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Kyoung Hoon Kim 1 - 1

Heat Transfer

2010

Kyoung Hoon Kim

School of Mechanical Engineering Kumoh National Institute of Technology


References

References

• Y.A. Cengel, Heat transfer, a practical approach, 2nd

Ed., McGraw-Hill, 2003

• M.J. Moran, H.N. Shapiro, B.R. Munson, D.R. Dewit,

Introduction to thermal system engineering, John

Wiley & Sons, 2003


Contents

• Basic of heat transfer

• Conduction

• Convection

• Boiling and condensation

• Thermal radiation

• Heat exchangers


CHAPTER 1

Basics of Heat Transfer


1.1 Thermodynamics and heat transfer

heat and heat transfer§heat: the form of energy that can be transferred from one

system to another as a result of temperature difference§heat transfer: the science that deals with the determination of

the rates of such energy transfer

thermodynamics§ it is concerned with the amount

of heat transfer as a system undergoes a process from one equilibrium state to another§ it gives no indication how long

the process will take

Heat flows in the direction of decreasing temperature.


Some application areas of heat transfer


1.2 Engineering heat transfer

rating and sizing problems§ rating problems: deal with the

determination of the heat transfer rate for a existing system at a specified temperature difference§sizing: deal with the determination of the

size of a system in order to transfer heat at a specified rate for a specified temperature difference


1.2 Engineering heat transfer

modeling in heat transfer§The descriptions of most

scientific problems involve expressions that relate the changes in some key variables to each other.§Modeling is a powerful

engineering tool that provides great insight and simplicity at the expense of some accuracy.

Modeling is a powerful engineering that provides great insight and simplicity at the expense of some accuracy.


1.3 Heat and other forms of energy

forms of energy§ total energy and internal energy§sensible and latent energy§chemical energy and nuclear energy§enthalpy: h = u + Pv

The internal energy urepresents the microscopic energy of a non-flowing fluid, whereas enthalpy hrepresents the microscopic energy of a flowing fluid.



Specific heats1) ideal gas: Pv = RT or P = rRT2) specific heat3) specific heat at constant volume Cv, kJ/kgK4) specific heat at constant pressure Cp, kJ/kgK5) ideal gas case: Cp - Cv = R6) Du = Cv DT, DU = mCv DT7) Dh = Cp DT, DH = mCp DT

Specific is the energy required to raise the temperature of a unit mass of a substance by one degree in a specific way.


Energy transfer

1) power = work done per unit time W, W or hp

2) heat: kJ, kcal, BTU

3) heat transfer rate, W, kcal/hr, BTU/hr

4) heat flux = heat transfer rate per unit area: q = Q / A



<Ex 1-1> Heating of a copper ball

A D-cm diameter ball is to be heated from Ti

oC to an average temperature Tf

oC in Dt minutes. Taking the average density and specific heat in this temperature range to be r kg/m3 and Cp

kJ/kgoC, respectively. Determine a) the total amount of heat transfer to the ball, b) the average rate of heat transfer to the ball, and c) the average heat flux.

Input Data

daimeter D 10.0 cm×:=

initial temperature Ti 100C×:=

final temperature Tf 150C×:=

density r 8950kg

m3

×:=

specific heat Cp 0.395kJ

kg K××:=

time interval Dt 30 min×:=

Solution

surface area As p D2

×:= As 0.0314m2

=

mass of the ball mb rp

6× D

3×:= mb 4.6862kg=

total amount of heat Qtot mb Cp× Tf Ti-( )×:= Qtot 92.5526kJ=

average rate of heat transfer QavQtot

Dt:= Qav 51.4181W=

average heat flux qQav

As:= q 1.6367

kW

m2

=


1.4 The first law of thermodynamics

the first law of thermodynamics1) 1st law: conservation of energy principle

total energy entering the system - total energy leaving the system= changing in the total energy of the system

2) energy balance: Ein - Eout = DEsystem

3) rate form: Ein - Eout = dEsystem / dt4) steady: Ein = Eout

5) heat balance: Qin - Qout + Egen = DEsystem

In steady operation, the rate of energy transfer to a system is equal to the rate of energy transfer from the system.


1.4 The first law of thermodynamics

Energy balance for closed systems1) Ein - Eout = DU = m Cv DT2) stationary closed system: Q = m Cv DT

Energy balance for steady flow systems1) mass flow rate: mdot = r V Ac

2) volume flow rate: Vdot = V Ac

3) net heat transfer: Qdot = mdot Cp DT

Surface energy balance1) Q1 = Q2 + Q3

In the absence of any work interactions, the change in the energy content of a closed system is equal to the net heat transfer.


<Ex 1-2> Heating of water in an electrical teapot

Liquid water of mw kg initially at Ti oC is to be heated to Tf oC in a teapot equipped with a P W electric heating element inside. The teapot is mp kg and has an average specific heat of Cpp

kJ/kgoC. Taking the specific heat of water to be Cpw kJ/kgoC and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated.

Input Data

mass of water mw 1.20 kg×:=initial temperature Ti 15 C×:=specific heat of water Cpw 4.18

kJ

kg K××:=

mass of tea pot mp 0.50 kg×:= final temperature Tf 95 C×:=

specific heat of teapot Cpp 0.70kJ

kg K××:= electric power of heat P 1200W×:=

Solution

input energy to water Ew mwCpw× Tf Ti-( )×:= Ew 401.2800kJ=

input energy to teapot Ep mp Cpp× Tf Ti-( )×:= Ep 28.0000kJ=

total input energy Etot Ew Ep+:= Etot 429.2800kJ=

heating time DtEtot

P:= Dt 357.7333s=


<Ex 1-3> Heat loss from heating ducts in a basement

A L-m-long section of an air heating system of a house through an unheated space in the basement. The cross section of the rectangular duct of the heating system is W*H. Hot air enters the duct at P kPa and Tin

oC at an average velocity of V m/s. The temperature of the air in the duct drops to Tout as a result of heat loss to cool the space in the basement under steady conditions. Also, determine the cost of heat loss per hour if the house is heated by a natural gas furnace that has an efficiency of h %, and the cost of the natural gas in this area is c $/therm (1 therm = 100,000 Btu = 105,500 kJ).

Input Datalength L 5 m×:= inlet temperature Tin 60 C×:=width a 0.20m×:= outlet temperature Tout 54 C×:=height b 0.25m×:= efficiency of furnace h 80 %×:=pressure P 100 kPa×:=

cost of natural gas c 0.60dollor

therm×:=velocity V 5 m× s

1-×:=

Solution

reference temperature TrTin Tout+

2273.15K×+:= Tr 330.1500K=

specific heat of air Cp Cp_air Tr( ):= Cp 1.0086kJ

kg K×=

density at inlet rP

Rair Tr×:= r 1.0554kg m

3-×=

mass flow rate mdot r V× a b×( )×:= mdot 0.2638kg

s=

heat transfer Q mdot Cp× Tin Tout-( )×:= Q 1.5967kW=

cost of heat loss C1 Q c×:= C1 0.0327dollor

hr=


<Ex 1-4> Electric heating of a house at high elevation

Consider a house that has a floor space of A ft2 and an average height of H ft at E ft elevation where the atmospheric pressure is Pat psia. Initially the house is at a uniform temperature of T1oC. Noe the electric heater is turned on, and the heater runs until the air temperature in the house rises to an average value of T2 oC. Determine the energy transferred to the air assuming (a) the house is air-tight and thus no air escapes during the heating process and (b) some air escapes through the cracks as the heated air in the house expands at constant pressure. Also determine the cost of this heat for each case if the cost of electricity in that area is c $/kWh

Input Dataarea of floor space A 2000ft

2×:= final temperature T2 70 F×:=average height of house H 9 ft×:= atmospheric pressure P 12.20psi×:=average elevation z 5000ft×:=

initial temperature T1 50 F×:= cost of electricity c 0.075dollor

kWh×:=

Solution

reference temperature TrT1 T2+

2459.67R×+:= Tr 519.6700R=

isobaric specific heat Cp Cp_air Tr( ):= Cp 1.0058kJ

kg K×=

isovolume specific heat Cv Cvair Tr( ):= Cv 0.1717BTU

lb R×=

volume of house V A H×:= V 1.8000 104

´ ft3

=

mass of air maP V×

Rair T1 459.67R×+( )×:= ma 1.1631 10

3´ lb=

a) model of constant volumeinput energy Ein ma Cv× T2 T1-( )×:= Ein 3.9936 10

3´ BTU=

energy cost C1 Ein c×:= C1 0.0878dollor=b) model of constant pressure

input energy Ein ma Cp× T2 T1-( )×:= Ein 5.5883 103

´ BTU=energy cost C2 Ein c×:= C2 0.1228dollor=


1.5 Heat transfer mechanisms

1) conduction2) convection3) radiation

1.6 Conduction

1) conduction- liquids & gases: collisions and diffusion- solids: vibration and transport by free

electrons2) Fourier's law

Qdot = - k DT / Dx , Wwhere k = thermal conductivity, W/mK


<Ex 1-5> The cost of heat loss through a roof

The roof of an electrically heated home is H m long, W m wide, and L m thick, and is made of a flat layer whose thermal conductivity is k W/moC. The temperatures of the inner and outer surfaces of the roof one night are measured to be T1 and T2, respectively, for a period of Dt. Determine the rate of heat loss through the roof that night and b) the cost of that heat loss to the home owner if the cost of electricity is c $/kWh.

Input Dataouter temperature T2 15 C×:=

length a 8 m×:= time interval Dt 10 hr×:=width b 6 m×:=

k 0.8W

m K××:=thermal conductivity

thickness L 0.25m×:=

inner temperature T1 4 C×:= electricity cost c 0.08dollor

kWh×:=

Solutionarea A a b×:= A 48.0000m

2=

heat transfer Q k A×T2 T1-

L×:= Q 1.6896kW=

total input energy Ein Q Dt×:= Ein 16.8960kWh=

electricity cost C1 Ein c×:= C1 1.3517dollor=


1.6 Conduction

A simple experimental setup to determine the thermal conductivity of a material


1.6 Conduction

The range of thermal conductivity of various materials at room temperature.


1.6 Conduction

The variation of the thermal conductivity of various solids, liquids, and gases with temperature.


<Ex 1-6> Measuring the thermal conductivity of a material

A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one dimensional. Two thermocouples are embedded into each sample some distance L apart, and a differential thermometer reads the temperature drop DT across this distance along each sample. When steady operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, cylindrical samples of diameter D cm and length L cm are used. The two thermocouples are placed d cm apart. After initial transient, the electric heater is observed to draw I A at V V, and both differential thermometers read a temperature difference of DT C. Determine the thermal conductivity of the sample.

Input Data

diameter D 5 cm×:= electric current I 0.40 amp×:=length L 10 cm×:= voltage DE 110 volt×:=distance of thermocouple d 3 cm×:= temperature difference DT 15 C×:=

Solution

electric power consumed We DE I×:= We 44.0000W=

heat transfer QWe

2:= Q 22.0000W=

cross sectional area Acp

4D2

×:= Ac 19.6350cm2

=

thermal conductivity kQ d×

Ac DT×:=

k 22.4090W

m K×=


<Ex 1-7> Conversion between SI and English units

An engineer who is working on the heat transfer analysis of a brick building in English units needs the thermal conductivity of brick. But the only value he can find from his handbook is k W/moC. To make matters worse, the engineer does not have a direct conversion factor

between the two unit systems for thermal conductivity. Can you help him out?

Input Data

thermal conductivity k 0.72W

mK××:=

Solution

unit conversionsW

mK×

W hr×

BTU

ft

m×

R

K×

BTU

hr ft× R××

conversion factors c1W hr×

BTU:= c2

ft

m:= c3

R

K:= c1 3.4121= c2 0.3048= c3 0.5556=

English unit ke km K×

W× c1× c2× c3×

BTU

hr ft× R××:= ke 0.4160

BTU

hr ft× R×=

automatic conversion with MathCAD k 0.6191kcal

hr m× K×= k 0.4160

BTU

hr ft× R×=


1.6 Convection

1) natural(free) convection and forced convection2) Newton's law of cooling: Q = As h DT

h = heat transfer coefficient, W/m2 K


<Ex 1-8> Measuring heat transfer coefficient

A L-m-long, D-cm-diameter electrical wire extends across a room at Ta oC. Heat is generated in the wire as a result of resistance heating and the surface temperature of the wire is measured to be Ts oC in steady operation. Also, the voltage drop and electrical current through the wore are measured to be V V and I A, respectively. Disregarding any heat transfer by radiation, determine the convection heat transfer coefficient for heat transfer between the outer surface of the wire and the air in the room.

Input Data

length of wire L 2 m×:= wire surface temperature Ts 152 C×:=

diameter of wire D 0.3 cm×:= voltage drop DE 60 volt×:=

room temperature Ta 15 C×:= electric current I 1.5 amp×:=

Solution

heat generation Q DE I×:= Q 90.0000W=

surface area As p D× L×:= As 188.4956cm2

=

heat transfer coefficient hQ

As Ts Ta-( )×:= h 34.8514

W

m2K×

=


1.9 Radiation

1) energy transport by electromagnetic wave form (or photons)

2) volumetric phenomena: generally surface phenomena: for solids that

are opaque to thermal radiation

3) blackbody: Qrad = s As Ts4, W

Ts = surface temperature, K

s = Stephan-Boltzmann constant = 5.67*10-8 W/m2 K4

4) incident radiation: a + r + t = 1

a = absorptivity

r = reflectivity

t = transmissivity

5) emmision: Qemit = e s As Ts4 , W

e = emissivity

6) Kirchhoff's law: e = a for same temperature and wavelength

7) completely enclosed by a much larger surface at Tsurr:

Qrad = e s As (Ts4 - Tsurr

4) , W

8) combined heat transfer coefficient

Qtotal = hcombined A (Ts - Tinf)


<Ex 1-9> Radiation effect on thermal comfort

It is a common experience to feel "chilly" in winter and "warm" in summer in our homes even when the thermostat setting is kept the same. This is due to the so called "radiation effect" resulting from radiation heat exchange between our bodies and the surrounding surfaces of the walls and the ceiling.

Consider a person standing in a room maintained at TaoC at all times. The inner surfaces of

the walls, floors, and the ceiling of the house are observed to be at an average temperature of Twinter

oC in winter and TsummeroC in summer. Determine the rate of radiation heat transfer

between this person and the surrounding surfaces if the exposed surface area and the average outer surface temperature of the person are As m2 and Ts

oC, respectively.

Input Data

room temperature Ta 22 C×:= Ta Ta Tk0+:=

average winter temperature Twinter 10 C×:= Twinter Twinter Tk0+:=

average summer temperature Tsummer 25 C×:= Tsummer Tsummer Tk0+:=

surface area of the person As 1.40m2

×:=

surface temperature of the person Ts 30 C×:= Ts Ts Tk0+:=

emissivity of the person e 0.95:=

Solution

winter time Qrw e s× As× Ts4

Twinter4

-( )×:= Qrw 152.1596W=

summer time Qrs e s× As× Ts4

Tsummer4

-( )×:= Qrs 40.9900W=

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