# Heat Transfer

HEAT TRANSFER

Heat can be transferred from place to place in measurable quantities. Heat can be transferred in three ways, which are; by conduction, convection and radiation.

Heat transfer by Conduction.

Heat conduction occurs as hot, rapidly moving or vibrating atoms and molecules interact with neighbouring atoms and molecules, transferring some of their energy (heat) to these neighbouring particles. In other words, heat is transferred by conduction when adjacent atoms vibrate against one another, or as electrons move from one atom to another.

NOTE: The rate of heat flow is proportional to Difference in temperature. This means that no flow of heat if there is no difference in temperature.

It is known that the rate of heat flow, in Joules per second(J/s)through a substance is proportional to the difference in temperature between its ends. Thus,

ΔQ/Δt = kA (T2 – T1)/L

Therefore, K = QL/A(T2 – T1)

Where ΔQ/Δt is the rate of heat flow,

A is the cross-sectional area,

T2 is the higher Temperature, T1 is the lower temperature

L is the distance(m) between the two ends at temperatures T1 and T2,

K is the constant of proportionality, known as the thermal conductivity. The S.I unit of thermal conductivity is Joule per seconds, per meter, per degree Celsius(J/s./m/C) or Watt per meter, per Kelvin (Wm-1K-1)

Note: materials with high thermal conductivity(k) value are known as good conductors while materials with a small thermal conductivity value, are known as insulators.

Note: The temperature gradient is equal to (T2 – T1)/L

U values and R-Value

The R-Value combines the thickness l, and the thermal conductivity k, in one number. It is directly proportional to the material thickness. R is a property attributed to a slab of a specified thickness

R = L/k

The lower the thermal conductivity of the material of which a slab is made, the higher the R-value of the slab.

The U value of a building element is the inverse of the total thermal resistance of that element. The U-value is a measure of how much heat is lost through a given thickness of a particular material.

U = k/L

Inserting this equation into the equation above,

ΔQ/Δt = UA(T2 – T1)

CALCULATIONS ON HEAT CONDUCTION

Question 1: Calculate the thermal conductivity through a conductor when 30 kW of heat flows through it having a length of 4 m and area of 12 m2m2 if the temperature gradient is 40 K.
Solution:

Given :
Heat flow Q = 30 kW, length L = 4 m, Area A = 12 m2, temperature difference ΔT = 40 K.

The thermal conductivity is given by,
k = QL/AΔT
= (30000×4)/(12×40)
= 250 W/m K.

2.A major source of heat loss from a house is through the windows. Calculate the rate of heat flow through a glass window 2.0m x 1.5m in area and 3.2mm thick, if the temperatures at the inner and outer sufaces are 15.00C and 14.00C, respectively. ( k = 0.84J/s./m/C)

Solution:

Given Values;

A = 2.0 x 1.5 = 3.0m2, l = 3.2mm = 3.2 x 10-3m, T1 = 15.00C, T2 = 14.00C

Rate of heat flow ΔQ/Δt = kA (T1 – T2)/L

Therefore, ΔQ/Δt = 0.84 x 3.0 (15.0 – 14.0)/3.2 x 10-3

= 2.52 (1.0)/3.2 x 10-3

= 2.25/3.2 x 10-3

= 787.5J/s. 