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Heat Transfer

8% of exam

Conductive, convective (natural and forced), and radiation heat transfer, overall, local, and fouling heat-transfer coefficients, and heat-transfer equipment design (double-pipe and shell-and-tube exchangers, log-mean temperature difference, the effectiveness-NTU method, and flow configuration).

4 concepts

A. Conductive heat transfer

Conduction, Thermal Resistance and the Overall U

Fourier conduction through walls and cylinders, the series-parallel resistance network, fouling, and the overall heat-transfer coefficient U.

Almost every heat-transfer point on the FE Chemical exam is won or lost in one habit: turning a temperature drop into a resistance and a resistance into a heat rate. Conduction is where that habit is built. Heat flows down a temperature gradient, and the material's resistance to that flow is as concrete as an electrical resistance — wire several layers in series, add the convective films on each face, and the whole assembly collapses to a single number you can divide a temperature difference by. The NCEES FE Reference Handbook (Heat Transfer) gives you Fourier's law, the plane-wall and cylindrical-wall resistances, the convection resistance, fouling factors, and the overall coefficient

U

; this concept makes them one fluent calculation. Convection films and radiation each get their own concept — here we treat the film coefficient

h

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as a given resistance

1/(hA)

and focus on building the network and extracting

U

.

Fourier's law and the driving force

Conduction obeys Fourier's law: the heat rate is proportional to the cross-sectional area, the thermal conductivity

k

, and the temperature gradient. The minus sign says heat flows from hot to cold, down the gradient. For one-dimensional steady flow through a slab of thickness

L

with no internal generation, the gradient is constant and the law integrates to a simple ratio of temperature drop over thickness. Think of

k\,A/L

as a conductance and its reciprocal as a resistance — that single mental move powers everything below.

\dot{Q} = -kA\,\frac{dT}{dx} \quad\Longrightarrow\quad \dot{Q} = \frac{kA}{L}\,(T_1 - T_2)

Thermal resistance: the unifying idea

Define resistance by analogy to Ohm's law,

\dot{Q} = \Delta T / R

, so temperature is the potential and heat rate is the current. A plane wall has

R = L/(kA)

; a convective film has

R = 1/(hA)

. Resistances in series add, exactly like resistors carrying the same current — the same

\dot{Q}

passes through every layer in a one-dimensional stack, and the temperature drops add up to the total. Once you write the network you never solve simultaneous equations; you sum resistances and divide once.

R_{\text{plane}} = \frac{L}{kA}, \qquad R_{\text{conv}} = \frac{1}{hA}, \qquad \dot{Q} = \frac{\Delta T_{\text{total}}}{\sum R}

The cylindrical wall

Pipes and tubes are the chemical engineer's daily geometry, and a cylinder is not a slab: the area for heat flow grows with radius, so the conduction resistance carries a logarithm of the radius ratio, not a simple thickness. For a tube of length

L

between inner radius

r_1

and outer radius

r_2

the resistance uses

\ln(r_2/r_1)

. Stack tube wall plus insulation the same way you stack plane layers — series resistances — but be careful which radius pairs with each convective film: the inside film uses the inner area

2\pi r_1 L

and the outside film the outer area

2\pi r_2 L

.

R_{\text{cyl}} = \frac{\ln(r_2/r_1)}{2\pi k L}

Critical insulation radius

Insulation on a small pipe has a counterintuitive trap. Adding insulation raises the conduction resistance but also enlarges the outer surface, which lowers the convective resistance — for a thin pipe the second effect can win, so a little insulation actually increases heat loss until the outer radius reaches the critical value

r_{cr} = k/h

. Below

r_{cr}

you are heating, not insulating. For most lagged process lines

r_{cr}

is only a few millimetres and you are safely past it, but the FE likes to test whether you know the rule exists.

r_{cr} = \frac{k_{\text{ins}}}{h_o}

Fouling and the overall coefficient U

Real exchangers accumulate scale, biofilm, and corrosion products that add a fouling resistance

R_f

(tabulated per unit area, so divide by area to get K/W) on each fluid side. Roll the whole series chain — inside film, inside fouling, wall conduction, outside fouling, outside film — into one overall coefficient

U

defined so that

\dot{Q} = UA\,\Delta T

. Because the areas differ between the inside and outside of a tube,

U

must be quoted against a stated reference area (

U_i A_i = U_o A_o = 1/\sum R

); the handbook's

UA

form sidesteps the ambiguity by carrying the area inside. The smallest conductance (largest resistance) — usually a gas film or heavy fouling — dominates

U

, which tells you exactly where to spend money to improve an exchanger.

\frac{1}{UA} = \frac{1}{h_i A_i} + \frac{R_{f,i}}{A_i} + \frac{\ln(D_o/D_i)}{2\pi k L} + \frac{R_{f,o}}{A_o} + \frac{1}{h_o A_o}

Exam strategy

Build the resistance list before you touch a calculator: write

R

for every film, every solid layer, and every fouling allowance, in order from hot fluid to cold. Sum them, divide the end-to-end

\Delta T

once to get

\dot{Q}

, then walk back through the chain (

T_{\text{next}} = T_{\text{prev}} - \dot{Q}R

) to find any interface temperature. Watch three things: use

\ln(r_2/r_1)

for cylinders (never

L/kA

), pair each film with the correct area (

A_i

vs

A_o

), and keep fouling factors per-area until you divide by the matching area. Temperature differences are identical in K or °C, so a

\Delta T

never needs a kelvin conversion — but an absolute temperature in radiation always does.

Key equations

Fourier's law (1-D)

\dot{Q} = -kA\,\dfrac{dT}{dx}

Conduction heat rate (W). $k$ = thermal conductivity [W/(m·K)], $A$ = area normal to flow (m²); minus sign = flow down the gradient.

Plane-wall conduction

\dot{Q} = \dfrac{kA}{L}\,(T_1 - T_2)

Thermal-resistance (Ohm) form

\dot{Q} = \dfrac{\Delta T_{\text{total}}}{\sum R}

Plane-wall resistance

R = \dfrac{L}{kA}

Conduction resistance of a slab (K/W). $L$ = thickness, $A$ = face area.

Convection resistance

R = \dfrac{1}{hA}

Film resistance (K/W). $h$ = convection coefficient [W/(m²·K)], $A$ = wetted area on that side.

Cylindrical-wall resistance

R = \dfrac{\ln(r_2/r_1)}{2\pi k L}

Critical insulation radius

r_{cr} = \dfrac{k_{\text{ins}}}{h_o}

Fouling resistance

R_{\text{foul}} = \dfrac{R_f}{A}

$R_f$

Overall coefficient (UA)

\dfrac{1}{UA} = \dfrac{1}{h_i A_i} + \dfrac{R_{f,i}}{A_i} + \dfrac{\ln(D_o/D_i)}{2\pi k L} + \dfrac{R_{f,o}}{A_o} + \dfrac{1}{h_o A_o}

Overall U vs reference area

U_i A_i = U_o A_o = \dfrac{1}{\sum R}

Interface temperature

T_{n+1} = T_n - \dot{Q}\,R_n

Biot number (lumped check)

Bi = \dfrac{h V}{k A_s} = \dfrac{h L_c}{k}

Worked examples

Composite furnace wall with films

Problem. A 1 m² furnace wall has, from inside out: a $0.20\,\text{m}$ firebrick layer ( $k = 1.2\,\text{W/(m·K)}$ ), a $0.05\,\text{m}$ insulating layer ( $k = 0.07$ ), and a $0.005\,\text{m}$ steel sheet ( $k = 45$ ). Hot gas at $1000\,^\circ\text{C}$ has $h_i = 25\,\text{W/(m}^2\text{·K)}$ ; outside air at $30\,^\circ\text{C}$ has $h_o = 12$ . Find the heat loss per m², the overall $U$ , and the temperature at the brick/insulation interface.

Solution. Build the series resistances (per 1 m², so divide by

A=1

):

R_{conv,i}=1/25=0.0400

;

R_{brick}=0.20/1.2=0.1667

\dot{Q} = \dfrac{1000-30}{0.0400+0.1667+0.7143+0.000111+0.0833} = 966\ \text{W}

Insulated steam pipe

Problem. A $10\,\text{m}$ steel pipe ( $k=45$ , $r_1=0.050\,\text{m}$ , $r_2=0.055\,\text{m}$

Common pitfalls

•Using $L/(kA)$ for a pipe wall. Radial conduction needs $\ln(r_2/r_1)/(2\pi k L)$ — the area grows with radius, so a logarithm appears.
•Pairing both convective films with the same area. The inside film uses $A_i=2\pi r_1 L$ (or the inner face) and the outside film uses $A_o$
•Quoting $U$ without saying which area it references. Only $UA$ (or $U_iA_i=U_oA_o$
•Forgetting to divide a tabulated fouling factor $R_f$ [(m²·K)/W] by area before adding it to a K/W network — it is per unit area, not already a resistance.
•Assuming insulation always helps a small pipe. Below the critical radius $r_{cr}=k/h$ , adding insulation increases heat loss; check it on small-diameter, low- $h$ lines.
•Converting a temperature difference to kelvin. A $\Delta T$ is identical in K and °C; only absolute temperatures (radiation, ideal-gas) require the $+273.15$ .
•Treating parallel paths as series. Two materials side-by-side (e.g., studs and insulation in a wall) carry separate currents — add conductances $1/R$ , not resistances.

References

NCEES FE Reference Handbook — Heat Transfer — Fourier's law, plane/cylindrical resistance, critical radius, overall UA.
Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — Resistance-network method and composite walls.
Perry's Chemical Engineers' Handbook — Heat Transfer — Tabulated fouling factors and conductivities.

B. Convective heat transfer (natural and forced)

Convective Heat Transfer: Natural and Forced

Newton's law of cooling, the film coefficient h, Nusselt-Reynolds-Prandtl correlations for forced flow, and Grashof/Rayleigh natural convection.

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C. Radiation heat transfer

Radiation Heat Transfer: Stefan-Boltzmann and View Factors

Blackbody emission and the Stefan-Boltzmann law, emissivity and gray surfaces, view factors, and net radiation exchange between surfaces.

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E. Heat-transfer equipment, operation, and design

Heat Exchangers: LMTD and Effectiveness-NTU

Double-pipe and shell-and-tube exchangers, the log-mean temperature difference with its F-correction, and the effectiveness-NTU method.

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Steady 1-D slab of thickness $L$ (m); $T_1>T_2$ are the two face temperatures (K or °C — a difference is the same in both).

Heat rate equals total temperature drop over the sum of series resistances (K/W). The same $\dot{Q}$ flows through each.

Radial conduction through a tube of length $L$ between radii $r_1$ (inner) and $r_2$ (outer).

Outer radius that maximizes heat loss from an insulated cylinder; below it, adding insulation increases loss.

is the tabulated per-area fouling factor [(m²·K)/W]; divide by the fouled-side area to get K/W.

Series sum of all tube-side resistances. $UA$ (W/K) carries the area, so $\dot{Q}=UA\,\Delta T$ needs no reference-area choice.

$U$ must be quoted against a stated area (inside or outside); only the product $UA$ is unambiguous.

March the same $\dot{Q}$ through each resistance to recover any intermediate temperature.

If $Bi < 0.1$ a body is nearly isothermal and the lumped-capacitance transient model applies; $L_c=V/A_s$ .

;

R_{ins}=0.05/0.07=0.7143

;

R_{steel}=0.005/45=0.000111

;

R_{conv,o}=1/12=0.0833

(all K/W).

\sum R = 1.0044\,\text{K/W}

.

\dot{Q}=\Delta T/\sum R=(1000-30)/1.0044=\mathbf{966\ \text{W}}

(to 3 s.f.).

U = 1/(\sum R\cdot A)=1/1.0044=\mathbf{0.996\ \text{W/(m}^2\text{·K)}}

. Interface temperatures by marching:

T_1=1000-966(0.0400)=961\,^\circ\text{C}

(brick inner face);

T_2=961-966(0.1667)=\mathbf{800\,^\circ C}

(brick/insulation). Sanity: the insulating layer holds 71% of the total resistance, so it should take ~71% of the

970\,^\circ\text{C}

drop (

\approx 690\,^\circ\text{C}

) — and indeed

T

falls from

800

to

111\,^\circ\text{C}

across it. Units:

\text{K}/(\text{K/W})=\text{W}

. ✓

) carries steam at

200\,^\circ\text{C}

(

h_i=500\,\text{W/(m}^2\text{·K)}

) and is lagged with insulation (

k=0.05

) out to

r_3=0.105\,\text{m}

. Outside air is

25\,^\circ\text{C}

with

h_o=10

. Find the heat loss. Is the insulation past its critical radius?

Solution. Inside film:

R_{conv,i}=1/(h_i\,2\pi r_1 L)=1/(500\cdot2\pi\cdot0.05\cdot10)=0.000637\,\text{K/W}

. Steel:

R_{steel}=\ln(0.055/0.050)/(2\pi\cdot45\cdot10)=3.37\times10^{-5}

. Insulation:

R_{ins}=\ln(0.105/0.055)/(2\pi\cdot0.05\cdot10)=0.2058

. Outside film:

R_{conv,o}=1/(10\cdot2\pi\cdot0.105\cdot10)=0.01516

.

\sum R = 0.2217\,\text{K/W}

.

\dot{Q}=(200-25)/0.2217=\mathbf{789\ \text{W}}

(to 3 s.f.). Critical radius:

r_{cr}=k_{ins}/h_o=0.05/10=0.005\,\text{m}=5\,\text{mm}

. The insulation outer radius (

105\,\text{mm}

) is far beyond

r_{cr}

, so the lagging genuinely reduces loss. Sanity: the insulation is 93% of the resistance, as expected for low-

k

lagging. ✓

\dot{Q} = \dfrac{200-25}{R_{conv,i}+R_{steel}+R_{ins}+R_{conv,o}} = \dfrac{175}{0.2217} = 789\ \text{W}

; mismatching them corrupts

U

.

) is unambiguous; a bare

U

for a tube is meaningless until

A_i

or

A_o

is named.