# Conduction through slab

#### Electrical analogy of Heat Transfer

Rele. = ΔV/ i  Ω

• ##### Rthermal,  → thermal resistance

Rthermal = ΔT/ q   K/Watt

#### We will use electrical analogy and two adjacent slabs will act like resistance in series

##### 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

#### We will use electrical analogy and two adjacent slabs will act like resistance in series

##### 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)

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