4.5 Heat Exchanger Analysis: Log Mean Temperature Difference (LMTD) & Effectiveness-NTU (ε-NTU)

4.5 Heat Exchanger Analysis: Log Mean Temperature Difference (LMTD) & Effectiveness-NTU ($\varepsilon\text{-NTU}$)

Heat exchangers transfer thermal energy between two moving fluid streams at different temperatures across a solid conducting boundary. In HVAC and refrigeration systems, heat exchangers include air handling heating/cooling coils, shell-and-tube chiller evaporators and condensers, plate-and-frame waterside economizers, air-to-air energy recovery ventilators (ERVs), and cooling towers. Analyzing heat exchangers requires selecting between two standard analytical approaches:

1. Log Mean Temperature Difference (LMTD) Method: Preferred for sizing/design problems, where all inlet and outlet fluid temperatures are specified and the required heat transfer surface area ($A$) or overall $U$-factor must be determined.
2. Effectiveness-NTU ($\varepsilon\text{-NTU}$) Method: Preferred for rating/performance problems, where the heat exchanger geometry ($A$) and fluid inlet conditions are known, and the outlet temperatures and total heat transfer rate ($\dot{q}$) must be calculated without iterative solving.

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1. Flow Classifications & Temperature Profiles

+-----------------------------------------------------------------------------------------+
| HEAT EXCHANGER FLOW CONFIGURATIONS AND TEMPERATURE PROFILES                             |
+-----------------------------------------------------------------------------------------+
| PARALLEL FLOW (Co-Current):                     COUNTER FLOW (Counter-Current):         |
|   Hot:  T_h,in  ================> T_h,out         Hot:  T_h,in  ================> T_h,out |
|   Cold: T_c,in  ================> T_c,out         Cold: T_c,out <================ T_c,in  |
|                                                                                         |
| Temp ^                                          Temp ^                                  |
|      | T_h,in                                        | T_h,in                                  |
|      |   \                                           |   \                                     |
|      |    \---------> T_h,out                        |    \---------> T_h,out                  |
|      |                ------- T_c,out                |               /                         |
|      |    /--------->                                |   T_c,out <--/                          |
|      |   /                                           |              \                          |
|      | T_c,in                                        |               \<--------- T_c,in        |
|      +-------------------------> Area                +-------------------------> Area   |
| Limit: T_c,out CANNOT exceed T_h,out!          Advantage: T_c,out CAN EXCEED T_h,out!   |
+-----------------------------------------------------------------------------------------+

Primary Flow Configurations

1. Counter-Flow (Counter-Current): Fluids flow in opposite directions. Counter-flow provides the highest theoretical thermal efficiency and the maximum effective mean temperature difference. Crucially, counter-flow allows the cold fluid outlet temperature to exceed the hot fluid outlet temperature ($T_{c,out} > T_{h,out}$).
2. Parallel-Flow (Co-Current): Fluids flow in the same direction. Parallel-flow has the lowest thermal efficiency; the cold fluid outlet temperature can never exceed the hot fluid outlet temperature ($T_{c,out} < T_{h,out}$).
3. Cross-Flow: Fluids flow perpendicular to each other (e.g., air flowing across finned hydronic tubes in AHUs). Categorized as unmixed/unmixed (both streams channeled by fins/tubes), mixed/unmixed (one stream mixes freely in the plenum), or mixed/mixed.
4. Shell-and-Tube: One fluid flows inside tubes while the other flows across baffles inside the outer shell (designated by TEMA standards, such as 1-shell pass, 2-tube passes or 2-shell passes, 4-tube passes).

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2. Heat Capacity Rates & Overall Energy Balance

Under steady-state adiabatic conditions (neglecting heat loss from the outer shell to surroundings):

$$\dot{q} = \dot{m}_h c_{p,h} (T_{h,in} - T_{h,out}) = \dot{m}_c c_{p,c} (T_{c,out} - T_{c,in})$$

Defining the heat capacity rate ($C$) for hot and cold streams:

$$C_h = \dot{m}_h c_{p,h} \quad \text{and} \quad C_c = \dot{m}_c c_{p,c} \quad \left[\frac{\text{Btu}}{\text{hr}\cdot{}^\circ\text{F}} \text{ or } \frac{\text{W}}{\text{K}}\right]$$ $$\dot{q} = C_h (T_{h,in} - T_{h,out}) = C_c (T_{c,out} - T_{c,in})$$

Capacity Rate Extremes & Capacity Ratio ($C_r$)

$$C_{min} = \min(C_h, C_c) \quad \text{and} \quad C_{max} = \max(C_h, C_c)$$ $$C_r = \frac{C_{min}}{C_{max}} \quad (0 \le C_r \le 1.0)$$

• Pure Phase Change (Boilers & Condensers): For an evaporating or condensing fluid (such as steam or refrigerant), phase change occurs at constant temperature ($c_p \to \infty$), meaning $C_{max} \to \infty$ and the capacity ratio is identically $C_r = 0$.
• Balanced Flow: When hot and cold mass capacity rates are equal ($C_h = C_c$), $C_r = 1.0$.

Maximum Theoretical Heat Transfer Rate ($\dot{q}_{max}$)

The absolute thermodynamic maximum heat transfer rate achievable in an infinitely long counter-flow heat exchanger is limited by the fluid stream with the smaller capacity rate ($C_{min}$), which undergoes the maximum possible temperature difference ($T_{h,in} - T_{c,in}$):

$$\dot{q}_{max} = C_{min} (T_{h,in} - T_{c,in})$$

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3. Log Mean Temperature Difference (LMTD) Method

The fundamental heat transfer rate equation using the LMTD method is:

$$\dot{q} = U A F \Delta T_{lm}$$

Where:

• $U = \text{overall heat transfer coefficient } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \text{or W/(m}^2\cdot\text{K)})$
• $A = \text{total heat transfer surface area } (\text{ft}^2 \text{ or m}^2)$
• $F = \text{LMTD geometry correction factor } (F = 1.0 \text{ for true pure counter-flow or parallel-flow})$
• $\Delta T_{lm} = \text{Log Mean Temperature Difference } ({}^\circ\text{F or } {}^\circ\text{C})$

Definition of $\Delta T_{lm}$

$$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)}$$

Where the terminal temperature differences $\Delta T_1$ and $\Delta T_2$ are defined based on flow orientation:

Special Balanced Case ($\Delta T_1 = \Delta T_2$): When $\Delta T_1 = \Delta T_2$ (which occurs in counterflow when $C_h = C_c$), $\frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)}$ yields the indeterminate form $0/0$. Applying L'Hôpital's Rule confirms that $\Delta T_{lm} = \Delta T_1 = \Delta T_2$.

4. LMTD Correction Factor ($F$) for Complex Geometries

For multi-pass shell-and-tube heat exchangers and cross-flow coils, flow paths deviate from pure counter-flow, reducing the effective mean temperature difference ($F \le 1.0$).

+-----------------------------------------------------------------------------------------+
| LMTD CORRECTION FACTOR DIMENSIONLESS PARAMETERS (P and R)                               |
+-----------------------------------------------------------------------------------------+
| Temperature Effectiveness P:    P = (t_out - t_in) / (T_in - t_in)                      |
| Capacity Rate Ratio R:          R = (T_in - T_out) / (t_out - t_in) = C_tube / C_shell  |
|                                                                                         |
| Where: T = Shell-side fluid temperatures (T_in, T_out)                                  |
|        t = Tube-side fluid temperatures  (t_in, t_out)                                  |
+-----------------------------------------------------------------------------------------+

Analytical Form for 1-Shell Pass, $2n$-Tube Passes

For a standard 1-shell-pass, 2-tube-pass (or 4, 6 tube passes) shell-and-tube exchanger:

$$F = \frac{\sqrt{R^2 + 1} \ln\left(\frac{1 - P}{1 - P R}\right)}{(R - 1) \ln\left(\frac{2 - P(R + 1 - \sqrt{R^2 + 1})}{2 - P(R + 1 + \sqrt{R^2 + 1})}\right)}$$

Critical Design Rules for $F$ on the PE Exam

• The $F \ge 0.75-0.80$ Threshold: In professional HVAC engineering, heat exchangers are never selected with $F < 0.75$. Low $F$ values indicate severe temperature cross and steep performance drops where tiny changes in fluid flow cause large capacity losses. If $F < 0.75$, add more shell passes in series (e.g., switch to a 2-shell-pass, 4-tube-pass exchanger).
• Phase Change Equivalence ($F = 1.0$): When one fluid is condensing or boiling ($C_r = 0$, $R = 0$ or $R \to \infty$), $F = 1.00$ for all heat exchanger configurations (shell-and-tube, cross-flow, etc.).

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5. The Effectiveness-NTU ($\varepsilon\text{-NTU}$) Method

When sizing an existing heat exchanger with known surface area $A$ and entering fluid conditions, solving via LMTD requires tedious iterations because outlet temperatures appear inside logarithms. The $\varepsilon\text{-NTU}$ method provides direct, explicit closed-form solutions.

1. Heat Exchanger Effectiveness ($\varepsilon$)

Effectiveness is the ratio of actual heat transfer rate $\dot{q}$ to the maximum theoretical heat transfer rate $\dot{q}_{max}$:

$$\varepsilon = \frac{\dot{q}}{\dot{q}_{max}} = \frac{C_h (T_{h,in} - T_{h,out})}{C_{min}(T_{h,in} - T_{c,in})} = \frac{C_c (T_{c,out} - T_{c,in})}{C_{min}(T_{h,in} - T_{c,in})}$$

Once $\varepsilon$ is known, total heat transfer is calculated immediately:

$$\dot{q} = \varepsilon \dot{q}_{max} = \varepsilon C_{min} (T_{h,in} - T_{c,in})$$

Fluid outlet temperatures follow directly:

$$T_{h,out} = T_{h,in} - \frac{\dot{q}}{C_h} \quad \text{and} \quad T_{c,out} = T_{c,in} + \frac{\dot{q}}{C_c}$$

2. Number of Transfer Units ($\text{NTU}$)

$\text{NTU}$ is a dimensionless measure of the physical heat transfer size of the exchanger:

$$\text{NTU} = \frac{U A}{C_{min}}$$

Summary of Analytical $\varepsilon\text{-NTU}$ Formulas (NCEES Reference Handbook)

6. Step-by-Step Worked Example 1: Sizing a Plate-and-Frame Waterside Economizer (LMTD)

Problem Statement

A plate-and-frame heat exchanger serves as a waterside economizer (free cooling) in a central chiller plant. The system cools $200.0\text{ GPM}$ of building chilled water from $T_{h,in} = 56.0^\circ\text{F}$ down to $T_{h,out} = 44.0^\circ\text{F}$ using cold tower water entering at $T_{c,in} = 38.0^\circ\text{F}$ with a flow rate of $250.0\text{ GPM}$. Fluid properties:

• Standard water specific heat: $c_p = 1.000\text{ Btu/(lbm}\cdot{}^\circ\text{F)}$
• Chilled water multiplier: $C_h = 500 \times 200.0\text{ GPM} = 100,000\text{ Btu/(hr}\cdot{}^\circ\text{F)}$
• Tower water multiplier: $C_c = 500 \times 250.0\text{ GPM} = 125,000\text{ Btu/(hr}\cdot{}^\circ\text{F)}$
• Heat exchanger overall heat transfer coefficient: $U = 450.0\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$
• True counter-flow configuration ($F = 1.00$)

Calculate:

1. Total heat duty $\dot{q}$ ($\text{Btu/hr}$ and Tons of Refrigeration)
2. Leaving cooling tower water temperature $T_{c,out}$ (${}^\circ\text{F}$)
3. Counter-flow Log Mean Temperature Difference $\Delta T_{lm}$ (${}^\circ\text{F}$)
4. Required heat transfer plate surface area $A$ ($\text{ft}^2$)

Solution Steps

Step 1: Calculate total heat duty $\dot{q}$

$$\dot{q} = C_h (T_{h,in} - T_{h,out}) = 100,000\text{ Btu/(hr}\cdot{}^\circ\text{F)} \times (56.0^\circ\text{F} - 44.0^\circ\text{F}) = 100,000 \times 12.0 = 1,200,000\text{ Btu/hr}$$ $$\text{Cooling Capacity} = \frac{1,200,000\text{ Btu/hr}}{12,000\text{ Btu/(hr}\cdot\text{Ton)}} = 100.0\text{ Tons}$$

Step 2: Calculate leaving cooling tower water temperature $T_{c,out}$

$$\dot{q} = C_c (T_{c,out} - T_{c,in}) \implies T_{c,out} = T_{c,in} + \frac{\dot{q}}{C_c}$$ $$T_{c,out} = 38.0^\circ\text{F} + \frac{1,200,000\text{ Btu/hr}}{125,000\text{ Btu/(hr}\cdot{}^\circ\text{F)}} = 38.0 + 9.60 = 47.60^\circ\text{F}$$

Notice that $T_{c,out} = 47.60^\circ\text{F} > T_{h,out} = 44.0^\circ\text{F}$ (temperature cross, demonstrating counterflow capability).

Step 3: Calculate counter-flow LMTD

$$\Delta T_1 = T_{h,in} - T_{c,out} = 56.0^\circ\text{F} - 47.60^\circ\text{F} = 8.40^\circ\text{F}$$ $$\Delta T_2 = T_{h,out} - T_{c,in} = 44.0^\circ\text{F} - 38.0^\circ\text{F} = 6.00^\circ\text{F}$$ $$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{8.40 - 6.00}{\ln(8.40 / 6.00)} = \frac{2.40}{\ln(1.400)} = \frac{2.40}{0.33647} = 7.133^\circ\text{F}$$

Step 4: Calculate required plate surface area $A$

$$\dot{q} = U A F \Delta T_{lm} \implies A = \frac{\dot{q}}{U F \Delta T_{lm}}$$ $$A = \frac{1,200,000\text{ Btu/hr}}{450.0\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \times 1.00 \times 7.133^\circ\text{F}} = \frac{1,200,000}{3,209.85} = 373.85\text{ ft}^2$$

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7. Step-by-Step Worked Example 2: Rating an Air-to-Air Energy Recovery Ventilator ($\varepsilon\text{-NTU}$)

Problem Statement

An air-to-air fixed plate energy recovery ventilator (ERV, counter-flow arrangement) handles an outdoor fresh airstream ($3,000\text{ CFM}$, $T_{c,in} = 10.0^\circ\text{F}$) and an building exhaust airstream ($3,000\text{ CFM}$, $T_{h,in} = 72.0^\circ\text{F}$). Standard air properties apply ($C_s = 1.08\text{ Btu/(hr}\cdot\text{CFM}\cdot{}^\circ\text{F)}$). The total heat transfer surface area is $A = 1200.0\text{ ft}^2$ and the overall $U$-factor is $U = 2.70\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$.

Calculate:

1. Heat capacity rates $C_h$ and $C_c$, capacity ratio $C_r$, and maximum possible heat transfer $\dot{q}_{max}$
2. Number of Transfer Units ($\text{NTU}$)
3. Exchanger effectiveness $\varepsilon$
4. Actual heat recovery rate $\dot{q}$ ($\text{Btu/hr}$)
5. Supply air temperature delivered to the building ($T_{c,out}$)

Solution Steps

Step 1: Calculate heat capacity rates and $\dot{q}_{max}$

$$C_h = C_c = 1.08 \times 3,000\text{ CFM} = 3,240\text{ Btu/(hr}\cdot{}^\circ\text{F)}$$ $$C_{min} = C_{max} = 3,240\text{ Btu/(hr}\cdot{}^\circ\text{F)} \implies C_r = \frac{C_{min}}{C_{max}} = 1.00$$ $$\dot{q}_{max} = C_{min} (T_{h,in} - T_{c,in}) = 3,240 \times (72.0^\circ\text{F} - 10.0^\circ\text{F}) = 3,240 \times 62.0 = 200,880\text{ Btu/hr}$$

Step 2: Calculate Number of Transfer Units ($\text{NTU}$)

$$\text{NTU} = \frac{U A}{C_{min}} = \frac{2.70\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \times 1200.0\text{ ft}^2}{3,240\text{ Btu/(hr}\cdot{}^\circ\text{F)}} = \frac{3,240}{3,240} = 1.00$$

Step 3: Calculate effectiveness $\varepsilon$ for counter-flow with $C_r = 1.0$

$$\varepsilon = \frac{\text{NTU}}{1 + \text{NTU}} = \frac{1.00}{1 + 1.00} = \frac{1.00}{2.00} = 0.500 \quad (50.0\%\text{ effectiveness})$$

Step 4: Calculate actual heat transfer rate $\dot{q}$

$$\dot{q} = \varepsilon \dot{q}_{max} = 0.500 \times 200,880\text{ Btu/hr} = 100,440\text{ Btu/hr}$$

Step 5: Calculate delivered supply air temperature $T_{c,out}$

$$T_{c,out} = T_{c,in} + \frac{\dot{q}}{C_c} = 10.0^\circ\text{F} + \frac{100,440\text{ Btu/hr}}{3,240\text{ Btu/(hr}\cdot{}^\circ\text{F)}} = 10.0 + 31.0 = 41.0^\circ\text{F}$$

The ERV preheats incoming outdoor ventilation air from $10^\circ\text{F}$ to $41^\circ\text{F}$ for zero heating fuel expenditure.