hm 3

10
Why We Need It? Fourier Law is for steady flow, in one dimension, and without heat generation. The cases other then this can not be solved by this equation. Cartesian Coordinates : T(r, , z) Derivation : Assumptions : K (conductivity), c (pecific heat) and (denisity) do not vary with position. Heat generation is uniform. Consider a small volume whose dimensions are dx, dy, dz. Material is homogenous and isotropic. Means its properties (density, h, k ) are same everywhere. Temperature is indicated by T Temperature is a function of distance do T changes as distance changes ( T changes as dx changes). So rate of change of temperature T/ x

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HM 3

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Why We Need It?

Fourier Law is for steady flow, in one dimension, and without heat generation. The cases other then this can not be solved by this equation.

Cartesian Coordinates: T(r, , z)

Derivation:

Assumptions:

K (conductivity), c (pecific heat) and (denisity) do not vary with position.

Heat generation is uniform.

Consider a small volume whose dimensions are dx, dy, dz. Material is homogenous and isotropic. Means its properties (density, h, k ) are same everywhere.

Temperature is indicated by T

Temperature is a function of distance do T changes as distance changes ( T changes as dx changes).

So rate of change of temperature T/ x

Change of temperature at a distance dx=> Tx - Tx+dx => T/x dx

In d s Kumar it is => T/x dx

Now

heat inflow during time dt

+ heat generated Eg during time dt

=

heat outflow during time dt

+ change in internal energy during time dt Est

ENERGY BALANCE EQ.

(dqx+dqy+dqz)dt

+Eg(dxdydz)dt

=

(dqx+dx + dqy+dy +dqz+dz)dt

c(dx.dy.dz)dT

Est =mcdT

m= X volume

Now consider single direction x and apply Fourier eq

Heat inflow per unit time:

dqx = -k(dydz) T/x where dy.dz is area

Heat outflow in x direction per unit time: it is increased by dx

dqx+dx = dqx + /x (dqs ).dx = - /x (-k(dydz) T/x ) + /x { (-k(dydz) T/x) dx}

dqx+dx dqx =qx= -k (dx dydz) /x { (T/x ) }

dqx+dx dqx = -k (dx dydz) 2T/x2

Heat outflow in y direction

dqy+dy dqy = -k (dx dydz) 2T/y2

Heat outflow in z direction

dqy+dy dqy = -k (dx dydz) 2T/y2

Putting them in energy balance eq and solving we get GENERAL HEAT CONDUCTION EQ

2T/x2 + 2T/y2 + 2T/z2 + Eg/k = cT/kt = (c/k) (T/t)

(Change in thermalenergy storage) (Net transfer of thermal energy into the control volume (inflow-outflow)control volume (inflow-outflow)) (Thermal energygeneration)

SIGNIFICANCE OF GENERAL HEAT CONDUCTION EQ

This eq. tells us about temp. distribution and heat flow in a solid homogeneous and isotropic material Via conduction.

Thermal Diffusivity

k/ c = is called thermal diffusivity and is property of the material science it consist of all property terms. Greater it is greater is the ability to store or conduct heat.

Thermal diffusivity is ratio of conductivity (k) to thermal storage capacity(c) .

Liquids have low conductivity but high heat storage capacity. Metals have low c and high k.

This also tells us how fast temperature change can occur in a material if surrounding temperature is changed.

Temperature distribution in unsteady state depend on conductivity and storage capacity / but in unsteady state only on conductivity.

If heat generation is nil Eg = 0 this eq. is Fouriers eq. in three dimension.

If system is in steady state but with heat generation this eq. is called poissons eq.

If no heat source and in steady state this is called laplace,s eq.

For one dimension / without generation / steady state the equation is

d2T/dx2 = 0

HEAT CONDUCTION EQUATION (RADIAL SYSTEMS)

When conduction occurs in shapes of radial geometries it is more convinent to work in cylindrical systems.

Cylindrical Coordinates: T(r, , z)

Derivation:

Assumptions: same

Now consider single direction x and apply Fourier eq

Net Heat flow in r direction per unit time

dqr+dr dqr = -k (dr ddz) 2T/r2

Net Heat flow in direction per unit time

dq+d dq = -k (dr ddz) 2T/2

Net Heat flow in z direction per unit time

dqz+dz dqz = -k (dr ddz) 2T/z2

Net heat generated per unit time

Eg r(dr ddz)

Net heat generated per unit time

c r (dr ddz) dT/dt

Putting in energy balance eqn and solving we get

2T/r2 + (1/r) T/r +(1/r2) 2T/2 + 2T/z2 + Eg/k = (c/k) (T/t) =(1/) (T/t)

(Thermal energygeneration) (Change in thermalenergy storage) (Net transfer of thermal energy into the control volume (inflow-outflow)control volume (inflow-outflow))

** Every other thing remains same as derivation before

For one dimension / without generation / steady state the equation is

2T/r2+(1/r) T/r = 0

1/r2 d/dr ( r dT/dr ) = 0

SPHERICAL COORDINATES: T(r, , )

Derivation:

Assumptions: same

Volume = (dr.rd.rsin.d)

Heat flow r- plan, direction per unit time

Inflow = dq = -k(dr . r.d)()

Heat stored or change in heat energy

dq+d - dq = .dt

dq+d - dq = k(dr.rd.rsin.d) []

Heat flow r- plan, direction

Inflow = dq = -k(dr . sin.d)()

dq+d - dq = (q) . rd

solving for time dt

dq+d - dq = k(V) ()().dt

Heat flow - plan, r direction

Similarly

dqr+dr - dqr = (q) . dr

solving for time dt

dqr+dr - dqr = k(V) () . dt

Putting in energy balance eqn and solving +

+ ()() + () + Eg/k =(1/) (T/t)

For one dimension / without generation / steady state the equation is

1/r2 d/dr ( r2 dT/dr ) = 0

OVER ALL

Steady-state conduction, no internal generation of energy

For one-dimensional, steady-state transfer by conduction without heat generation

i = 0 rectangular coordinates

i = 1 cylindrical coordinates

i = 2 spherical coordinates

THERMOPHYSICAL PROPERTIES

Thermal Conductivity: A measure of a materials ability to transfer thermal energy by conduction.

Thermal Diffusivity: A measure of a materials ability to respond to changes in its thermal environment.

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