INTRODUCTION

Energy can be converted from one form to another. (Second law) Heat is energy in transition. And this heat transfer is caused by temperature difference. Rate of heat transfer is temp difference and 1/ to resistance by material.

DIFFERENCE BETWEEN THERMODYNAMICS AND HEAT TRANSFER they are complimentary

Thermodynamics tells us:

• How much heat is transferred (dQ)

• How much work is done (dW)

• Final state of the system

Heat transfer tells us:

• How (with what modes) dQ is transferred

• At what rate dQ is transferred

• Temperature distribution inside the body

Classification

Transit: example radiator.

Periodic: example of engine.

Modes of Heat Transfer

Conduction:

An energy transfer across a system boundary due to a temperature difference by the mechanism of intermolecular interactions. Conduction needs matter and does not require any bulk motion of matter.

By Oscillation: in non conductor

By electron and oscillation: in conductor

By impact: in fluids

Examples:

Conduction rate equation is described by the Fourier Law: this law is based on observations rate of heat flow is to area of flow (A) and temperature difference (T) and 1/ to thickness (dx) So ..

We assume case of one dimensional heat flow and k remains constant.q Ax dT/dx = kAx dT/dx

k = thermal conductivity of material (W/m K) (which varies with temperature but we assume it to be constant..)

q = heat flow vector, (W) or (J/s)

A = Cross sectional area in direction of heat flow. (m2 )

dT = temp difference in one direction.

Conduction rate equation rectangular coordinate system

q = kAx (dT/dx i + dT/dy j + dT/dz k)

Conduction rate equation radial system

qr = kAr dT/dr

Conduction rate equation if temperature gradient is constant

q = kA T2-T1

x2-x1

** Here area will remain constant in direction of flow, later we will derive other cases.

One end of the block of 1mx1mx.5m is at 100oC, another is maintained at 0oC. (k= 385W/m K). find (1)rate of heat transfer (2) thermal resistance.

A=1m2 l= .05mk= 385W/m KT= 385

(i) q = kA T = 385X1X100/0.05 = 770 kW

l(ii) R= l/kA = 0.05/(385X1)=1.3 X 10-4

Convection: heat transfer is by mixing of fluid

An energy transfer across a system boundary due to a temperature difference by the combined mechanisms of intermolecular interactions and bulk transport. Convection needs fluid matter.

Type :Natural convection: the mixing is carried out by difference in density of cold and hot partials(induced by buoyancy forces)..

Ex:

Hot plate to atmosphere. Water heating system. Heating of room by stove.

Forced convection: the mixing is carried out by pump, fan etc. here heat transfer rate speeds up.

Cooling of I.C engine.

Heat transfer through wall to fluid or fluid to wall is very important in engineering heat transfer.

Convection rate equation is described by the Newton’s Law of Cooling :

q = h As Tbutq = kA dT/dx = kA(Tw –Ts) = h As (Tw –Ts)/ so k=h/

q = h As T

Where:

q = heat flow from surface, (W)

h = heat transfer coefficient (which is not a thermodynamic property of the material, but may depend on geometry of surface, flow characteristics, thermodynamic properties of the fluid, etc.

(W/m 2 K)

As = Surface area from which convection is occurring. (m 2 )

T = TS –Tw Temperature Difference between surface and coolant. (K)

Q. An air cooler has surface area 0.12m2 and temp 65 oC. atmospheric temp is 30 oC surface coefficient

of heat transfer 45.5 W/m 2 K. calculate heat transfer.

Sol.

q = h As (TS –Tatm)

= 45.4 X (0.12)(65-30) =190W

Q. Water is heated up to boiling by a wire (rod) of 10cm X 1mm , 23.5 watt of power is consumed.

h=5000 W/m 2 K find temperature of wire for steady state.

Sol.

q = h As (Twire –Twater)

23.5=5000 x ( x d x l) (Twire –Twater)= 5000 x ( x 0.001x 0.1) (Twire –100)= 1.57(Twire –100)

Twire= 23.5/1.57 + 100 = 115 oC