Thermodynamics- Heat Transfer Process

Heat Transfer Process

Heat can be transferred from one place to another by three different methods: conduction, convection and radiation. Conduction usually occurs in solids convection in liquids and gases. No medium is required for radiation.

Thermal conduction if A is area of cross-section of a conductor I its length K thermal conductivityT₁andT₂temperatures at two ends, ten rate of transfer of heatdQ / dtor thermal current is given by

dQ / dt = KA(T₂– T₁) /1 = - KA dT / dx

Temperature gradientdT / dxis negative in the direction of heat flow.

Comparing it with ohm’s law in electricityI = V/R

I thermal = dQ / dt Vthermal = T₁– T₂and Rthermal = 1 / KA

Laws of resistance in series and parallel as in electricity are valid in thermal resistance s also, it is believed that current carriers (free electrons) and heat carriers are same because all electrical conductors are also thermal conductors. In general metals are better thermal conductors than liquids and gases as metals have large number of free electrons.

Thermometric conductivity(D)it is the ratio of thermal conductivity to thermal capacity per unit volume. Thus thermometric conductivity or diffusivity is

D= K/pC where K -----> thermal conductivity

P ----> density

C -----> specific heat Cv

Thermal conductivityKof gasesK = 1 / 3 V av λp c_V

= Dp Cv where D = 1 / 3 v av λis diffusion coefficient and

λ = 1/2 πd²nis mean free path

{d--->effective diameter of a molecule

{n -->number of molecules / volume

Wiedemann-franz law Tiedemann and Franz have shown that at a given temperatureT, the ratio of thermal conductivityK) to electrical conductivity(σ)is constant

That is,

K/σT = constant

Ingen’s –housz experiment:Ingen Housz showed that if the number of identical rods of different metals are coated with wax an one of their ends is put in baling water, then in steady state the square of length of the bar over which wax melts is directly proportional to the thermal conductivity of the metal. That is,

K/L²= constant

Thermal resistances in series

R = R₁+ R₂

I₁+ I₂/ K_effA_eff+ I₁/ K₁A₁+ I₂/ K₂A₂

If A₁= A₂and I₁= I₁

Then 2/K_eff= 1/K₁+ 1/K₂

Thermal resistances in parallel

1/R = 1/R₁+ 1/R₂

K eff A_eff/1 = K₁A₁/ 1 + K₂A₂/ 1

Or K_effA_eff= K₁A₁+ K₂A₂

And K_eff= K₁+ k₂/2

If the area of cross-section is equal then

A_eff= A₁+ A₂+ 2A

增长啊f ice in a pond

dQ₁= KA 0 – ( - θ) / y dy = dQ₂= mL = PAdyL

Or dy / dt = Kθ / pL x 1 / y or 1 = pL y²/ Kθ²

The ratio of times for thicknesses0 to y : y to 2y : 2y to 2y : : 1 : 3 : 5

In a shell of radiusr₁and r₂

dQ / dt = K 4πr₁r₂/ (r₂– r₂) (θ₁– θ₂)

Thickness of the shell(r₂– r₁) = K4πr²θ / dQ/ dt

Wherer₁= r₂= r and θ₁– θ₂= θ

In a cylinder of lengthI, radii r₁and r₂

dQ / dt = 2πK I ( θ₁–n θ₂)

log e r₂/ r₁

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