Heat Transfer | What is conduction| Conduction| Heat| serise and parallel combination of rod

There are three types modes of heat transfer.

(1) Conduction

(2) Convention

(3) Radiation

*** Conduction :--

Conduction is a method of heat transfer in which the molecules of a substance transfer heat by vibrating without changes its position .

Conduction is a mode of transfer of heat between two adjacent part of body because of their temperature difference.

Length of the rod =L

Temperature difference(Δ T)= (T₁ - T₂)

**Heat current(H) :-

Rate of flow of energy from hotter end to colder end is called heat current.

[ H = d Q/dt]

:- S I unit of heat current is j/s = watt(W).

:- Dimensional formula of current is [ML²T⁻³]

:- It is scalar quantity .

***Temperature gradient (T. g) :-

The ratio of temperature difference between two end of rod and length of rod is called temperature gradient.

T.g = T₁ - T₂/L

:-S I unit of T.g is ° C /m.

:-Dimensional formula of T. g is [M⁰L⁻¹T⁰K].

:- It is scalar quantity.

***Heat current for conduction rod:-

A=Cross section area of rod ,L=Length of rod

Change in temperature(Δ Τ) = T₁ - T₂

Heat current ∝ Change in temperature

H ∝ Δ T

H ∝ T₁ - T₂ --------(1)

Heat current ∝ Cross-section area of rod

H ∝ A -------------(2)

Heat current ∝ 1/ Length of rod

H ∝ 1 /L --------------(3)

From eqⁿ (1) , (2) and (3)

:- S I unit of k is J m⁻¹ °C ⁻¹.

:- practical unit of k is Cal m⁻¹ °C⁻¹ .

: - Dimensional formula of k is [MLT⁻²K⁻¹].

Example (1) One end of a rod of length 4m and cross-section area 20 × 10⁻²m² maintained at 120°C and other end of rod is at 30°C. Heat flow through the rod is 100W then find thermal conductivity of rod.

:- S I unit of R is °C W⁻¹ .

:-Dimensional formula of R is [M⁻¹L⁻²T³K]

Q₂ A copper rod 2m long has a circular cross-section of radius 1cm. one end is kept at 100°C and the other at 0°C, and the surface insulated so that negligible heat is lost through the surface. Find

(1) the thermal resistance the rod.

(2) the thermal current .

(3) the temperature gradient.

(4)the temperature at 25cm from the hot end.( K= 401Wm⁻¹k⁻¹)

***Connection of rod with different value of thermal conductivity.

T₁ T₂ T₃ Tⁿ⁻¹ Tₙ

Let us consider that n rod each cross - sectional area A , Length L₁ , L₂ ------ L ₙ , thermal conductivities K₁ , K₂ , K ₙ and thermal resistance R₁ , R₂ , R ₙ respectively are connected in series as shown in the figure.

(1) Heat current is the same in all the rod in the series connection in study state.

H₁ =H₂ = ---------=H ₙ

(2) Equivalent thermal resistance:-

[ Rₑᵥ = R₁ + R₂ + -------+R ₙ ]

(3) Equivalent thermal conductivity(K)