4.4 Overall Heat Transfer Coefficients, Fouling Factors & Transient/Thermal Mass Effects

4.4 Overall Heat Transfer Coefficients, Fouling Factors & Transient/Thermal Mass Effects

In real-world HVAC equipment—such as shell-and-tube water chillers, finned-tube cooling coils, plate-and-frame heat exchangers, and boiler firetubes—heat flows across multi-component thermal barriers involving convection, conduction, surface extended fins, and fouling scale deposits. Furthermore, equipment startup, thermostat cycling, and building envelope thermal storage exhibit time-dependent transient heat transfer. Mastering the complete overall heat transfer coefficient formulation and transient lumped capacitance criteria is essential for the PE Mechanical exam.

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1. Overall Heat Transfer Coefficient Formulation ($U$)

The steady-state heat transfer rate across any composite thermal barrier is expressed in terms of the overall heat transfer coefficient ($U$) and a defined reference area $A$:

$\dot{q} = U A \Delta T = U_i A_i \Delta T = U_o A_o \Delta T = \frac{\Delta T}{R_{total}}$

Where:

• $U_i = \text{overall heat transfer coefficient based on inside surface area } A_i \quad [\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \text{ or W/(m}^2\cdot\text{K)}]$
• $U_o = \text{overall heat transfer coefficient based on outside surface area } A_o \quad [\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)} \text{ or W/(m}^2\cdot\text{K)}]$
• Because heat rate $\dot{q}$ is identical regardless of reference datum: $U_i A_i = U_o A_o = \frac{1}{R_{total}}$

+-----------------------------------------------------------------------------------------+
| COMPREHENSIVE THERMAL RESISTANCE NETWORK FOR A TUBULAR HEAT EXCHANGER                   |
+-----------------------------------------------------------------------------------------+
| T_hot o---[ 1/(h_i A_i) ]---[ R_fi/A_i ]---[ ln(r_o/r_i)/(2 pi k L) ]---[ R_fo/A_o ]---[ 1/(eta_o h_o A_o) ]---o T_cold
|           Inside            Inside          Tube Wall Conduction       Outside           Outside Finned Surface   |
|           Convection        Fouling                                    Fouling           Convection               |
+-----------------------------------------------------------------------------------------+

General Resistance Equation for Finned Tubes

Expanding the total thermal resistance network across a cylindrical tube of length $L$, inner radius $r_i$, outer radius $r_o$, with extended fins on the exterior surface:

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

Multiplying through by outside reference area $A_o$ gives the outside overall $U$-factor ($U_o$):

$\frac{1}{U_o} = \frac{1}{h_i} \left(\frac{A_o}{A_i}\right) + R_{f,i} \left(\frac{A_o}{A_i}\right) + \frac{r_o \ln(r_o / r_i)}{k} + R_{f,o} + \frac{1}{\eta_o h_o}$

Where:

• $h_i, h_o = \text{inside and outside convective heat transfer coefficients } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)})$
• $R_{f,i}, R_{f,o} = \text{inside and outside fouling factors } (\text{hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu})$
• $k = \text{tube wall material thermal conductivity } (\text{Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)})$
• $\eta_o = \text{overall surface temperature efficiency of the finned array}$

For thin-walled bare planar heat exchanger plates ($A_i \approx A_o = A$, thickness $t$, no fins):

$\frac{1}{U} = \frac{1}{h_i} + R_{f,i} + \frac{t}{k} + R_{f,o} + \frac{1}{h_o}$

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2. Extended Surfaces (Fins) & Surface Temperature Efficiency

In gas-to-liquid heat exchangers (such as hydronic cooling coils and air-cooled condensers), the air-side convective coefficient is much lower than the liquid-side coefficient ($h_{air} \approx 10-20\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$ vs $h_{water} \approx 800-1500\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$). To balance thermal resistances, thin aluminum fins are attached to the outside of the copper tubes to multiply the effective surface area ($A_o \gg A_i$).

+-----------------------------------------------------------------------------------------+
| FIN TEMPERATURE PROFILE AND EFFICIENCY                                                  |
+-----------------------------------------------------------------------------------------+
| Fin Base: Temp T_b                                                                      |
|   |                                                                                     |
|   |=========\                                                                           |
|   |          \                                                                          |
|   |           \-----> Temp drops along fin: T(x)                                        |
|   |                  Convective heat loss to surrounding air (T_inf)                    |
|   |                   \                                                                 |
|   |                    \========= Fin Tip: Temp T_L                                     |
|   +------------------------------> Fin Length L                                         |
| Fin Efficiency: eta_f = Actual Heat Transfer / Ideal Heat Transfer (if fin was at T_b)  |
+-----------------------------------------------------------------------------------------+

Fin Parameter ($m$) & Straight Fin Efficiency ($\eta_f$)

As heat conducts along a fin, convection continuously dissipates energy into the airstream, causing the fin temperature to drop from the base temperature $T_b$ toward fluid temperature $T_\infty$. For a straight rectangular fin of length $L$, thickness $t$, perimeter $P$, and cross-sectional area $A_c$:

$m = \sqrt{\frac{h P}{k A_c}} = \sqrt{\frac{h (2w + 2t)}{k (w \cdot t)}} \approx \sqrt{\frac{2 h}{k t}}$

For an adiabatic (insulated) tip fin, the fin efficiency ($\eta_f$) is:

$\eta_f = \frac{\dot{q}_{actual}}{\dot{q}_{ideal}} = \frac{\tanh(m L)}{m L}$

When accounting for convective heat loss from the tip, use the corrected fin length $L_c = L + \frac{t}{2}$:

$\eta_f = \frac{\tanh(m L_c)}{m L_c}$

Total Extended Surface Efficiency ($\eta_o$)

An HVAC coil exterior surface consists of unfinned tube area ($A_u$) at base temperature $T_b$ (efficiency $= 1.0$) and fin area ($A_f$) at efficiency $\eta_f$. The total outside area is $A_t = A_u + A_f$. The overall surface efficiency is:

$\eta_o = \frac{A_u(1.0) + A_f \eta_f}{A_t} = \frac{(A_t - A_f) + A_f \eta_f}{A_t} = 1 - \frac{A_f}{A_t}(1 - \eta_f)$

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3. Fouling Factors (Thermal Resistances)

During operation, mineral scaling (calcium carbonate), biological growth (algae/biofilms), particulate sedimentation, and corrosion deposit an insulating layer on heat exchanger surfaces.

Typical Design Fouling Factors (TEMA & AHRI Standards)

Impact on Water Chiller Performance

In a centrifugal water chiller, cooling tower water flows through condenser tubes. As scale deposits increase $R_{f,i}$:

1. Overall $U$-factor drops, requiring a higher temperature difference $\Delta T$ to reject heat.
2. Chiller condensing pressure and saturated condensing temperature ($T_{cond}$) rise.
3. Compressor lift increases, increasing compressor electrical consumption by approximately $2%$ to $3%$ per $0.0001\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$ increase in fouling resistance.

4. Transient Heat Conduction & Lumped Capacitance Method

When thermal boundary conditions change abruptly (e.g., thermostat temperature setback, hot water pipe startup, refrigerated food pull-down), temperature varies with both location and time: $T = T(x, y, z, t)$.

+-----------------------------------------------------------------------------------------+
| LUMPED CAPACITANCE VALIDITY CRITERION: BIOT NUMBER Bi < 0.10                            |
+-----------------------------------------------------------------------------------------+
| Biot Number: Bi = (h * L_c) / k_solid  where L_c = Volume / Surface Area                |
|                                                                                         |
| Case A: Bi < 0.10 (Lumped Capacitance Valid)    Case B: Bi >= 0.10 (Spatial Gradients)  |
|   Conduction resistance << Convection           Conduction resistance is significant    |
|   Solid temperature is uniform: T(t) only       T = T(x, t) requires Fourier / Heisler  |
|   +---------------------------------------+     +-------------------------------------+ |
|   | T_solid = constant across volume      |     |  T_center > T_surface (steep curve) | |
|   +---------------------------------------+     +-------------------------------------+ |
+-----------------------------------------------------------------------------------------+

The Biot Number Criterion ($\text{Bi}$)

The Biot number evaluates whether internal conduction resistance within the solid is negligible compared to external surface convection resistance:

$\text{Bi} = \frac{R_{cond}}{R_{conv}} = \frac{L_c / (k_s A_s)}{1 / (h A_s)} = \frac{h L_c}{k_s}$

Where:

• $h = \text{convective heat transfer coefficient } (\text{Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)})$
• $k_s = \text{thermal conductivity of the solid material } (\text{Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)})$
• $L_c = \frac{V}{A_s} = \text{characteristic length of the solid } (\text{ft})$:
  • Plane slab of thickness $2L$ (exposed both sides): $L_c = \frac{2L A}{2A} = L$
  • Long cylinder of radius $r_o$: $L_c = \frac{\pi r_o^2 L}{2 \pi r_o L} = \frac{r_o}{2}$
  • Sphere of radius $r_o$: $L_c = \frac{\frac{4}{3}\pi r_o^3}{4 \pi r_o^2} = \frac{r_o}{3}$

Governing Rule for Lumped Capacitance: If $\mathbf{\text{Bi} < 0.10}$, temperature gradients inside the body are negligible ($<5%$ error). The body can be treated as a single uniform "lumped" temperature $T(t)$.

Lumped Capacitance Formulation & Thermal Time Constant ($\tau$)

Performing an energy balance equating rate of internal energy change to surface convection:

$\rho V c_p \frac{dT}{dt} = -h A_s (T - T_\infty)$

Separating variables and integrating from initial temperature $T(0) = T_0$ at $t = 0$:

$\frac{T(t) - T_\infty}{T_0 - T_\infty} = \exp\left(-\frac{h A_s}{\rho V c_p} t\right) = \exp\left(-\frac{t}{\tau}\right) = e^{-t / \tau}$

Where $\tau$ is the thermal time constant of the system:

$\tau = R_{th} C_{th} = \left(\frac{1}{h A_s}\right)(\rho V c_p) = \frac{\rho V c_p}{h A_s} = \frac{\rho c_p L_c}{h} \quad [\text{hours or seconds}]$

• At $t = 1\tau$: $T(t)$ has completed $63.2%$ of its total temperature change.
• At $t = 3\tau$: $T(t)$ has completed $95.0%$ of change.
• At $t = 5\tau$: $T(t)$ has completed $99.3%$ of change (essentially steady-state).

Spatial Transient Conduction ($\text{Bi} \ge 0.10$) & Fourier Number ($\text{Fo}$)

When $\text{Bi} \ge 0.10$, internal temperature varies with spatial position, governed by the dimensionless Fourier number:

$\text{Fo} = \frac{\alpha t}{L_c^2}$

Where $\alpha = \frac{k}{\rho c_p}$ is the material's thermal diffusivity ($\text{ft}^2/\text{hr}$ or $\text{m}^2/\text{s}$). For $\text{Fo} > 0.2$, one-term approximate analytical solutions or Heisler charts provide exact temperature profiles.

Building Thermal Mass Dynamics (Decrement Factor & Time Lag)

In high-thermal-mass building envelopes (concrete, masonry, adobe):

• Decrement Factor ($f$): Ratio of the inside surface peak temperature fluctuation to the outside ambient sol-air peak temperature fluctuation ($f < 1.0$). High thermal mass damps outdoor temperature swings.
• Time Lag ($\phi$): Time delay (in hours) between the peak outdoor sol-air temperature and the resulting peak indoor cooling load, shifting peak cooling demand to late evening off-peak hours.

5. Step-by-Step Worked Example: Condenser Tube Sizing & Fouling Derating

Problem Statement

A water-cooled shell-and-tube refrigeration condenser uses copper tubes ($k = 225\text{ Btu/(hr}\cdot\text{ft}\cdot{}^\circ\text{F)}$) with outer diameter $D_o = 0.750\text{ in.} = 0.06250\text{ ft}$ and inner diameter $D_i = 0.652\text{ in.} = 0.05433\text{ ft}$. Operating parameters:

• Saturated refrigerant condensing on shell outside: $h_o = 1200.0\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$
• Cooling tower water flowing inside tubes: $h_i = 850.0\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$
• Inside tube fouling factor (cooling tower scale): $R_{f,i} = 0.00050\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$
• Outside fouling factor (clean condensing refrigerant): $R_{f,o} = 0.0$
• Bare smooth tubes (no fins, $\eta_o = 1.0$)

Calculate:

1. Ratio of outer-to-inner surface area ($A_o / A_i$)
2. Clean overall heat transfer coefficient based on outside area ($U_{o,clean}$)
3. Fouled overall heat transfer coefficient based on outside area ($U_{o,fouled}$)
4. Percentage reduction in heat transfer capacity caused by water-side tube fouling

Solution Steps

Step 1: Calculate area ratio and radii $r_o = \frac{0.06250}{2} = 0.03125\text{ ft}, \quad r_i = \frac{0.05433}{2} = 0.027165\text{ ft}$ $\frac{A_o}{A_i} = \frac{\pi D_o L}{\pi D_i L} = \frac{D_o}{D_i} = \frac{0.750\text{ in.}}{0.652\text{ in.}} = 1.1503$

Step 2: Calculate individual resistance components based on outside area $A_o$

• Inside Convection Resistance per unit $A_o$: $R_{conv,i} = \frac{1}{h_i} \left(\frac{A_o}{A_i}\right) = \frac{1}{850.0} \times 1.1503 = 0.0011765 \times 1.1503 = 0.0013533\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$
• Tube Wall Conduction Resistance per unit $A_o$: $R_{cond} = \frac{r_o \ln(r_o / r_i)}{k} = \frac{0.03125 \times \ln(0.750 / 0.652)}{225} = \frac{0.03125 \times 0.14002}{225} = 0.00001945\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$
• Outside Convection Resistance per unit $A_o$: $R_{conv,o} = \frac{1}{h_o} = \frac{1}{1200.0} = 0.0008333\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$

Step 3: Calculate clean overall $U$-factor ($U_{o,clean}$) $R_{clean,total} = R_{conv,i} + R_{cond} + R_{conv,o}$ $R_{clean,total} = 0.0013533 + 0.0000195 + 0.0008333 = 0.0022061\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$ $U_{o,clean} = \frac{1}{R_{clean,total}} = \frac{1}{0.0022061} = 453.29\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$

Step 4: Calculate fouled overall $U$-factor ($U_{o,fouled}$) Inside fouling resistance per unit outside area: $R_{f,i,o} = R_{f,i} \left(\frac{A_o}{A_i}\right) = 0.00050 \times 1.1503 = 0.0005752\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$ $R_{fouled,total} = R_{clean,total} + R_{f,i,o} = 0.0022061 + 0.0005752 = 0.0027813\text{ hr}\cdot\text{ft}^2\cdot{}^\circ\text{F/Btu}$ $U_{o,fouled} = \frac{1}{R_{fouled,total}} = \frac{1}{0.0027813} = 359.54\text{ Btu/(hr}\cdot\text{ft}^2\cdot{}^\circ\text{F)}$

Step 5: Calculate percentage reduction in heat transfer capacity $\%\text{ Reduction} = \frac{U_{o,clean} - U_{o,fouled}}{U_{o,clean}} \times 100\% = \frac{453.29 - 359.54}{453.29} \times 100\% = \frac{93.75}{453.29} \times 100\% = 20.68\%$

Water-side scale buildup of $R_{f,i} = 0.00050$ reduces the condenser heat rejection capability by over $20.7%$, demanding higher water flow rates and increasing chiller operating costs.