9.2 Heat Transfer Coefficients, Fouling, and Overall U

9.2 Heat Transfer Coefficients, Fouling, and Overall U

In process plants, heat transfer rarely occurs through a single, isolated material layer. More commonly, heat is transferred from a hot fluid through a solid wall (which may be multi-layered or insulated) and then to a cold fluid. To analyze such composite systems, chemical engineers utilize local and overall heat transfer coefficients along with the thermal resistance network method. This section covers the formulation of overall heat transfer coefficients ($U$), series and parallel thermal resistances, cylindrical geometries, and the impacts of fouling factors.

Convection Heat Transfer Coefficients

The convective heat transfer rate between a solid surface and a fluid is governed by Newton's Law of Cooling, written as $\dot{Q} = h A (T_s - T_\infty)$. The term $h$ is the convective heat transfer coefficient.

• Local convective coefficient ($h_{local}$): Represents the heat transfer rate at a specific point on the surface, which varies due to the development of velocity and thermal boundary layers.
• Average convective coefficient ($h$): The integrated average coefficient over the entire heat transfer area, used for macroscopic energy balances.

Thermal Resistance Network

The thermal resistance analogy is a powerful framework modeled after electrical circuits. The temperature difference ($\Delta T$) acts as the driving potential (voltage), the heat transfer rate ($\dot{Q}$) represents the thermal current (electric current), and the thermal resistance ($R_{th}$) represents the resistance.

$$\dot{Q} = \frac{\Delta T}{R_{total}}$$

Depending on the system configuration, thermal resistances can combine in series, parallel, or a combination of both.

Series Resistances

For a series heat transfer path (such as a composite plane wall), the heat must pass through each layer sequentially. The total thermal resistance is the sum of the individual resistances:

$$R_{total} = \sum R_i = R_1 + R_2 + R_3 + \dots$$

Under steady-state conditions, the rate of heat transfer is constant through all layers, allowing interface temperatures to be solved by equating heat rates across individual barriers.

Parallel Resistances

For a parallel heat transfer path (such as a wall with structural studs or different materials side-by-side), the heat can flow through multiple pathways simultaneously. The overall thermal resistance is combined reciprocally:

$$\frac{1}{R_{total}} = \sum \frac{1}{R_i} \implies R_{total} = \left( \frac{1}{R_1} + \frac{1}{R_2} + \dots \right)^{-1}$$

Expressions for Common Thermal Resistances

1. Conduction Resistance (Plane Wall): $$R_{cond} = \frac{L}{k A}$$
2. Conduction Resistance (Cylindrical Wall): $$R_{cyl} = \frac{\ln(r_o/r_i)}{2 \pi k L}$$
3. Convection Resistance: $$R_{conv} = \frac{1}{h A}$$

Overall Heat Transfer Coefficient (U)

The overall heat transfer coefficient ($U$) is defined as the reciprocal of the total thermal resistance scaled by the heat transfer area:

$$\dot{Q} = U A \Delta T_{overall}$$ $$U A = \frac{1}{R_{total}} \implies U = \frac{1}{A R_{total}}$$

For flat walls where the area ($A$) is constant across all layers, $U$ is independent of position. For a flat plate with convection on both sides and $N$ solid layers in series, the overall coefficient is:

$$\frac{1}{U} = \frac{1}{h_i} + \sum_{j=1}^N \frac{L_j}{k_j} + \frac{1}{h_o}$$

Cylindrical Geometries and Area-Dependent Overall U

In radial coordinates (e.g., pipes and tubes), the heat transfer area increases with radius ($A = 2\pi r L$). Consequently, the overall heat transfer coefficient must be defined with respect to a specific reference area—typically the inner surface area ($A_i$) or the outer surface area ($A_o$).

Because the total heat transfer rate $\dot{Q}$ is constant, the relation is:

$$\dot{Q} = U_i A_i \Delta T = U_o A_o \Delta T \implies U_i A_i = U_o A_o = \frac{1}{R_{total}}$$

Consider a pipe with inner radius $r_i$, outer radius $r_o$, thermal conductivity $k$, length $L$, and convective coefficients $h_i$ (internal) and $h_o$ (external). The total thermal resistance is:

$$R_{total} = R_{conv, i} + R_{cond, pipe} + R_{conv, o} = \frac{1}{h_i A_i} + \frac{\ln(r_o/r_i)}{2\pi k L} + \frac{1}{h_o A_o}$$

Thus, the overall heat transfer coefficient based on the outer area $A_o = 2\pi r_o L$ is:

$$\frac{1}{U_o} = A_o R_{total} = \frac{A_o}{h_i A_i} + \frac{A_o \ln(r_o/r_i)}{2\pi k L} + \frac{A_o}{h_o A_o}$$ $$\frac{1}{U_o} = \frac{r_o}{r_i h_i} + \frac{r_o \ln(r_o/r_i)}{k} + \frac{1}{h_o}$$

Similarly, based on the inner area $A_i = 2\pi r_i L$:

$$\frac{1}{U_i} = A_i R_{total} = \frac{1}{h_i} + \frac{r_i \ln(r_o/r_i)}{k} + \frac{r_i}{r_o h_o}$$

Fouling Factors

Over time, heat exchanger surfaces accumulate deposits such as mineral scale, rust, soot, or biological slime. This layer of deposits introduces an additional thermal resistance, known as the fouling resistance.

The fouling resistance is expressed using a fouling factor ($R_f$), which has units of $\text{m}^2\cdot \text{K/W}$ (or $\text{h}\cdot \text{ft}^2\cdot \text{°F/Btu}$):

$$R_{fouling} = \frac{R_f}{A}$$

When fouling occurs on both the inner and outer surfaces of a tube, these resistances are added in series to the overall resistance network:

$$R_{total} = \frac{1}{h_i A_i} + \frac{R_{f,i}}{A_i} + \frac{\ln(r_o/r_i)}{2\pi k L} + \frac{R_{f,o}}{A_o} + \frac{1}{h_o A_o}$$

Based on the outer area $A_o$, the dirty overall heat transfer coefficient ($U_{o,dirty}$) is related to the clean overall coefficient ($U_{o,clean}$) by:

$$\frac{1}{U_{o,dirty}} = \frac{1}{U_{o,clean}} + R_{f,i}\left(\frac{r_o}{r_i}\right) + R_{f,o}$$

Worked Examples

Worked Example 1: Composite Wall Interface Temperatures

A furnace wall is constructed of $0.12 \text{ m}$ of firebrick ($k_1 = 1.4 \text{ W/(m}\cdot \text{K)}$) and $0.08 \text{ m}$ of insulating brick ($k_2 = 0.2 \text{ W/(m}\cdot \text{K)}$). The hot gas inside the furnace is at $900^\circ\text{C}$ with a convective coefficient of $h_i = 30 \text{ W/(m}^2\cdot \text{K)}$. The ambient air outside is at $30^\circ\text{C}$ with a convective coefficient of $h_o = 10 \text{ W/(m}^2\cdot \text{K)}$. Calculate the steady-state heat loss per unit area ($q''$) and the temperature at the interface between the two bricks.

Step 1: Calculate the individual thermal resistances per unit area ($R'' = R \cdot A$).

• Inner convection: $R''_{conv, i} = \frac{1}{h_i} = \frac{1}{30} \approx 0.0333 \text{ m}^2\cdot \text{K/W}$
• Firebrick conduction: $R''_{cond, 1} = \frac{L_1}{k_1} = \frac{0.12}{1.4} \approx 0.0857 \text{ m}^2\cdot \text{K/W}$
• Insulation conduction: $R''_{cond, 2} = \frac{L_2}{k_2} = \frac{0.08}{0.2} = 0.4000 \text{ m}^2\cdot \text{K/W}$
• Outer convection: $R''_{conv, o} = \frac{1}{h_o} = \frac{1}{10} = 0.1000 \text{ m}^2\cdot \text{K/W}$

Step 2: Calculate the total thermal resistance per unit area.

$$R''_{total} = R''_{conv, i} + R''_{cond, 1} + R''_{cond, 2} + R''_{conv, o}$$ $$R''_{total} = 0.0333 + 0.0857 + 0.4000 + 0.1000 = 0.6190 \text{ m}^2\cdot\text{K/W}$$

Step 3: Calculate the heat flux ($q''$).

$$q'' = \frac{T_{\infty, i} - T_{\infty, o}}{R''_{total}} = \frac{900 - 30}{0.6190} \approx 1405.5 \text{ W/m}^2$$

Step 4: Solve for the interface temperature ($T_{int}$).

The temperature drop from the hot gas to the interface is:

$$T_{\infty, i} - T_{int} = q'' (R''_{conv, i} + R''_{cond, 1})$$ $$900 - T_{int} = 1405.5 \times (0.0333 + 0.0857) = 1405.5 \times 0.1190 \approx 167.3^\circ\text{C}$$ $$T_{int} = 900 - 167.3 = 732.7^\circ\text{C}$$

Worked Example 2: Overall U of a Condenser Tube with Fouling

Water flows inside a copper tube ($k_{wall} = 400 \text{ W/(m}\cdot \text{K)}$, $r_i = 9.0 \text{ mm}$, $r_o = 10.0 \text{ mm}$) while steam condenses on the outer surface. The convective heat transfer coefficients are $h_i = 4000 \text{ W/(m}^2\cdot \text{K)}$ and $h_o = 10000 \text{ W/(m}^2\cdot \text{K)}$. The inner fouling factor is $R_{f,i} = 0.0001 \text{ m}^2\cdot \text{K/W}$ and the outer fouling factor is neglected. Calculate the overall heat transfer coefficient based on the outer area ($U_o$).

Step 1: Set up the equation for $1/U_o$.

$$\frac{1}{U_o} = \frac{r_o}{r_i h_i} + R_{f,i}\left(\frac{r_o}{r_i}\right) + \frac{r_o \ln(r_o/r_i)}{k_{wall}} + \frac{1}{h_o}$$

Step 2: Substitute the known values (convert radii to meters).

• $r_i = 0.009 \text{ m}$
• $r_o = 0.010 \text{ m}$
• $r_o / r_i = 10 / 9 \approx 1.111$

$$\frac{r_o}{r_i h_i} = \frac{1.111}{4000} \approx 0.000278 \text{ m}^2\cdot\text{K/W}$$ $$R_{f,i}\left(\frac{r_o}{r_i}\right) = 0.0001 \times 1.111 \approx 0.000111 \text{ m}^2\cdot\text{K/W}$$ $$\frac{r_o \ln(r_o/r_i)}{k_{wall}} = \frac{0.010 \times \ln(1.111)}{400} = \frac{0.010 \times 0.10536}{400} \approx 2.63 \times 10^{-6} \text{ m}^2\cdot\text{K/W}$$ $$\frac{1}{h_o} = \frac{1}{10000} = 0.000100 \text{ m}^2\cdot\text{K/W}$$

Step 3: Sum the terms to find $1/U_o$.

$$\frac{1}{U_o} = 0.000278 + 0.000111 + 0.00000263 + 0.000100 \approx 0.0004916 \text{ m}^2\cdot\text{K/W}$$

Step 4: Take the reciprocal to find $U_o$.

$$U_o = \frac{1}{0.0004916} \approx 2034 \text{ W/(m}^2\cdot\text{K)}$$

The overall heat transfer coefficient based on the outer surface area is $2034 \text{ W/(m}^2\cdot \text{K)}$.