 # Conduction through slab

#### Assumptions:→  Steady state Heat Transfer → Temperature, T ≠ f(Time) at a particular point.→ One dimensional conduction→Uniform thermal conductivity ‘k’

##### Case1: Conduction through a slab #### T = f(x)Boundary Conditionsat x = 0, T = T1at x = b , T = T2from fourier’s law of conduction q = -kA dT/dxqdx = -kA dT #### To satisfy steady state conditions q ≠ f(x)qx = q(x+dx)solving integrationq * b = kA (T1-T2) #### Electrical analogy of Heat Transfer

Rele. = ΔV/ i  Ω

• ##### Rthermal,  → thermal resistance

Rthermal = ΔT/ q   K/Watt

##### For a single slab ##### Case2: Conduction through a composite slab #### We will use electrical analogy and two adjacent slabs will act like resistance in series  ##### Rate of Heat Transfer through composite slab ##### Heat Transfer per unit area  or Heat Flux = q/A  by this equation we can find intermediate temperature T2

##### Till now we have read about conductive thermal resistance. In some cases there is convective thermal resistance.

Convective thermal resistance

Newton’s law of cooling

qconv. = hA (Tw – T) Watt Since, h value is associated with water (liquid) being more the conventional thermal resistance with water shall be lesser

##### Case3: Conduction-Convection Heat Transfer through a composite slab #### We will use electrical analogy and two adjacent slabs will act like resistance in series  ##### Rate of Heat Transfer between gases and ambient fluid through composite slab ##### Heat Transfer per unit area  or Heat Flux = q/A  Overall Heat Transfer Coefficient (U)

##### q = UAΔT watt

where ΔT Total temp. differnce (TG – T)

q = UA (TG – T)

##### U and h have the same units (watt/m2k) ### I hope you like this section. Please share with friends and like my Facebook page and never miss an update.

##### Share the knowledge
Share the knowledge