Heat Transfer Notes

Unit 1: Introduction to Heat Transfer

• Heat: Energy in transit.
  • Aristotle & Plato: Fire was one of the four fundamental elements.
  • Modern understanding: Heat is related to the motion of particles.
• Two main theories regarding heat transfer:
  1. Heat exchange: Heat lost by a hot body equals heat gained by a cold body.
  2. Heat Flow: Always occurs from higher to lower temperatures.
  3. Substances expand when heated.
  4. Phase Change: Heat can change a substance's state (solid to liquid or liquid to gas) without changing its temperature.
  5. Heating a body doesn't alter its weight.

Importance of Heat Transfer

• Heat Transfer: Transmission of energy from one region to another due to a temperature difference.
• Temperature difference is the driving potential.
• Mass Transfer (MT):
  • Driving potential: Concentration difference.
  • Involves mass motion, resulting in composition changes, initialed variations of concentration of various constituent species and diffusion

Purposes of Studying Heat Transfer

1. Estimate the rate of energy flow as heat through system boundaries.
2. Determine temperature fields under steady-state and transient conditions.

Areas of Heat Transfer Application

1. Design of thermal and nuclear power plants:
   • Heat engines
   • Steam generators
   • Condensers
   • Heat exchange equipment
   • Catalytic converters
   • Heat shields for space vehicles
   • Furnaces
   • Cryogenic equipment
2. Internal combustion engines.
3. Refrigeration and air conditioning systems.
4. Cooling systems for:
   • Electronic resistors
   • Generators
   • Transformers

Further Applications

• Heating and cooking processes.
• Chemical and biochemical operations.
• Construction of dams and structures.
• Minimizing building heat losses using infrared insulation techniques.
• Thermal control of space vehicles.
• Heat treatment of metals.
• Dispersion of atmospheric pollutants.

Heat Transfer in Biotechnology

1. Sterilization:

   • Moist heat sterilization: Uses conduction and convection of steam (e.g., autoclaving at $121^\circ C$ for 15-30 minutes).
   • Dry heat sterilization: Performed in dry ovens (e.g., $160^\circ C$ for 2 hours or $170^\circ C$ for 1 hour). Used for glassware, metal instruments, and heat-stable powders.
     • Example of heat transfer through conduction.

2. Lyophilization (Freeze-drying):

   • Involves freezing a product and removing water under vacuum.
   • Heat transfer provides the energy needed for the sublimation of ice.

3. Bioreactor Temperature Control:

   • Utilizes jackets or internal coils.
     • Convection: Heating or cooling fluid circulates in the jacket/coil.
     • Conduction: Heat is conducted through the bioreactor wall to the culture media.
   • Fluid movement enhances convective heat transfer.
   • Heat is conducted across the solid boundary to influence the temperature of bioreactor contents.

4. Pasteurization of Liquids (using heat exchangers):

   • Convection: Heat transfer occurs from the liquid food product and the heating/cooling fluid.
   • Conduction: Heat transfer occurs through the walls of the heat exchangers.
   • Rapid fluid flow on either side of the heat exchanger surfaces enhances convective heat transfer across the surface.

5. Microwave Sterilization/Heating:

   • Radiation (EM waves).
   • Microwaves interact directly with water molecules in biological materials, causing vibration and internal heat generation.
   • This is a form of radiative heat transfer where energy is absorbed and converted into thermal energy within the sample.

Modes of Heat Transfer

• Heat transfer is the transmission of energy from one region to another due to a temperature gradient.
• Three primary modes:
  1. Conduction
  2. Convection
  3. Radiation
• In real-world scenarios, heat transfer often occurs through a combination of these modes.
• Example: Water in a boiler shell receives heat from the firebed through conduction, convection, and radiation.
• Heat always flows in the direction of lower temperatures.

1. Conduction

• Heat transfer from one part of a substance to another, or from one substance to another in physical contact.
• Occurs without appreciable displacement of the substances.
• Mechanisms in Solids:
  1. Lattice vibration: Molecules in hotter areas transfer energy to adjacent molecules through impact.
  2. Transport of free electrons: Free electrons facilitate energy flow.
• Simple Explanation:
  • Kinetic energy of a molecule is a function of temperature.
  • Molecules are in constant random motion, exchanging energy and momentum.
• Liquids:
  • Molecules are more closely spaced, so intermolecular forces play a significant role.

2. Convection

• Heat transfer within a fluid by mixing one portion of the fluid with another.
• Only possible in a fluid medium.
• Directly linked to the transport of the medium itself (macroform of heat transfer).
• Microscopic particles of the fluid moving in space cause heat exchange.
• Heat transfer depends on the motion of the fluid.
• Effectiveness depends on fluid properties.
• Situations involve energy transfer as heat to a surface over which fluid is flowing.
  • Conduction in a thin fluid layer at the surface.
  • Mixing caused by the flow.
• Heat flow depends on:
  1. Properties of the fluid.
  2. Properties of the surface material.
  3. Shape of the surface (influences flow).
• Two types:
  1. Free/Natural convection
  2. Forced convection

Free or Natural Convection

• Occurs when the fluid circulates due to natural density differences caused by temperature variations.
• Denser portions of the fluid move downward under the influence of gravity.

Forced Convection

• Occurs when external work is done to force the fluid.

3. Radiation

• Heat transfer through space or matter by electromagnetic waves (EMWs) or quanta.
• Similar in nature to light and radio waves.
• All bodies radiate heat.
• Heat transfer occurs as:
  • Hot bodies emit radiant energy.
  • Cold bodies receive radiant energy.
• Hot bodies emit more heat than they receive and cold bodies receive more heat than they emit.
• Electromagnetic radiation requires no medium for propagation and can pass through a vacuum.

Heat Transfer by Conduction: Fourier's Law

• Empirical law stating that the rate of heat flow through a homogeneous solid is:
  • Directly proportional to the area of the section at right angles to the direction of heat flow.
  • Directly proportional to the change of temperature with respect to the length of the path of flow.
• Equation: $Q = -kA \frac{dt}{dx}$
  • Q = Heat flow rate (W)
  • A = Surface area perpendicular to heat flow ($m^2$)
  • $dt$ = Temperature difference (°C or K)
  • $dx$ = Thickness (m)
  • k = Thermal conductivity (W/mK)
• The negative sign indicates the decreasing temperature along the direction of increasing thickness.

Assumptions for Fourier's Law

1. Steady-state conditions.
2. Unidirectional heat flow.
3. Constant temperature gradient and linear temperature profile.
4. No internal heat generation.
5. Isothermal bounding surfaces.
6. Homogeneous and isotropic material (constant thermal conductivity in all directions).

Essential Features of Fourier's Law

1. Applicable to all matter (solid, liquid, gas).
2. Based on experimental evidence.
3. Vector expression.
4. Helps define thermal conductivity k.

Thermal Conductivity (k) of Materials

• From Fourier's Law, $k = \frac{Q}{A \frac{dt}{dx}}$
• If $Q = 1$, $A = 1$, and $\frac{dt}{dx} = -1$; then $k = Q$
• Thermal conductivity can be defined as the amount of energy conducted through a body of unit area and unit thickness in unit time when the temperature difference between the faces carrying heat flow is one unit.
• High k = Good conductor
• Low k = Good thermal insulator
• Metals have high k, alloys have lower k, and non-metals have the lowest k.

Factors Affecting Thermal Conductivity

1. Material structure.
2. Moisture content.
3. Density of the material.
4. Temperature and pressure.

Examples of Thermal Conductivity Values

Example Problem 1

• Calculate the rate of heat transfer per unit area through a copper plate of 45mm thick, with one face maintained at $350^\circ C$ and the other at $50^\circ C$. Take thermal conductivity of copper as 370 W/m°C.
• Solution:
  • Given:
    • Temperature difference, $dt = 50^\circ C - 350^\circ C = -300^\circ C$
    • Thickness of Copper plate, $L = 45mm = 0.045m$
    • Thermal conductivity of Copper, $K = 370 W/m^[circ]C$
  • Rate of Heat transfer per unit area, $q = \frac{Q}{A}$
  • From Fourier's Law:
    $q = -K \frac{dt}{dx} = -370 \times \frac{50-350}{0.045} = 2.466 \times 10^6 W/m^2 = 2.466 MW/m^2$

Example Problem 2

• A plane wall is 150mm thick and its wall area is $4.5m^2$. If its conductivity is 9.35 W/m°C and surface temperatures are steady at $150^\circ C$ and $45^\circ C$, determine:
  • Heat flow across the plane wall
  • Temperature gradient in the flow direction.
• Solution:
  • Given:
    $L = 150mm = 0.15 m$
    $A = 4.5 m^2$
    $dt = t<em>2 - t</em>1 = 45 - 150= -105^\circ C$
    $k = 9.35 W/m^\circ C$
  • Heat flow across the plane wall:
    $Q = -kA \frac{dt}{dx} = -9.35 \times 4.5 \times \frac{-105}{0.15} = 29452.5 W$
  • Temperature gradient in the flow direction:
    $\frac{dt}{dx} = \frac{Q}{kA} = \frac{-29452.5}{9.35 \times 4.5} = -700^\circ C/m$

Example Problem 3

• The following data relates to an oven:
  • Thermal conductivity of wall insulation: 0.044 W/m°C
  • Thickness of side wall of oven = 82.5 mm
  • Temperature on the inside of the wall = 175°C
  • Energy dissipated by the electrical coil within the oven = 40.5 W
• Determine the area of wall surface perpendicular to heat flow so that temperature on the other side of the wall does not exceed 75°C.
  • Given:
    • $x = 82.5 mm = 0.0825 m$
    • $k = 0.044 W/m^\circ C$
    • $t_1 = 175^\circ C$
    • $t_2 = 75^\circ C$
    • $Q = 40.5 W$
  • Assuming one-dimensional steady-state heat conduction: Rate of energy dissipated in the oven = Rate of heat transfer (conduction) across the wall$Q = -kA \frac{dT}{dx}$
    $Q = \frac{kA(t<em>1 - t</em>2)}{x}$
    $A = \frac{Qx}{k(t<em>1-t</em>2)} = \frac{40.5 \times 0.0825}{0.044(175-75)} = 0.759 m^2$

Steady-State Heat Conduction Through a Plain Wall

• Steady-state heat conduction means the temperature does not vary with time.
• One-dimensional heat conduction implies temperature gradient exists only in one direction, making the heat flow unidirectional.
• Examples:
  • Heat flow through a slab (plane wall)
  • Heat flow through a circular cylinder
  • Heat flow through a sphere
  • Heat flow along long fins
  • Heat flow through the wall of a stirred tank containing hot or cold fluid
  • Heat flow through the wall of a cryogenic furnace

Conduction Through a Plane Wall

• Considerations:
  • Wall of uniform thickness x and made of material with thermal conductivity k.
  • Cross-sectional area A is constant.
  • Area A is large compared to the thickness, so heat losses from edges are negligible.
  • Time-invariant face temperatures (T1, T2).
  • Heat flow is perpendicular to the wall, varying only in the x-axis direction.
  • Steady-state: No accumulation or depletion of heat within the plain wall.
  • Q = constant along the path of heat flow.

Mathematical Formulation

• From Fourier's Law:$Q = -kA\frac{dT}{dx}$
• Rearranging: $Q \cdot dx = -kA \cdot dT$
• Integrating: $\int<em>{0}^{x} Q \cdot dx = -kA \int</em>{T<em>1}^{T</em>2} dT$
• Then, $Q \cdot x = kA(T<em>1-T</em>2)$
• Define a temperature difference \[AT=T1-T2\]
• Then rearrange to be, $Q = \frac{AT}{x/kA}$
• We can rewrite this as, $Q=\frac{AT}{R}$
• Where, $R=\frac{x}{kA}$ is defined as Thermal Resistance
• Thus,
  $Q = \frac{Driving Force}{Thermal Resistance}$

Thermal Resistance and Conductance

• Thermal resistance: $R = \frac{x}{kA}$
• Its reciprocal is the heat conductance which can be determined as: $1/R = \frac{kA}{x}$

Analogy to Ohm's Law

Heat Transfer Through Composite Plane Wall

• Composite wall: A series of layers of different materials.
• Steady-state conduction through a composite wall is analogous to conduction through resistances in series.

Assumptions

1. Layers are in excellent thermal contact.
2. Area of the composite wall is at right angles to the plane of illustration.
3. Overall temperature drop is related to individual temperature drops:
   $\Delta T = \Delta T<em>1 + \Delta T</em>2 + \Delta T_3$
4. Thermal Conductivities of the materials for the layers K1, K2, K3.
5. Thickness of the Materials layers.
6. Temp drop accross each layer \[Delta T1, Delta T2,Delta T3].
7. Over all Temp Drop for the entire composite is \[Delta T]

Heat Flow Equations

*Rate of heat flow through layer 1: Q1=\frac{K1A}{x1} \[\Delta T1=(T1-T')\]
*Rearranging : \Delta T1=Q\frac{x1}{K1A}

*Rate of heat flow through layer 2Q2=\frac{K2A}{x2} \[\Delta T2=(T'-T'')\]
*Rearranging : \Delta T2=Q\frac{x2}{K2A}

*Rate of heat flow through layer 3Q3=\frac{K3A}{x3} \[\Delta T3=(T''-T2)\]
*Rearranging : \Delta T3=Q\frac{x3}{K3A}

Adding The equations yields,
\Delta T1+\Delta T2+\Delta T3=Q\frac{x1}{K1A}+Q\frac{x2}{K2A}+Q\frac{x3}{K3A}
\Delta T=Q( \frac{x1}{K1A}+\frac{x2}{K2A}+\frac{x3}{K3A})
Q=\frac{\Delta T}{(\frac{x1}{K1A}+\frac{x2}{K2A}+\frac{x3}{K3A})}

Thermal Resistance

• R1 = \frac{x1}{K1 A}$,$ R2 = \frac{x2}{K2 A}$,$ R3 = \frac{x3}{K_3 A}
• We can write our expression as Q = \frac{AT}{R1 + R2 + R_3} = \frac{AT}{R}

Rates and Interface Temperatures

*Temp at face as the hot face = T1.
*The driving force for Heat Transfer Heat \[\Delta T]
*R = The overall resitance for Hear Transfer over the composite wall
*So Q,$The~Rate~of~Heat~Transfer = \frac{overall~driving~force}{The~Overall~resistance}$
*Also, Temp @ the face interface of face of the wall can be expressed as the following,
$\Delta T1/T= \frac{R1}{ R<em>{overall}}$$\Delta T/T= \frac{R2}{ R</em>{overall}}$
$\Delta T3/T= \frac{R3}{ R_{overall}}$

Example: Furnace Wall

Problem: A furnace wall is constructed with a 229mm thick layer of fire brick, a 115mm thick layer of insulation brick, and a 229mm thick layer of building brick. The temperature at the inside is 1233 K and the outermost wall is 323 K. Find the heat lost per unit area and the temperature at the interfaces.

*Given:
FireBrick.
Kl=6.05 W/(m.k)
x1= 229/1000 m =0.229 m
Insulating Brick
K2= .581 W? (m.k)
x2 = 115/1000 m=0.115m
Building Brick
K3= 2.33 W(m.k)
x3= 229/1000 m =0.229 m
Overall Temp Drop \[\Delta T = T - T2 = 1233 - 323 = 900 k]
R1 = (x1)/(K1 A) =>R1=(0.229)/(6.05A)
R2 = (x2)/(K2 A) => R2=(0.11)5/(.581 A)
R3 = (x3)/(K1 A) => R3=(0.229)/(2.33 A)
Q= \frac{\Delta T}{ R_{overall}} =A ((2884)W/m2)
Given that the Heat transfer does not depend on A, heat transfer/m2 is 2900 W.
Now for Tb which occurs between the Interface can be determined as the following using the formulas mentioned earlier,
$T-TB=Q/(A(1/K2)\frac{x2}{K3})$

Which gives us
Tb = 588k
Therefore ,
Between FB and IB (1100K) , and between IB and BB(~500K).

Example: Furnace Wall with Multiple Layers

• Problem:A furnace wall is made up of a steel plate 10mm thick lined on the inside with silica brick 150mm thick and on the outside with magnesite brick 150mm thick. The temperature on the inside surface of the wall is 973 K and on the outside is 288 K. Calculate the quantity of heat lost watt per $m^2$. It is required to reduce the heat flow to 1163 W/$m^2$ by means of an air gap between the steel plate and magnesite brick. Estimate the width of this gap.
  • Thermal Conductivities in W/cm. K) are 16.86, 1.75, 5.23, & 0.33 Respectively for steel,silica brick, magnerite brick & air.
    *HeatLoss,
    $\frac{\Delta T}{\sum R}=Q$
    *ThermalResistance(steel plate, silica brick , magnerite brick) ,respectively.
  • $15/ (1.75e2 W/K ) = 0.085\frac { K } { watts }$(
  • $0.01 / ( 16.86e2 W/K ) = 0.0006 \frac { K } { watts }$
  • $15 / ( 5.23e2 W/K ) = 0.028 \frac { K } { watts }$
    *HeatLoss calculation based on area of 1 M^2,
    **Heatloss = (288-973K)/.114 = -6009 Watts/ M^2
    *NewHeatLoss , 1163 w/M^2, and the newThermal Resistance can be calculated as follows,
    1163 = (973-288) / RNew(
    RNew -ThermalResisanceSteel+ThermalReistanceSilica+ ThermalReistanceAirGap+ThermalResisanceMagnasite
    *Solving for thermalResistance AirGap,
    NewResisitance = 288 = QNEW- ThermalReistanceSilica- ThermalReistanceSteel -ThermalReistanceMagnesite
    *The ThermalResistance Air =.0589, Solving for the Length which achieves that thermalResistance , we plug iinto ( l/ (k A )). Which nets us ~15mm of air to achieve the reducedQ.

Heat Loss Through a Gold Room Wall

• Problem: A Gold Room has one of the walls which measures 4.6m \times 2.3m$constructed of bricks 115 mm thick insulated by cork sheeting 75mm thick. The cork is protected externally by wood 25mm thick. Estimate the heat leakage through the wall per day if the temperatures of the interior 271 K surface and exterior surface is 291K. Conductivities of brick, cork and wood are$0.113, 0.005, & 0.021 W/(m.K). \frac{\Delta T}{\sum R}=Q *ThermalResistance( wood, cork, brick) ,respectively.
  • 15 / ( 113e2 W/K ) = 0.102 \frac { K } { watts }(
  • 0.075 / ( .005e2 W/K ) = 1.41 \frac { K } { watts }
  • .025 / ( .021e2 W/K ) =1.19 \frac { K } { watts }
    TotalArea, Area = 4.62.3=10.58M^2
    Delta-T= (291-W71)=20 K$$\Delta T / R = 325
    M^2/ 2.76 =100 KJ/ Day##

Example: Reactor Wall with Insulation

Problem: A reactor's wall, 320mm thick, is made up of an inner layer of fire brick (k = 0.84 W/m°C) covered with a layer of insulation (k = 0.16 W/m°C). The reactor operates at a temperature of 1325°C, while