Engineering task and calculation objective
This module performs the thermal and hydraulic design of plate heat exchangers according to the VDI Heat Atlas (VDI-Wärmeatlas). From the mass or volume flows, inlet and outlet temperatures and the properties of both media, it determines the heat duty, the heat transfer coefficients of both sides, the overall heat transfer coefficient, the log mean temperature difference with its correction factor, and the required heat transfer area. The performance factor shows at a glance whether a selected unit fulfills the task. Fouling allowances and the viscosity at wall temperature are taken into account.
Calculating a plate heat exchanger is a task found in almost every industry: in building services and district heating, in breweries and dairies, in the chemical industry and in refrigeration. The range of media extends from water, brines and thermal oils to acids, alkalis and refrigerants. Compared with shell-and-tube units, corrugated plates provide high turbulence even at low Reynolds numbers, large specific transfer areas and a compact design — sizing to the VDI Heat Atlas ensures that heat duty and pressure drop match.
Standard and calculation basis: VDI Wärmetlas
Calculation workflow
- Set up the heat balance: From the mass flow, specific heat capacity and temperature change of one side follows the heat duty; via the balance, the missing quantity of the opposite side (outlet temperature or mass flow) is determined. Both sides must deliver the same heat duty.
- Evaluate properties at the mean temperature: Density, heat capacity, thermal conductivity, viscosity and Prandtl number of both media are determined at their respective mean temperatures; in addition, the dynamic viscosity at wall temperature is needed to correct the heat transfer for temperature-dependent viscosity.
- Calculate the heat transfer coefficients of the plate channels: From the flow velocities in the plate gaps, the hydraulic diameter and the chevron angle of the plates, the heat transfer coefficients of both sides are determined using the correlations of the VDI Heat Atlas.
- Form the overall heat transfer coefficient including fouling: The individual resistances — heat transfer on both sides, conduction through the plate wall and the fouling resistances — are combined into the overall heat transfer coefficient.
- Determine the driving temperature difference and area: From the four inlet and outlet temperatures follows the log mean temperature difference (LMTD); the correction factor accounts for the actual flow arrangement and pass configuration. The required area results from the heat duty, the U-value and the corrected temperature difference.
- Select and evaluate the unit: For the selected heat exchanger, the available area is compared with the required one (performance factor); in parallel, the pressure drops of both sides are checked so that the design also works hydraulically.
Input quantities
| Quantity | Symbol | Unit |
|---|---|---|
| Fouling resistance | f | m²·K/W |
| Inlet temperature | ϑe | °C |
| Inlet temperature | ϑe | °C |
| Outlet temperature | ϑa | °C |
| Outlet temperature | ϑa | °C |
| Mean temperature | ϑm | °C |
| Mean temperature | ϑm | °C |
| Pressure (abs.) | p | Pa |
| Pressure (abs.) | p | Pa |
| Mass flow | m | kg/s |
| Mass flow | m | kg/s |
| Wall temperature | ϑw | °C |
| Wall temperature | ϑw | °C |
| Specific heat capacity | cp | J/(kg·K) |
| Specific heat capacity | cp | J/(kg·K) |
| Density | ρ | kg/m³ |
| Density | ρ | kg/m³ |
| Dynamic viscosity | η | mPa·s |
| Dynamic viscosity | η | mPa·s |
| Prandtl number | Pr | - |
| Prandtl number | Pr | - |
| Thermal conductivity | λ | W/(m·K) |
| Thermal conductivity | λ | W/(m·K) |
| Dynamic viscosity at ϑwall | ηw | mPa·s |
Calculated results
| Quantity | Symbol | Unit |
|---|---|---|
| Heat transfer coefficient | αi | W/(m²·K) |
| Heat transfer coefficient | α | W/(m²·K) |
| Overall heat transfer coefficient | k | W/(m²·K) |
| Log. mean temperature difference (LMTD) | ΔTlog | K (diff) |
| LMTD correction factor | F | - |
| Pressure drop in the internal coil | ΔpWi | Pa |
| Pressure drop in the external coil ΔpWe | Außenwendel | Pa |
| Total pressure drop in the internal tube | Δpi | Pa |
| Total pressure drop in the concentric annulus | Δpa | Pa |
| Flow velocity internal coil vi | Zentralrohr | m/s |
| Flow velocity external coil ve | Ringspalt | m/s |
Worked example
In a plate heat exchanger, 20,000 kg/h of hot water are to be cooled from 80 °C to 50 °C. Cooling water with an inlet temperature of 15 °C is available and may warm up to 40 °C. This worked example determines the heat duty, the required cooling water flow, the log mean temperature difference for counter-current flow and the required heat transfer area for an assumed overall heat transfer coefficient of k = 3,500 W/(m²·K) (typical water/water value including fouling; correction factor F = 1 assumed).
Given values
| Mass flow hot side | 20,000 kg/h = 5.556 kg/s |
| Temperatures hot side | 80 °C → 50 °C |
| Temperatures cold side | 15 °C → 40 °C |
| Specific heat capacity of water | cp = 4.19 kJ/(kg·K) |
| Overall heat transfer coefficient (assumed) | k = 3,500 W/(m²·K) |
| Flow arrangement | Counter-current, F = 1 |
Solution
Heat duty from the balance of the hot side
Q̇ = ṁh · cp · (ϑh,in − ϑh,out) = 5.556 · 4.19 · (80 − 50) = 698.3 kW
Required cooling water flow
ṁc = Q̇ / (cp · Δϑc) = 698.3 / (4.19 · 25) = 6.667 kg/s = 24,000 kg/h
Log mean temperature difference (counter-current)
Δϑ1 = 80 − 40 = 40 K, Δϑ2 = 50 − 15 = 35 K
Δϑm = (Δϑ1 − Δϑ2) / ln(Δϑ1/Δϑ2) = (40 − 35) / ln(40/35) = 37.4 K
Required heat transfer area
A = Q̇ / (k · F · Δϑm) = 698,300 / (3,500 · 1 · 37.4) = 5.3 m²
The overall heat transfer coefficient is specified here as an empirical value; in the module it is calculated from the plate correlations of the VDI Heat Atlas using the chevron angle, gap geometry and fouling resistances.
Result
| Heat duty | Q̇ ≈ 698 kW |
| Cooling water flow | ṁ_c = 24,000 kg/h |
| Log mean temperature difference | Δϑ_m ≈ 37.4 K |
| Required area | A ≈ 5.3 m² |
All values are illustrative. The applicable standard and project-specific boundary conditions remain authoritative.
Frequently asked questions
Why do plate heat exchangers achieve such high overall heat transfer coefficients?
The chevron-corrugated plates enforce a strongly swirled flow even at Reynolds numbers of a few hundred; at the same time, the gap widths are small and the wall thicknesses low (typically 0.4 to 0.8 mm). For water/water applications, U-values of 3,000 to 7,000 W/(m²·K) are common — several times what shell-and-tube units achieve. The price is a higher specific pressure drop, which is why heat duty and pumping power must always be evaluated together.
What is the correction factor of the log mean temperature difference for?
The LMTD is strictly valid only for pure counter-current or co-current flow. Real plate units deviate from this as soon as multiple passes, unequal channel numbers or end-plate effects occur. The correction factor (≤ 1) reduces the driving temperature difference accordingly. If it falls significantly below about 0.75, the chosen arrangement is thermodynamically unfavorable and should be changed — for instance by a different pass configuration.
How do I handle fouling in plate heat exchangers?
Fouling resistances are included as allowances in the overall heat transfer resistance. Important: the fouling resistances tabulated for shell-and-tube exchangers are usually too high for plate units, because the high wall shear stress partly removes deposits by itself; excessive allowances lead to unnecessarily large units that exceed their target outlet temperatures in the clean state. Reduced values or a percentage area margin are common practice.
What are the application limits of gasketed plate heat exchangers?
The limiting factor is the elastomer gaskets: depending on the material, about 150 to 180 °C and 16 to 25 bar are the usual upper limits. For higher pressures and temperatures, brazed or fully welded plate units come into consideration. Media heavily laden with solids or fibers are also critical, since the narrow gaps can clog; wide-gap plates or a change to other designs help here.