Heat Transfer Assignment Help

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 conductivity T1 and T2temperatures at two ends, ten rate of transfer of heat dQ / dt or thermal current is given by

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

Temperature gradient dT / dx is negative in the direction of heat flow.

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

I thermal = dQ / dt Vthermal = T1 – T2 and 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 conductivity K of gases K = 1 / 3 V av λp cV

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

λ = 1/2 πd2n is 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 temperature T, the ratio of thermal conductivity K) 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 / Keff Aeff + I1 / K1 A1 + I2 / K2 A2

If A1 = Aand I1 = I1

Then 2/Keff = 1/K1 + 1/K2

Thermal resistances in parallel

1/R = 1/R+ 1/R2

K eff Aeff/1 = K1 A1 / 1 + K2 A2 / 1

Or Keff Aeff = K1 A1 + K2 A

And Keff = K1 + k2/2

If the area of cross-section is equal then

Aeff = A1 + A2 + 2A

Growth of ice in a pond

dQ= KA 0 – ( – θ) / y dy = dQ= mL = PAdyL

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

The ratio of times for thicknesses 0 to y : y to 2y : 2y to 2y : : 1 : 3 : 5 

In a shell of radius rand r2

dQ / dt = K 4πr1 r2 / (r2 – r2) (θ1 – θ2)

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

Where r1 = r2 = r and θ1 – θ= θ

In a cylinder of length I, radii r1 and r2

dQ / dt = 2πK I ( θ1 –n θ2)

log e r2 / r1

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