HMT Short Notes

Basic

1. Thermal Conductivity

Variation

Thermal conductivity is very high for Non-metallic crystals

2. Specific Heat: ρ c p

3. Thermal Diffusivity: How fast the heat propagates

α= kρ c p

4. Convection heat transfer coefficient

Forced>Free , Liquid>Gas , Boiling∧condensationmeinmax

Solid

Gases

Water

Liquid

Conduction Governing PDE

1. General Formula for 1D Heat Conduction

1

rn∂∂r (rn k ∂T

∂r )+ g= ρ cp∂T∂ t

r={0 ; for longwall∧r→ x1 ; for long cylinder2; for sphere

2. Heat flux varies with r in case of infinitely long cylinder and sphere, as area perpendicular to heat flow varies with r.

3. General Formula for 3D Heat Conduction (Cartesian coordinate)

∂∂ x (k ∂T

∂ x )+ ∂∂ y (k ∂T

∂ y )+ ∂∂ z (k ∂T

∂ z )+ g=ρ cp∂T∂ t

For constant k = Fourier-Biot Eq For steady State = Poisson’s eq For No heat gen = Diffusion eq No heat gen and steady = Laplace

4. General Formula for 3D Heat Conduction (Polar coordinate)

1r

∂∂r (rk ∂T

∂r )+ 1r2∂∂θ (k ∂T

∂θ )+ ∂∂ z (k ∂T

∂ z )+ g=ρ c p∂T∂ t

5. Boundary conditions for solving diff eq 1D:

At insulated surface or at thermal symmetry

( ∂T∂ x )=0Rest all are easy, simply Q will remain same

6. Heat generation in solids

T s−T ∞=f ( x )={gLhgr2hgr3h

T o−T s=f ( x )={g L2

2kgr2

4 kgr3

6k

If temperature distribution is to be found out then:

Take small element and then apply energy balance, then solve diff eq and put given boundary conditions.

7. Variable thermal conductivity:

k (T )=k o(1+βT )

β=temperature coefficient of thermal conductivity

In a temperature range T 1¿T 2 average value of thermal conductivity is equal to value of function k (t ) at T=T avg, that is

k avg=k o(1+βT 1+T 22 )

Steady State Heat Conduction

1. Conduction Thermal Resistances (ALKA)

Rwall=LkA

Rcyl=ln ¿¿

2. Convection Thermal Resistances

Rconv=1hA

3. Radiation Thermal Resistances

Rrad=1

hrad A

where ,hrad=εσ (T s2+T surr

2 )(T s+T surr)

4. Combined Convection Radiation heat transfer coefficient.

If both convection and radiation are present then use:

hcombined=hconv+hrad

Then

Req=1

hcombined A

Note that mostly you do not know T s in that case you take any suitable T s and solve the problem, then find check T s again, if error is significant then re-iterate.

5. Overall heat transfer coefficient

Express equation in form on Newton’s law of cooling

Q=UA ∆T

Here U is overall heat transfer coefficient.

UA= 1Rtotal

For hollow cylinder and sphere you need to define the area with respect to which you are calculating the overall heat transfer coefficient, because surface area is different. We say surface area w.r.t inner radius or outer radius

6. Critical Thickness of insulation

For r<r cr as you increase insulation heat transfer rate increases

For r>r cr as you increase insulation heat transfer rate decreases

rcr=khfor cyl∧rcr=

2kh

for sph

Wires vagerah mein zyada significant order of rcr is of mm

Fins (HAPKA)

1. Pin-Fin equation

θ−m2θ=0

Where, θ=excess temperature ( thanambient )=T−T ∞∧¿

m=√ hpk Ac

P is perimeter and Ac is cross-sectional area.

2. Basic Eq for various boundary conditions

Infinitely Long Fin Adiabatic Fin Tip

Temperature Distribution:θθb

=e−mx θθb

=cosh [m (l−x )]coshmL

Fin Efficiency: η= 1mL

η= tanhmLmL

Heat Transfer: Q=M θb Q=M θb tanhmL

M=√hpk Ac

3. Fin efficiency:

η=Q actual

Qmax

Qmax=h A fin(Tb−T∞ )

Qactual=h (η A fin) (T b−T ∞ )

By the above relation we find that we can replace the actual fin by the extra surface equals to ηA fin

.Add this area to unfinned outer surface area and we get new area to put in newton’s law of cooling to give heat transfer from a finned (fin+unfin) surface.

4. Fin effectiveness

For a single fin

ε=Q fin

Q nofin

=Qfin

h ( Ab ) (T b−T∞ )=h (η A fin ) (T b−T ∞ )h ( Ab ) (T b−T ∞ )

ε=η A fin

Ab

εη=

A fin

Ab

For a finned surface: Overall Fin effectiveness

ε=Q fin+Qun fin

Qnofin

ε=A fin+Aun fin

Anofin

For increase decrease problems use: (KP Sir bht effective the)

ε=√ kph Ac

Unsteady state conduction

1. Lumped System

Temperature does not vary with location, uniform temperature thought-out.

Condition of applicability:

Biot number < 0.1, implies temperature gradient inside body is negligible

Bi=h Lc

k

Lc=VA s

=charachteristic length=L ,r2,r3forwall cyl , sph

Biot number provides measure of temperature drop of solid relative to temperature difference between solid surface and the fluid

Also interpreted as ratio of thermal resistance to convection resistance.

2. Governing equation

θθo

=e−b x

Or alternate form -

θθo

=e−FoBi

Where, b= time constant

b= h Aρ cpV

And Fo=ℱ number

Fo=αt

Lc2

Effect of high/low b

High b implies equilibrium will be achieved quicker. Jitna bada b utni fast process

3. Un-lumped Analysis

We use Hiesler Charts for this

Non dimentionalised solution-

θ¿=f (Fo ,Bi , X¿)

X ¿=x /L

Bi=h Lk

θ¿=T ( x ,t )−T ∞

T o−T ∞

Fo=αt

L2

Lumped and un-lumped ke L different hai

For un-lumped analysis

L = L , r , r for wall, cyl, sph