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HEAT TRANSFER

 

 

HEAT TRANSFER MODES      - conduction          - convection       - radiation 

Conduction:

Transfer of energy from molecule to molecule, e.g. thru opaque solids. Substances that touch transmit heat by conduction.

Thermal conductivity is the proportionality constant relating heat flux by conduction and the temperature. gradient:

      q = -Ak(dt/dx) = UA(dt)            where k = thermal conductivity 

From the above it can be seen that the rate of heat transfer is dependent on the temperature difference, the area, the material (conductivity), and the time.

Thermal conductivities:

Btu/(hr-sqft-F/in.)

glass wool
wood
concrete
steel
aluminum

0.29
0.80
12.5
326
1536



 

Example

 Determine the heat transfer through a 2 in. steel plate that is 4 ft by 6 ft if the temperature on one side is 450 F and the other side is 70 F.

  q = Ak/x(dt)

  = (4 ft x 6 ft)*326 Btu/hr(sqft)F/in./2 in. (450 - 70) F

= 1.49e6 Btu/hr

 

What if the above were wood instead of steel?

  q = ((4 x 6) x 0.80/2) x 380 = 3648 Btu/hr

 

Convection:      (a) from object:             q = hA(ts - tf)          where              h
= heat transfer coefficient            ts = surface temperature            tf = fluid
temperature       (b) transfer by mass movement of a media (e.g. air)               q =
mc(dt)      Factors affecting convective heat transfer:             thermal conductivity   
       coefficient of expansion           viscosity (drag effect)           density
buoyance           specific heat     

Natural Convection

Cold air is warmed by surface, becomes less dense and rises, cold (dense) air falls. Vertical movement is opposed by viscosity and inertial effects. Air movement across a surface enhances heat transfer.

 

 

Example

Given: A 2.4 m x 3.0 m (8 ft x 10 ft) plate glass with a convection heat transfer coefficient on the outside of h= 6.8 W/K-m2 (1.2 Btu/hrFsqft). The surface temperature is 24 C (75 F) and the outside ambient temperature is -4.0 C(25 F).

Find: The rate of heat loss

Solution:
q = hA(t2 - t1)
q = 6.8 W/Km2 x 2.4 m x 3.0 m x (24 - (-4)) K = 1370 W
( q = 1.2 x 8 x 10 x (75 - 25) = 4800 Btu.hr)

Radiation:

Transmission of energy by electromagnetic waves thru space from a body:

q = AEsT4

From a body to a wall:

q = A(Fa)(Fe)s(T14- T24)

     where          E = emissivity          s = Stefan-Boltzmann constant (5.67e-8 W/m2K4 or 0.1714 x 10-8
Btu/(h-ft2-R4)          Fa = shape factor determined by geometry          Fe = a factor related to
emissivity and absorptivity of surfaces 

Example

Given a 2.4 m x 3.0 m (7.9' x 9.8') window with a surface temperature of 24 C (75 F) and an emissivity of 0.8. The sky temperature is - 6 C (21 F).

Find: The radiant heat loss.

Solution:

q = 5.673EA[(TH/100)4- (TL/100)4]
= 5.673 x 2.4 x 3.0 x 0.8 x [(297/100)4 - (267/100)4]

= 882 W

{0.1714 x 10-8(0.8)(7.9 x 9.8)[(75+460)4 - (21+460)4] = 3015 Btu/h}