🏗️ Double-Pipe Exchanger Sizing
Size a concentric double-pipe heat exchanger. Input convection coefficients, pipe diameters, and fouling factors to calculate required clean and fouled lengths, areas, and overall coefficients.
Calculation Domain Inputs
Specify design temperatures, convective heat transfer coefficients (HTC), fouling factors, and concentric tube diameters to determine required heat exchanger size:
- Fluid Streams: Hot and cold flow rates ($\dot{m}$), heat capacities ($C_p$), and temperatures.
- Heat Transfer Coefficients ($h_i, h_o$): Tube-side and annulus-side convective properties.
- Fouling ($R_{fi}, R_{fo}$): Thermal resistance layers built up on internal/external surfaces.
- Pipes Geometry: Tube internal/external diameters ($D_{i,in}, D_{i,out}$) and shell internal diameter ($D_{o,in}$).
📝 Sizing Inputs
Thermal Sizing Equations:
Q = U_fouled × A_fouled × LMTD
L = A_fouled / (π × D_tube,out)
Overall Resistance:
1/U_fouled = 1/h_o + R_fo + ln(D_out/D_in)·D_out/(2·k_w) + R_fi·(D_out/D_in) + 1/h_i·(D_out/D_in)
Overdesign Percentage:
% Overdesign = (A_fouled - A_clean) / A_clean × 100
Q = U_fouled × A_fouled × LMTD
L = A_fouled / (π × D_tube,out)
Overall Resistance:
1/U_fouled = 1/h_o + R_fo + ln(D_out/D_in)·D_out/(2·k_w) + R_fi·(D_out/D_in) + 1/h_i·(D_out/D_in)
Overdesign Percentage:
% Overdesign = (A_fouled - A_clean) / A_clean × 100
📊 Results & Sizing Details
Configure the inputs and click "Size Double-Pipe Exchanger" to see calculation results.
ℹ️ About Double-Pipe Heat Exchangers
A double-pipe heat exchanger consists of one pipe concentrically placed inside another pipe of larger diameter. One fluid flows through the inner pipe, while the other flows through the annular space between them.
Key Sizing Principles:
• **Clean vs Fouled Condition**: Over time, scale and deposit layers build up on pipe surfaces. This fouling resistance decreases heat transfer rate. Designing with fouling factors ($R_f$) ensures the exchanger performs adequately over its operating life.
• **Convection Coefficients ($h_i, h_o$)**: The convective resistance depends heavily on flow velocities, fluid viscosities, and turbulence. Higher coefficients reduce required size.
• **Log Mean Temperature Difference (LMTD)**: Sizing calculations use the LMTD based on temperature profiles. Counter-flow offers higher LMTD, resulting in shorter pipe lengths.
A double-pipe heat exchanger consists of one pipe concentrically placed inside another pipe of larger diameter. One fluid flows through the inner pipe, while the other flows through the annular space between them.
Key Sizing Principles:
• **Clean vs Fouled Condition**: Over time, scale and deposit layers build up on pipe surfaces. This fouling resistance decreases heat transfer rate. Designing with fouling factors ($R_f$) ensures the exchanger performs adequately over its operating life.
• **Convection Coefficients ($h_i, h_o$)**: The convective resistance depends heavily on flow velocities, fluid viscosities, and turbulence. Higher coefficients reduce required size.
• **Log Mean Temperature Difference (LMTD)**: Sizing calculations use the LMTD based on temperature profiles. Counter-flow offers higher LMTD, resulting in shorter pipe lengths.
📘 Calculation Methodology
Mathematical Model & Theory
Double-pipe heat exchangers (hairpin) are sized by calculating tube-side and annulus-side heat transfer coefficients and incorporating fouling resistances. The design heat transfer coefficient is:
$$\frac{1}{U_{fouled}} = \frac{1}{h_o} + R_{fo} + \frac{D_{o,out} \ln(D_{o,out}/D_{o,in})}{2 k_w} + \frac{D_{o,out}}{D_{o,in}} R_{fi} + \frac{D_{o,out}}{D_{o,in} h_i}$$
$$A_{req} = \frac{Q}{U_{fouled} \Delta T_{lm}}, \quad L = \frac{A_{req}}{\pi D_{o,out}}$$
Assumptions & Boundary Conditions:
- Steady-state operation (constant inlet temperatures and flow rates over time).
- Negligible heat loss to surroundings (perfectly insulated outer shell).
- Constant fluid physical properties (density, specific heat, thermal conductivity, viscosity).
- One-dimensional axial fluid flow and temperature variations.
- Uniform overall heat transfer coefficient ($U$) and fouling resistances along the heat exchanger length.
Academic References:
- Incropera, F. P., Dewitt, D. P., Bergman, T. L., & Lavine, A. S. (2011). Fundamentals of Heat and Mass Transfer (7th ed.). John Wiley & Sons.
- Kakaç, S., Liu, H., & Pramuanjaroenkij, A. (2012). Heat Exchangers: Selection, Rating, and Thermal Design (3rd ed.). CRC Press.
Worked Engineering Example
Problem Statement:
Determine the required length of a double-pipe heat exchanger cooling hot organic fluid ($Q = 20,000$ W, $\Delta T_{lm} = 40$ K). Outer pipe outer diameter $D_o = 60$ mm, inner pipe coefficients: $h_i = 1000$ W/m²·K (tube side), $h_o = 800$ W/m²·K (shell side). Fouling factors: $R_{fi} = R_{fo} = 0.0002$ m²·K/W. Neglect wall conduction resistance.
Step-by-step Solution:
1. Calculate design $U$ (neglecting tube thickness for approximation):
$$\frac{1}{U_f} = \frac{1}{1000} + \frac{1}{800} + 0.0002 + 0.0002 = 0.001 + 0.00125 + 0.0004 = 0.00265 \text{ m}^2\text{K/W}$$ $$U_f = 1 / 0.00265 = 377.36 \text{ W/m}^2\text{K}$$ 2. Calculate required surface area $A$:
$$A = \frac{Q}{U_f \Delta T_{lm}} = \frac{20,000}{377.36 \times 40} = 1.325 \text{ m}^2$$ 3. Calculate required pipe length $L$:
$$L = \frac{A}{\pi D_o} = \frac{1.325}{\pi \times 0.060} = 7.03 \text{ m}$$
Final Result:
The required length is 7.03 meters.
Determine the required length of a double-pipe heat exchanger cooling hot organic fluid ($Q = 20,000$ W, $\Delta T_{lm} = 40$ K). Outer pipe outer diameter $D_o = 60$ mm, inner pipe coefficients: $h_i = 1000$ W/m²·K (tube side), $h_o = 800$ W/m²·K (shell side). Fouling factors: $R_{fi} = R_{fo} = 0.0002$ m²·K/W. Neglect wall conduction resistance.
Step-by-step Solution:
1. Calculate design $U$ (neglecting tube thickness for approximation):
$$\frac{1}{U_f} = \frac{1}{1000} + \frac{1}{800} + 0.0002 + 0.0002 = 0.001 + 0.00125 + 0.0004 = 0.00265 \text{ m}^2\text{K/W}$$ $$U_f = 1 / 0.00265 = 377.36 \text{ W/m}^2\text{K}$$ 2. Calculate required surface area $A$:
$$A = \frac{Q}{U_f \Delta T_{lm}} = \frac{20,000}{377.36 \times 40} = 1.325 \text{ m}^2$$ 3. Calculate required pipe length $L$:
$$L = \frac{A}{\pi D_o} = \frac{1.325}{\pi \times 0.060} = 7.03 \text{ m}$$
Final Result:
The required length is 7.03 meters.