Processteknik / Process Engineeringusers.abo.fi/rzevenho/PRC181920-OHpart0rz.pdf · Processteknik...

Processteknik / Process Engineering424104.0 v. 2018-2019 + 2019-2020

0. Introduktion / IntroductionTermodynamik / ThermodynamicsVärmeöverföring / Heat transfer

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering Laboratory / Värme- och strömningsteknikProcess and Systems Technology / Laboratoriet för process- och systemteknik

tel. 3223 ; [email protected]

NOT PARTOF THE EXAM

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 2/66

0.1 This course(new as of 2016 - 2017)

Processteknik 424104.0 5 sp

RZ An intro / brief summary of Thermodynamicsand Heat transfer from earlier course 414101 Processteknikens grunder (PTG) (not part of exam)

RZ Fluid mechanics from PTG (#6) PMA Chemical reaction processing from earlier

course 421102 Kemiteknikens grunder (KTG) (#1,4) RZ Dimensional analysis & Mass transfer from earlier course

424302 Massöverföring & separationteknik (MÖF-ST) (#1-8) PMA Chemical and thermal separation processes

and Process synthesis from KTG (#3,5) 5 sp = 133 h = 52 h class teaching & exam + 81 h self study Course exam: 2 x 2h during the course + several re-exams /yr

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland 3

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What is process engineering ?“Process engineering is often a synonym for chemical engineering and focuses on the design, operation and maintenance of chemical and material manufacturing processes” (source: wikipedia)

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland

Natural gas + CO2

Liquid solventfor example

alkanol amine

Natural gas

Liquid solvent+ CO2

A typicalexample

Socolow, R.H. ”Can we buryglobal warming” ScientificAmerican, July 2005, 49-55

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What is a process?”A process occurswhenever some propertyof a system changes or if there is an energy or mass flow across the boundary of a system”(source: KJ05)


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Systems and boundaries Any engineering device can be

denoted as a system, separatedfrom the rest of the universe by a system boundary.

Interactions between the system and the rest of the universe can be exchange of mass and/or energy.

The part of the universe that interactswith the system is called the surroundings, or environment(sv: omgivning, närområde)

Picture: SEHB06

Engineering systems: Isolated systems: no exchange

of mass or energy with the surroundingsfor example: thermos bottle (Dewarflask)

Closed systems: exchange ofenergy with the surroundings, but no exchange of massfor example: light bulb, space station

Open systems: exchange ofenergy and mass with the surroundingsfor example: combustion engines, pumps, distillation columns, livingorganisms



0.2 Brief re-wrap of thermodynamics

taken from Processteknikens grunder (PTG) #3 : see Z13

http://users.abo.fi/rzevenho/PTG14-OH3.pdf http://users.abo.fi/rzevenho/PTGkap3-aug2013.pdf

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What is energy?

”Energy is any quantity that changes the state of a closedsystem when crossing the system boundary” (SEHB06)

”Energy is the capacity to do work” (A83)

”Energy is the capacity to do work or produce heat” (ZZ03)

Energy cannot be produced or destroyed(First Law of Thermodynamics *) but it can be convertedfrom one form into another, and vice versa.

Energy can be degraded to lower quality, as a consequence of producing and converting heat (Second Law of Thermodynamics *)

* sv: Termodynamikens Första och Andra Huvudsatsen



Types of energy In a closed system, energy is

present (”stored”) as:

potential energy sv: lägesenergi

Ep and kinetic energy sv:

rörelseenergi Ek (mechanical energy) and internal energy sv: inre energi

U (thermal energy)

For a mass m, with verticalposition z in a gravity field and velocity v defined as

Ep = m· g· z

Ek = ½· m· v2

= ½· m· v2

U = m· u

These energies are relative to a reference frame reference state:

– potential energy relative to a position z = 0 in a gravity field

– kinetic energy relative to a non-moving reference frame with velocity v = 0

– internal energy relative to zero temperature T = 0, or surroundings temperature T = T°.

Pictures: SEHB06

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Internal energy, U Internal energy U is a consequence of the motion of

constituent molecules of a substance and their interactions (T06)

Unit: J ; Specific internal energy u = U/m (J/kg)

Internal energy of a material or substance is related to temperature via the specific heat, also referred to as heat capacity, (sv: värmekapacitet) c (unit: J kg-1K-1 or J mol-1K-1)

Defining u = 0 at reference temperature T° (for example 0°C or 25°C) gives u = T° ∫T c dT ≈ c· (T – T°). In general: u = T ∫T+ΔT c dT ≈ c· ΔT

Picture: SEHB06



Total energy of a system

Total energy of a system E = Ek + Ep + U (J)

= m· ek + m· ep + m· u (J)

with specific energies ek, ep and u (J/kg)

A mass stream m (kg/s) corresponds to an energy stream Ė

Ė = Ėk + Ėp + U = m· (ek + ep + u) (J/s).

.

.


Energy transfer: heat & work For a closed system ENERGY can be transferred across

the boundary by WORK or by HEAT transfer

For an open system, MASS may cross the bounday and the ENERGY associated with that can enter the system, in addition to WORK or by HEAT transfer (as for a closedsystem)

Generally, for solid matter and objects, work changes kinetic energy and potential energy of a system, while heat changesinternal energy. For gases and liquids it is morecomplicated: friction in fluid flow gives viscous heat, compression of a gas at constant volume increases itstemperature, etc.

Picture: KJ05


”The energy of an isolated system is constant” (A83)

Energy can be converted from one form into another, and vice versafor example heat Q can be converted into work W and vice versa - but there are limitations !

Closed system ΔE = W + Q (or: dE = δW + δQ)Open system ΔE = W + Q + min· ein - mout· eout

Energy cannot be produced and speaking about ”energyproduction” is wrong and silly (but it allows for distinguishingamateurs from specialists)

First Law of ThermodynamicsAlso known as the ”conservation of energy” principle

..


Enthalpy /1

Consider the addition of heat to a gas (assume an ideal gas for the moment) in two different situations(a) a rigid tank, i.e. a constant volume process(b) a cylinder-piston assembly, i.e. a constant pressureprocess

picture: KJ05

Or: why we talk about energybut calculate with enthalpy.....

REMINDER:Ideal gas lawp· V = n· R· T

Pressure p, PaTemperature T, KVolumeV, m3,Number of moles nGas constant R = 8.314 J/(mol· K)

(a) constant volumeThe energy balance (no potential energy or kinetic energy effects):ΔU = Q + W, and since no work is done: ΔU = Q

This implies that all heat is used to increasethe temperature of the gas at constant volume.With specific heat at constant volume, cv (J kg-1K-1), gas mass m and temperature T:ΔU = m· cv· ΔT, or Δu = cv· ΔT

which can be written as cv = Δu / ΔT or cv = du / dT

IF u is a function of T only !


Enthalpy /2

picture: KJ05


Enthalpy /3

(b) constant pressureThe energy balance (no potential energy or kinetic energy effects):ΔU = Q + W, and some work is done to increase the volume as to keep pressure constant: W = - V∫ p dV, with p = external pressure = internal pressure (at equilibrium: assume a slow or ”quasi-equilbrium” process !) As p is constant, W = - p· ΔV, and thusΔU = Q - p· ΔV, or Q = ΔU + p· ΔV= (at constant pressure) Q = Δ (U + p· V)

Part of the heat Q is used to increase volume V !

picture: KJ05


(b) constant pressure (continued)The quantity U + p· V is referred to as enthalpy, H (unit: J)→ the energy balance givesQ = ΔH = m· Δh with specific enthalphy h

With specific heat at constant pressure, cp (J kg-1K-1), gas mass m and temperature T:Q = ΔH = m· cp· ΔT, or Δh = cp· ΔT

which can be written as cp = Δh / ΔT or cp = dh / dT

IF h is a function of T only !Simular to u, h must be fixed to a referencevalue h = h° at T = T°

Enthalpy /4

picture: KJ05

Enthalpy takes intoconsideration that a constant pressure process may give a volume changewhich implies exchange ofwork with the environment

Like U, p, T and V, enthalpy is a thermodynamic state property


The total energy balance The total energy of a system with mass m equals

E = mgz + ½mv2 + U + pV = mgz + ½mv2 + H or as energy concentration (and using m = ρ· V)e = gz + ½v2 + u + p/ρ = gz + ½v2 + h with gravity g

For a process that involves energy transfer to the system as heat and/or work:

ΔE = Q + W

Δ(mgz + ½mv2)+ ΔU + Δ(pV) = Δ(mgz + ½mv2) + ΔH = Q + W

Δe = q + w

Δ(gz + ½v2)+ Δu + Δ(p/ρ) = Δ(gz + ½v2)+ Δh = q + w


Entropy /1 A spontaneous process occurs (fast or slow) without

outside intervention Earlier it was thought that processes are spontaneous if

exothermic (= producing heat, ΔH < 0), but, for example, ice melting at > 0°C is endothermic (= consuming heat, ΔH > 0) and spontaneous!

The driving force for a spontaneous process is an increase of the ENTROPY of the universe

Nature spontaneously proceeds towards the states that have the highest probability

http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter17/Text_Images/FG17_01.JPG


Entropy /2 The entropy of the universe must increase:

ΔSuniverse = ΔSsystem + ΔSsurroundings

ΔSuniv > 0 : spontaneous processΔSuniv < 0: spontaneous process in reverse directionΔSuniv = 0: equilibrium

http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter17/Text_Images/FG17_03.JPG


Entropy /3

ΔSuniv = ΔSsys + ΔSsurr , > 0 if ΔSsys > -ΔSsurr

ΔSsurr is primarily determined by the flow of energy in or out the system as heat

Exothermic processes: Q to surroundings, ΔSsurr > 0Endothermic processes: Q into system, ΔSsurr < 0

If ΔSsurr or ΔSsys < 0, temperature determines what willhappen, for example: water → ice if T < 0°C (at 1 bar)

ΔSsurr depends on temperature:

0ΔG T

ΔHΔS

T

ΔH ΔH)- Q- done, is workno (if

T

Q

etemperatur

gssurroundin to system from heat of quantity ΔS

syssys

sys

system

surr

Note: Q = heat into the system !


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Second Law of Thermodynamics ”The entropy of an isolated system will

always increase until it reaches a stateof equilibrium at which the entropy is maximised” (SEHB06)

Clausius: ”Heat cannot movespontaneously from cold to hot bodies” (KJ05); ”The entropy of the universe tends toward a maximum” (T06)

Kelvin-Planck: ”Although all work canalways be converted completely to heat, heat cannot be completely and continuously converted into work” (T06)

RZ: ”Producing work by extractingheat from 1 kg of hot substance willresult in 1 kg of cold substance”

vS91: ”Endast sådana processer och förändringar är möjliga som ökar den totala oordningen”

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Some implications: 1) Every process has losses; 2) We cannot build a perpetuum mobile machine;3) Heat cannot be converted intowork with 100% efficiency.


Heat → work: power cycles /1

An air power cycle and its p-V diagram1→2: Heat-up at constant volume: pressure ↑2→3: Heat-up at constant pressure: volume ↑, work is done3→4: Cooling at constant volume: pressure ↓4→1: Compression (volume ↓) and cooling at constant pressure

pictures: KJ05

”A thermodynamic cycle consists of a sequence of processes in which the working fluid returns to it’s original thermodynamic state” (T06)



The cycle generates work Wnet = Wout – Win, or power Wnet = Wout – Win,

The First Law of Thermodynamics gives

Qin = Wnet + Qout, or Qin = Wnet + Qout

Assuming that the goal is to produce work (i.e. generate power) using the incoming heat streams, the thermal efficiency or Carnotefficiency of the cycle, ηcycle, can be defined.

pictures: KJ05. ..

in

out

in

outin

in

net

in

out

in

outin

in

net

thermal

Q

Q

Q

QQ

Q

W

Q

Q

Q

QQ

Q

W

in put must energy we

use to wantenergy weη

. . .



Example of a power cycle Example of a refrigeration cycle

pictures: KJ05


Environment

at T = T°

Energy conversion heat ↔ work

Conversion of heat (to work) is limited

Heat at T=T1

p,V work

Heat at T=T2

Non - p,V work100% conversion

100% conversion

100% conversion

100% conversion

Only if T1 > T2

Only if T1 < T2

Conversion partly

Fraction T°/T2

is lost

Con

vers

ion

part

ly

Fraction T°/T1

is lost


0.3 Brief re-wrap of heat transfer

taken from Processteknikens grunder (PTG) #5 - see Z13

http://users.abo.fi/rzevenho/PTG14-OH5.pdf http://users.abo.fi/rzevenho/PTGkap5-aug2013.pdf

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What is Heat? /1 (sv: värme)

”Heat is energy transferred, without transfer of mass, across a boundary of a system (or across a control surface)because of a temperature difference between the system and the surroundings or a temperature gradient at the boundary” (T06)

”Heat involves the transfer of energy betweenobjects due to a temperature difference” (ZZ03)

”Heat is thermal energy in transit” (BB10)

(Similar: ”work is mechanical energy in transit ”)

”Heat stimulates random motion (thermal motion)”(A83)

(Similar: ”Work stimulates organised motion”)

Note (a detail): heat is a path function (like work) and is not a system property -but temperature is! (Therefore, differential dQ has no meaning, and δQ is used.)


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What is Heat? /2

.

.

Picture: SEHB06

Heat occurs at the boundary of a system; a system cannot ”contain” heat.

Heat is an energy transfer process(similar to ”work”) and ”heat transfer” is the same as ”heat” ! (”After all, ”work transfer” is not used either !)

Nomenclature for heat transfer and rate:heat (transfer) (sv: värmemängd) Q unit: Jheat transfer rate (sv: värmeström) Q unit: J/s = Wheat flux (sv: värmeströmtäthet) Q” unit: J/(s· m2) = W/m2

A process without heat transfer is called adiabatic(which is not the same as isothermal !!)


Three heat transfer mechanismsConduction Convection Radiation

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Pic: BÖ88


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Heat conduction: Fourier’s Law In a non-moving medium (i.e. a solid, or

stagnant fluid) in the presence of a temperature gradient, heat is transferred from high to low temperature as a result of molecular movement: heat conduction(sv: värmeledning)

For a one-dimensional temperature gradient ΔT/Δx or dT/dx, Fourier’s Law gives the conductive heat transfer rate Q through a cross-sectional area A (m2). If λ is a constant:

with thermal conductivity λ, unit: W/(mK)

(sv: termisk konduktivitet eller värmeledningsförmåga)

)(W/m (W) 2

dx

dT

A

Q"Q

dx

dTAQ

Pictures: T06

.


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Material conductivity data

r

Pictures: T06

Thermalconductivity is temperature-dependent!

Table: KJ05

λ


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Fourier’s Law /2

Alternative notation based on energy instead of temperature: use the heat concentration (heat per volume) ρ· cp·T(with density ρ, specific heat cp) with unit J/m3

This gives (for 1-dimensional conduction):

/s)(m cρ

λ ay diffusivit heat with

)(W/m dx

)Tcρ(da

dx

)Tcρ(d

cρ

λ"Q

2

p

2pp

p

Typical values λ (W/mK) a (m2/s)

Gases ~ 0.02 ~ 20×10-6

Liquids ~ 0.2 (water ~ 0.6) ~ 0.1×10-6

Non-metallic solids ~ 2 ~ 1×10-6

Metals 20 – 200 5 - 50×10-6

There is no suchthing as a

temperature balance


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Heat conduction: plane surfaces /1/2

The heat flux Q” through a layered planar wall of materials with thickness di

and conductivity λi is foundby considering the material as a series of heat resistances di/λi

Note that Q” is the same for every location, so λ·ΔT/Δx is constant: Q” = -λ1· ΔT1/Δx1 = -λ2· ΔT2/Δx2 for a two-layermaterial

Thus, for each layer Q”· (di/λi) = - ΔTi = (Ti-1 – Ti)

T1T0=Tin T2=Tout

λ1 λ2Tin Tout

Q”

d1 d2

x

.

.

.

.

.


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Heat conduction: plane surfaces /2/2

Thus, for each layer

Q”· (di/λi) = - ΔTi = (Ti-1 – Ti)

For two layers ”1” and ”2”, the (unknown?) T1 can be eliminated: 1) Q”· (d1/λ1) = (T0 – T1) and 2) Q”· (d2/λ2) = (T1 – T2) 1) into 2) gives Q”· (d2/λ2) = (T0 – Q”· (d1/λ1) –T2)

→ Q”· (d2/λ2+ d1/λ1) = (T0 – T2) = Q”· d1+2/λ1+2

Similarly for N layers: Q”· (d1/λ1+ d2/λ2+ d3/λ3+ ... dN/λN) = ΔTtotal

T1T0=Tin T2=Tout

λ1 λ2Tin Tout

Q”

d1 d2

x

.

.

.

.

.

..

.

.

.

.


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Example: a brick wall /1

An oven wall consists in succession of – a layer of firebricks λf = 1.21 W/(m· K), – a layer of insulation material λi = 0.08 W/(m· K) and– a layer of outside bricks λb = 0.69 W/(m· K).

Each layer is 10 cm thick. Inside the oven the temperature Tin = 872°C, outside the temperature is Tout = 32°C. The surface area of the wall is 42 m2.

a. Calculate the heat loss by conduction during 24 h.b. Calculate the centretemperature Tm in the middleof the layer of insulation

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Example: a brick wall /2answer a:The heat transfer equation isQ = - A· λf/df·∆Tf

= - A· λi/di·∆Ti

= - A· λb/db·∆Tb

= - A· (λ/d)total·∆Ttotal

1/(λtotal/dtotal) = df/λf + di/λi + db/λb

→ λtotal/dtotal= 0.677 W/(m2K)

with ∆Ttotal = - 840°C (or K)gives Q = 42 m2· 0.677 W/(m2K)· 840 K = 23.88 kW (or kJ/s) which after 24 h means 23.88 kJ/s·24 h·3600 s/h = 2063 MJ

answer b:Thot-Tcold = (Q/A)· (d/λ)→ T1 = Tin - (Q/A)· (df/λf)= 872°C – 47°C = 825°Cand T2 = Tout + (Q”/A)· (db/λb)= 32°C + 82°C = 114°Ccheck: = T1 - (Q/A)· (di/λi)= 825°- 711°C = 114°C→ in centre of middle layerTm = (T1 + T2)/2 = 469°C

T1 T2T0 T3

λf λbλiTin Tout

Q”

.

.

.

.

.

.


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Cylindrical, spherical geometry /1

For stationary conduction in radial direction in cylindrical or spherical geometries, Fourier’s Law becomes:

) sphericalD-(1 )(m

l)cylindrica D-(1 )(m

with(W)

2

2

2r4A(r)

rL2A(r)

dr

dT)r(A)r(Q

Picture: KJ05

Picture: T06

see the literature for cases with angulargradients dT/dθ, dT/dφ etc.


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Heat convection /1

In convection (sv: konvektion)heat transfer, heat is entrained with a movingconducting medium.

The medium flow may be the result of external forces: forced convection(sv: påtvungen konvektion); or is the result of densitydifferences caused by temperature differences: free, or natural convection(sv: fri, eller naturlig konvektion)

Forced convection is usuallymuch more important thannatural convection Pictures: T06

Forced convection Free convection

wall

medium dy

dTλconv"Q :Law sFourier'

to according medium flowing

the into surface the from away

conducted is heat the:Note


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The general expression for heat transfer by convection is

Q = h· A· ΔT (W)

for heat exchange surface A (m2), temperature difference ΔT (K, °C) between the media or materials, and heat transfer coefficient h, unit: W/(m2·K)

The heat transfer coefficientdepends on

– Geometry– Flow velocity (or velocities,

if both media are fluids)– Type of flowing media

(gas, liquid)– Temperatures

The heat transfer coefficient(sv: värmeövergångs-koefficient) is a purely emperical and phenomenological quantitythat embodies ”all we do not know” about the heat transfer process

Picture: T06

.


Heat convection /2

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Heat transfer coefficient

Some typical values:

Table & pictures: KJ05

(a) Forced and (b) Natural convectionover a cylinder.


For circular tube flow, the laminar → turbulent flow transition occurs at Reynolds number Re = 2100 – 2300 (fully turbulent at Re = 4000) with dimensionless Re defined as Re = ρ· <v>· d / η

for ρ = fluid density (kg/m3), <v> = fluid average velocity (m/s), d = tube diameter (m) and η = fluid dynamic viscosity (Pa· s)

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Laminar ↔ turbulent fluid flow

Pictures: T06

Osborne Reynolds’s dye-streakexperiment (1883) for measuring laminar → turbulent flow transition

laminar: Re < 2100

laminar → turbulent

turbulent: Re > 4000


𝑅𝑒𝜌 · 𝑣

𝜂 ·𝑣𝑑

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Viscosity Viscosity (sv: viskositet) is a

measure of a fluid's resistance to flow; it describes the internal friction of a moving fluid. More specifically, it defines the

rate of momentum transfer in a fluid as a result of a velocitygradient. Dynamic viscosity η

(unit: Pa·s) is related to a kinematic viscosity, ν(unit: m2/s) via fluid density ρ (kg/m3) as: ν = η/ρ

η

Picture T06 Picture: KJ05


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Boundary layers

Growth of the velocity boundary layeron a flat surface.* This can be a solid surface or

another flowing medium

At the interface of a surface* and a flowing medium, a thinlayer (0.01 – 1 mm) of fluid is created in which the velocityincreases from v = 0 at the interface to the free-flowvelocity v = v∞ (or 0.99·v∞ )

In this boundary layer(sv: gränsskikt) all the thermaland/or viscous effects of the surface are concentrated

The boundary layer candevelop from laminar to turbulent flow

Pictures: KJ05


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Thermal boundary layers /1

Growth of the thermal boundary layeron a flat surface.

The boundary layer is a thinmore or less stagnant region where the fluid flow velocitydecreases until reachingv = 0 (”no slip”) at the surface The heat transfer resistance on

the flow side can be assumedconcentrated in and limited to the boundary layer A thermal boundary layer

is also formed as a result of the temperature difference BUT: the thicknesses the of the

hydrodynamic (δH) and thermalboundary (δT) layer are not the same! Pictures: KJ05

Growth of the velocity boundary layeron a flat surface.


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Thermal boundary layers /2

Typically the heat conductivity of the (flowing) fluid medium is (much) lower than that of the surfacematerial, and the very thin fluid layer may be the largest heat transfer resistance

For the boundary layer, two heat balances hold, which gives someinsight into the heat transfer coefficient:

Q” = Q/A = h·(Ts-Tf) = -λfluid·dT/dy│surface ≈ -λ fluid·ΔT/Δy│boundary layer

with ΔT = (Tf-Ts) and Δy = δthermal boundary layer = δT this gives

h·ΔT = λfluid·ΔT/δT → h ≈ λfluid/δT

This shows that thin boundary layers promote (heat) transfer.

Pictures: KJ05

. .


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Prandtl number The relative thicknesses of the thermal and hydrodynamic boundary

layers depends on: - Thermal conductivity λ

- Specific heat cp

- Dynamic viscosity η

(or kinematic viscosity v = η/ρ with density ρ )

defining the dimensionless Prandtl number Pr

Typical Prandtl numbers

for common fluids.

For a typical(di-atomic) gas

Pr ~ 0.7; Pr⅓ ~ 0.9

3

13

13

1

3

1

Pr)/(

a

cc pp

T

H

Picture: KJ05


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Nusselt number Consider convective heat transfer Qconv = h·A·ΔT

through the boundary layer with thickness δT at surfaceA with h = λfluid/δT

Can be compared with the conduction of heat througha stagnant layer of the same thickness of the same material Qconductive = - λfluid·A·ΔT/δT

→ the ratio of convective and conductive heat transfer: Qconvective / Qconductive = h·δT/λfluid = Nu

This defines the dimensionless Nusselt number Nu

For a more general geometry with acharacteristic size Lchar (for example, Lchar

= volume / surface = V/A), Nu is defined as:Pr)(Re,f

LhNu

fluid

char

.

.

.

.


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Nusselt number: flow around obstacles

Convection over a sphere

term last via T T or T T for allowing

380Pr0.7 and 80000 Re3.5 for

flowsurfaceflowsurface

D

41403221 060402 /.// )(Pr)Re.Re.(surface

DDDhD

Nu

Picture T06


Viscositycorrection

term

50/66

Example: convective heat transfer /1

Birds fluff their feathers to stay warm during winter. Estimate the convective heat transfer from a small bird in a v = 9 m/s wind, treating the bird as a d = 6 cm diameter sphere.

Data: the surface temperature at the feathers is -7°C, the temperature of the surrounding air is -10°C, pressure is 1 bar.The convective heat transfer coefficient h is related to windvelocity v by:

for air viscosity η = 1.68×10-5 Pa· s; air density ρ = 1.33 kg/m3; air thermal conductivity λ = 0.024 W/(m· K)

0.50.53Re Nu :or ;

50

530.

vd.

hd

Picture: T06


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Example: convective heat transfer /2

The heat flux Q (W) for convective heat transfer: Q = h· A· (Tsurface – Tsurroundings)

with A = πd2 = 0.0113 m2

ΔT = (Tsurface – Tsurroundings) = 3°C

and h follows from the relation betweenNu and Re:

Re = 42750 → Re0.5 = 207 → Nu = 110

h = (λ/d)· 0.53· Re0.5 = 43.9 W/(m2K)

This gives for the heat flux:

Q = 43.9· 0.0113· 3 = 1.49 W

Note: using the expression for a sphereNu = 2 + ......, with Pr = 0.71 and Pr ≈ Prsurface or η≈ ηsurface gives Nu = 121

.

.

.

Pictures: T06


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Nusselt number: tube flow

For turbulent flows in pipes (with diameter D) with large (or at least significant) temperature differencesbetween wall and flow – and this is usually the case - the following expression is usedwith a viscosity correction for near-wall fluid viscosity

ηwall = η at wall temperature:

NuD = 0.027· ReD0.8· Prn· (η/ηwall)0.14

with ReD > 10000; 0.7 < Pr ≤ 16700; L/D ≥ 10 and

n = 0.3 for cooling (Twall < Tflow) or n = 0.4 for heating (Twall > Tflow)


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Example: Conduction + convection /1

A 2 cm×2 cm×0.3 cm computer chip is cooled by a forced air flow with heat transfer coefficient h = 152 W/(m2· K).

The electronic component is located as a thin layer at the bottom surface of the chip, generating 1.6 W heat.

Air temperature Tf is 20°C Calculate the top and bottom

surface temperatures of the chip T2 and T1

For silicon, λ= 148 W/(m· K)Pictures: KJ05


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Example: Conduction + convection /2

Two mechanisms: For the convective heat

transfer at the surfaceQ = h·A· (T2-Tf) givesT2 = Tf + Q/(h·A) = 46.32 °C

For the conductive heat transfer through the chipQ = -λ·A· (T2-T1)/L givesT1 = T2 + Q· L/(λ·A) =

46.40 °CPictures: KJ05

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Natural convection

• A heated surface can cause a temperature gradient in a surrounding medium, and the resulting density differences Δρ induce (buoyancy driven) convection. Can be important in situations without forced flow or mixing.

• Natural convection (or ”free convection”) can also be induced by concentration differences Δc, and sometimes temperature and concentrationeffects must both be considered.

Picture: BMH99


More detail: see courseVärmeteknik, or # 5.2.3 ofProcessteknikens grunder

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Radiation heat transfer /1

Radiative heat transfer (sv: värme-överföringgenom strålning) involves the transfer of heat between surfaces of different temperature separated by a transparent (”diathermal”) medium, by electromagnetic waves

Radiant energy can be exchangedwithout any intervening medium and across (very) long distances.

For heat transfer, most important is the wavelength range 0.3 ≤ λ≤ 100 µm (300 ≤ λ≤ 105 nm), primarily the infrared region

A complicating factor is that radiant energyand also radiant properties of materials are dependent on wavelength λ

Picture: T06


The radiation QR (W) from a surfacewith emissivity ε (-) (sv: emissivitet), surface A (m2) and temperature T(K) equals

QR = ε· σ·A·T4 = ε·A· Eb

withStefan-Boltzmann coefficientσ = 5.67×10-8 W/m2K4

For a blackbody surface– ε =1 in the Stefan-Boltzmann Law– all incident radiation is absorbed– radiation is maximum for its temperature

at each wavelength– the intensity of the emitted radiation is

independent of direction: diffuse emitter57/66


Pictures: T06

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0

4TdEb ħ = h/2π, h = Planck’s constant 6.626×10-34 J.s c0= vacuum speed of light 2.998×108 m/skB = Boltzmann’s constant 1.381 ×10-23 J/KT = temperature K, λ = wavelength m

Blackbody radiation Planck’s

radiation Lawgives the spectraldistributionof the radiationemitted by a blackbody(sv: svart kropp, svart strålning)

The area under the curve equalsthe radiation(W/m2):


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Emissivity, gray surfaces

)(W/m 2

0

4Td)T,(E

Note:radiation heat transfer can be important also at low temperaturessuch as room temperature !

For a blackbody, the wavelength λmax for whichthe intensity is maximal is related to temperature T byλmax· T = 2898 µm· K

λmax = 10 µm @ T = 300 K

λmax = 0.5 µm @ T = 6000 K

Emissivity of a gray (sv: grå) surface is defined as ε = Eλ / Eλb = constant


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Tables and pictures: KJ05

Radiation interaction with (a) a general surface; (b) a black surface.

Gray bodies - emissivity data


ε

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The radiant heat transfer of a surface at temperature T with the surroundings can be expressed as

Qrad = ε· A· σ· (T4 – T4surroundings)

assuming that– the surface and the surroundings are isothermal– the whole surface A is incident with the

surroundings– the surroundings behave as a blackbody or

the area of the surroundings >> A– the surface is a diffusive emitter and reflector

The effect of an incident angle θ with respect to the normal to the surface:

Lambert’s cosine Law : Q(θ) = Q(0)·cos(θ)


Picture: T06

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. .


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Example: radiation from the sun /1

Consider the sun as a blackbody radiator at 5800 K. The diameter of the sun is ~ 1.39×109 m, the diameter of earth is ~ 1.29×107 m; the distance between earth and sun is ~ 1.5×1011 m.

Calculate:1. The total energy emitted by the sun (W)2. The amount of the sun’s energy

intercepted by earth (W)3. The solar radiation flux (W/m2) that

strikes reaches the earth’s atmosphereat a right angle. Compare it with a measured solar flux of 720 W/m2

measured on a clear day in Colorado, USA. Pictures: T06


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Example: radiation from the sun /2

1. The solar radiation output equals

Qsun = (π· d2sun)· (σ·T4

sun ) = 3.89×1026 W

2. The fraction of Qsun that is intercepted by earthequals, with S = distance earth-sun

Qsun-earth = Qsun × (π/4)· d2earth /(4π·S2)

= Qsun × d2earth / (16·S2) = 1.80×1017 W

3. Heat flux to spherical earth equalsQ”sun-earth = Qsun / (4πS2)= 1376 W/m2

which is ~ 2× the value measured in Colorado.

Picture: T06

Picture: KJ05

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Emission, reflection, transmission An energy balance for a

surface depends on reflected, absorbed, transmitted and of course emitted energy

Reflectivity ρ (-), absorptivity α (-) and transmittivity τ (-) are defined as

Energy balance → ρ + α + τ = 1For an opaque (sv: opak, ogenomskinlig) material τ = 0 → ρ + α = 1More exact, for incident angle θ, temperature T and wavelength λ

ρ = ρ(λ,θ,T), α = α(λ,θ,T), τ = τ(λ,θ,T), as also ε = ε(λ,θ,T)

Picture: KJ05

energy incident

energy dtransmitte ;

energy incident

energy absorbed ;

energy incident

energy reflected

Incident angleθ with respectto normal


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Example: convection and radiationfrom an outdoor grill The outside surface temperature of an

outdoor grill is Ts = 50°C. It looses heat to the surroundings by natural convection withh = hconv= 5.4 W/m2K and by heat radiation with emissivity = 0.87.The grills surface area is A = 0.63 m2. For surroundings temperatureTsurr = 20°C, calculate the heat transfer from the grill to the environment. Picture: KJ05

• Answer: the heat transfer as a result of the two mechanisms is

Q = Qconv + Qrad = h· A· (Ts-Tsurr) + ε· σ· A· (Ts4-Tsurr

4)

gives Qconv = 102 W and Qrad = 109 W, total Q = 211 W

..

.. .

.



Literature A83: P.W. Atkins ”Physical chemistry”, 2nd ed., Oxford Univ. Press (1983) BMH99: W.J. Beek, K.M.K Muttzall, J.W. van Heuven ”Transport phenomena” Wiley,

2nd edition (1999) BSL60: R.B. Bird, W.E. Stewart, E.N. Lightfoot ”Transport phenomena” Wiley (1960) BÖ88: Y. Bayazitoglu, M. Necati Özisik “Elements of heat transfer“ McGraw-Hill (1988) ÇB98: Y.A. Çengel, M.A. Boles “Thermodynamics. An Engineering Approach”, McGraw-Hill (1998) IdWBL07: F.P. Incropera, D.P. DeWitt, T.L. Bergman, A.S. Lavine ”Fundamentals of heat and

mass transfer”, Wiley, 6th ed. (2007) KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids Engineering”, Wiley (2005) SEHB06: P.S. Schmidt, O. Ezekoye, J. R Howell, D. Baker “Thermodynamics: An Integrated

Learning System” (Text + Web) Wiley (2006) T06: S.R. Turns ”Thermal – Fluid Sciences”, Cambridge Univ. Press (2006) vS91: R. von Schalien ”Teknisk termodynamik och modellering”, 6. ed, Åbo Akademi

University (1991)

Z13 R. Zevenhoven, course compendium ”Processteknikens grunder” / ”Principles ofprocess engineering (PTG), Åbo Akademi, Värme- och strömningsteknik (2013)

Available on - line (22 Mb): http://users.abo.fi/rzevenho/PTG%20Aug2013.pdf

ÖS96: G. Öhman, H. Saxén ”Värmeteknikens grunder”, Åbo Akademi University (1996)Available on - line (28 Mb): http://users.abo.fi/rzevenho/VTG -1996.pdf

See ÅA course library

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