9.2 Effectiveness-NTU Method and Exchanger Rating

9.2 Effectiveness-NTU Method and Exchanger Rating

In chemical plant operations and on the NCEES PE Chemical Exam, engineers face two fundamentally distinct categories of heat exchanger problems: sizing (design) and rating (performance evaluation).

• In a sizing problem, fluid flow rates and all four terminal temperatures are specified; the engineer uses the Log Mean Temperature Difference (LMTD) method to directly calculate the required heat transfer surface area ($A$).
• In a rating problem, an existing heat exchanger with a fixed surface area ($A$) and known overall heat transfer coefficient ($U$) receives streams at specified inlet temperatures ($T_{h,in}, T_{c,in}$) and flow rates. The outlet temperatures ($T_{h,out}, T_{c,out}$) and heat duty ($Q$) are unknown.

Attempting to solve a rating problem with the LMTD method leads to tedious, circular trial-and-error: you must guess an outlet temperature, compute LMTD and $F$, calculate $Q$, update the outlet temperatures, and repeat until convergence. To eliminate iteration entirely, W. M. Kays and A. L. London formulated the Effectiveness-NTU ($\epsilon$-NTU) method, which solves rating problems analytically in a single direct sequence.

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1. Fundamental Definitions of the $\epsilon$-NTU Method

1. Heat Capacity Rates ($C_h, C_c$)

The heat capacity rate ($C$) represents the thermal energy carried by a flowing stream per degree of temperature change:

$C_h = \dot{m}_h C_{p,h}, \quad C_c = \dot{m}_c C_{p,c}$

Where:

• $C_h, C_c$ = heat capacity rates of hot and cold fluids ($\text{W/K}$, $\text{kW/K}$, or $\text{Btu/(hr}\cdot^\circ\text{F)}$).
• $\dot{m}$ = mass flow rate ($\text{kg/s}$ or $\text{lb/hr}$).
• $C_p$ = specific heat capacity ($\text{J/(kg}\cdot\text{K)}$ or $\text{Btu/(lb}\cdot^\circ\text{F)}$).

Identify the minimum and maximum heat capacity rates:

$C_{min} = \min(C_h, C_c)$ $C_{max} = \max(C_h, C_c)$

2. Heat Capacity Ratio ($C_r$)

The dimensionless capacity ratio is bounded between $0$ and $1$:

$C_r = \frac{C_{min}}{C_{max}} \quad (0 \le C_r \le 1.0)$

3. Maximum Theoretical Heat Transfer Rate ($Q_{max}$)

The maximum possible heat duty that could be transferred in an idealized, infinitely long countercurrent heat exchanger is governed strictly by the fluid with the minimum heat capacity rate ($C_{min}$):

$Q_{max} = C_{min} (T_{h,in} - T_{c,in})$

[!IMPORTANT] Why $C_{min}$ Limits Heat Transfer:
The maximum temperature change any fluid can experience is the difference between fluid inlet temperatures: $\Delta T_{overall} = T_{h,in} - T_{c,in}$. From energy balances, $Q = C_h \Delta T_h = C_c \Delta T_c$. Because energy is conserved, the fluid with the smaller heat capacity rate ($C_{min}$) experiences the larger temperature change ($\Delta T_{max} = Q / C_{min}$). If the fluid with $C_{max}$ were to undergo the full temperature span $(T_{h,in} - T_{c,in})$, the required heat duty would force the temperature of the $C_{min}$ fluid to exceed the inlet temperature of the opposing stream, violating the Second Law of Thermodynamics! Therefore, $C_{min}$ always dictates $Q_{max}$.

4. Heat Exchanger Effectiveness ($\epsilon$)

Effectiveness ($\epsilon$) is the dimensionless ratio of the actual heat transfer rate ($Q$) to the maximum theoretical heat transfer rate ($Q_{max}$):

$\epsilon \equiv \frac{Q}{Q_{max}} = \frac{C_h (T_{h,in} - T_{h,out})}{C_{min} (T_{h,in} - T_{c,in})} = \frac{C_c (T_{c,out} - T_{c,in})}{C_{min} (T_{h,in} - T_{c,in})}$

• If $C_h = C_{min}$ (hot fluid is limiting): $\epsilon = \frac{T_{h,in} - T_{h,out}}{T_{h,in} - T_{c,in}}$
• If $C_c = C_{min}$ (cold fluid is limiting): $\epsilon = \frac{T_{c,out} - T_{c,in}}{T_{h,in} - T_{c,in}}$

Effectiveness is bounded by $0 \le \epsilon < 1.0$. Once $\epsilon$ is determined from the exchanger geometry and operating conditions, actual duty is calculated immediately as $Q = \epsilon Q_{max}$.

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

The Number of Transfer Units (NTU) is a dimensionless parameter that characterizes the physical thermal size of the heat exchanger relative to the stream's heat capacity:

$\text{NTU} \equiv \frac{U \cdot A}{C_{min}}$

Where:

• $U$ = overall heat transfer coefficient ($\text{W/(m}^2\cdot\text{K)}$).
• $A$ = total surface area ($\text{m}^2$).
• $C_{min}$ = minimum heat capacity rate ($\text{W/K}$).

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2. Explicit Analytical Effectiveness Formulas

Effectiveness is a function of flow geometry, $\text{NTU}$, and $C_r$:

$\epsilon = f(\text{NTU}, C_r, \text{Flow Arrangement})$

1. Pure Countercurrent Flow

• For $C_r < 1.0$: $\epsilon = \frac{1 - \exp\left[-\text{NTU}(1 - C_r)\right]}{1 - C_r \exp\left[-\text{NTU}(1 - C_r)\right]}$
• For $C_r = 1.0$ (Balanced Exchanger, $C_h = C_c$): $\epsilon = \frac{\text{NTU}}{1 + \text{NTU}}$
• Inverse Formula (Sizing Area from Desired Effectiveness): $\text{NTU} = \frac{1}{1 - C_r} \ln\left[ \frac{1 - \epsilon C_r}{1 - \epsilon} \right] \quad (C_r < 1.0)$ $\text{NTU} = \frac{\epsilon}{1 - \epsilon} \quad (C_r = 1.0)$

2. Pure Cocurrent / Parallel Flow

$\epsilon = \frac{1 - \exp\left[-\text{NTU}(1 + C_r)\right]}{1 + C_r}$

$\text{NTU} = -\frac{\ln\left[ 1 - \epsilon(1 + C_r) \right]}{1 + C_r}$

[!NOTE] The Parallel Flow 50% Limit:
As surface area approaches infinity ($\text{NTU} \to \infty$), the exponential term $\exp[-\text{NTU}(1+C_r)] \to 0$. Consequently, the maximum possible effectiveness for parallel flow is: $\epsilon_{max, \text{parallel}} = \frac{1}{1 + C_r}$ When fluid heat capacities are balanced ($C_r = 1.0$), $\epsilon_{max} = 1 / (1 + 1) = 0.50$ ($50%$). In parallel flow, fluid streams can never exchange more than half the maximum theoretical heat because they approach a common intermediate equilibrium temperature!

3. Shell-and-Tube Exchangers (1 Shell Pass, $2, 4, 6, \dots$ Tube Passes)

For a TEMA 1-2 shell-and-tube exchanger:

$\epsilon_1 = 2 \left[ 1 + C_r + \sqrt{1 + C_r^2} \cdot \frac{1 + \exp\left(-\text{NTU} \sqrt{1 + C_r^2}\right)}{1 - \exp\left(-\text{NTU} \sqrt{1 + C_r^2}\right)} \right]^{-1}$

For $n$ identical shell passes in series (each shell containing an even number of tube passes):

$\epsilon = \frac{\left[ \frac{1 - \epsilon_1 C_r}{1 - \epsilon_1} \right]^n - 1}{\left[ \frac{1 - \epsilon_1 C_r}{1 - \epsilon_1} \right]^n - C_r}$

4. Crossflow Exchangers (Air Coolers, Fin-Fan Exchangers)

• Both Fluids Unmixed (e.g., plate-fin or finned-tube radiators): $\epsilon = 1 - \exp\left\{ \frac{\text{NTU}^{0.22}}{C_r} \left[ \exp\left(-C_r \cdot \text{NTU}^{0.78}\right) - 1 \right] \right\}$
• $C_{max}$ Mixed, $C_{min}$ Unmixed: $\epsilon = \frac{1}{C_r} \left( 1 - \exp\left\{ -C_r \left[ 1 - \exp(-\text{NTU}) \right] \right\} \right)$
• $C_{min}$ Mixed, $C_{max}$ Unmixed: $\epsilon = 1 - \exp\left\{ -\frac{1}{C_r} \left[ 1 - \exp(-C_r \cdot \text{NTU}) \right] \right\}$

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3. Phase Change Heat Exchangers ($C_r = 0$)

In condensers, reboilers, evaporators, and steam heaters, one process stream undergoes an isothermal phase transition ($T = \text{constant}$). Thermodynamically, the effective heat capacity of a boiling or condensing fluid is infinite:

$C_p = \left( \frac{\partial H}{\partial T} \right)_P \to \infty \implies C_{max} \to \infty$

Therefore, the heat capacity ratio is identically zero:

$C_r = \frac{C_{min}}{C_{max}} = \frac{C_{min}}{\infty} = \mathbf{0}$

Substituting $C_r = 0$ into the effectiveness formulas for all heat exchanger configurations (counterflow, parallel flow, shell-and-tube, and crossflow) collapses the equations into a single universal relationship:

$\epsilon = 1 - \exp(-\text{NTU})$

$\text{NTU} = -\ln(1 - \epsilon)$

[!TIP] The $C_r = 0$ Shortcut on the PE Exam:
Whenever an exam question involves steam condensation or liquid vaporization, flow arrangement is completely irrelevant! Counterflow, parallel flow, and shell-and-tube all yield the identical effectiveness: $\epsilon = 1 - e^{-\text{NTU}}$.

4. Summary Table of Effectiveness-NTU Relations across Flow Configurations

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5. Systematic Exchanger Rating Algorithm (Step-by-Step)

To rate an existing heat exchanger on the PE Chemical exam, follow this rigorous 7-step sequence:

[Step 1: Calculate C_h & C_c] ---> [Step 2: Find C_min, C_max, C_r] ---> [Step 3: Compute Q_max]
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[Step 6: Compute Actual Q]    <--- [Step 5: Calculate epsilon]     <--- [Step 4: Compute NTU]
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            v
[Step 7: Solve Outlet Temps (T_h,out & T_c,out)]

1. Compute heat capacity rates: $C_h = \dot{m}h C{p,h}$ and $C_c = \dot{m}c C{p,c}$.
2. Identify extremes: Determine $C_{min}, C_{max}$, and compute $C_r = C_{min} / C_{max}$.
3. Calculate maximum heat duty: $Q_{max} = C_{min} (T_{h,in} - T_{c,in})$.
4. Evaluate Number of Transfer Units: $\text{NTU} = U A / C_{min}$.
5. Evaluate effectiveness $\epsilon$ using the analytical relation corresponding to the exchanger flow arrangement.
6. Determine operating heat duty: $Q = \epsilon \cdot Q_{max}$.
7. Solve for stream outlet temperatures using straightforward energy balances: $T_{h,out} = T_{h,in} - \frac{Q}{C_h}$ $T_{c,out} = T_{c,in} + \frac{Q}{C_c}$

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6. Comprehensive Worked Numerical Example: Rating an Industrial Effluent Heat Recovery Exchanger

Problem Statement

A chemical synthesis plant utilizes an existing countercurrent plate-and-frame heat exchanger to recover waste heat from a hot aqueous effluent stream to preheat cold boiler feedwater.

Exchanger and Stream Operating Data:

• Exchanger surface area: $A = 45.0\text{ m}^2$
• Overall heat transfer coefficient: $U = 1,250\text{ W/(m}^2\cdot\text{K)}$
• Hot Effluent Stream: enters at $T_{h,in} = 95.0^\circ\text{C}$ with mass flow rate $\dot{m}h = 12.0\text{ kg/s}$, $C{p,h} = 3.80\text{ kJ/(kg}\cdot\text{K)}$
• Cold Boiler Feedwater: enters at $T_{c,in} = 20.0^\circ\text{C}$ with mass flow rate $\dot{m}c = 8.0\text{ kg/s}$, $C{p,c} = 4.19\text{ kJ/(kg}\cdot\text{K)}$

Calculate:

1. The heat capacity rates $C_h$ and $C_c$, identifying $C_{min}$ and capacity ratio $C_r$.
2. The maximum theoretical heat transfer rate ($Q_{max}$) in $\text{kW}$.
3. The Number of Transfer Units ($\text{NTU}$).
4. The thermal effectiveness ($\epsilon$) of the countercurrent exchanger.
5. The actual steady-state operating heat transfer rate ($Q$) in $\text{kW}$.
6. The outlet temperatures of both process streams ($T_{h,out}$ and $T_{c,out}$).
7. Verify the solution by calculating $\Delta T_{lm}$ and confirming that $Q = U A \Delta T_{lm}$.

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Step 1: Heat Capacity Rates and $C_r$

Hot fluid capacity rate: $C_h = \dot{m}_h C_{p,h} = (12.0\text{ kg/s}) \times (3.80\text{ kJ/(kg}\cdot\text{K)}) = \mathbf{45.60\text{ kW/K}} = 45,600\text{ W/K}$

Cold fluid capacity rate: $C_c = \dot{m}_c C_{p,c} = (8.0\text{ kg/s}) \times (4.19\text{ kJ/(kg}\cdot\text{K)}) = \mathbf{33.52\text{ kW/K}} = 33,520\text{ W/K}$

Comparing rates: $C_{min} = C_c = \mathbf{33.520\text{ kW/K}} = \mathbf{33,520\text{ W/K}}$ $C_{max} = C_h = \mathbf{45.600\text{ kW/K}} = \mathbf{45,600\text{ W/K}}$

Heat capacity ratio: $C_r = \frac{C_{min}}{C_{max}} = \frac{33,520\text{ W/K}}{45,600\text{ W/K}} = \mathbf{0.73509}$

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Step 2: Maximum Theoretical Heat Duty

$Q_{max} = C_{min} (T_{h,in} - T_{c,in})$ $Q_{max} = (33.520\text{ kW/K}) \times (95.0^\circ\text{C} - 20.0^\circ\text{C}) = 33.520 \times 75.0 = \mathbf{2,514.0\text{ kW}}$

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Step 3: Number of Transfer Units ($\text{NTU}$)

$\text{NTU} = \frac{U \cdot A}{C_{min}} = \frac{(1,250\text{ W/(m}^2\cdot\text{K)}) \times (45.0\text{ m}^2)}{33,520\text{ W/K}} = \frac{56,250\text{ W/K}}{33,520\text{ W/K}} = \mathbf{1.6781}$

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Step 4: Exchanger Thermal Effectiveness ($\epsilon$)

For pure countercurrent flow with $C_r = 0.73509 < 1.0$:

$\epsilon = \frac{1 - \exp\left[-\text{NTU}(1 - C_r)\right]}{1 - C_r \exp\left[-\text{NTU}(1 - C_r)\right]}$

Evaluate exponent argument: $\text{NTU}(1 - C_r) = 1.6781 \times (1 - 0.73509) = 1.6781 \times 0.26491 = \mathbf{0.44455}$ $\exp(-0.44455) = \mathbf{0.64111}$

Numerator: $\text{Num} = 1 - 0.64111 = \mathbf{0.35889}$

Denominator: $\text{Den} = 1 - C_r \exp(-0.44455) = 1 - (0.73509 \times 0.64111) = 1 - 0.47127 = \mathbf{0.52873}$

Effectiveness: $\epsilon = \frac{0.35889}{0.52873} = \mathbf{0.67878} \approx \mathbf{0.6788} \quad (67.88\%)$

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Step 5: Actual Operating Heat Duty ($Q$)

$Q = \epsilon \cdot Q_{max} = 0.67878 \times 2,514.0\text{ kW} = \mathbf{1,706.45\text{ kW}} \approx \mathbf{1,706.5\text{ kW}}$

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Step 6: Stream Outlet Temperatures

Hot effluent exit temperature: $T_{h,out} = T_{h,in} - \frac{Q}{C_h} = 95.0^\circ\text{C} - \frac{1,706.45\text{ kW}}{45.60\text{ kW/K}} = 95.0 - 37.42 = \mathbf{57.58^\circ\text{C}}$

Cold boiler feedwater exit temperature: $T_{c,out} = T_{c,in} + \frac{Q}{C_c} = 20.0^\circ\text{C} + \frac{1,706.45\text{ kW}}{33.52\text{ kW/K}} = 20.0 + 50.91 = \mathbf{70.91^\circ\text{C}}$

Notice that $T_{c,out} (70.91^\circ\text{C}) > T_{h,out} (57.58^\circ\text{C})$—a significant temperature cross of $13.33^\circ\text{C}$, which is fully achievable because plate-and-frame exchangers operate in pure countercurrent flow!

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Step 7: LMTD Cross-Verification

Terminal temperature differences: $\Delta T_1 = T_{h,in} - T_{c,out} = 95.0^\circ\text{C} - 70.91^\circ\text{C} = 24.09^\circ\text{C}$ $\Delta T_2 = T_{h,out} - T_{c,in} = 57.58^\circ\text{C} - 20.0^\circ\text{C} = 37.58^\circ\text{C}$

$\Delta T_{lm} = \frac{37.58 - 24.09}{\ln(37.58 / 24.09)} = \frac{13.49}{\ln(1.5600)} = \frac{13.49}{0.44469} = \mathbf{30.336^\circ\text{C}}$

Recalculating heat duty via LMTD ($F = 1.00$ for countercurrent): $Q = U \cdot A \cdot \Delta T_{lm} = (1,250\text{ W/(m}^2\cdot\text{K)}) \times (45.0\text{ m}^2) \times (30.336\text{ K})$ $Q = 56,250 \times 30.336 = 1,706,400\text{ W} = \mathbf{1,706.4\text{ kW}}$

(The calculation closes with $100.00%$ precision, proving exact analytical consistency between the two methods).

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7. Critical PE Exam Traps & Pitfalls

Trap 1: Calculating $Q_{max}$ Using $C_{max}$ Instead of $C_{min}$
A catastrophic exam error is evaluating $Q_{max} = C_{max} (T_{h,in} - T_{c,in})$. Doing so violates the First Law of Thermodynamics: if you transfer that much heat, the temperature change of the $C_{min}$ stream would exceed $(T_{h,in} - T_{c,in})$, meaning a cold stream would exit hotter than the entering hot stream could ever supply. Always use $C_{min}$ to establish $Q_{max}$.

Trap 2: Neglecting the $C_r = 0$ Simplification in Phase Change Operations
When an exam question specifies a boiler, evaporator, or condenser, do not attempt to calculate the mass flow rate times liquid heat capacity for the phase-changing stream! Because latent heat transfer occurs at constant temperature, $C_{max} \to \infty$ and $C_r = 0$. The effectiveness collapses immediately to $\epsilon = 1 - \exp(-\text{NTU})$ for all configurations.

Trap 3: The Economic Law of Diminishing Returns for $\text{NTU} > 3.0$
Heat exchanger surface area scales linearly with $\text{NTU}$ ($A = \text{NTU} \cdot C_{min} / U$), but effectiveness scales asymptotically ($,1 - e^{-\text{NTU}}$). For $\text{NTU} = 1.0$, $\epsilon \approx 63%$; for $\text{NTU} = 3.0$, $\epsilon \approx 95%$. Increasing $\text{NTU}$ from $3.0$ to $6.0$ doubles the capital cost of the heat exchanger but yields less than $4%$ in additional heat recovery. On process optimization exam questions, sizing an exchanger for $\text{NTU} > 3.5$ is rarely economically justified.