Heat Transfer Mechanisms

President University Erwin Sitompul Thermal Physics 3/1

Lecture 3Thermal Physics

Dr.-Ing. Erwin SitompulPresident University

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Heat Transfer MechanismsChapter 18 First Law of Thermodynamics

Conduction Conduction is the transfer of heat from one

substance to another by direct contact. Denser substances are better conductors.

The direction of heat is always from the warmer to the colder substance.

Consider a slab of face A and thickness L, whose faces are maintained at temperature TH and TC by a hot reservoir and a cold reservoir.

Let Q be the energy that is transferred as heat through the slab, from its hot face to its cold face, in time t.

Experiment shows that the conduction rate Pcond (the amount of energy transferred per unit time) is:

condH CT TQP kA

t L

k : Thermal conductivity [W/m·K]


Heat Transfer Mechanisms A material that readily transfers energy by

conduction is a good thermal conductor and has a high value of k.

For the purpose of insulation, we are more concerned with poor heat conductors than with good ones.

For this reason, the concept of thermal resistance R is introduced. The R-value of a slab of thickness L is defined as:

Chapter 18 First Law of Thermodynamics

LRk



Conduction Through a Composite Slab Assume a composite slab, consisting of two

materials having different thickness L1 and L2 and different thermal conductivities k1 and k2.

Each face of the slab has area A. In the steady state, the conduction rates

through the two materials must be equal. Let TX be the temperature of the interface between the two materials, so that:

cond 2 12 1

X CH X T TT TP k A k AL L

After a little algebra,

1 2 2 1

1 2 2 1

C HX

k L T k LTT

k L k L



Substituting TX into either equality yields:

Extending the equation above to apply to any number n of materials making up a slab:

cond1 1 2 2

( )H CA T TP

L k L k

cond

1

( )H Cn

n ni

A T TP

L k


CheckpointChapter 18 First Law of Thermodynamics

The figure shows the face and interface temperatures of a composite slab consisting of four materials, of identical thicknesses, through which the heat transfer is steady. Rank the materials according to their thermal conductivities, greatest first.

ΔTc > ΔTa > ΔTb = ΔTd

kb and kd ties, then ka, then kc.



Convection Convection is the transfer of heat from one substance to another

through the movement of molecules within fluids (liquid or gas). As example, convection occurs when water comes in contact with

an object whose temperature is higher than that of the water. The temperature of the part of the water that is in contact with the

hot object increases, and (in most cases) water expands and thus becomes less dense.

The expanded fluid is now lighter than the surrounding cooler fluid, and the buoyant forces cause it to rise.

Some of the surrounding cooler fluid then flows down to take the place of the rising warmer fluid, and the process can then continue.


Heat Transfer Mechanisms Radiation Radiation is the transfer of heat through electromagnetic waves or

photons in the form of rays, waves, or particles. Radiation does not need a propagating medium and it moves at

the speed of light. The heat radiated by the sun can be exchanged between the solar

surface and the Earth’s surface without heating the transitional space.

The rate Prad at which an object emits energy via electromagnetic radiation depends on the object’s surface A and the temperature T of that area in kelvins and is given by:

Chapter 18 First Law of Thermodynamics

4radP AT σ : Stefan-Boltzman constant

= 5.6704×10–8 W/m2·K4

ε : emissivity of the object’s surface [0..1]


ProblemChapter 18 First Law of Thermodynamics

The cross section of a wall is made of white pine of thickness La and brick of thickness Ld (=2.0La), sandwiching two layers of unknown material with identical thicknesses and thermal conductivities.The thermal conductivity of the pine is ka and that of the brick is kd (=5.0ka). The face area A of the wall is unknown.Thermal conduction through the wall has reached the steady state; the only known interface temperatures are T1 = 25°C, T2 = 20°C, T5 =–10°C. What is the interface temperature T4?

4 51 2cond,steady state a d

a d

T TT TP k A k AL L

4 ( 10)25 20 (5 )2a a

a a

Tk A k AL L

4525 20 ( 10)2T 4 8 CT

?If Lb = Lc but kb = 2kc, can you find T3?


ProblemChapter 18 First Law of Thermodynamics

(a) A cube, 10 cm on a side, of rough steel is heated in a furnace to a temperature of 400.0°C. If its total emissivity is 0.97, determine the rate at which it radiates energy from each face.

(b) A black body is a body that completely absorbs all the electromagnetic radiation falling on it. A small hole in the wall of an object behaves like a black body. At what rate does radiation escape from a hole 10 cm2 in area in the wall of a furnace whose interior is at temperature of 700oC?

(a)

(b)

4radP AT

8 2 4(5.6704 10 )(0.97)(10 )(673.15) 112.94 W

4radP AT

8 3 4(5.6704 10 )(1.00)(10 )(973.15) 50.85 W ● A black body absorbs

all the radiation falling on it


Chapter 19Kinetic Theory

Thermal Physics


Physics of Gases As one of the main subjects in thermodynamics, a gas consists of

atoms or molecules that fill their container’s volume and exert pressure on the container’s wall.

There are three variables associated with a gas: volume, pressure, and temperature. All these can be related with the motion of the atoms.

The kinetic theory of gases relates the motion of the atoms to the volume, pressure, and temperature of the gas.

Chapter 19 Kinetic Theory


Avogadro’s Number In working with gas consisted of atoms and molecules, it makes

sense to measure the size of the samples in mole. By doing so, we can be certain that we are comparing samples

that contain the same number of atoms or molecules. One mole is the number of atoms in a 12 g sample of carbon-12.


The number of atoms or molecules in a mole is determined experimentally and is given by the Avogadro’s number:

23 1A 6.02 10 molN

Amedeo Avogadro suggested that all gases occupying the same volume under the same conditions of temperature and pressure contain the same number of atoms or molecules.


Avogadro’s NumberChapter 19 Kinetic Theory

The number of moles n contained in a sample of any substance is equal to the ratio of the number of molecules N in the sample to the number of molecules NA in 1 mol:

A

NnN

We can also find the number of moles n in a sample from the mass Msam of the sample and either the molar mass M (the mass of 1 mol) or the molecular mass m (the mass of one molecule):

sam sam

A

M MnM mN


Ideal GasesChapter 19 Kinetic Theory

The objective now is to explain the macroscopic properties of a gas –pressure, volume, temperature– in terms of the behavior of the atoms or molecules that make it up.

Although there are so many different gases, experiments show that if we confine 1 mol sample of various gases in boxes of identical volume and hold the gases at the same temperature, then their measured pressures are nearly the same.

Further experiments show that, at low enough densities, all real gases tend to obey the relation:

Ideal Gas LawpV nRTp : absolute pressuren : number of moles of gasT : temperature (K)R : gas constant

= 8.31 J/mol·K An alternative form of the ideal gas law is:

pV NkT Ideal Gas LawN : number of molecules of gask : Boltzmann constant

= R/NA= 1.38×10–23 J/K



Work Done by an Ideal Gas at Constant Temperature (Isothermal Process)

Suppose we put an ideal gas in a piston-cylinder arrangement. Suppose that the gas expands or contracts from an initial volume

Vi to a final volume Vf while we keep the temperature T of the gas constant.

Such a process, at constant temperature, is called an isothermal process.

On p-V diagram, an isotherm is a curve that connects points that have the same temperature.

For n moles of an ideal gas, the equation of the graph is:

1 1(a constant)p nRTV V



The work done by an ideal gas during an isothermal process is:f

i

V

V

W pdV1f

i

V

V

nRT dVV

lnf

i

V

VnRT Vf

Completing the calculation:

ln f

i

VW nRT

V Ideal Gas, Isothermal Process

For an expansion, Vf > Vi the natural logarithm of the ratio is positive the work W is positive.

For compression, Vf < Vi the natural logarithm of the ratio is negative the work W is negative.



Work Done by an Ideal Gas at Constant Volume (Isochoric Process) and at Constant Pressure (Isobaric Process)

To find the work W done by an ideal gas when the temperature varies, we can go back to the equation:

f

i

V

V

W pdV

For a constant-volume process (isochoric process), then the equation above yields:

0W Ideal Gas, Isochoric Process

For a constant-pressure process (isobaric process), then the equation above becomes:

( )f iW p V V p V Ideal Gas, Isobaric Process


CheckpointsAn ideal gas has an initial pressure of 3 pressure units and an initial volume of 4 volume units. The table next to the right gives the final pressure and volume of the gas (in those same units) in five processes.Which processes start and end on the same isotherm?

All but c.At isothermal process, pV = constant



ProblemChapter 19 Kinetic Theory

A cylinder contains 12 l of oxygen at 20°C and 15 atm. The temperature is raised to 35°C, and the volume is reduced to 8.5 l.What is the final pressure of the gas in atmospheres (atm)? Assume that the gas is ideal.

pV nRT (a constant)pV nRT

f fi i

i f

p VpVT T

fif i

f i

TVp p

V T (12) (35 273.15)(15)

(8.5) (20 273.15)

22.26 atm


ProblemChapter 19 Kinetic Theory

One mole of oxygen (assume it to be an ideal gas) expands at a constant temperature T of 310 K from an initial volume Vi of 12 l to a final volume Vf of 19 l.How much work is done by the gas during the expansion?

ln f

i

VW nRT

V

(19)(1)(8.31)(310) ln(12)

1183.8 J

?Is there any heat transfer Q along the expansion?


Class Group Assignments1. A styrofoam (k = 0.010 W/m2·K) box used to keep drinks cold at a picnic has total

wall area (including the lid) of 0.80 m2 and wall thickness of 2.0 cm. It is filled with ice and water at 0°C. If the temperature of the outside wall is 30°C, how much ice melts in one day?(a) 1.178 kg (b) 4.210 kg (c) 2.562 kg (d) 6.228 kg (e) 3.114 kg

2. A certain scuba main tank is designed to hold 66 ft3 of air when it is at atmospheric pressure at 22°C. When the volume of air is compressed to an absolute pressure of 3 000 psi (pound per square inch) and stored in a 10-liter portable tank, the air becomes so hot that the tank must be allowed to cool before it can be used. Determine the temperature of the air just after the compression process.(a) 27°C (b) 55°C (c) 40°C (d) 310°C (e) 48°C

3. A 2.00-liter container holds half a mole of an ideal gas at a pressure of 12.5 atm. If R = 8.31 J/mol·K, what is the gas temperature?(a) 1980 K (b) 1190 K (c) 965 K (d) 609 K (e) 578 K


4) An air bubble doubles in volume as it rises from the bottom of a lake (ρwater = 1000 kg/m3). Assuming that the process is isothermal, the depth of the lake is:(a) 21 m (b) 0.76 m (c) 4.9 m (d) 10 m (e) 0.99


Homework 31. (18-57) In part (a) of the next figure, two identical rectangular

rods of metal are welded end to end, with a temperature of T1 = 0°C on the left side and a temperature of T2 = 100°C on the right side. In 2.0 min, 10 J is conducted at a constant rate from the right side to the left side. How much time would be required to conduct the same 10 J if the rods were welded side to side as in part (b)?

Deadline: Wednesday, 30 April 2014 at 10:30. Quiz 1: Thursday, 1 May 2014 (International Labor Day)

Wednesday, 30 April 2014. We have make-up class at 10:30-13:00

2. (19-13) Air that initially occupies 0.140 m3 at a gauge pressure of 103.0 kPa is expanded isothermally to an absolute pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. (a) Draw the p-V diagram of the process

(b) Compute the net work done by the air. (Gauge pressure is the difference between the absolute/actual pressure an atmospheric pressure.)


Heat Transfer Mechanisms

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