Introduction & Context
Regeneration section heat recovery efficiency is a critical performance metric in process engineering, particularly within continuous thermal processing systems such as pasteurizers and heat exchangers. In these systems, a counter-current plate heat exchanger is utilized to transfer sensible heat from a hot product stream exiting a holding section to a cold incoming product stream. By preheating the raw feed using the energy already present in the processed product, the system significantly reduces the external energy required for final heating. This calculation is essential for monitoring thermal performance, identifying fouling accumulation, and ensuring the system operates within its design specifications.
Methodology & Formulas
The efficiency of the regeneration section is determined by comparing the actual temperature rise of the cold fluid to the maximum theoretical temperature rise possible, which is limited by the inlet temperature of the hot fluid. The following formulas define the thermal performance of the system:
First, the mass flow rate (\(\dot{m}\)) is derived from the volumetric flow rate (\(\dot{V}\)) and the fluid density (\(\rho\)):
\[ \dot{m} = \frac{\dot{V} \cdot \rho}{3600} \]
The regeneration efficiency (\(\eta\)) is calculated as the ratio of the cold side temperature increase to the total available temperature gradient:
\[ \eta = \left( \frac{T_{\text{co}} - T_{\text{ci}}}{T_{\text{hi}} - T_{\text{ci}}} \right) \cdot 100 \]
The energy savings (\(Q_{\text{saved}}\)) achieved through this heat recovery process is determined by the sensible heat gain of the cold fluid:
\[ Q_{\text{saved}} = \dot{m} \cdot c_{p} \cdot (T_{\text{co}} - T_{\text{ci}}) \]
Finally, the temperature approach (\(\Delta T_{\text{app}}\)), which indicates the proximity of the system to the thermodynamic limit, is defined as:
\[ \Delta T_{\text{app}} = T_{\text{hi}} - T_{\text{co}} \]
| Parameter |
Condition/Regime |
Engineering Significance |
| Efficiency (\(\eta\)) |
85% – 95% |
Typical range for modern, well-maintained systems. |
| Efficiency (\(\eta\)) |
< 60% |
Indicates severe fouling, flow imbalance, or design mismatch. |
| Temperature Approach (\(\Delta T_{\text{app}}\)) |
< 5 °C |
Required for high-efficiency (>90%) heat recovery. |
| Temperature Approach (\(\Delta T_{\text{app}}\)) |
\(\leq\) 0 °C |
Physical impossibility; violates the Second Law of Thermodynamics. |
| Thermal Gradient |
\(T_{\text{hi}} > T_{\text{ci}}\) |
Mandatory condition for positive heat transfer. |
Worked Example: Regeneration Section Heat Recovery Efficiency
A milk pasteurisation plant uses a counter-current plate heat exchanger (PHE) for regeneration. The raw milk enters the regeneration section at a cold inlet temperature, and the pasteurised milk returns from the holding tube at a hot inlet temperature. The system operates under steady-state conditions with balanced flow. The goal is to determine the regeneration efficiency, the energy savings, and the temperature approach.
Scenario: Raw milk at 5.0 °C is preheated by hot pasteurised milk at 72.0 °C. The preheated milk leaves the cold side at 55.3 °C. The flow rate is 10 m³/h, with milk density 1030 kg/m³ and specific heat 3850 J/(kg·K).
Knowns (Input Parameters)
- Cold fluid inlet temperature, \( T_{\text{ci}} = 5.0 \; ^{\circ}\text{C} \)
- Hot fluid inlet temperature, \( T_{\text{hi}} = 72.0 \; ^{\circ}\text{C} \)
- Cold fluid outlet temperature, \( T_{\text{co}} = 55.3 \; ^{\circ}\text{C} \)
- Volumetric flow rate, \( \dot{V} = 10.0 \; \text{m}^3/\text{h} \)
- Density, \( \rho = 1030.0 \; \text{kg/m}^3 \)
- Specific heat, \( c_{p} = 3850.0 \; \text{J/(kg·K)} \)
Step-by-Step Calculation
- Compute the regeneration efficiency.
The efficiency is defined as the ratio of the actual cold‑side temperature rise to the maximum possible rise:
\[
\eta = \frac{T_{\text{co}} - T_{\text{ci}}}{T_{\text{hi}} - T_{\text{ci}}} \times 100
\]
Substituting the given values:
\[
\eta = \frac{55.3 - 5.0}{72.0 - 5.0} \times 100 = \frac{50.3}{67.0} \times 100 = 75.075 \%
\]
The efficiency is 75.075%.
- Convert the volumetric flow rate to mass flow rate.
Mass flow rate is given by:
\[
\dot{m} = \frac{\dot{V} \cdot \rho}{3600}
\]
Using the provided numbers:
\[
\dot{m} = \frac{10.0 \cdot 1030.0}{3600} = 2.861 \; \text{kg/s} \quad (\text{to three decimal places})
\]
The exact value \( \dot{m} = 10300/3600 = 2.861111\dots \; \text{kg/s} \) is used in subsequent calculations to avoid rounding error.
- Calculate the energy saved in the regeneration section.
The heat recovered is:
\[
Q_{\text{saved}} = \dot{m} \cdot c_{p} \cdot (T_{\text{co}} - T_{\text{ci}})
\]
Insert the exact mass flow and the temperatures:
\[
Q_{\text{saved}} = \frac{10.0 \times 1030.0}{3600} \cdot 3850.0 \cdot (55.3 - 5.0) = \frac{10300}{3600} \times 3850.0 \times 50.3 = 554.068 \; \text{kW}
\]
- Determine the temperature approach.
The approach temperature is the difference between the hot inlet and cold outlet:
\[
\Delta T_{\text{app}} = T_{\text{hi}} - T_{\text{co}} = 72.0 - 55.3 = 16.7 \; ^{\circ}\text{C}
\]
Final Answer
- Regeneration Efficiency: 75.075 %
- Mass Flow Rate: 2.861 kg/s (approx.)
- Energy Saved: 554.068 kW
- Temperature Approach: 16.7 °C
