Heat Transfer

Transfer of energy in the form of HEAT is a part of the many processes in the chemical industries.Design problems require estimation and control of the heat flowrate.

Mechanism

The driving force of heat transfer id the temperature difference.Heat transfers from a hot system to a colder system.Three mechanisms are involved:1. Conduction2. Convection, and3. Radiation

Conduction

Heat flows through conduction results from transfer of kinetic and internal energy of the molecules in fluid or solid from the difference in temperature.

Conduction

The basic heat transfer through conduction is;

q = -k q = heat flux (rate of heat transfer per unit area. = temperature gradient in the x direction.k = thermal conductivity (natural thermophysics property)

Conduction

Steady state conduction through a slab:

T0 Temperature

profile

q T1 q

area A x q = -k = k

Rate of heat transfer

• The rate of heat transfer is the multiple of heat flux, q, and area ,A:

Q = qA = (T0 –T1)

Thermal resistances in series

k1 k2 k3

T1

T2

T3

T4

x1 x2 x3

The rate of heat transfer must be the same otherwise the temperature of the surfaces would either increase or decrease.

Q = (T1 –T2) = (T2 –T3)= (T3 –T4)

(T1 –T4)= (T1 –T2) + (T2 –T3) + (T3 –T4)

= + +

= Q( + + )

Q= (T1 –T4)/ ( + + )

=DF/RT (driving force/thermal resistance)

Example

• Oven: 0.2 m brick (kb=3.0 Wm-1K-1)

0.1 m asbestos (ka=0.15 Wm-1K-1)

Inner temperature is 250oC and the outer wall is 50oC. Calculate the amount of heat lost through 1 m2 in 1 hour.

Solution

• Q= (T1 –T2)/ ( + )

Q= (250 – 50)/ ( + )

Heat lost in 1 hour =[(250 – 50)/ ( + )] x 60x60= 981.82 kJ

Assignment (group)

1. Derive the rate of heat transfer, Q, through 3 different slabs in parallel as below;

k1