flow rates

and energy transfers by work and heat.

Energy forms Fluids and solids can possess severalforms of energy. All fluids possess energy due totheir temperature and this is referred to as ‘internalenergy’. They will also possess ‘ potential energy’(PE) due to distance (z) above a datum level and ifthe fluid is moving at a velocity (v), it will alsopossess ‘kinetic energy’. If the fluid is pressurised, itwill possess ‘flow energy’ (FE). Pressure andtemperature are the two governing factors andinternal energy can be added to FE to produce asingle property called ‘enthalpy’.

Internal energy The molecules of a fluid possess bothkinetic energy (KE) and PE relative to an internal datum.Generally, this is regarded simply as the energy due to itstemperature and the change in internal energy in a fluid thatundergoes a temperature change is given by

ΔU = mcΔT

The total internal energy is denoted by the symbol ‘U’, whichhas values of J, kJ or MJ; also the specific internal energy ‘u’has the values of kJ/kg. Note: The change in temperature is ineither degrees Celsius or Kelvin.

potential energy When a mass ‘m’ kg is raised to

a height of ‘z’ metres above a datum level, a

lifting force is required. The work done in raising

the mass is force × distance moved, so

PE = mgz = Work where g is the gravity constant

Kinetic energy

Energy has been expended in doing this work and

has therefore been stored in the mass and will be

carried along with it. This energy is called ‘kinetic

energy’.

KE = m.v2/2

Enthalpy, Enthalpy (H) requires both pressure

and temperature; it therefore must possess

both flow (FE) and internal energy (U). These

two energies are added together

FE=PV

H = FE + U

The units are J for the total enthalpy or kJ/kg

for the specific enthalpy.

Boyle’s law

Boyle’s law states that provided the temperature‘T’ of a perfect gas remains constant, the volume‘V’ of a given mass of gas is inverselyproportional to its pressure ‘P’ of the gas, that is,P∝1/V or P × V = constant if the temperatureremains constant. If the gas experiences a changein state during an isothermal process, then

P1V1 = P2V2 = constant

EXAMPLE 1

A certain perfect gas is heated at a constant temperature from an initial state of 0.22 m3 and 325 kN/m2 to a final state of 170 kN/m2. Determine the final volume of the gas.

solution State 1: P1 = 325 kN/m2 and V1 = 0.22 m3.

State 2: P2 = 170 kN/m2 and V2 = ? From the equation, P1 ⋅ V1 = P2 ⋅ V2

P1 ⋅ V1 = P2 ⋅ V2

P2= P1. V1/V2

V2 = 0.4206 m3

Charles’s law Charles’s law states that provided

the pressure ‘P’ of a given mass of gas remains

constant, the volume ‘V’ of the gas will be directly

proportional to the absolute temperature ‘T’ of the

gas, that is, V ∞ T, or V = constant × ‘T’.

Therefore, V/T constant = for a constant pressure

‘P’. If the gas experiences a change in state during

a constant pressure process, then

V1/T1 = V2/T2 = constant

EXAMPLE 2

A quantity of gas is subjected to a constant

pressure process causing the volume of gas to

reduce from 0.54 m3 at a temperature of 345°C

to 0.32 m3. Calculate the final temperature of the

gas at the end of this process.

solution

From the question V1 = 0.54 m3 T1 = 345 +

273 K V2 = 0.32 m3

V1/T1 = V2/T2 = constant

T2= 366.22K

Universal gas law The universal gas equation

combines pressure, volume and temperature and

the relation between Boyle’s and Charles’s laws

is expressed in Equation

P . V/ T = R (constant)

where R is known as the universal gas constant.

That is

P1V1/T1 = P2V2/T2

or for ‘m’ kg, occupying V m3

PV = mRT

From the definition of the kilogramme-mole, for

‘m’ kg of the gas

m = nM

where n is the number of moles

where M is the molecular weight of the gas.

Oxygen has a molecular weight of 32

EXAMPLE

A volume of gas at a pressure of 325 kN/m2 is 0.22 m3 and

temperature of 618 K is compressed to a volume 0.16 m3 and a

pressure of 380 kn/m2. Determine the final temperature of the gas.

Solution State 1:

P1 = 325 kN/m3, V1 = 0.22 m3 and T1 = 618 K. State 2: P2 = 380

kN/m3, V2 = 0.16 m3 and T2 = ?

From Equation P1V1/T1 = P2V2/T2

T2 = 525.52 K.

The specific heat capacity of any substance isdefined as the amount of energy required toraise a unit mass through one degreetemperature rise. In thermodynamics, there aretwo specified conditions used:

1. Constant volume (Cv) 2. Constant pressure(Cp)

The two specific heat capacities do not havethe same value, and it is very important todistinguish between them.

SPECIFIC HEAT CAPACITY

specific heat capacity at constant volume (cv)

Consider one kg of a gas supplied with an

amount of heat energy sufficient to raise its

temperature by 1 K while the volume of the gas

constant, the amount of heat energy supplied,

is known as the specific heat capacity at

constant volume and is denoted by Cv. The

basic unit of Cv is J/kg K. For a reversible non-

flow process at constant volume:

dQ = mCvdT

For a perfect gas, the value of Cv will be constant for any one gas atall pressures and temperatures. From equation can be expandedas follows. Heat flow in a constant volume process between twostates:

Q12 = mCv(T2 − T1)

From the non-flow energy equation:

Q − W = (U2 − U1)

mCv(T2 − T1) − 0 = (U2 − U1)

(U2 − U1) = mCv(T2 − T1)

that is

dU = Q

Note: In a reversible constant volume process, there will be no work energy transfer as the piston will be unable to move; therefore, W = 0

EXAMPLE

A quantity of 4.5 kg of gas is heated at aconstant volume of 1.5 m3 and temperature20°C until the temperature rose to 150°C. Ifthe gas is assumed to be perfect, determine

1. The heat flow during the process 2. Thepressure at the beginning of the cycle 3. Thefinal pressure

Given:

Cv = 0.72 kJ/kg K and R = 0.287 kJ/kg K.

solution

m = 4.5 kg

V1 = 1.5 m3 V2 = 1.5 m3

T1 = 20 + 273 = 293 K T2 = 150 + 273 = 423

K

Cv = 0.72 kJ/kg K R = 0.287 kJ/kg K

Q12 = mCv(T2 − T1)

= 4.5 kg × 0.72 kJ/kg K × (423 – 293) K

= 421.2 kJ

2. From Equation, PV = mRT

For state 1:

P1V1 = mRT

P1=mRT 1/V1

4.5kg0.287kJ/kg K 293K/ 1.5m1 3=

P1 = 252.27 kN/m2

3. For state 2:

P2V2 = mRT2

P2=

4.5kg0.287kJ/kg K 423K/ 1.5m3

P2 = 364.20 kN/m2

specific heat capacity at constant pressure (cp)

When 1 kg of a gas is supplied with an amount

of heat energy sufficient to raise the

temperature by 1 K while the pressure of the

gas remains constant, the amount of heat

energy that is supplied is

known as the specific heat capacity at constantpressure and is denoted by Cp. The unit of Cp is J/kg K. For a reversible non-flow process at constantpressure:

dQ = mCpdT

For a perfect gas, the value of Cp is constant forany one gas at all pressures and temperatures.Equation dQ = mCpdT can be expanded asfollows: In a reversible constant pressure process,the heat flow

Q = mCp(T2 − T1)

relationship Between the specific heats Consider aperfect gas being heated at constant pressure from T1to T2. Referring to the non-flow equation, Q = U2 – U1 +W and the equation for a perfect gas, U2 – U1 = mCv(T2– T1), combining will give

Q = mCv(T2 – T1) + W

In a constant pressure process, the work done by thefluid is given by

W = P . ΔV

that is

W = P(V2 – V1)

Using the equation PV = mRT:

W = mR(T2 – T1)

Substituting:

Q = mCv(T2 – T1) + mR(T2 – T1)

= m(Cv + R)(T2 – T1)

mCp(T2 – T1) = m(Cv + R)(T2 – T1)

Therefore,

Cp = Cv + R

This equation may also be written as

R = Cp − Cv

specific heat ratio ‘γ’

The ratio of specific heat at constant pressure to thespecific heat at constant volume is given by thesymbol ‘γ’ (gamma).

γ= Cp/ Cv

From Equation Cp = Cv + R , it is clear that Cp has tobe greater than Cv for a perfect gas. It follows,therefore, that the ratio Cp/Cv = γ is always greaterthan unity. In general, ‘γ’ is 1.4 for diatomic gasessuch as carbon monoxide (CO), hydrogen (H2),nitrogen (N2) and oxygen (O2).

EXAMPLE

A particular perfect gas has a specific heat asfollows:

Cp = 0.846 kJ/kg K and Cv = 0.657 kJ/kg K

Determine the gas constant and the molecularweight of the gas.

solution

From Equation

R = Cp – Cv

R = Cp – Cv

that is

R = 0.846 – 0.657 = 0.189 kJ/kg K or R =

189 Nm/kg K.

M=Ro/R that is

M=8314.4 /189

= 44.0 kg/kmol