# hmt chap 1

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HEAT AND MASS TRANSFERSYLLABUSIt is branch of science which deals with the study of heat transfer rate and the mechanism of heat transfer.Thermodynamics Vs Heat transfer Thermodynamics tells us:How much heat is transferred (Q)How much work is done (W)Final state of the systemHeat transfer tells us:How (with what modes) Q is transferredAt what rate Q is transferredTemperature distribution inside the bodyMODE OF HEAT TRANSFERConductionConduction is the flow of heat in a substance due to exchange of energy between molecules having more energy and molecules having less energy.

[ solids Lattice vibrations, (ii) motion of free electrons][ fluids conduction is due to collision between the molecules caused by the random motion]

-needs matter (solid liquid, gas)-molecular phenomenon (diffusion process)-without bulk motion of matter

ConvectionThe transfer of energy from one region to another region due to macroscopic motion in fluid, added on to the energy transfer by conduction is called heat transfer by conduction.

-heat carried away by bulk motion of fluid-needs fluid matter Forced Convection :: fluid motion is caused by an external agency.Natural Convection:: fluid motion occurs due to density variations caused by temperature differences.RadiationAll the physical matter emits thermal radiation in the form of electromagnetic waves because of vibrational and rotational movements of the molecules and atoms which make up the matter.

-does not needs matter-transmission of energy by electromagnetic waves Radiation increases with the temperature level

MODE OF HEAT TRANSFERCONDUCTION- MECHANISMCONDUCTION:Conduction is the transfer of energy from more energetic particles of a substance to the adjacent less energetic one as a result of interactions between the particles.In gases and liquids, conduction is due to the collisions and diffusion of the molecules during their random motion.In solids, it is due to the combination of vibrations of the molecules in a lattice and the energy transport by free electrons.Micro form of heat transfer

FOURIERS LAW

q =-KA( dT / dX)CONCECTION-MECHANISMConvection: Heat transfer between a solid surface and an adjacent gas or liquid. It is the combination of conduction and flow motion. Heat transferred from a solid surface to a liquid adjacent is conduction. And then heat is brought away by the flow motion.Newton's law of cooling: whereh = Convection heat transfer coefficientTs = Temperature of the solid surfaceTf = Temperature of the fluidMacroform of heat transfer

CONVECTION NEWTONS LAW OF COOLINGq = hA (Ts-Tf)RADIATION-MECHANISMRadiation: The energy emitted by matter in the form of electromagnetic waves as a result of the changes in the electronic configurations of the atoms or molecules.Stefan - Boltzmann law:

where = Stefan - Boltzmann constant = emissivityTs = Surface temperature of the object q = AT4Conduction- Convection- RadiationThe three modes of heat transfer always exist simultaneously. For example, the heat transfer associated with double pane windows are:Conduction: Hotter (cooler) air outside each pane causes conduction through solid glass. Convection: Air between the panes carries heat from hotter pane to cooler pane. Radiation: Sunlight radiation passes through glass to be absorbed on other side.

Examples for different mode of heat transfer:

Conduction:Heat loss through thermal insulation on steam pipe

Convection:Heat transfer to water flowing through a pipe in condenser

Radiation:Heat transfer in an electric furnace

Conduction, Convection and Radiation:Solar energy used water heater

Fouriers Law of Heat Conduction

Newtons Law of Cooling

Laws of Thermal Radiation

LAWS FOR HEAT TRANSFERCONDUCTION

CONVECTION

NEWTONS LAW OF COOLING

RADIATION

STEFAN BOLTZMAN LAW

q=-KA( dT / dX)Assumptions:Steady stateUnidirectional heat flowConst. temperature gradientLinear profileNo internal heat generationIsothermal bounding surfaceHomogeneousIsotropic Applicable to solid, liquid, gasq=AT4q= hA (Tw-T)

FOURIERS LAWq = AT4ELECTRICAL &THERMAL ANALOGY THERMAL RESISTANCE(ELECTRICAL ANALOGY)

OHMs LAW :Flow of Electricity

V=I R elect

Voltage Drop = Current flowResistance

Temp Drop=Heat FlowResistance

T= q RthermThermal Analogy To OHMS Law

PLANE WALL-UNIDIRECTIONAL

PLANE WALL -SERIESPLANE WALL IN SERIESPLANE WALL -PARALLELPLANE WALL-PARALLELPLANE WALL-COMBINATIONPLANE WALL-FILM RESISTANCEPLANE WALL- THERMAL CONTACT RESISTANCE PLANE WALL- THERMAL CONTACT RESISTANCE At the joining surface of the two slabs air is trapped in voids due to surface irregularities.CYLINDER-SOLIDHEAT TRANSFER IN SOLID CYLINDERHEAT TRANSFER RATE IN HOLLOW CYLINDERHEAT TRANSFER RATE IN HOLLOW CYLINDER-COMPOSITEHEAT TRANSFER RATE IN HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER HOLLOW CYLINDER-MULTY LAYERHEAT TRANSFER HOLLOW CYLINDER-MULTY LAYER

HEAT TRANSFER IN SPHERE HEAT TRANSFER IN SPHERE -MULTY LAYERHEAT TRANSFER IN SPHERE -MULTY LAYERProblem 1. The wall in a furnace consists of 125 mm thick refractory bricks and 125 mm thick insulating fire bricks separated by an air gap. A 12 mm thick plaster covers the outer wall. The inner surface of wall is at 1100oC and the ambient temperature 25oC. the heat transfer coefficient on the outer wall to the air is 17 W/m2K and the resistance to heat flow of the air gap is 0.16 K/W. the thermal conductivities of refractory brick, insulating firebrick and plaster are 1.6, 0.3 and 0.14 W/mK, respectively. Calculate,(a). the rate of heat loss per unit area of the wall surface,(b). the interface temperatures throughout the wall(c). the temperature at the outside surface of the wall.Problem 2:Two large aluminum plates (k=240 W/mK) each 2 cm thick, with 10 mm surface roughness are placed in contact under 105 N/m2 pressure in air. The temperature at the outside surfaces are 390 oC and 406 oC.Calculate : A) the heat flux, B) the temperature drop due to the contact resistance C) the contact temperatures. Thermal contact resistance with air as the interfacial fluid for 10 mm roughness is 2.75*10-4 m2K/W.Problem 3:Steam at 350 oC flowing in a pipe (k=80 W/mK) of 5 cm inner diameter and 5.6 cm outer diameter is covered with 3 cm thick insulation of k=0.05 W/mK. Heat is lost to surroundings at 5 oC by natural convection and radiation, the combined h being 20 W/m2K. Taking the heat transfer coefficient inside the pipe as 60 W/m2K,DetermineA) the rate of heat loss from the steam per unit length of the pipeB) the temperature drop across the pipe and the insulation.

PROBLEM:4A nuclear reactor has a sperical pressure vessel 750 mm inside diameter 80 mm wall thickness. The temperature at the inner surface is 500 oC . If the tempertarue at the outer surface is 495 oC.Calculate:The rate of heat loss through the metal wallIf the rate of heat loss is to limited to 305W,how much inulation thickness needs to be applied?Thermal conductivity of steel wall=46 W/moCThermal conductivity of insulation=0.04 W/moCFilm coefficient of heat tranfer at inner and outer surface are 200 W/m2 oC and 15 W/m2 oC . Ambient air temperature=28 oCCRITICAL RADIUSThe radius upto which heat flow increases and after decreases. At critical radius q becomes qmax.The addition of insulation always increases (Rth)cond.Rtotal= (Rth)cond + (Rth)conv The addition of insulation may reduce the (Rth)conv due to increase in surface area.r1 rc- heat transfer decreases by adding insulation above the rc

Plane wall- Insulation thickness -studyCylinder & Sphere- Insulation thickness -studyProblem#5A 3mm diameter and 5 mm long electric wire is tightly wrapped with a 2 mm thick plastic cover whose K=0.15 w/moC. Electrical measurements indicate that a current of 10 A passes through the wire and there is a voltage drop of 8 V along the wire. If the insulated wire is exposed to a medium at Ta=30oC with a heat transfer coefficient of h=12 w/m2oC. Determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine whether doubling the thickness of the plastic cover will increase or decrease this interface temperature.

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATESConsider an unsteady (transient) three dimensional heat conduction with internal heat generation in heterogeneous & anisotropic PLANE WALL.Cartesian co-ordinates: x, y, zElemental length: dx,dy,dzConsider an elemental volume ABCDEFGH = (dx.dy.dz) Let, Temperature distribution is a function of T=T(x,y,z,t)Let,

=internal heat generation per unit volume per unit time (W/ m3)q= rate of heat flow (W)Q= q.dt=total heat flow (J)r=mass density of the material element(kg/m3)Cp=specific heat of the material element (J/kg K)

Plane (y - z) : x direction:Quantity of heat leaving from the face EFGH in X direction,

Quantity of heat flowing into the face ABCD in X direction,

Heat accumulation in the element due to heat flow by conduction in X direction,

Plane (x - z) : y direction:Quantity of heat flowing into the face ABFE in Y direction,

Heat accumulation in the element due to heat flow by conduction in Y direction,

Plane (x - y) : z direction:Quantity of heat flowing into the face ADHE in Z direction,

Heat accumulation in the element due to heat flow by conduction in Z direction,

Quantity of heat leaving from the face DCGH in Y direction,

Quantity of heat leaving from the face BCGF in Z direction,

Total amount of internal heat generation in the element ,Ein + Egen = Eout+Est(Ein- Eout)+ Egen=Est

stgzyxQQdQdQdQ= +++)(

Total heat energy stored in the element ,(i.e, thermal energy of the