🥞 Plate Heat Exchanger

Chevron Plate Heat Exchanger (PHE) rating and sizing. Model enlargement factor, convection coefficients, and port/channel drops.

⚡ Fortran 90 Engine Double Precision (IEEE 754) ✓ ISO / ASME Validated
Plate Heat Exchanger Heat Transfer
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📅 Released Jun 2026
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Plates Pack Stack: N_plates Chevron beta (30° - 60°), Pressing b, Pitch pc

PHE Design Parameters

Configure Chevron plates stack parameters and check flows or size the stack count:

  • chevron Angle ($\beta$): Angle relative to vertical axis. Affects Nusselt and friction heavily.
  • Pressing depth ($b$): Spacing between plates forming flow channels.
  • Rating mode vs Sizing mode: Verify existing stack parameters or design the stack to fit target heat and pressure drop constraints.

📝 Configuration

Symmetric flow requires odd count (e.g. 11, 21, 31...)

🚀 Process Streams Configuration

🔴 Hot Fluid (Heating/Cooling)
🔵 Cold Fluid (Heating/Cooling)

🥞 Plates Geometry

📊 Results & Visualization

Configure parameters and click "Sizing Plate Heat Exchanger" to view channels stacks, temperature lines, and results.

📘 Calculation Methodology: Plate Heat Exchanger (PHE) Thermal & Hydraulic Sizing

Mathematical Model & Theory

Plate heat exchangers use corrugated chevron plates with angle $\beta$ to generate intense turbulence at low Reynolds numbers ($Re > 50$), maximizing Nusselt numbers while controlling pressure drops:

$$Nu = C \cdot Re^m Pr^{1/3} \left(\frac{\mu}{\mu_w}\right)^{0.14}, \quad \Delta p = 4 f \left(\frac{L}{D_h}\right) \frac{\rho V^2}{2} N_{pass}$$
$$D_h = \frac{2 b}{\Phi}, \quad q = U A \cdot LMTD \cdot F$$

Assumptions

  • Equal channel mass flow distribution across parallel plates.
  • Constant fluid properties at bulk mean temperature.

Academic References

  1. Kakac, S., Liu, H., & Pramuanjaroenkij, A.: Heat Exchangers: Selection, Rating, and Thermal Design, CRC Press.
  2. Shah, R. K., & Sekulic, D. P.: Fundamentals of Heat Exchanger Design, Wiley.

Worked Engineering Example

Problem Statement:
A PHE with 40 chevron plates transfers $q = 250\text{ kW}$ with $LMTD = 8.5^\circ\text{C}$, $F = 0.98$, and $U = 3500\text{ W/m}^2\cdot\text{K}$. Calculate required total plate heat transfer area $A$.

Step-by-step Solution:
1. Calculate effective temperature difference $\Delta T_{eff} = 8.5 \times 0.98 = 8.33^\circ\text{C}$.
2. Solve for area $A = q / (U \Delta T_{eff}) = 250,000 / (3500 \times 8.33) = 250,000 / 29,155 \approx 8.575\text{ m}^2$.
Final Result:
Required active plate surface area is $A = \mathbf{8.58\text{ m}^2}$ (approx. $0.22\text{ m}^2/\text{plate}$).