Heat Transfer · Chapter Thirteen

Heat Exchangers

A heat exchanger moves energy between two fluids without mixing them. Designing one means relating the heat duty to the surface area, the flow arrangement, and an overall coefficient that bundles every resistance in the path.

The central engineering choice is the model: use an energy balance first, then LMTD when terminal temperatures are known or effectiveness–NTU when outlets are unknown. Property temperatures, fouling, pressure drop, and correlation limits determine whether the result is credible.

Learning objectives

Classify exchanger construction and flow arrangement from a physical description.
Build an area-consistent overall thermal-resistance model including fouling.
Apply fluid energy balances with heat capacities evaluated at defensible mean temperatures.
Select and execute LMTD or effectiveness–NTU analysis.
Audit temperature ordering, units, effectiveness bounds, and design assumptions.

13–1

Types of Heat Exchangers

Double-pipe units provide simple counter or parallel flow; shell-and-tube units tolerate pressure and temperature; plate units offer high area density and cleanability; compact cross-flow cores favor gas service. Arrangement changes the temperature field, not the first-law balance.

Flow Arrangements

Parallel flow has the largest approach at the inlet and a rapidly collapsing driving force. Counter flow sustains a more uniform difference and can produce a cold outlet hotter than the hot outlet without violating the second law. Cross-flow effectiveness depends on whether each stream mixes transverse to its mean motion.

Physical meaning

The local heat flux is \(q''=U\Delta T(x)\). Flow arrangement reshapes \(\Delta T(x)\); it does not create energy. A temperature cross is feasible in counter flow but not in a single-pass parallel-flow exchanger.

13–2

The Overall Heat Transfer Coefficient

Heat crosses a hot-side film, deposits, wall, deposits, and cold-side film. The conductance \(UA\) is unambiguous, whereas \(U\) depends on the chosen reference area. For a cylindrical wall, inside and outside areas differ and must not be mixed.

Overall coefficient (thin wall) and resistance series

\[ \frac{1}{UA} = \frac{1}{h_i A_i} + R_{\text{wall}} + \frac{1}{h_o A_o} \]

Effect of fouling

\[ \frac{1}{UA} = \frac{1}{h_i A_i} + \frac{R_{f,i}}{A_i} + R_{\text{wall}} + \frac{R_{f,o}}{A_o} + \frac{1}{h_o A_o} \]

Fouling factors \(R_f\) account for deposits that build up over time and degrade performance.

Use local or mean fluid temperatures appropriate to the property variation. If viscosity or \(c_p\) changes substantially, segment the exchanger rather than relying on one averaged value. Correlations used to obtain \(h_i\) and \(h_o\) must match geometry, Reynolds regime, entrance condition, and roughness.

Course Note

In most exchangers, one side's film coefficient dominates the resistance — usually the gas side. There is little point polishing the water-side coefficient if the air-side film is the bottleneck. Find the largest resistance first, then spend your design effort there.

13–3

Analysis of Heat Exchangers

For steady operation with negligible heat loss, kinetic and potential changes, hot-side loss equals cold-side gain. The heat-capacity rate \(C=\dot m c_p\) has units W/K and measures how resistant a stream is to temperature change. Evaluate \(c_p\) near the stream mean temperature, or integrate enthalpy when variation is important.

Energy balances and heat capacity rate

\[ \dot{Q} = C_h(T_{h,i} - T_{h,o}) = C_c(T_{c,o} - T_{c,i}), \qquad C = \dot{m}c_p \]

Problem-solving strategy

Recognize sizing (known terminals) versus rating (unknown outlets), and identify the flow arrangement.
Draw both streams, label inlets/outlets, and write the hot and cold energy balances before a transfer equation.
Evaluate properties at stated mean temperatures; determine \(UA\) on one declared area basis.
Use LMTD for known terminals or the arrangement-specific \(\varepsilon\)-NTU relation for unknown terminals.
Audit energy closure, positive end differences, \(0\le\varepsilon\le1\), and correlation validity.

13–4

The Log Mean Temperature Difference Method

Because the temperature difference between the fluids varies along the exchanger, the correct average is logarithmic, not arithmetic. The LMTD method is ideal when inlet and outlet temperatures are known and area is sought.

Log Mean Temperature Difference

The appropriate mean driving temperature difference for a heat exchanger is the logarithmic mean of the end differences, not their average. Using the arithmetic mean overestimates performance, especially when the end differences differ greatly.

\[ \dot{Q} = U A\,\Delta T_{\text{lm}}, \qquad \Delta T_{\text{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1/\Delta T_2)} \]

Correction factor for multipass / cross flow

\[ \Delta T_{\text{lm}} = F\,\Delta T_{\text{lm,CF}} \]

\(F \le 1\) is read from charts using the dimensionless ratios \(P\) and \(R\); \(\Delta T_{\text{lm,CF}}\) is the counter-flow value.

EXAMPLE 1Sizing a Lubricant Cooler

Problem. A counter-flow cooler reduces biodegradable lubricant from 92 to 54 °C while cooling water warms from 18 to 31 °C. The measured duty is 48.0 kW and the clean overall coefficient is 310 W/m²·K. Determine required clean area.

Strategy

All four terminal temperatures are known, so LMTD closes the sizing problem directly.

Givens

\(T_{h,i}=92^\circ C,\ T_{h,o}=54^\circ C,\ T_{c,i}=18^\circ C,\ T_{c,o}=31^\circ C\)
\(\dot Q=48.0\) kW, \(U=310\) W/m²·K; find \(A\).

Physical/model interpretation

Steady, adiabatic counter flow with constant mean \(U\); heat rate is positive from lubricant to water.

Numbered derivation steps

1. Define end approaches. \(\Delta T_1=T_{h,i}-T_{c,o}=61\) K and \(\Delta T_2=T_{h,o}-T_{c,i}=36\) K.

2. Form the logarithmic mean. \(\Delta T_{lm}=(61-36)/\ln(61/36)\).

3. Evaluate it. \(\Delta T_{lm}=47.4\) K.

4. Solve the transfer equation. \(A=\dot Q/(U\Delta T_{lm})=48{,}000/(310\times47.4)=3.27\ {\rm m^2}\).

Unit/sign/limit checks

Units: In \(A=\dot Q/(U\Delta T_{lm})\), W divided by \(({\rm W/m^2K})({\rm K})\) leaves m², so the computed 3.27 is dimensionally an area.

Signs: Both terminal differences are positive, so the assumed hot-to-cold direction is consistent.

Limits: The logarithmic mean 47.4 K lies strictly between the 36 K and 61 K end differences, as every positive two-point mean must; this independently checks the logarithm and terminal pairing.

Final answer: \(\boxed{A=3.27\ {\rm m^2}}\).

Engineering insight

This is clean area; a specified fouling resistance would lower \(U\) and increase installed area.

13–5

The Effectiveness–NTU Method

When outlet temperatures are unknown (a rating problem), the effectiveness–NTU method is more direct. Effectiveness compares actual heat transfer to the maximum thermodynamically possible.

Effectiveness, maximum rate, and NTU

\[ \varepsilon = \frac{\dot{Q}}{\dot{Q}_{\max}}, \quad \dot{Q}_{\max} = C_{\min}(T_{h,i} - T_{c,i}), \quad NTU = \frac{UA}{C_{\min}} \]

Counter-flow effectiveness relation

\[ \varepsilon = \frac{1 - \exp[-NTU(1-c)]}{1 - c\,\exp[-NTU(1-c)]}, \qquad c = \frac{C_{\min}}{C_{\max}} \]

For \(c=0\) (one fluid changing phase), all arrangements reduce to \(\varepsilon = 1 - e^{-NTU}\).

EXAMPLE 2Rating a Glycol Heat-Recovery Unit

Problem. A counter-flow process exchanger receives hot water at 78 °C with \(C_h=3.60\) kW/K and glycol at 12 °C with \(C_c=2.40\) kW/K. If \(UA=3.00\) kW/K, find duty and both outlets.

Strategy

Outlet temperatures are unknown, so use the counter-flow effectiveness–NTU relation and then two energy balances.

Givens

\(C_h=3.60,\ C_c=2.40,\ UA=3.00\) kW/K.
\(T_{h,i}=78^\circ C,\ T_{c,i}=12^\circ C\).

Physical/model interpretation

Steady adiabatic exchanger; the glycol is \(C_{\min}\) and therefore undergoes the larger temperature change.

Numbered derivation steps

1. Form parameters. \(C_{\min}=2.40\), \(C_r=2.40/3.60=0.667\), \(NTU=3.00/2.40=1.25\).

2. Find effectiveness. \(\varepsilon=[1-e^{-1.25(1-0.667)}]/[1-0.667e^{-1.25(1-0.667)}]=0.608\).

3. Find duty. \(\dot Q_{\max}=2.40(78-12)=158.4\) kW, so \(\dot Q=0.608(158.4)=96.3\) kW.

4. Recover outlets. \(T_{h,o}=78-96.3/3.60=51.3^\circ C\); \(T_{c,o}=12+96.3/2.40=52.1^\circ C\).

Unit/sign/limit checks

Units: \(C_{\min}\Delta T\) has \(({\rm kW/K})({\rm K})={\rm kW}\); subsequently, \(\dot Q/C_h\) and \(\dot Q/C_c\) each have units K, so both outlet-temperature corrections are dimensionally consistent.

Signs: Hot stream cools and cold stream warms by equal 96.3 kW.

Limits: \(0<0.608<1\); a cold outlet slightly above the hot outlet is permissible in counter flow, while each remains between inlet extremes.

Final answer: \(\boxed{\dot Q=96.3\ {\rm kW},\ T_{h,o}=51.3^\circ C,\ T_{c,o}=52.1^\circ C}\).

Engineering insight

The temperature cross is a useful arrangement check: applying a parallel-flow relation would be physically incompatible with this result.

13–6

Selection of Heat Exchangers

Selecting an exchanger balances heat duty against cost, size, weight, pressure drop, and maintainability. The "best" choice is rarely the one with the highest \(U\) alone.

Key Idea A bigger \(U\) or area always raises heat transfer but also raises cost and pressure drop. Good selection meets the required duty at acceptable pumping power and capital cost — counter-flow when temperature approach is tight, compact cross-flow when space and weight dominate.

EXAMPLE 3Quantifying Fouling Margin

Problem. A thin-wall food-process plate exchanger has \(h_h=900\) and \(h_c=1800\) W/m²·K. Its combined area-based fouling resistance is \(3.0\times10^{-4}\) m²·K/W. Neglect wall resistance. Find clean and fouled \(U\), and the area penalty for unchanged duty and LMTD.

Strategy

All resistances share one plate area, so add them per unit area and compare conductances.

Givens

\(h_h=900,\ h_c=1800\) W/m²K; \(R_f''=3.0\times10^{-4}\) m²K/W.
Thin wall; unchanged \(\dot Q\) and \(\Delta T_{lm}\).

Physical/model interpretation

Steady one-dimensional series resistance; fouling adds resistance rather than subtracting a percentage from \(U\).

Numbered derivation steps

1. Clean resistance. \(1/U_c=1/900+1/1800=1.667\times10^{-3}\ {\rm m^2K/W}\).

2. Clean coefficient. \(U_c=600\ {\rm W/m^2K}\).

3. Fouled coefficient. \(1/U_f=1.667\times10^{-3}+0.300\times10^{-3}=1.967\times10^{-3}\), hence \(U_f=508\ {\rm W/m^2K}\).

4. Area ratio. Since \(UA\) must remain fixed, \(A_f/A_c=U_c/U_f=600/508=1.18\).

Unit/sign/limit checks

Units: Every resistance is m²K/W and its reciprocal is W/m²K.

Signs: Positive deposit resistance lowers \(U\).

Limits: As \(R_f''\to0\), \(U_f\to600\) and area ratio tends to one.

Final answer: \(\boxed{U_c=600,\ U_f=508\ {\rm W/m^2K};\ 18.0\%\ {\rm more\ area}}\).

Engineering insight

Installed margin must also consider cleaning interval, allowable pressure drop, uncertainty, and sanitary design; this resistance calculation is not a complete equipment specification.

Chapter recap

Begin with two stream energy balances and a declared property basis.
Use known terminal temperatures to recognize LMTD sizing; use unknown outlets to recognize \(\varepsilon\)-NTU rating.
Carry film, wall, and fouling resistances on a consistent area basis.
Before moving on: be able to sketch temperature profiles, select the method, solve outlets or area, and audit energy closure and bounds.

← Chapter 12 · Radiation Heat Transfer Chapter 14 · Mass Transfer →