9.3 Heat Exchanger Design

9.3 Heat Exchanger Design

Heat exchangers are devices that facilitate the transfer of thermal energy between two or more fluids at different temperatures. They are ubiquitous in chemical processes, ranging from simple double-pipe heat exchangers to complex shell-and-tube units. When designing or rating heat exchangers, chemical engineers rely on two key methods: the Log Mean Temperature Difference (LMTD) method and the Effectiveness-Number of Transfer Units (${\epsilon}$-NTU) method. This section covers double-pipe and shell-and-tube configurations, co-current vs. counter-current flows, and the applications of both methods using formulas consistent with the FE Reference Handbook.

Heat Exchanger Configurations

Heat exchangers are classified based on their construction and flow configuration.

Double-Pipe Heat Exchangers

The simplest design consists of two concentric pipes. One fluid flows through the inner pipe while the other flows through the annular space.

• Parallel Flow (Co-current): Both fluids enter at the same end, flow in the same direction, and leave at the same end. The outlet temperature of the cold fluid can never exceed the outlet temperature of the hot fluid.
• Counter-Current Flow: The fluids enter at opposite ends and flow in opposite directions. This maintains a more uniform temperature difference along the exchanger length. The cold fluid outlet temperature can exceed the hot fluid outlet temperature, resulting in higher thermodynamic efficiency and lower thermodynamic irreversibility (entropy generation).

Shell-and-Tube Heat Exchangers

For larger thermal loads, shell-and-tube exchangers are used. These consist of a bundle of tubes mounted inside a cylindrical shell. Baffles are installed in the shell to support the tubes, direct the shell-side fluid flow across the tube bundle to increase turbulence, and enhance the convective heat transfer coefficient.

• Pass Configurations: Described by the number of shell passes and tube passes. A "1-2 exchanger" has one shell pass and two tube passes. Multipass designs increase the fluid velocity and heat transfer rate, though they increase pressure drop.

Log Mean Temperature Difference (LMTD) Method

The heat transfer rate in a heat exchanger is expressed as:

$\dot{Q} = U A \Delta T_{lm}$

where $U$ is the overall heat transfer coefficient, $A$ is the heat transfer area, and $\Delta T_{lm}$ is the Log Mean Temperature Difference (LMTD).

The LMTD is defined as:

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

The definition of the terminal temperature differences $\Delta T_1$ and $\Delta T_2$ depends on the flow direction:

If the temperature differences at the two ends are equal ($\Delta T_1 = \Delta T_2$), the expression for LMTD is mathematically indeterminate ($0/0$). LMTD is then evaluated as the limit, which simplifies to $\Delta T_{lm} = \Delta T_1 = \Delta T_2$.

LMTD Correction Factor (F)

For multipass shell-and-tube and cross-flow heat exchangers, the flow is not purely counter-current. To account for this, a dimensionless correction factor ($F$) is introduced:

$\dot{Q} = U A F \Delta T_{lm,CF}$

where $\Delta T_{lm,CF}$ is the LMTD calculated assuming pure counter-current flow.

• $F$ is always less than or equal to $1.0$ ($F = 1.0$ for pure counter-current flow).
• $F$ is determined using charts in the FE Reference Handbook based on two dimensionless ratios:
  1. Temperature Effectiveness ($P$): $P = \frac{t_{out} - t_{in}}{T_{in} - t_{in}}$
  2. Capacity Ratio ($R$): $R = \frac{T_{in} - T_{out}}{t_{out} - t_{in}}$ (Here, uppercase $T$ represents the shell-side fluid and lowercase $t$ represents the tube-side fluid).
• If $F < 0.75$, the design is generally inefficient due to potential temperature cross, and a multi-shell-pass design should be selected.

Effectiveness-NTU (ε-NTU) Method

When the fluid outlet temperatures are unknown, the LMTD method requires iterative calculations. In such cases, the Effectiveness-NTU ($\epsilon$-NTU) method is preferred.

Key Parameters

1. Heat Capacity Rate ($C$): $C = \dot{m} C_p$ Identify $C_{min} = \min(C_h, C_c)$ and $C_{max} = \max(C_h, C_c)$.
2. Maximum Possible Heat Transfer Rate ($\dot{Q}_{max}$): $\dot{Q}_{max} = C_{min} (T_{h,in} - T_{c,in})$ This represents the heat transfer in an infinitely long counter-current exchanger.
3. Heat Exchanger Effectiveness (${\epsilon}$): $\epsilon = \frac{\dot{Q}}{\dot{Q}_{max}}$ The actual heat transfer rate is then: $\dot{Q} = \epsilon C_{min} (T_{h,in} - T_{c,in})$
4. Number of Transfer Units ($NTU$): $NTU = \frac{U A}{C_{min}}$
5. Capacity Ratio ($C_r$): $C_r = \frac{C_{min}}{C_{max}} \qquad (0 \le C_r \le 1)$

Effectiveness Relations

The effectiveness is a function of NTU and $C_r$ depending on flow configuration:

• Parallel Flow: $\epsilon = \frac{1 - \exp[-NTU(1 + C_r)]}{1 + C_r}$
• Counter-Current Flow: $\epsilon = \frac{1 - \exp[-NTU(1 - C_r)]}{1 - C_r \exp[-NTU(1 - C_r)]} \qquad (\text{for } C_r < 1)$ $\epsilon = \frac{NTU}{1 + NTU} \qquad (\text{for } C_r = 1)$
• Condensers/Boilers ($C_r = 0$): When one fluid undergoes a phase change (e.g., steam condensing or refrigerant boiling), its specific heat is effectively infinite, meaning $C_{max} \to \infty$ and $C_r = 0$. For all flow configurations: $\epsilon = 1 - \exp(-NTU)$

Worked Examples

Worked Example 1: LMTD Method for Exchanger Sizing

A double-pipe, counter-current heat exchanger is used to cool $0.5 \text{ kg/s}$ of an oil ($C_{p} = 2200 \text{ J/(kg·K)}$) from $120^\circ\text{C}$ to $70^\circ\text{C}$. Cooling water ($C_p = 4180 \text{ J/(kg·K)}$) enters at $20^\circ\text{C}$ and leaves at $55^\circ\text{C}$. The overall heat transfer coefficient is $U = 350 \text{ W/(m}^2\text{·K)}$. Calculate the required heat transfer area.

Step 1: Calculate the rate of heat transfer ($\dot{Q}$).

$\dot{Q} = \dot{m}_{oil} C_{p,oil} (T_{h,in} - T_{h,out}) = 0.5 \times 2200 \times (120 - 70) = 55,000 \text{ W}$

Step 2: Calculate the terminal temperature differences for counter-current flow.

• $\Delta T_1 = T_{h,in} - T_{c,out} = 120 - 55 = 65^\circ\text{C}$
• $\Delta T_2 = T_{h,out} - T_{c,in} = 70 - 20 = 50^\circ\text{C}$

Step 3: Calculate the LMTD ($\Delta T_{lm}$).

$\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} = \frac{65 - 50}{\ln(65 / 50)} = \frac{15}{0.2624} \approx 57.17^\circ\text{C}$

Step 4: Solve for the heat transfer area ($A$).

$\dot{Q} = U A \Delta T_{lm} \implies A = \frac{\dot{Q}}{U \Delta T_{lm}} = \frac{55,000}{350 \times 57.17} \approx 2.75 \text{ m}^2$

The required heat transfer area is $2.75 \text{ m}^2$.

Worked Example 2: ε-NTU Method for Exchanger Rating

A counter-current heat exchanger with an area of $5.0 \text{ m}^2$ and an overall heat transfer coefficient of $U = 280 \text{ W/(m}^2\text{·K)}$ is used to heat water ($C_p = 4180 \text{ J/(kg·K)}$) flowing at $0.2 \text{ kg/s}$. The heating fluid is an oil stream ($C_p = 2000 \text{ J/(kg·K)}$) flowing at $0.3 \text{ kg/s}$. The oil enters at $150^\circ\text{C}$ and the water enters at $20^\circ\text{C}$. Calculate the effectiveness of the exchanger and the heat transfer rate.

Step 1: Calculate the heat capacity rates.

• $C_c = \dot{m}c C{p,c} = 0.2 \times 4180 = 836 \text{ W/K}$
• $C_h = \dot{m}h C{p,h} = 0.3 \times 2000 = 600 \text{ W/K}$

Thus:

• $C_{min} = 600 \text{ W/K}$ (oil)
• $C_{max} = 836 \text{ W/K}$ (water)
• $C_r = \frac{C_{min}}{C_{max}} = \frac{600}{836} \approx 0.718$

Step 2: Calculate the Number of Transfer Units ($NTU$).

$NTU = \frac{U A}{C_{min}} = \frac{280 \times 5.0}{600} \approx 2.333$

Step 3: Calculate the effectiveness (${\epsilon}$) for counter-current flow.

$\epsilon = \frac{1 - \exp[-NTU(1 - C_r)]}{1 - C_r \exp[-NTU(1 - C_r)]}$ $NTU(1 - C_r) = 2.333 \times (1 - 0.718) = 2.333 \times 0.282 \approx 0.658$ $\exp[-0.658] \approx 0.518$ $\epsilon = \frac{1 - 0.518}{1 - 0.718 \times 0.518} = \frac{0.482}{1 - 0.372} = \frac{0.482}{0.628} \approx 0.768 \quad (76.8\%)$

Step 4: Calculate the actual heat transfer rate ($\dot{Q}$).

$\dot{Q}_{max} = C_{min} (T_{h,in} - T_{c,in}) = 600 \times (150 - 20) = 78,000 \text{ W}$ $\dot{Q} = \epsilon \dot{Q}_{max} = 0.768 \times 78,000 = 59,904 \text{ W} \approx 59.9 \text{ kW}$

The heat transfer rate is $59.9 \text{ kW}$.