Heat Transfer · Chapter 1
0%
Heat Transfer · Chapter One

Basics of Heat Transfer

Thermodynamics tells us how much heat moves between two equilibrium states. Heat transfer tells us how fast — the rate at which that energy crosses a boundary, and the mechanisms that carry it.

This chapter sets the foundation: the relationship between heat and the other forms of energy, the first law as an energy-balance tool, and the three physical mechanisms — conduction, convection, and radiation — that transport thermal energy. It closes with how these mechanisms act together and a disciplined approach to solving problems.

Learning objectives

  • Distinguish heat, stored energy, heat rate, and heat flux.
  • Apply closed- and open-system energy balances with a declared sign convention.
  • Recognize conduction, convection, radiation, and simultaneous mechanisms.
  • Select introductory rate laws and state their assumptions and property reference temperatures.
  • Audit units, directions, limits, and physical plausibility.
1–1

Thermodynamics and Heat Transfer

Both sciences rest on the conservation of energy, but they ask different questions. Thermodynamics is concerned with equilibrium end states and the net energy exchanged in going between them. Heat transfer is concerned with the journey: the rate of energy flow and how long a process takes.

Heat

The form of energy that crosses a system boundary solely because of a temperature difference. There is no heat transfer between two regions at the same temperature.

Heat Transfer

The study of the rates at which thermal energy moves, and the temperature distributions that result. It supplies the time dimension that equilibrium thermodynamics omits.

Key Idea A thermodynamic analysis can tell you a cup of coffee will eventually reach room temperature and how much energy it sheds. Only a heat transfer analysis tells you whether that takes five minutes or an hour.
Course Note

When you read a problem, first decide which question is being asked — "how much" (thermodynamics) or "how fast" (heat transfer). That single decision tells you which tools to reach for and saves you from over-complicating a simple energy balance.

1–2

Engineering Heat Transfer

Equipment such as heaters, radiators, heat exchangers, and electronic cooling systems is designed around heat transfer rates. Engineering analysis can be experimental, analytical, or — most often — a combination of both, guided by modeling.

Modeling in Engineering

Replacing a real, complex situation with a simplified representation that keeps its essential features. The right model is the simplest one that still yields acceptable accuracy; results are never more reliable than the assumptions behind them.

Modeling vs. Experiment

Experiments measure real systems but are costly and time-consuming. Analytical modeling is fast and cheap but only as trustworthy as its assumptions. Practical engineering blends the two — using experiment to validate the model.

Course Note

Resist the urge to model everything at maximum fidelity. A model you can actually solve and interpret beats an "exact" one you cannot. Add complexity only when a simpler model disagrees with reality.

1–3

Heat and Other Forms of Energy

Energy appears in many forms — thermal, mechanical, kinetic, potential, electrical, chemical, nuclear — whose sum is the total energy of a system. The forms tied to molecular activity make up the internal energy U.

Internal Energy and Its Parts

Sensible energy is associated with molecular kinetic energy and rises with temperature. Latent energy is associated with phase. Chemical and nuclear energy reside in atomic and nuclear bonds. In daily speech the sensible and latent forms together are loosely called "heat."

Enthalpy

The convenient combination u + Pv that arises for flowing fluids. The Pv term is the flow work needed to push the fluid; enthalpy bundles it with internal energy so a moving stream can be treated by a single property.

Specific Heat

The energy required to raise the temperature of a unit mass by one degree. Two versions matter: at constant volume (c_v) and at constant pressure (c_p). For solids and liquids the two are essentially equal and written simply as c.

Ideal-gas relation
\[ Pv = RT \qquad\text{or}\qquad P = \rho R T \]
Internal energy & enthalpy change (ideal gas)
\[ \Delta U = m\,c_{v,\text{avg}}\,\Delta T \qquad \Delta H = m\,c_{p,\text{avg}}\,\Delta T \]
For incompressible substances (solids, liquids): \( \Delta U = m\,c_{\text{avg}}\,\Delta T \), with \( c_p \approx c_v \approx c \).
Heat transfer rate, total heat, and heat flux
\[ Q = \int_{0}^{\Delta t}\dot{Q}\,dt \quad\Rightarrow\quad Q = \dot{Q}\,\Delta t \ \ (\dot{Q}=\text{const}) \qquad \dot{q} = \frac{\dot{Q}}{A} \]
\( \dot{Q} \) is the heat transfer rate (W), and \( \dot{q} \) is heat flux — rate per unit area (W/m²).
EXAMPLE 1 Thermal storage in an instrument enclosure
Problem. A 6.0-kg aluminum instrument enclosure warms uniformly from 18°C to 46°C in 14 min. Using the supplied mean specific heat \(c=0.900\ \mathrm{kJ/(kg\,K)}\), find its sensible-energy increase and average net heat rate.

Strategy

The target is stored-energy change and its time average, not a surface mechanism; use a closed-system sensible-energy balance.

Givens

  • \(m=6.0\) kg, \(T_1=18^\circ\)C, \(T_2=46^\circ\)C, \(\Delta t=840\) s, supplied \(c_{\rm avg}=0.900\) kJ/(kg·K).

Physical/model interpretation

Lumped uniform temperature and negligible kinetic, potential, phase, and deformation energy; positive heat enters the enclosure.

Step-by-step solution
1. Temperature change. \(\Delta T=46-18=28\) K.
2. Symbolic balance. With no work, \(Q_{\rm net,in}=\Delta U=mc_{\rm avg}\Delta T\).
3. Energy. \(\Delta U=(6.0)(0.900)(28)=151.2\) kJ.
4. Average rate. \(\dot Q_{\rm avg}=151200/840=180\) W.

Unit/sign/limit checks

Units: In \(mc\Delta T\), kg·kJ/(kg·K)·K reduces to kJ; dividing \(151{,}200\) J by 840 s gives J/s = W, as required for a heat rate.

Signs: Because \(T_2-T_1=+28\) K under the heat-in-positive convention, both \(\Delta U\) and the net heat input are positive; a negative result would contradict the observed warming.

Limit: If the temperature change tends to zero at fixed mass and heat capacity, stored-energy change and average net rate both tend to zero; doubling the heating time would halve only the average rate.

Final answer: \(\boxed{\Delta U=151\ \mathrm{kJ},\ \dot Q_{\rm avg}=180\ \mathrm W\ inward}\)

Engineering insight

This is net input; heater power must be larger if the enclosure simultaneously loses heat. The uniform-temperature assumption should later be checked with a Biot number.

1–4

The First Law of Thermodynamics

The first law — the conservation of energy principle — is the central accounting tool of heat transfer. Energy is neither created nor destroyed; every joule that enters, leaves, or is generated must be accounted for.

First Law of Thermodynamics (Energy Balance)

For any system undergoing any process, the net energy transferred in (by heat, work, and mass) minus the net energy transferred out equals the change in the system's total energy.

\[ E_{\text{in}} - E_{\text{out}} = \Delta E_{\text{system}} \]

Rate form of the energy balance
\[ \dot{E}_{\text{in}} - \dot{E}_{\text{out}} = \frac{dE_{\text{system}}}{dt} \]

Closed-System Energy Balance

With no mass flow, the balance reduces to heat and work only: \( Q_{\text{net,in}} - W_{\text{net,out}} = \Delta U \). For a stationary closed system heated at constant pressure or volume, \( Q = m c \,\Delta T \).

Steady-flow energy balance (one inlet, one exit)
\[ \dot{Q} = \dot{m}\,c_p\,(T_{\text{out}} - T_{\text{in}}) \]
With \( \dot{m} = \rho \dot{V} \); neglecting kinetic and potential energy changes.
Surface energy balance
\[ \dot{E}_{\text{in}} = \dot{E}_{\text{out}} \]
A surface stores no energy, so all heat arriving must leave — true even under transient conditions.
EXAMPLE 2 Energy balance on an electronics cooling stream
Problem. Air cools a sealed power cabinet at steady state, entering at 22°C and leaving at 31°C. The supplied mean properties are \(\rho=1.16\ \mathrm{kg/m^3}\) and \(c_p=1.007\ \mathrm{kJ/(kg\,K)}\); volumetric flow is \(0.085\ \mathrm{m^3/s}\). If 70 W escapes through the cabinet walls, determine the internal electrical dissipation.

Strategy

Mass crosses the control volume, so use a steady-flow energy balance and include both outgoing paths.

Givens

  • \(\dot V=0.085\) m³/s, \(22\to31^\circ\)C, \(\rho=1.16\) kg/m³, \(c_p=1007\) J/(kg·K), wall loss 70 W.

Physical/model interpretation

Steady control volume around cabinet; negligible kinetic/potential changes; electrical work enters and heat leaves in warmed air and through walls.

Step-by-step solution
1. Mass flow. \(\dot m=\rho\dot V=(1.16)(0.085)=0.0986\ \mathrm{kg/s}\).
2. Air removal. \(\dot Q_{air}=\dot mc_p(T_o-T_i)\).
3. Calculate. \(\dot Q_{air}=(0.0986)(1007)(9)=893.7\) W.
4. Balance. \(\dot W_{elec}=\dot Q_{air}+\dot Q_{wall}=893.7+70=963.7\) W.

Unit/sign/limit checks

Units: \((\mathrm{kg/s})[\mathrm{J/(kg\,K)}](\mathrm K)=\mathrm{J/s}=\mathrm W\); adding the 70 W wall loss is therefore dimensionally valid.

Signs: Both air enthalpy rise and wall heat loss remove energy from the cabinet, so both appear positive on the output side and require positive electrical input.

Limit: With zero wall loss the electrical dissipation becomes 893.7 W, exactly the air removal; with zero air temperature rise only the 70 W wall path remains.

Final answer: \(\boxed{\dot W_{elec}=0.964\ \mathrm{kW}}\)

Engineering insight

Air properties are mean-temperature values. Fan heat, leakage, altitude, and nonuniform component temperatures must be included in a qualified thermal design.

1–5

Heat Transfer Mechanisms

Heat is transported by three physically distinct mechanisms. Each requires a temperature difference, and in real situations they frequently act at once.

The Three Mechanisms

Conduction — energy passes from more energetic particles to adjacent less energetic ones; it needs a medium. Convection — energy moves between a surface and a moving fluid; conduction plus bulk fluid motion. Radiation — energy travels as electromagnetic waves and needs no medium at all.

Conduction hot → cold (solid) Convection surface ↔ fluid flow Radiation EM waves, no medium
Figure 1–1The three mechanisms of heat transfer. Conduction needs a medium; convection couples a surface to a moving fluid; radiation crosses a vacuum.
1–6

Conduction

Conduction is energy transfer by molecular interaction and, in metals, by mobile electrons. Fourier’s law is local and broadly applicable to continuum media; the plane-wall form below additionally requires steady one-dimensional flow, constant area, no internal generation, and a representative constant \(k\). Evaluate \(k\) over the wall’s actual temperature range.

Fourier's Law of Heat Conduction

The conduction heat rate through a layer is proportional to the temperature gradient across it and to the area normal to the flow, with the proportionality constant being the material's thermal conductivity. The minus sign makes heat flow in the direction of decreasing temperature.

\[ \dot{Q}_{\text{cond}} = -k A \frac{dT}{dx} \]

Steady conduction through a plane wall
\[ \dot{Q}_{\text{cond}} = k A \frac{T_1 - T_2}{L} = k A \frac{\Delta T}{L} \]

Thermal Conductivity, \(k\)

A material property (W/m·K) measuring how readily a substance conducts heat. It is high for metals (copper ≈ 401, silver ≈ 429), low for insulators and gases (air ≈ 0.026). It generally depends on temperature.

Thermal Diffusivity, \(\alpha\)

How fast heat spreads through a material versus how much it stores: \( \alpha = k/(\rho c_p) \), in m²/s. A large \(\alpha\) means the material responds quickly to changes in its thermal environment.

Thermal diffusivity
\[ \alpha = \frac{k}{\rho c_p} \quad \left(\text{m}^2/\text{s}\right) \]
1–7

Convection

Convection is heat transfer between a solid surface and an adjacent moving fluid. Externally driven motion gives forced convection; buoyancy-driven motion gives natural convection. Newton’s law defines \(h\), but predicting \(h\) requires a correlation matching geometry, flow regime, thermal boundary condition, and validity range; fluid properties are commonly evaluated at a film temperature when the correlation specifies it.

Newton's Law of Cooling

The rate of convective heat transfer from a surface is proportional to the area and to the temperature difference between the surface and the bulk fluid, through the convection heat transfer coefficient \(h\).

\[ \dot{Q}_{\text{conv}} = h A_s (T_s - T_\infty) \]

Convection Coefficient, \(h\)

Not a material property — it lumps together the effects of geometry, fluid properties, and flow conditions (W/m²·K). Typical ranges: free convection of gases 2–25, forced convection of gases 25–250, forced convection of liquids 50–20,000, and boiling/condensation 2500–100,000.

Course Note

Students often treat \(h\) like a constant from a table. It isn't — it shifts with velocity, orientation, and surface temperature. Later chapters give correlations (via the Nusselt number) to actually predict it; for now, just know it bundles all the messy flow physics into one number.

1–8

Radiation

Radiation is energy emitted by matter as electromagnetic waves due to changes in the electronic configuration of atoms. Unlike conduction and convection, it requires no intervening medium and is fastest in a vacuum.

Stefan–Boltzmann Law

The maximum radiation a surface at absolute temperature \(T_s\) can emit is proportional to the fourth power of that temperature. A real (gray) surface emits a fraction \(\varepsilon\) of this blackbody maximum.

\[ \dot{Q}_{\text{emit,max}} = \sigma A_s T_s^4, \qquad \dot{Q}_{\text{emit}} = \varepsilon \sigma A_s T_s^4 \]

Stefan–Boltzmann constant
\[ \sigma = 5.67 \times 10^{-8}\ \text{W/m}^2\text{·K}^4 \]

Emissivity and Absorptivity

Emissivity \(\varepsilon\) (0–1) is how efficiently a surface emits relative to a blackbody. Absorptivity \(\alpha\) (0–1) is the fraction of incident radiation a surface absorbs. By Kirchhoff's law, at the same temperature \(\varepsilon = \alpha\).

Net radiation exchange with large surroundings
\[ \dot{Q}_{\text{rad}} = \varepsilon \sigma A_s \left(T_s^4 - T_{\text{surr}}^4\right) \]
Temperatures must be absolute (K). Valid when a small surface is enclosed by much larger surroundings.
Key Idea Radiation scales with \(T^4\): it dominates many furnace and solar applications and can remain comparable to natural convection near room temperature. Always convert to kelvin before raising temperature to the fourth power.
1–9

Simultaneous Heat Transfer Mechanisms

In practice the three mechanisms rarely act alone. A surface losing heat to air, for instance, does so by convection and radiation at the same time, and the two simply add.

Combined convection and radiation from a surface
\[ \dot{Q}_{\text{total}} = \dot{Q}_{\text{conv}} + \dot{Q}_{\text{rad}} = h A_s (T_s - T_\infty) + \varepsilon \sigma A_s (T_s^4 - T_{\text{surr}}^4) \]

Which Mechanisms Coexist

Conduction and radiation can occur within a transparent solid; convection always brings conduction with it at the wall; radiation accompanies the other two whenever surfaces are exposed. A solid opaque medium transmits heat only by conduction; a vacuum, only by radiation.

EXAMPLE 3 Combined loss from an oven access panel
Problem. The outer surface of an industrial oven access panel has area 0.72 m², temperature 78°C, and emissivity 0.82. Room air is 24°C with a measured \(h=7.5\ \mathrm{W/(m^2K)}\); large surrounding surfaces are at 20°C. Find convective, radiative, and total heat loss.

Strategy

The exposed surface loses heat through parallel convection and radiation paths. Use Newton’s law and the small-gray-surface/large-surroundings relation.

Givens

  • \(A=0.72\) m², \(T_s=351.15\) K, \(T_\infty=297.15\) K, \(T_{surr}=293.15\) K, \(\varepsilon=0.82\), \(h=7.5\), \(\sigma=5.67\times10^{-8}\).

Physical/model interpretation

Steady uniform gray surface with view factor approximately one to large isothermal surroundings; positive heat leaves the panel.

Step-by-step solution
1. Convection. \(\dot Q_{conv}=hA(T_s-T_\infty)=7.5(0.72)(54)=291.6\) W.
2. Radiation setup. \(\dot Q_{rad}=\varepsilon\sigma A(T_s^4-T_{surr}^4)\); absolute temperatures are required.
3. Radiation. Substitution gives \(0.82(5.67\times10^{-8})(0.72)(351.15^4-293.15^4)=261.8\) W.
4. Combine paths. \(\dot Q_{tot}=291.6+261.8=553.4\) W.

Unit/sign/limit checks

Units: \(hA\Delta T\) gives [W/(m²·K)]m²K = W; \(\varepsilon\sigma A\Delta(T^4)\) gives [W/(m²·K⁴)]m²K⁴ = W.

Signs: Since the panel is hotter than both room air and surrounding surfaces, both calculated terms are positive outward losses under the declared convention.

Limit: Each path vanishes when its own temperature difference vanishes; radiation is 47.3% of the total, so omitting it would underpredict loss by nearly one-half.

Final answer: \(\boxed{\dot Q_{conv}=292\ \mathrm W,\ \dot Q_{rad}=262\ \mathrm W,\ \dot Q_{tot}=553\ \mathrm W\ outward}\)

Engineering insight

The supplied \(h\) is configuration-specific and emissivity depends on finish and aging. Guarding, touch-temperature limits, and code compliance need separate review.

1–10

Problem-Solving Technique

A disciplined solution begins with recognition, not equation hunting. Decide system, steady/transient behavior, geometry, active mechanisms, knowns, and target before selecting the minimum equation set.

Model before arithmetic

State geometry and boundary conditions; separate stored energy from energy crossing the boundary; identify conduction, forced or natural convection, radiation, and phase change; evaluate properties at the method’s reference temperature; and verify any correlation’s geometry, Reynolds/Rayleigh/Prandtl range, and boundary condition.

Key Idea Always carry units through every step and sanity-check magnitudes. An answer that says a coffee cup loses 50 kW, or a wall temperature of 5000 °C, is signalling a setup error — not a surprising result.

Physical meaning

Conservation closes the energy accounting; mechanism laws determine rates. A result is credible only when both descriptions agree.

Problem-solving strategy

  1. Recognize whether the target is energy, rate, flux, temperature, or time and sketch the system.
  2. Declare steady/transient behavior, active mechanisms, coordinates, and positive heat direction.
  3. Select the minimum independent balances and constitutive laws.
  4. Use traceable properties at appropriate mean or film temperatures and check correlation validity.
  5. Solve symbolically where useful, then audit units, signs, limiting cases, and expected magnitude.

Chapter recap

  • Heat is boundary-crossing energy driven by temperature difference; rate and flux describe how fast and how concentrated.
  • Energy balances account for storage, mass flow, heat, and work.
  • Conduction, convection, and radiation have distinct models and often act together.

Before moving on

  • Can you choose a system and sign convention?
  • Can you distinguish properties from coefficients and state reference temperatures?
  • Can you check units, direction, limits, and plausibility?
← All chapters Chapter 2 · Heat Conduction Equation →