Introduction & Basic Concepts of Thermodynamics

Reading Problems2-1→ 2-8 2-58, 2-61, 2-84, 2-85

Introduction to Thermal SciencesWe will study the basic laws and principles that govern the three disciplines of thermal sciences,namely, thermodynamics, heat transfer and fluid mechanics.

Th

erm

od

yn

am

ics

Heat T

ran

sfer

Fluids Mechanics

Thermal

Systems

Engineering

Thermodynamics

Fluid Mechanics

Heat TransferConservation of mass

Conservation of energy

Second law of thermodynamics

Properties

Fluid statics

Conservation of momentum

Mechanical energy equation

Modeling

Conduction

Convection

Radiation

Conjugate

Thermodynamics: the study of energy, energy conversion and its relation to matter.

Heat Transfer: the study of energy in transit including the relationship between energy, matter,space and time.

Fluid Mechanics: the study of fluids at rest or in motion.

1

Thermodynamic Systems

Isolated Boundary

System

Surroundings

Work

Heat

System Boundary

(real or imaginary

fixed or deformable)

System

- may be as simple as a melting ice cube

- or as complex as a nuclear power plant

Surroundings

- everything that interacts

with the system

SYSTEM:

• thermodynamic systems are classified as either closed or open

Closed System:

– consists of a fixed mass

– NO mass crosses the boundary

– energy crosses the boundary as workand heat

Work

Heat

Mass is fixed

Open System:

– consists of a fixed volume

– mass crosses the boundary

– energy crosses the boundary as workand heat

Heat

Work

Mass out

Mass in

2

WORK & HEAT TRANSFER:

• a system interacts with its surroundings through work and heat transfer

• work and heat transfer are NOT properties

Thermodynamic Properties of Systems

Basic Definitions

Thermodynamic Property: Any observable or measurable characteristic of a system or anymathematical combination of measurable characteristics

Intensive Properties: Properties that are independent of the size (or mass) of the system

Extensive Properties: Properties that are dependent of the size (or mass) of the system

Intensive Properties

Pressure P N/m2 or PaTemperature T K

Extensive Properties

Volume V m3

Mass m kg

Other Properties (Intensive and Extensive)

Specific VolumeA V/m = v m3/kgTotal EnergyB E/m = e J/kgInternal EnergyB U/m = u J/kgEnthalpyB H/m = h = u+ Pv J/kgEntropyB S/m = s J/kg ·KSpecific Heat Cp, Cv J/kg ·KIdeal Gas Constant R = Cp− Cv J/kg ·KDensity ρ = 1/v = m/V kg/m3

Notes: A The term “specific” denotes that the property is independent of massand therefore intensive

B Internal energy, enthalpy and entropy are specific properties.The term “specific” is dropped.

3

Measurable Properties

• P, V, T, andm are important because they are measurable quantities

– pressure (P ) and temperature (T ) are easily measured intensive properties.Note: They are not always independent of one another.

– volume (V ) and mass (m) are easily measured extensive properties

• other thermodynamic quantities are determined using the Equations of State based on knownvalues of P, V, T, andm

Pressure

• Pressure =Force

Area; →

N

m2≡ Pa

absolute

pressure

gauge

pressure

vacuum

pressure

absolute

vacuum pressure

ABSOLUTE

ATMOSPHERIC

PRESSURE

Pre

ssure

Patm

• gauge pressure = absolute pressure - atmospheric pressure

Pgauge = Pabs − Patm when Pabs > Patm

• vacuum pressure

Pvac = Patm − Pabs when Pabs < Patm

• thermodynamics properties depend on absolute pressure

4

Temperature

• temperature is a pointer for the direction of energy transfer as heat

Q QT

A

TA

TA

TA

TB

TB

TB

TB

> <

0th Law of Thermodynamics: if system C is in thermal equilibrium with system A,and also with system B, then TA = TB = TC

• the 0th law makes a thermometer possible

CC

A

T = TA C

B

T = TB C

T = TA B

• experimentally obtained temperature scales: The Celsius and Fahrenheit scales are based onthe melting and boiling points of water

T (K) = T (◦C) + 273.15

T (R) = T (◦F ) + 459.67

• for temperature differences between two states, note that

∆T (K) = ∆T (◦C)

5

Other Properties

• Internal energy U [kJ ]:

– the energy of molecules within the system

– specific internal energy is an intensive property u = U/m [kJ/kg]

• Kinetic energy KE:

– the energy of motion relative to some reference frame

– KE =1

2m(V)2

• Potential energy PE:

– the energy of position within a gravitational field

– PE = mgz

State and Equilibrium

• the state of a system is its condition as described by a set of relevant energy related properties.

• a system at equilibrium is in a state of balance

Definitions

Simple Compressible Material: One only open to the compression-expansion reversible workmode.

State Postulate:

State Postulate (for a simple compressible system): The state of a simplecompressible system is completely specified by 2 independent and intensive properties.

• only PdV work is done

• in a single phase system, T, v, and P are independent and intensive (in a multiphasesystem however, T and P are not independent)

• if any two properties i.e. (P, T ), (P, v), (v, T ) are known, then all other propertiescan be determined

• if the system is not simple, for each additional effect, one extra property has to beknown to fix the state.

6

Example 1-1: A glass tube is attached to a water pipe. If the water pressure at the bottomof the glass tube is 115 kPa and the local atmospheric pressure is 92 kPa, determine howhigh the water will rise in the tube, in m. Assume g = 9.8 m/s2 at that location and takethe density of water to be 1000 kg/m3.

Step 1: Draw a clearly labeled diagram to represent the system

Step 2: State what the problem is asking you to determine.

Step 3: State all assumptions used during the solution process.

Step 4: Prepare a table of properties.

Step 5: Solve (start by writing a force balance on the system)

Step 6: Clearly identify your final answer

7

Properties of Pure Substances

Reading Problems4-1→ 4-7 4-41, 4-42, 4-79, 4-84, 4-92, 4-955-3→ 5-5

Pure Substances

• a Pure Substance is the most common material model used in thermodynamics.

– it has a fixed chemical composition throughout (chemically uniform)

– a pure substance is not necessarily physically uniform (different phases)

Phases of Pure Substances• 3 principal phases:

solid→ liquid→ gas

Solids:

– strong molecular bonds– molecules form a fixed (but vibrating) structure (lattice)

Liquids:

– molecules are no longer in a fixed position relative to one another

Gases:

– there is no molecular order– intermolecular forces≈ 0

Behavior of Pure Substances (Phase Change Processes)• all pure substances exhibit the same general behavior.

• Critical point: is the point at which the liquid and vapor phases are not distinguishable (wewill talk about this more)

• Triple point: is the point at which the liquid, solid, and vapor phases can exist together

1

sublimatio

n

vaporizatio

n

melti

ng

meltin

g

triple point

critical point

SOLID

VAPOR

LIQUID

vaporization

condensation

sublimation

sublimation

freezingmelting

P

T

substances that

expand on freezing

substances that

contract on freezing

Phase Diagram

• phase change processes:

Critical Point1 Triple Point2P (MPa) T (K / ◦[C]) P (MPa) T (K / ◦[C])

H2O 22.06 647.10/[373.95] 0.00061 273.16 [0.01]

O2 5.08 154.80/[−118.35] 0.000152 54.36/[−218.79]

CO2 7.39 304.2/[31.05] 0.517 216.55/[−56.6]

N 3.39 126.2/[−146.95] 0.0126 63.18/[−209.97]

He 0.23 5.3/[−267.85] 0.0051 2.19/[−270.96]

1 - see Table A-1 for other substances2 - see Table 4.3 for other substances

2

T − v Diagram for a Simple Compressible Substance

• consider an experiment in which a substance starts as a solid and is heated up at constantpressure until it all becomes as gas

P0

P0

P0

P0

P0

P0

dQ dQ dQ dQ dQ

S S

L LL

G G

S + G

L + G

SS + L

G

L

T

v

critical point

sta urate

dvap

orline

triple point line

satu

rate

dliq

uid

line

fusio

nlin

e

• depending on the prevailing pressure, the matter will pass through various phase transforma-tions. At P0:

3

The Vapor Dome

• the general shape of a P − v diagram for a pure substance is very similar to that of a T − vdiagram

v

critical point

vfg

subcooled

liquid

saturated

liquid line

L+G

vfvg

P

P (T)sat

saturated

vapor line

gas (vapor)

two-phase region

(saturated liquid &

saturated vapor)

Tcr

T

T

• we will express values of properties for conditions on the vapor dome as:

specific volume: vf , vg and vfg = vg − vf

internal energy: uf , ug and ufg = ug − uf

specific enthalpy: hf , hg and hfg = hg − hf

specific entropy: sf , sg and sfg = sg − sf

• in the two phase region, pressure and temperature cannot be specified independently

Psat = P (Tsat) ⇔ Tsat = T (Psat)

this only holds true under the vapor dome in the two-phase region.

The Two-Phase RegionQuestion: How can we describe the mixture? P and T are obviously not enough since they are

inter-dependent.

Answer: We must also specify how much of each phase is in the mixture, for instance at C inthe above sketch.

4

LL

G G

A BC

saturated

liquid

mixture of

saturated liquid

and vapor

saturated

vapor

all three systems have the same T and P

We need to define a new thermodynamic property: Quality≡ x =mg

m=

mg

mg +mf

L

Gx = m /mg

1-x = m /mf

total mass of mixture ⇒ m = mg +mf

mass fraction of the gas ⇒ x = mg/m

mass fraction of the liquid ⇒ 1− x = mf/m

Properties of Saturated Mixtures

• P and T are not independent in the wet vapor region.To fix the state we require (T, x), (T, v), (P, x) . . .

• use Table A-4 or Table A-5 if water mixture:1) is saturated, 2) has a quality, 3) liquid & vapor are present

• to define a state

– identify the phase, i.e. sub cooled, 2-phase, superheated

– find two independent properties

v =V

m=Vf + Vg

m=mfvf +mgvg

m= (1− x) vf + x vg

= vf + x(vg − vf)

= vf + x vfg

x =v − vfvfg

5

Properties of Superheated Vapor

• To fix the state we require (T, v), (P, T ) . . .

v

L

L+GG

T

T (P)sat

P

P

P = 100 kPa

state: 200 C and 100 kPao

T = 200 Co

Properties of Sub-cooled (Compressed) Liquid• can be treated as incompressible (independent of P )

v = v(T, P )⇒ v ≈ v(T ) = vf(T )

u = u(T, P )⇒ u ≈ u(T ) = uf(T )

s = s(T, P )⇒ s ≈ s(T ) = sf(T )

• for enthalpy:

h(T, P ) = hf(T ) + vf(T )[P − Psat(T )]

Example 2-1: A saturated mixture of R-134a with a liquid volume fraction of 10% and apressure of 200 kPa fills a rigid container whose volume is 0.5 m3. Find the quality of thesaturated mixture and the total mass of the R134-a.

6

Using the “Equation of State” to Calculate the Properties ofGaseous Pure Substances

• a better alternative for gases: use the Equation of State which is a functional relationshipbetween P, v, and T (3 measurable properties)

Ideal Gases

• gases that adhere to a pressure, temperature, volume relationship are referred to as idealgases

Pv = RT or PV = mRT

Real Gases

• experience shows that real gases obey the following equation closely:

Pv = ZRT (T and P are in absolute terms)

Calculation of the Stored Energy

• for most of our 1st law analyses, we need to calculate ∆E

∆E = ∆U + ∆KE + ∆PE

• for stationary systems, ∆KE = ∆PE = 0

• in general: ∆KE =1

2m(V2

2 − V21) and ∆PE = mg(z2 − z1)

Specific Heats: Ideal Gases• for an ideal gas

Cv =du

dT⇒ Cv = Cv(T ) only

Cp =dh

dT⇒ Cp = Cp(T ) only

• the calculation of ∆u and ∆h for an ideal gas is given as

∆u = u2 − u1 =∫ 2

1Cv(T ) dT (kJ/kg)

∆h = h2 − h1 =∫ 2

1Cp(T ) dT (kJ/kg)

7

Specific Heats: Solids and Liquids• for solids and liquids it can be shown (mathematically) that for incompressible materials:

Cp = Cv = C

• as with ideal gases, for incompressible materials C = C(T )

∆u =∫ 2

1C(T ) dT ≈ Cavg ∆T

∆h = ∆u+ ∆(Pv) = ∆u+ v∆P + P∆v↗0≈ Cavg∆T + v∆P

Example 2-2: Using water as the working fluid, complete the table shown below. Usinggraph paper, create a separate T − v diagram for each state point that clearly shows how youhave located the state point (approximately to scale). For instance, using State Point 1 as anexample, show a curve of T = 400 ◦C and a curve of P = 0.8 MPa. State Point 1 willbe at the intersection of these two curves. The intersection should have the correct relationshipto the vapor dome (i.e. inside or outside).

For each case, show all calculations necessary to establish the actual location of the statepoint. Identify the relevant charts or tables used to perform your calculations

In the table, under the column for quality (x), indicate the quality of the state point is underthe dome, CL if the state point is in the compressed liquid region and SV if the state pointis in the superheated vapor region.

State T P v x◦C MPa m3/kg

1 400 0.82 200 0.0053 3.5 0.054 0.01 20.05 0.2 0.306 100 5.07 300 0.58 70.0 0.609 250 0.125

8

First Law of Thermodynamics

Reading Problems3-2→ 3-7 3-39, 3-755-1→ 5-2 5-25, 5-29, 5-35, 5-42, 5-67, 5-72, 5-966-1→ 6-5 6-23, 6-37, 6-52, 6-61, 6-78, 6-87, 6-156

Control Mass (Closed System)In this section we will examine the case of a control surface that is closed to mass flow, so that nomass can escape or enter the defined control region.

Conservation of Mass: states that mass cannot be created or destroyed, is implicitly satisfied bythe definition of a control mass.

Conservation of Energy: which is the underlying principle for the 1st Law of Thermodynamics,states energy cannot be created or destroyed it can only change forms

energy entering − energy leaving = change of energy within the system

A Gas Compressor

1

Performing a 1st law energy balance:

InitialEnergyE1

+−

{Energy gainW1−2

Energy loss Q1−2

}=

FinalEnergyE2

E1 +W1−2 −Q1−2 = E2 ⇒ ∆E = Q−W

The differential form of the energy balance can be written as a rate equation by dividing throughby dt, a differential time, and then letting dt→ 0 in the limit to give

dE

dt=δQ

dt−δW

dt⇒

dE

dt= Q− W

Example 3-1: A fluid with an energy content of 800 kJ is contained in an uninsulated rigidcontainer that is equipped with a stirring device. 100 kJ of work is added to the tank usingthe stirring process and 500 kJ of heat is dissipated to the surroundings. If kinetic andpotential energy are assumed to be negligible, determine the final energy state of the fluid inthe tank.

Forms of Energy Transfer

Work Versus Heat• Work is macroscopically organized energy transfer.

• Heat is microscopically disorganized energy transfer.

Heat Energy• heat is defined as a form of energy that is transferred solely due to a temperature difference

(without mass transfer)

• modes of heat transfer:

2

– conduction: diffusion of heat in a stationary medium (Chapters 10, 11 & 12)

– convection: it is common to include convective heat transfer in traditional heat transferanalysis. However, it is considered mass transfer in thermodynamics. (Chapters 13 &14)

– radiation: heat transfer by photons or electromagnetic waves (Chapter 15)

Work Energy• work is a form of energy in transit. One should not attribute work to a system.

Mechanical Work

• force (which generally varies) times displacement

W12 =∫ 2

1F ds

Moving Boundary Work

• consider compression in a piston/cylinder,whereA is the piston cross sectional area (fric-tionless)

• the work is:

– +ve for compression

– −ve for expansion

• sometimes called P dV work orcompression /expansion work

W12 = −∫ 2

1F ds = −

∫ 2

1P ·A ds = −

∫ 2

1P dV

• a decrease in the volume, dV → −ve results in work addition (+ve) on the system

• polytropic processes: where PV n = C

3

Gravitational Work Work is defined as force through a distance

W12 =∫ 2

1Fds

Acceleration Work

• if the system is accelerating, the work associated with the change of the velocity can becalculated as follows:

W12 =∫ 2

1F ds =

∫ 2

1ma ds =

∫ 2

1mdVdt

ds =∫ 2

1m dV

ds

dt︸︷︷︸V

and we can then write

W12 =∫ 2

1mV dV = m

(V2

2

2−V2

1

2

)

Example 3-2: A pneumatic lift as shown in the figure below undergoes a quassi-equilibriumprocess when the valve is opened and air travels from tank A to tank B.

A AB

Bair

atm.

valve closed valve open

control mass

system

Ap

mp

state 1 state 2

The initial conditions are given as follows and the final temperatures can be assumed to be thesame as the initial temperatures.

Patm = 100 kPa 1 Pa = 1 N/m2

mp = 500 kg Ap = 0.0245 m2

VA,1 = 0.4 m3 VB,1 = 0.1 m3

PA,1 = 500 kPa PB,1 = 100 kPaTA,1 = 298 K TB,1 = 298 K = 25 ◦C

Find the final pressures PA,2 and PB,2 and the work, W12, in going from state 1 to state 2.

4

Control Volume (Open System)The major difference between a Control Mass and and Control Volume is that mass crosses thesystem boundary of a control volume.

CONSERVATION OF MASS:

m

mIN

OUT

mcv

cv

{rate of increase ofmass within the CV

}=

{net rate of

mass flow IN

}−{

net rate ofmass flow OUT

}

CONSERVATION OF ENERGY:

5

The 1st law states:

ECV (t) + δQ+ δWshaft + (∆EIN −∆EOUT )+

(δWIN − δWOUT ) = ECV (t+ ∆t)

Some Practical Assumptions for Control Volumes

Definitions

1. Steady: no change with respect to time

2. Uniform: no change with respect to position

Steady State Process: The properties of the material inside the control volume do not changewith time. For example

P1 P

2

P > P2 1

V < V2 1

Diffuser: P changes inside the control volume, butthe pressure at each point does not change withtime. [steady but not uniform]

Steady Flow Process: The properties of the material crossing the control surface do not changewith time. For example

y

x

T = T(y)

T ≠ T(t)

Inlet Pipe: T at the inlet may be different at differ-ent locations, but temperature at each bound-ary point does not change with time. [steadybut not uniform]

Uniform State Process: The properties of the material inside the control volume are uniform andmay change with time. For example

50 Co

50 Co

50 Co

50 Co

60 Co

60 Co

60 Co

60 Co

Q

t=0

t=10

Heating Copper: Cu conducts heat well, so that itheats evenly. [uniform but not steady]

6

Uniform Flow Process: The properties of the material crossing the control surface are spatiallyuniform and may change with time. For example

y

x

P ≠ P(y) Inlet Pipe: P at the inlet is uniformacross y. [uniform but not steady]

Example 3-3: Determine the heat flow rate, Q, necessary to sustain a steady flow processwhere liquid water enters a boiler at 120 ◦C and 10 MPa and exits the boiler at 10 MPaand a quality of 1 for a mass flow rate is 1 kg/s. The effects of potential and kinetic energyare assumed to be negligible.

L

G

H O ( )2

l

in

out

steam

Q

Example 3-4: Steam with a mass flow rate of 1.5 kg/s enters a steady-flow turbine with aflow velocity of 50 m/s at 2 MPa and 350 ◦C and leaves at 0.1 MPa, a quality of 1,and a velocity of 200 m/s. The rate of heat loss from the uninsulated turbine is 8.5 kW .The inlet and exit to the turbine are positioned 6 m and 3 m above the reference position,respectively. Determine the power output from the turbine.Note: include the effects of kinetic and potential energy in the calculations.

Q

W

in

out

7

Systems of Steady Flow ComponentsSteady flow devices can be assembled into closed-loop thermodynamic cycles. Examples include:liquid-vapour power plants (Rankine cycle), vapour-compression refrigeration systems and jet en-gines (Brayton cycles).

The Ideal Rankine Cycle

T

s

Q

Q

W

W

P = P

P = P

H

L

1

2

4

3

Q

Q

L

H

T

P

Boiler

Turbine

CondenserPump

• the water at state point 1 can be conveniently pumped to the boiler pressure at state point 2

• energy in the form of heat must be added to change the water at state point 2 to saturatedwater before it undergoes a phase transformation under the vapour dome

• saturated vapour or steam enters the turbine at state point 3

• as the vapour is expanded this energy is converted to turbine work

• heat is extracted from the condenser (heat exchanger) to ensure that only liquid water entersthe pump at state point 1

8

Example 3-5: For the steam power plant shown below,

T

s

Q

Q

W

W

P = P

P = P

H

L

1

2

4

3

Q

Q

L

H

T

P

Boiler

Turbine

CondenserPump

find: QH, QL, Wnet = WT − WP , and the overall cycle efficiency, ηR given the followingconditions:

P1 = 10 kPa P2 = 10 MPaP3 = 10 MPa P4 = 10 kPaT1 = 40 ◦C T3 = 530 ◦CT2 = 110 ◦Cx4 = 0.9 m = 5 kg/s

9

Entropy and the Second Law of Thermodynamics

Reading Problems7-1→ 7-3 8-29, 8-43, 8-58, 8-68, 8-997-6→ 7-10 8-148, 8-149, 8-159,8-1→ 8-10, 8-13 8-168, 8-183, 8-189

Second Law of ThermodynamicsThe second law of thermodynamics states:

The entropy of an isolated system can never decrease. When an isolated systemreaches equilibrium, its entropy attains the maximum value possible under theconstraints of the system

Entropy of a system plus its surroundings (i.e. an isolated system) can never decrease (2nd law).

System

Surroundings

Work

Heat

Non-Isolated Systems

- their entropy may decrease

Isolated System

- its entropy may

never decrease

• we have conservation of mass and energy, but not entropy. Entropy is not conserved.

• the 2nd law dictates why processes occur in a specific direction i.e., Sgen cannot be−ve

• Sgen is created by irreversibilities in a system, and therefore can be used to quantify theefficiency of a process. The smaller Sgen is, the more efficient the process.

(∆S)system + (∆S)surr. ≥ 0

1

Entropy

1. Like mass and energy, every system has entropy.

Entropy is a measure of the degree of microscopic disorder and represents our uncertaintyabout the microscopic state.

2. Reference: In a prefect crystal of a pure substance at T = 0K, the molecules are com-pletely motionless and are stacked precisely in accordance with the crystal structure.

3. Relationship to Work: For a given system, an increase in the microscopic disorder (that isan increase in entropy) results in a loss of ability to do useful work.

4. Work: Energy transfer by work is microscopically organized and therefore entropy-free.

5. Heat: Energy transfer as heat takes place as work at the microscopic level but in a random,disorganized way. This type of energy transfer carries with it some chaos and thus results inentropy flow in or out of the system.

6. Reversibility: According to the 2nd law, no real process can be reversed without leavingany trace on the surroundings. A reversible process is the theoretical limit for an irreversible(real) process.

Reversible ProcessA process 1 → 2 is said to be reversible if the reverse process 2 → 1 restores the system to itsoriginal state without leaving any change in either the system or its surroundings.

Example: Slow adiabatic compression of a gas

2

→ idealization where S2 = S1 ⇒ Sgen = 0

Reversible︸ ︷︷ ︸Sgen=0

+ Adiabatic Process︸ ︷︷ ︸Q=0

⇒ Isentropic Process︸ ︷︷ ︸S1=S2

But does:

Isentropic Process⇒ Reversible + Adiabatic

NOT ALWAYS - the entropy increase of a substance during a process as a result of irreversibilitiesmay be offset by a decrease in entropy as a result of heat losses.

Evaluating EntropyFor a simple compressible system, Gibb’s equations are given as

Tds = du+ Pdv

Tds = dh− vdP

• these relations are valid for open and closed, and reversible and irreversible processes

Tabulated Calculation of ∆s for Pure SubstancesCalculation of the Properties of Wet Vapor:

Use Tables A-4 and A-5 to find sf , sg and/or sfg for the following

s = (1− x)sf + xsg s = sf + xsfg

3

Calculation of the Properties of Superheated Vapor:

Given two properties or the state, such as temperature and pressure, use Table A-6.

Calculation of the Properties of a Compressed Liquid:

Use Table A-7. In the absence of compressed liquid data for a property sT,P ≈ sf@T

Calculation of ∆s for Incompressible Materials

Tds = du+ Pdv (the other T ds equation reduces the same way)

- for an incompressible substance, dv = 0, and Cp = Cv = C

ds =du

T= C

dT

T

s2 − s1 =∫ 2

1C(T )

dT

T

∆s = Cavg lnT2

T1

where Cavg = [C(T1) + C(T2)]/2

Calculation of ∆s for Ideal Gases

Ideal Gas Equation ⇒ Pv = RT

du = Cv dT ⇒ u2 − u1 = Cv(T2 − T1)

dh = Cp dT ⇒ h2 − h1 = Cp(T2 − T1)

There are 3 forms of a change in entropy as a function of T & v, T & P , and P & v.

s2 − s1 = Cv lnT2

T1

+R lnv2

v1

= Cp lnT2

T1

−R lnP2

P1

= Cp lnv2

v1

+ Cv lnP2

P1

4

If we don’t assume Cv and Cp to be constant i.e. independent of T , then we can use the ideal gastables (Table A-21) as follows

s2 − s1 = so(T2)− so(T1) +R ln(v2/v1)

s2 − s1 = so(T2)− so(T1)−R ln(P2/P1)

2nd Law Analysis for a Closed System (Control Mass)

MER TER

CM

dWdQ

TTER

The entropy balance equation is given as

(dS)CM︸ ︷︷ ︸≡ storage

=δQ

TTER︸ ︷︷ ︸≡ entropy flow

+ dSgen︸ ︷︷ ︸≡ production

Integrating gives

(S2 − S1)CM =Q1−2

TTER+ Sgen︸ ︷︷ ︸≥0

5

Example 4-1: A bicycle tire has a volume of 1200 cm3 which is considered to be con-stant during “inflation”. Initially the tire contains air at atmospheric conditions given asP0 = 100 kPa and T0 = 20 ◦C. A student then hooks up a bicycle pump and beginsto force air from the atmosphere into the tire. After pumping stops and a new equilibrium isreached, the tire pressure is 600 kPa and the air temperature in the tire is 20 ◦C.

a) Determine the mass [kg] of air added to the tire.b) Determine the minimum amount of work [kJ ] required to reach this

end state.c) If more than the minimum work is actually used, where does this

extra energy go? That is what becomes of the extra energy?

6

2nd Law Analysis for Open Systems (Control Volume)

TER

TER

MER

AA

B

B

FR

FR

CV

S= s m

S= -s m

B

A

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

1-2

B

A

A

B

A

B

B

A

A

B

S= - Qd

S= Qd

T

T

TER

TER

dQ

dQm

m

SCVdW

S=0

isolated S 0gen ≥

For the isolated system going through a process from 1→ 2

δSgen = (∆S)sys + (∆S)sur

δSgen = ∆SCV︸ ︷︷ ︸system

+

(−sAmA

1−2 + sBmB1−2 −

δQA1−2

TATER+δQB

1−2

TBTER

)︸ ︷︷ ︸

surroundings

7

or as a rate equation

Sgen =

(dS

dt

)CV

+

sm+Q

TTER

OUT

−

sm+Q

TTER

IN

This can be thought of as

generation = accumulation+ OUT − IN

Example 4-2: A vortex tube is a steady-state device that splits a high pressure gas streaminto two streams, one warm and one cool. During a controlled test, air entered the vortex tubeat 19.3 ◦C and 0.52 MPa. The warm air left the tube at 26.5 ◦C and the cool air left at−21.8 ◦C. Both exit streams were at 101.35 kPa. The ratio of the mass flow rate of thewarm air to that of the cool air was 5.39. There was no work transfer. The temperature of thesurroundings was 20 ◦C. For a compressed air flow rate of 1.5 kg/s, determine:

a) the rate of heat loss from the vortex tube to the surroundings, [kW ]b) the rate of entropy generation in the vortex tube, [kW/K]

warm

air

cool

air

compressed

air

1

2

3

T = 20 C0

o

8

Heat Engine

A heat engine is a device in which a working substance (control mass) undergoes a cyclic processwhile operating between two temperature reservoirs (TER).

combustion

gases

cooling

water

boiler

condenser

WP

WT

QH

QH

QL

QL

TER

TER

TH

TL

Wnet

heat

engines

isolated systems

S 0gen ≥

Wnet = QH −QL = QH

(1−

QL

QH

)

= QH

(1−

TL

TH−Sgen TL

QH

)

= QH

(1−

TL

TH

)︸ ︷︷ ︸Wmax possible

− TLSgen︸ ︷︷ ︸Wlost due to irreversibilties

The engine efficiency is defined as the benefit over the cost

η =benefit

cost=Wnet

QH

= 1−TL

TH−TLSgen

QH

9

The Carnot Cycle• based on a fully reversible heat engine - it does not include any of the irreversibilities asso-

ciated with friction, viscous flow, etc.

T

T

T

s s s

Q

Q

Q

W

P = P

P = P

H

H

L

L

1

1

2

4

4

3

in

out

where the efficiency is given as: η = 1−TL

TH⇐ Carnot efficiency

Practical Problems

• at state point 1 the steam is wet at TL and it is difficult to pump water/steam (two phase) tostate point 2

• it is difficult to devise a Carnot cycle to operate outside the wet vapor region

T

s s s

Q

Q

P = P

P

P

H

L

1

1

4

4

3

2

The net effect is that the Carnot cycle is not feasible for steam power plants.

10

Conduction Heat Transfer

Reading Problems10-1→ 10-6 10-20, 10-48, 10-59, 10-70, 10-75, 10-92

10-117, 10-123, 10-151, 10-156, 10-16211-1→ 11-2 11-14, 11-20, 11-36, 11-41, 11-46, 11-53, 11-104

Fourier Law of Heat ConductionWe can examine a small representative element of the bar over the range x → x + ∆x. Just asin the case of a thermodynamic closed system, we can treat this slice as a closed system where nomass flows through any of the six faces.

x=0

xx

x+ xD

x=Linsulated

Qx

Qx+ xD

gA

From a 1st law energy balance:

∂E

∂t= Qx − Qx+∆x + W

Using Fourier’s law of heat conduction→ Qx = −kA∂T

∂xwe can write the general 1-D con-

duction equation as

∂

∂x

(k∂T

∂x

)︸ ︷︷ ︸

longitudinalconduction

+ g︸︷︷︸internal

heatgeneration

= ρC∂T

∂t︸ ︷︷ ︸thermalinertia

1

Special Cases1. Multidimensional Systems: The general conduction equation can be extended to three dimen-

sional Cartesian systems as follows:

∂

∂x

(k∂T

∂x

)+

∂

∂y

(k∂T

∂y

)+

∂

∂z

(k∂T

∂z

)+ g = ρC

∂T

∂t

2. Constant Properties: If we assume that properties are independent of temperature, then theconductivity can be taken outside the derivative

∇2T +g

k=

1

α

∂T

∂t

3. Steady State: If t→∞ then all terms∂

∂t→ 0

∇2T = −g

k⇐ Poisson’s Equation

4. No Internal Heat Generation:

∇2T = 0 ⇐ Laplace’s Equation

Thermal Resistance NetworksThermal circuits based on heat flow rate, Q, temperature difference, ∆T and thermal resistance,R, enable analysis of complex systems. This is analogous to current flow through electrical circuitswhere, I = ∆V/R

Thermal Electrical Role

Q, heat flow rate I , current flow transfer mechanism

Tin − Tout, temperature difference Vin − Vout, voltage difference driving force

R, thermal resistance R, electrical resistance impedance to flow

Resistances in Series

The total heat flow across the system can be written as

Q =T∞1 − T∞2

Rtotal

where Rtotal =4∑i=1

Ri

2

Resistances in Parallel

For systems of parallel flow paths as shown below

L

T1

T2

QQ

R1

R2

R3

k1

k2

k3

In general, for parallel networks we can use a parallel resistor network as follows:

T1

T1

T2 T

2

R1

RtotalR

2

R3

=

1

Rtotal

=1

R1

+1

R2

+1

R3

+ · · · and Q =T1 − T2

Rtotal

3

Thermal Contact Resistance• the thermal interface between contacting materials offers resistance to heat flow

• the total heat flow rate can be written as

Qtotal = hcA∆Tinterface

The conductance can be written in terms of a resistance as

hcA =Qtotal

∆T=

1

Rc

Table 10-2 can be used to obtain some representative values for contact conductance.

Cylindrical SystemsFor steady, 1D heat flow from T1 to T2 in a cylindrical system as shown below

L

r1

r2

Qr

T1

T2

A=2 rLp

k

r

we can write

Qr =T1 − T2(

ln(r2/r1)

2πkL

) where R =

(ln(r2/r1)

2πkL

)

4

Composite Cylinders

Rtotal = R1 +R2 +R3 +R4

=1

h1A1

+ln(r2/r1)

2πk2L+

ln(r3/r2)

2πk3L+

1

h4A4

Example 5-1: Determine the temperature (T1) of an electric wire surrounded by a layer ofplastic insulation with a thermal conductivity if 0.15 W/mK when the thickness of the insu-lation is a) 2 mm and b) 4 mm, subject to the following conditions:

Given: Find:I = 10 A T1 = ???

∆ε = ε1 − ε2 = 8 V when:D = 3 mm δ = 2 mmL = 5 m δ = 4 mmk = 0.15 W/mK

T∞ = 30 ◦Ch = 12 W/m2 ·K

5

Critical Thickness of InsulationAlthough the purpose of adding more insulation is to increase resistance and decrease heat trans-fer, we can see from the resistor network that increasing ro actually results in a decrease in theconvective resistance.

Rtotal = R1 +R2 =ln(ro/ri)

2πkL+

1

h2πroL

Could there be a situation in which adding insulation increases the overall heat transfer?

rcr,cyl =k

h[m]

6

Spherical SystemsWe can write the heat transfer as

Q =4πkriro

(r0 − ri)(Ti − To) =

(Ti − To)R

where R =ro − ri4πkriro

ri

ro

Ti

To

Notes:

1. the critical thickness of insulation for a spherical shell is given as

rcr,sphere =2k

h[m]

Heat Transfer from Finned SurfacesThe conduction equation for a fin with constant cross section is

kAc

d2T

∂x2︸ ︷︷ ︸longitudinalconduction

−hP (T − T∞)︸ ︷︷ ︸lateral

convection

= 0

The temperature difference between the fin and the surroundings (temperature excess) is usuallyexpressed as

θ = T (x)− T∞

7

which allows the 1-D fin equation to be written as

d2θ

dx2−m2θ = 0

where the fin parameterm is

m =

(hP

kAc

)1/2

The solution to the differential equation for θ is

θ(x) = C1 sinh(mx) + C2 cosh(mx) [≡ θ(x) = C1emx + C2e

−mx]

Potential boundary conditions include:

Base: → @x = 0 θ = θbTip: → @x = L θ = θL [T -specified tip]

θ =dθ

dx

∣∣∣∣∣x=L

= 0 [adiabatic (insulated) tip]

θ → 0 [infinitely long fin]

8

Fin Efficiency and Effectiveness

The dimensionless parameter that compares the actual heat transfer from the fin to the ideal heattransfer from the fin is the fin efficiency

η =actual heat transfer rate

maximum heat transfer rate whenthe entire fin is at Tb

=Qb

hPLθb

An alternative figure of merit is the fin effectiveness given as

εfin =total fin heat transfer

the heat transfer that would haveoccurred through the base area

in the absence of the fin

=Qb

hAcθb

Transient Heat ConductionThe Biot number can be obtained as follows:

TH − TsTs − T∞

=L/(k ·A)

1/(h ·A)=

internal resistance to H.T.external resistance to H.T.

=hL

k= Bi

Rint << Rext: the Biot number is small and we can conclude

TH − Ts << Ts − T∞ and in the limit TH ≈ Ts

Rext << Rint: the Biot number is large and we can conclude

Ts − T∞ << TH − Ts and in the limit Ts ≈ T∞

9

Transient Conduction Analysis• if the internal temperature of a body remains relatively constant with respect to time

– can be treated as a lumped system analysis

– heat transfer is a function of time only, T = T (t)

• typical criteria for lumped system analysis→ Bi ≤ 0.1

Lumped System AnalysisFor the 3-D body of volume V and surface areaA, we can use a lumped system analysis if

Bi =hV

kA< 0.1 ⇐ results in an error of less that 5%

For an incompressible substance specific heat is constant and we can write

mC︸ ︷︷ ︸≡Cth

dT

dt= − Ah︸︷︷︸

1/Rth

(T − T∞)

where Cth = lumped capacitance andRth = thermal resistance

We can integrate and apply the initial condition, T = Ti @t = 0 to obtain

T (t)− T∞Ti − T∞

= e−t/(Rth·Cth) = e−t/τ = e−bt

10

where the time constant can be written as

1

b= τ = Rth · Cth = thermal time constant =

mC

Ah=ρV C

Ah

Example 5-2: Determine the time it takes a fuse to melt if a current of 3 A suddenly flowsthrough the fuse subject to the following conditions:

Given:D = 0.1 mm Tmelt = 900 ◦C k = 20 W/mK

L = 10 mm T∞ = 30 ◦C α = 5× 10−5 m2/s ≡ k/ρCp

Assume:

• constant resistance R = 0.2 ohms

• the overall heat transfer coefficient is h = hconv + hrad = 10 W/m2K

• neglect any conduction losses to the fuse support

11

Approximate Analytical and Graphical Solutions (Heisler Charts)The lumped system analysis can be used if Bi = hL/k ≤ 0.1 but what if Bi > 0.1 we mustfind a solution to the PDE where temperature is T (x, t) = f(x, L, t, k, α, h, Ti, T∞)

∂2T

∂x2=

1

α

∂T

∂t

The analytical solution to this equation takes the form of a series solution

T (x, t)− T∞Ti − T∞

=∞∑

n=1,3,5...

Ane

(−nλ

L

)2

αt

cos

(nλx

L

)

By using dimensionless groups, we can reduce the temperature dependence to 3 dimensionlessparameters

Dimensionless Group Formulation

temperature θ(x, t) =T (x, t)− T∞Ti − T∞

position x = x/L

heat transfer Bi = hL/k Biot number

time Fo = αt/L2 Fourier number

note: Cengel uses τ instead of Fo.

With this, two approaches are possible

1. use the first term of the infinite series solution. This method is only valid for Fo > 0.2

2. use the Heisler charts for each geometry as shown in Figs. 11-15, 11-16 and 11-17

12

First term solution: Fo > 0.2→ error about 2% max.

Plane Wall: θwall(x, t) =T (x, t)− T∞Ti − T∞

= A1e−λ2

1Fo cos(λ1x/L)

Cylinder: θcyl(r, t) =T (r, t)− T∞Ti − T∞

= A1e−λ2

1Fo J0(λ1r/ro)

Sphere: θsph(r, t) =T (r, t)− T∞Ti − T∞

= A1e−λ2

1Fosin(λ1r/ro)

λ1r/ro

λ1, A1 can be determined from Table 11-2 based on the calculated value of the Biot number (willlikely require some interpolation). The Bessel function, J0 can be calculated using Table 11-3.

Heisler Charts

• find T0 at the center for a given time (Table 11-15 a, Table 11-16 a or Table 11-17 a)

• find T at other locations at the same time (Table 11-15 b, Table 11-16 b or Table 11-17 b)

• findQtot up to time t (Table 11-15 c, Table 11-16 c or Table 11-17 c)

Example 5-3: An aluminum plate made of Al 2024-T6 with a thickness of 0.15 m isinitially at a temperature of 300 K. It is then placed in a furnace at 800 K with aconvection coefficient of 500 W/m2K.

Find: i) the time (s) for the plate midplane to reach 700 Kii) the surface temperature at this condition. Use both the Heisler charts

and the approximate analytical, first term solution.

13

Convection Heat Transfer

Reading Problems12-1→ 12-8 12-41, 12-48, 12-53, 12-68, 12-74, 12-78, 12-8113-1→ 13-6 13-39, 13-47, 13-5514-1→ 14-4 14-19, 14-25, 14-42, 14-48

Introduction

• the controlling equation for convection is Newton’s Law of Cooling

Qconv =∆T

Rconv

= hA(Tw − T∞) ⇒ Rconv =1

hA

A = total convective area, m2

h = heat transfer coefficient, W/(m2 ·K)

Tw = surface temperature, ◦C

T∞ = fluid temperature, ◦C

1

The following table gives the range of heat transfer coefficient expected for different convectionmechanisms and fluid types.

Process h [W/(m2 ·K)]

Natural

Convection

• gases 3 - 20

• water 60 - 900

Forced

Convection

• gases 30 - 300

• oils 60 - 1 800

• water 100 - 1 500

Boiling

• water 3 000 - 100 000

Condensation

• steam 3 000 - 100 000

Dimensionless Groups

It is common practice to reduce the total number of functional variables by forming dimensionlessgroups consisting of relevant thermophysical properties, geometry, boundary and flow conditions.

Prandtl number: Pr = ν/α where 0 < Pr < ∞ (Pr → 0 for liquid metals and Pr →∞ for viscous oils). A measure of ratio between the diffusion of momentum to the diffusionof heat.

Reynolds number: Re = ρUL/µ ≡ UL/ν (forced convection). A measure of the balancebetween the inertial forces and the viscous forces.

Peclet number: Pe = UL/α ≡ RePr

2

Grashof number: Gr = gβ(Tw − Tf)L3/ν2 (natural convection)

Rayleigh number: Ra = gβ(Tw − Tf)L3/(α · ν) ≡ GrPr

Nusselt number: Nu = hL/kf This can be considered as the dimensionless heat transfercoefficient.

Stanton number: St = h/(UρCp) ≡ Nu/(RePr)

Forced ConvectionThe simplest forced convection configuration to consider is the flow of mass and heat near a flatplate as shown below.

• as Reynolds number increases the flow has a tendency to become more chaotic resulting indisordered motion known as turbulent flow

– transition from laminar to turbulent is called the critical Reynolds number,Recr

Recr =U∞xcr

ν

– for flow over a flat plateRecr ≈ 500, 000

Boundary Layers

Velocity Boundary Layer

• the region of fluid flow over the plate where viscous effects dominate is called the velocityor hydrodynamic boundary layer

3

• the velocity of the fluid progressively increases away from the wall until we reach approxi-mately 0.99 U∞ which is denoted as the δ, the velocity boundary layer thickness.

Thermal Boundary Layer

• the thermal boundary layer is arbitrarily selected as the locus of points where

T − TwT∞ − Tw

= 0.99

Flow Over Plates

We will use empirical correlations based on experimental data where

Cf,x = C1 ·Re−m

Nux = C2 ·Rem · Prn =hx · xkf

4

1. Laminar Boundary Layer Flow, Isothermal (UWT)

The local values of the skin friction and the Nusselt number are given as

Cf,x =0.664

Re1/2x

Nux = 0.332Re1/2x Pr1/3 ⇒ local, laminar, UWT, Pr ≥ 0.6

An average value of the skin friction coefficient and the heat transfer coefficient for the full extentof the plate can be obtained by using the mean value theorem.

Cf =1.328

Re1/2L

NuL =hLL

kf= 0.664 Re

1/2L Pr1/3 ⇒ average, laminar, UWT, Pr ≥ 0.6

For low Prandtl numbers, i.e. liquid metals

Nux = 0.565Re1/2x Pr1/2 ⇒ local, laminar, UWT, Pr ≤ 0.6

2. Turbulent Boundary Layer Flow, Isothermal (UWT)

The local skin friction is given as

Cf,x =0.0592

Re0.2x

⇒ local, turbulent, UWT, Pr ≥ 0.6

Nux = 0.0296Re0.8x Pr1/3 ⇒

local, turbulent, UWT,0.6 < Pr < 100, 500, 000 ≤ Rex ≤ 107

Cf,x =0.074

Re0.2x

⇒ average, turbulent, UWT, Pr ≥ 0.6

NuL = 0.037Re0.8L Pr1/3 ⇒

average, turbulent, UWT,

0.6 < Pr < 100, 500, 000 ≤ Rex ≤ 107

5

3. Combined Laminar and Turbulent Boundary Layer Flow, Isothermal (UWT)

NuL =hLL

k= (0.037Re0.8

L − 871) Pr1/3 ⇒

average, combined, UWT,0.6 < Pr < 60,500, 000 ≤ ReL ≤ 107

4. Laminar Boundary Layer Flow, Isoflux (UWF)

Nux = 0.453Re1/2x Pr1/3 ⇒ local, laminar, UWF, Pr ≥ 0.6

5. Turbulent Boundary Layer Flow, Isoflux (UWF)

Nux = 0.0308Re4/5x Pr1/3 ⇒ local, turbulent, UWF, Pr ≥ 0.6

Example 6-1: Hot engine oil with a bulk temperature of 60 ◦C flows over a horizontal, flatplate 5 m long with a wall temperature of 20 ◦C. If the fluid has a free stream velocity of2 m/s, determine the heat transfer rate from the oil to the plate if the plate is assumed to beof unit width.

6

Flow Over Cylinders and Spheres

1. Boundary Layer Flow Over Circular Cylinders, Isothermal (UWT)

The Churchill-Bernstein (1977) correlation for the average Nusselt number forlong (L/D > 100) cylinders is

NuD = 0.3 +0.62Re

1/2D Pr1/3

[1 + (0.4/Pr)2/3]1/4

1 +

(ReD

282, 000

)5/84/5

⇒ average, UWT,ReD < 107, 0 ≤ Pr ≤ ∞, ReD · Pr > 0.2

All fluid properties are evaluated at Tf = (Tw + T∞)/2.

2. Boundary Layer Flow Over Non-Circular Cylinders, Isothermal (UWT)

The empirical formulations of Zhukauskas and Jakob given in Table 12-3 are commonly used,where

NuD ≈hD

k= C RemD Pr

1/3 ⇒ see Table 12-3 for conditions

3. Boundary Layer Flow Over a Sphere, Isothermal (UWT)

For flow over an isothermal sphere of diameterD

NuD = 2 +[0.4Re

1/2D + 0.06Re

2/3D

]Pr0.4

(µ∞

µs

)1/4

⇒

average, UWT,0.7 ≤ Pr ≤ 380

3.5 < ReD < 80, 000

where the dynamic viscosity of the fluid in the bulk flow, µ∞ is based on the free stream tem-perature, T∞ and the dynamic viscosity of the fluid at the surface, µs, is based on the surfacetemperature, Ts. All other properties are based on T∞.

7

Example 6-2: An electric wire with a 1 mm diameter and a wall temperature of 325 Kis cooled by air in cross flow with a free stream temperature of 275 K. Determine the airvelocity required to maintain a steady heat loss per unit length of 70 W/m.

Internal Flow

The mean velocity and Reynolds number are given as

Um =1

Ac

∫Ac

u dA =m

ρmAc

ReD =UmD

ν

8

Hydrodynamic (Velocity) Boundary Layer

• the hydrodynamic boundary layer thickness can be approximated as

δ(x) ≈ 5x

(Umx

ν

)−1/2

=5x√Rex

• the hydrodynamic entry length can be approximated as

Lh ≈ 0.05ReDD (laminar flow)

Thermal Boundary Layer

• the thermal entry length can be approximated as

Lt ≈ 0.05ReDPrD = PrLh (laminar flow)

• for turbulent flow Lh ≈ Lt ≈ 10D

Wall Boundary Conditions

1. Uniform Wall Heat Flux: Since the wall flux qw is uniform, the local mean temperature de-noted as

Tm,x = Tm,i +qwA

mCp

will increase in a linear manner with respect to x.

The surface temperature can be determined from

Tw = Tm +qw

h

9

2. Isothermal Wall: The outlet temperature of the tube is

Tout = Tw − (Tw − Tin) exp[−hA/(mCp)]

Because of the exponential temperature decay within the tube, it is common to present themean temperature from inlet to outlet as a log mean temperature difference where

Q = hA∆Tln

∆Tln =Tout − Tin

ln

(Tw − ToutTw − Tin

) =Tout − Tin

ln(∆Tout/∆Tin)

10

1. Laminar Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF)

For laminar flow whereReD ≤ 2300

NuD = 3.66 ⇒ fully developed, laminar, UWT, L > Lt & Lh

NuD = 4.36 ⇒ fully developed, laminar, UWF, L > Lt & Lh

NuD = 1.86

(ReDPrD

L

)1/3 (µbµs

)0.14

⇒

developing laminar flow, UWT,Pr > 0.5

L < Lh or L < Lt

For non-circular tubes the hydraulic diameter, Dh = 4Ac/P can be used in conjunction withTable 10-4 to determine the Reynolds number and in turn the Nusselt number.

In all cases the fluid properties are evaluated at the mean fluid temperature given as

Tmean =1

2(Tm,in + Tm,out)

except for µs which is evaluated at the wall temperature, Ts.

2. Turbulent Flow in Circular Tubes, Isothermal (UWT) and Isoflux (UWF)

For turbulent flow whereReD ≥ 2300 the Dittus-Boelter equation (Eq. 13-68) can be used

NuD = 0.023Re0.8D Prn ⇒

turbulent flow, UWT or UWF,0.7 ≤ Pr ≤ 160

ReD > 2, 300

n = 0.4 heatingn = 0.3 cooling

For non-circular tubes, again we can use the hydraulic diameter,Dh = 4Ac/P to determine boththe Reynolds and the Nusselt numbers.

In all cases the fluid properties are evaluated at the mean fluid temperature given as

Tmean =1

2(Tm,in + Tm,out)

11

Natural Convection

What Drives Natural Convection?• fluid flow is driven by the effects of buoyancy

Natural Convection Over Surfaces

• note that unlike forced convection, the velocity at the edge of the boundary layer goes to zero

12

Natural Convection Heat Transfer Correlations

The general form of the Nusselt number for natural convection is as follows:

Nu = f(Gr, Pr) ≡ CGrmPrn where Ra = Gr · Pr

• C depends on geometry, orientation, type of flow, boundary conditions and choice of char-acteristic length.

• m depends on type of flow (laminar or turbulent)

• n depends on the type of fluid and type of flow

• Table 14-1 should be used to find Nusselt number for various combinations of geometry andboundary conditions

– for ideal gases β = 1/T∞, (1/K)

– all fluid properties are evaluated at the film temperature, Tf = (Tw + T∞)/2.

Natural Convection From Plate Fin Heat Sinks

The average Nusselt number for an isothermal plate fin heat sink with natural convection can bedetermined using

NuS =hS

kf=

[576

(RaSS/L)2+

2.873

(RaSS/L)0.5

]−0.5

13

A basic optimization of the fin spacing can be obtained as follows: For isothermal fins with t < S

Sopt = 2.714

(S3L

RaS

)1/4

= 2.714

(L

Ra1/4L

)

with

RaL =gβ(Tw − T∞)L3

ν2Pr

The corresponding value of the heat transfer coefficient is

h = 1.307kf/Sopt

All fluid properties are evaluated at the film temperature.

Example 6-3: Find the optimum fin spacing, Sopt and the rate of heat transfer, Q for thefollowing plate fin heat sink cooled by natural convection.

Given:

W = 120 mm H = 24 mmL = 18 mm t = 1 mmTw = 80 ◦C T∞ = 25 ◦CP∞ = 1 atm fluid = air

Find: Sopt and the corresponding heat transfer, Q

14

Radiation Heat Transfer

Reading Problems15-1→ 15-7 15-27, 15-33, 15-49, 15-50, 15-77, 15-79,

15-86, 15-106, 15-107

IntroductionThe following figure shows the relatively narrow band occupied by thermal radiation.

An even narrower band inside the thermal radiation spectrum is denoted as the visible spectrum,that is the thermal radiation that can be seen by the human eye. The visible spectrum occupiesroughly 0.4− 0.7 µm.

Blackbody RadiationA blackbody is an ideal radiator that

• absorbs all incident radiation regardless of wavelength and direction

• emitted radiation is a function of wavelength and temperature but is independent of direction,i.e. a black body is a diffuse emitter (independent of direction)

1

Definitions1. Blackbody emissive power: the radiation emitted by a blackbody per unit time and per

unit surface area

Eb = σ T 4 [W/m2] ⇐ Stefan-Boltzmann law

2. Spectral blackbody emissive power: the amount of radiation energy emitted by a black-body per unit surface area and per unit wavelength about the wavelength λ. The followingrelationship between emissive power, temperature and wavelength is known as Plank’s dis-tribution law

Eb,λ =C1

λ5[exp(C2/λT )− 1][W/(m2 · µm)]

3. Blackbody radiation function: the fraction of radiation emitted from a blackbody at tem-perature, T in the wavelength band λ = 0→ λ

f0→λ =

∫ λ0Eb,λ(T ) dλ∫ ∞

0Eb,λ(T ) dλ

=

∫ λ0

C1

λ5[exp(C2/λT )− 1]dλ

σT 4

let t = λT and dt = T dλ, then

f0→λ =

∫ t0

C1T5(1/T )dt

t5[exp(C2/t)− 1]

σT 4=C1

σ

∫ λT0

dt

t5[exp(C2/t)− 1]

= f(λT )

f0→λ is tabulated as a function λT in Table 15.2

2

Radiation Properties of Real SurfacesThe thermal radiation emitted by a real surface is a function of surface temperature, T , wavelength,λ, direction and surface properties.

Eλ = f(T, λ, direction, surface properties) ⇒ spectral emissive power

while for a blackbody, the radiation was only a function of temperature and wavelength

Eb,λ = f(T, λ) → diffuse emitter ⇒ independent of direction

Definitions1. Diffuse surface: properties are independent of direction.

2. Gray surface: properties are independent of wavelength.

3. Emissivity: defined as the ratio of radiation emitted by a surface to the radiation emitted bya blackbody at the same surface temperature.

ε(T ) =radiation emitted by surface at temperature T

radiation emitted by a black surface at T

=

∫ ∞0Eλ(T ) dλ∫ ∞

0Ebλ(T ) dλ

=

∫ ∞0ελ(T )Ebλ(T ) dλ

Eb(T )=E(T )

σT 4

3

4. Irradiation,G: the radiation energy incident on a surface per unit area and per unit time

An energy balance based on incident radiation gives

G = ρG+ αG+ τG

If we normalize with respect to the total irradiation

α+ ρ+ τ = 1

In general ε 6= α. However, for a diffuse-gray surface (properties are independent of wave-length and direction)

ε = α diffuse-gray surface

5. Radiosity, J : the total radiation energy leaving a surface per unit area and per unit time.

For a surface that is gray and opaque, i.e. ε = α and α+ ρ = 1, the radiosity is given as

J = radiation emitted by the surface + radiation reflected by the surface

= ε Eb + ρG

= εσT 4 + ρG

Since ρ = 0 for a blackbody, the radiosity of a blackbody is

J = σT 4

4

Solar RadiationThe incident radiation energy reaching the earth’s atmosphere is known as the solar constant, Gs

and has a value of

Gs = 1353W/m2

For most calculations we can use

Esky ≈ σT 4sky

where the sky temperature is assumed to be

Tsky ≈ 230K (cold, clear sky) → 285K (warm, cloudy sky)

View Factor (Shape Factor, Configuration Factor)

• the radiative exchange between surfaces clearly depends on how well the surfaces “see” oneanother. This information is provided by using shape factors (or view factors or configurationfactors).

• Definition: The view factor, Fi→j is defined as the fraction of radiation leaving surface iwhich is intercepted by surface j. Hence

Fi→j =Qi→j

AiJi=

radiation reaching jradiation leaving i

5

Reciprocity relation: AiFi→j = AjFj→i

Summation rule:N∑j=1

Fi→j = 1 ; i = 1, 2, . . . , N

Note that Fi→i 6= 0 for a concave surface. For a plane or convex surface Fi→i = 0.

Superposition rule: if the surface is not available in the tables sometimes it can be treatedas the sum of smaller known surfaces to form the full extent of the surface of interest.

F1→(2,3) = F1→2 + F1→3

Symmetry rule: if the problem is symmetric, then the view factors will also be symmetric.

Hottel Crossed String Method

Can be applied to 2D problems where surfaces are any shape, flat, concave or convex. Note for a2D surface the area,A is given as a length times a unit width.

A1F12 = A2F21 =(total crossed)− (total uncrossed)

2

A1 andA2 do not have to be parallel

A1F12 = A2F21 =1

2[(ac+ bd)︸ ︷︷ ︸

crossed

− (bc+ ad)︸ ︷︷ ︸uncrossed

]

6

Radiation Exchange Between Diffuse-Gray SurfacesForming an Enclosure

We will assume that:

1. each surface of the enclosure is isothermal

2. radiosity, Ji, and irradiation,Gi are uniform over each surface

3. the surfaces are opaque (τi = 0) and diffuse-gray (αi = εi)

4. the cavity is filled with a fluid which does not participate in the radiative exchange process

• an energy balance on the i′th surface gives:

Qi = qiAi = Ai(Ji −Gi)

7

Qi = Ai(Ei − αiGi) (1)

Ji = Ei + ρiGi (2)

Ei = εiEb,i = εiσT4i (3)

ρi = 1− αi = 1− εi (4) ⇒ since αi + ρi + τi↗0= 1

and αi = εi

Combining Eqs. 2, 3 and 4 gives

Ji = εiEb,i + (1− εi)Gi (5)

Combining this with Eq. 1 gives

Qi =Eb,i − Ji(

1− εiεiAi

) ≡ potential differencesurface resistance

This surface resistance represents real surface behavior as opposed to a blackbody surface.Recall that for a blackbody εi = αi = 1 and Ji = E1 = Eb,i = σT 4

i .

• next consider radiative exchange between the surfaces.

By inspection it is clearly seen that

{irradiation on

surface i

}=

{radiation leaving theremaining surfaces

}

8

AiGi =N∑j=1

Fj→i(AjJj) =N∑j=1

AiFi→jJi

Therefore

Gi =N∑j=1

Fi→jJj

Combining this with Eq. 5 gives

Ji = εiσT4i + (1− εi)

N∑j=1

Fi→jJj

In addition we can write

Qi = AiJi −N∑j=1

AiFi→jJj

Since the summation rule statesN∑j=1

Fi→j = 1, the above equation becomes

Qi = Ai

N∑j=1

Fi→jJi −N∑j=1

Fi→jJj

Qi =

N∑j=1

AiFi→j(Ji − Jj) =N∑j=1

Qi→j

or

Qi =N∑j=1

Ji − Jj(1

AiFi→j

) ≡ potential differencespace resistance

9

Example 7-1: Consider two very large parallel plates with diffuse, gray surfaces, Determinethe net rate of radiation heat transfer per unit surface area, Q12/A, between the two surfaces.For Case 2, also determine T3, the temperature of a radiation shield, positioned midway be-tween surfaces 1 and 2.

Given:

ε1 = 0.2 T1 = 800 Kε2 = 0.7 T2 = 500 Kε3 = 0.02 A1 = A2 = A3 = A

Assume steady state conditions.

10

Example 7-2: A thermocouple is suspended between two parallel surfaces as shown in thefigure below. Find Tf , the temperature of the air stream by performing an energy balance onthe thermocouple.

Given:

Tw = 400 K Tth = 650 Kεth = 0.6 h = 80 W/(m2 ·K)

Assume steady state conditions.

11

Example 7-3: Consider a room that is 4 m long by 3 m wide with a floor-to-ceiling dis-tance of 2.5 m. The four walls of the room are well insulated, while the surface of the flooris maintained at a uniform temperature of 30 ◦C using an electric resistance heater. Heat lossoccurs through the ceiling, which has a surface temperature of 12◦C. All surfaces have anemissivity of 0.9.

a) determine the rate of heat loss, (W ), by radiation from the room.b) determine the temperature, (K), of the walls.

12