By the end of this chapter, you will be able to:
Mastering these skills will help you confidently handle fluid and heat transfer systems, making you a valuable technician in many industrial and engineering settings.
Heat transfer and heat exchangers are fundamental concepts in mechanical production, especially for technicians working in fluid systems. Understanding how heat moves through different media and how to control it is crucial when designing, operating, or maintaining equipment like boilers, condensers, and cooling systems. In Kenya’s manufacturing and agricultural processing industries, efficient heat transfer improves energy utilization, reduces costs, and enhances product quality. This chapter explores the mechanisms of heat transfer, the governing laws, and practical applications relevant to fluid systems.
Heat is energy that moves from a hotter place to a colder place because of a temperature difference. It never flows by itself from cold to hot. In a workshop, heat transfer decides how fast a welded plate cools, how far the heat of the arc spreads into the parent metal, how much heat a furnace wall loses and how well a heat exchanger works. A heat transfer medium is the material, or the empty space, through which the heat travels. Solids (a steel plate, a brick wall), liquids (cooling water, oil) and gases (air, steam) are all media, and radiation can cross even a vacuum. The rate of heat transfer is measured in watts (W), where 1 W = 1 J/s, and the heat flux is the rate per unit area, in W/m².
Heat moves in three ways: conduction, convection and radiation. In most real cases two or three act together. A hot weld, for example, loses heat by conduction into the plate beneath it, by convection to the air above it and by radiation to the surroundings.
Conduction is the transfer of heat through a material, or between materials in contact, without the material itself moving. In a solid, the heat passes from one particle to the next, and in metals the free electrons carry much of it. That is why metals conduct heat well. The rate of conduction is given by Fourier's law:
Q = k × A × (T₁ − T₂) ÷ L
where Q is the heat flow in W, k is the thermal conductivity in W/m·K, A is the area in m², T₁ − T₂ is the temperature difference across the material in K and L is its thickness in m.
Examples: heat flows along a steel rod from the hot end held in a fire to the end held in a hand; the arc heat flows from the weld pool into the plate; heat flows through the wall of a boiler from the flame side to the water side.
Convection is the transfer of heat between a surface and a moving fluid (liquid or gas), or within a fluid, by the movement of the fluid itself. There are two kinds:
The rate of convection from a surface is given by:
Q = h × A × (T_s − T_f)
where h is the convection heat transfer coefficient in W/m²·K, A is the surface area in m², T_s is the surface temperature and T_f is the temperature of the fluid far from the surface. The coefficient h depends on the fluid, its speed and the shape of the surface. It is small for natural convection in air (about 5 to 25 W/m²·K) and much larger for forced flow and for liquids.
Example: a flat plate of area 2 m² at 80 °C stands in still air at 30 °C. If h = 25 W/m²·K, then Q = 25 × 2 × (80 − 30) = 2500 W.
Radiation is the transfer of heat by electromagnetic waves. It needs no medium, so the heat from the sun reaches the earth through space. Every body above absolute zero radiates, and the rate rises steeply with temperature. The net rate between a surface at temperature T₁ and its surroundings at temperature T₂ (both in kelvin) is:
Q = ε × σ × A × (T₁⁴ − T₂⁴)
where σ = 5.67 × 10⁻⁸ W/m²·K⁴ is the Stefan-Boltzmann constant and ε is the emissivity of the surface, a number between 0 and 1. A dull dark surface has an ε near 1, and a polished metal surface has a low ε.
Example: a plate of area 0.5 m² and emissivity 0.8 is at 500 °C (773.15 K) in surroundings at 27 °C (300.15 K).
The plate loses about 7.9 kW by radiation alone. Radiation is why a person stands back from a hot workpiece, and why welders wear screens and protective clothing against the heat of the arc and the metal.
Newton's law of cooling says that the rate at which a hot body loses heat to a cooler surrounding fluid is proportional to the temperature difference between them. For the heat flow from the surface it is the convection equation:
Q = h × A × (T_s − T_f)
For a body that cools in surroundings at constant temperature T_f, the temperature falls at a rate proportional to the difference:
dT ÷ dt = −k × (T − T_f)
where k is a constant for the body, in 1/s or 1/min. Solving this equation, (T − T_f) falls exponentially with time:
T = T_f + (T₀ − T_f) × e^(−k × t)
where T₀ is the temperature at the start. It follows that k = ln[(T₀ − T_f) ÷ (T − T_f)] ÷ t, and the time to cool to a temperature T is t = ln[(T₀ − T_f) ÷ (T − T_f)] ÷ k.
Worked example: a steel plate at 300 °C cools in air at 30 °C. After 10 minutes it is at 200 °C.
The plate cools fast at first and slowly later, because the difference in temperature that drives the heat flow becomes smaller. The law holds best for small differences in temperature and a steady surrounding, so it is used for estimates. It does not include radiation, which matters at high temperatures. In welding, the cooling rate of a joint must follow the approved welding procedure, so this law is never a substitute for it.
Fourier's law is found from experiment. A slab of material with area A and thickness L has one face at temperature T₁ and the other at the lower temperature T₂. The heat flows steadily from the hot face to the cold face. The measurements show that:
Combining these gives Q ∝ A × (T₁ − T₂) ÷ L. A different material gives a different value for the same slab, and the constant of proportionality is the thermal conductivity k of the material:
Q = k × A × (T₁ − T₂) ÷ L
For a very thin layer of thickness dx across which the temperature changes by dT, the law is written in its general form:
Q = −k × A × dT ÷ dx
The negative sign shows that heat flows in the direction in which the temperature falls, because dT ÷ dx is negative when the temperature drops along the flow. For a slab with a uniform k, the temperature falls in a straight line, and dT ÷ dx = −(T₁ − T₂) ÷ L, which gives the first form again.
Application to a wall. A steel plate 20 mm thick has an area of 1.2 m², k = 50 W/m·K, one face at 120 °C and the other at 80 °C.
Q = 50 × 1.2 × (120 − 80) ÷ 0.02 = 50 × 1.2 × 40 ÷ 0.02 = 120 000 W = 120 kW.
Application to heat loss from a furnace. A brick wall of thickness 0.23 m and area 10 m² has k = 0.8 W/m·K. The inner face is at 600 °C and the outer face is at 60 °C.
Q = 0.8 × 10 × (600 − 60) ÷ 0.23 = 4320 ÷ 0.23 = 18 783 W, which is about 18.8 kW.
Method for any slab. The steps are the same every time: write down k, A, the two face temperatures and the thickness in metres and kelvin differences, then substitute into Q = kA(T₁ − T₂) ÷ L. A thicker wall, or a material of lower conductivity, reduces the heat loss. That is why furnace walls and boiler casings carry a layer of insulation.
Factors that affect conduction:
Limits of the equations: Q = kA(T₁ − T₂) ÷ L assumes steady heat flow in one direction, a uniform k and no heat generation inside the wall. The result is a good estimate for most engineering work.
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Create a free accountThis chapter explored the fundamental media and methods through which heat transfer occurs, focusing on conduction, convection, and radiation as the primary mechanisms. It examined Newton's Law of Cooling, which describes the rate of heat loss from a body to its surroundings, and detailed the derivation and practical application of Fourier's Law of Heat Conduction, providing the mathematical basis for quantifying heat flow through materials. The transfer of heat by conduction was analyzed in various geometries, including slabs, composite walls, thick cylinders, and thick spheres, highlighting the roles of thermal conductivity and temperature gradients in these contexts. Special attention was given to the concept of composite walls, which combine different materials and affect overall heat transfer. The chapter also introduced the overall coefficient of heat transfer, a critical parameter in assessing combined heat transfer processes. Finally, heat exchangers were discussed as practical devices that utilize these principles to transfer heat efficiently between fluids in diverse engineering and industrial applications.
Which of the following is NOT a mode of heat transfer? (2 marks)
a) Conduction
b) Convection
c) Radiation
d) Insulation
Explain Newton’s Law of Cooling and provide an example of its application in a mechanical workshop. (4 marks)
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