Introduction & Context
The calculation of the heat exchanger surface area for a bulk pasteurizer is a critical task in dairy and food process engineering. Specifically, the regeneration section utilizes the heat from the pasteurized product to pre‑heat the incoming raw product, significantly reducing the energy demand of the system. Accurate heat exchanger area sizing is essential to ensure the product reaches the required thermal treatment temperatures while maintaining economic efficiency. This methodology is typically employed during the preliminary design phase of Plate Heat Exchangers (PHE) to determine the number of plates required to achieve the desired thermal recovery.
Methodology & Formulas
The sizing follows a systematic approach based on energy balance and heat transfer principles. First, the mass flow rate is derived from the volumetric flow and the density of the fluid:
\[ \dot{m} = \frac{\dot{V} \cdot \rho}{3600} \]
where \(\dot{V}\) is the volumetric flow rate (L h⁻¹) and \(\rho\) the density (kg L⁻¹). The heat duty \(\dot{Q}\) required for the regeneration section is calculated from the temperature rise of the cold stream:
\[ \dot{Q} = \dot{m}_{\text{c}} \cdot C_{p} \cdot (T_{c,o} - T_{c,i}) \]
For a balanced energy system the hot‑stream duty must match; if the mass flow rates differ, the hot‑side flow must be adjusted accordingly. To determine the driving force for heat transfer, the Logarithmic Mean Temperature Difference (LMTD) is calculated using the temperature differences at both ends of the counter‑current exchanger:
\[ \Delta T_{1} = T_{h,i} - T_{c,o} \]
\[ \Delta T_{2} = T_{h,o} - T_{c,i} \]
\[ \mathrm{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln\!\left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)} \]
Finally, the required heat transfer area \(A_{\text{req}}\) is obtained from the overall heat transfer coefficient \(U\), and a safety factor \(S_{f}\) is applied to account for fouling over time:
\[ A_{\text{req}} = \frac{\dot{Q}}{U \cdot \mathrm{LMTD}} \]
\[ A_{\text{design}} = A_{\text{req}} \cdot S_{f} \]
| Parameter |
Condition/Threshold |
Engineering Significance |
| Pinch Point |
\(\Delta T_{1},\ \Delta T_{2} \ge 2.0\ \text{K}\) |
Prevents excessive area requirements and thermal instability. |
| Flow Regime |
\(\mathrm{Re} > 2000\) |
Ensures turbulent flow for effective heat transfer. |
| Velocity |
\(0.3\ \text{m s}^{-1} \le v \le 1.0\ \text{m s}^{-1}\) |
Balances heat transfer efficiency against pressure drop. |
| Correction Factor |
\(F \ge 0.8\) |
Validates the use of the counter‑current LMTD model. |
Worked Example: Bulk Pasteurizer Regeneration Section Heat Exchanger Area
Scenario: A bulk pasteurizer's regeneration section uses counter‑current flow to pre‑heat raw milk with hot pasteurized milk. The area must be sized to achieve the required thermal performance while accounting for fouling. The hot‑side flow is slightly higher than the cold‑side flow because of a small dilution stream, and the temperatures satisfy a closed energy balance.
Knowns (Input Parameters):
- Raw milk (cold side) inlet temperature: \(T_{c,i} = 4.0\ ^\circ\text{C}\)
- Raw milk outlet temperature: \(T_{c,o} = 70.0\ ^\circ\text{C}\)
- Pasteurized milk (hot side) inlet temperature: \(T_{h,i} = 72.0\ ^\circ\text{C}\)
- Pasteurized milk outlet temperature: \(T_{h,o} = 10.0\ ^\circ\text{C}\)
- Cold‑side volumetric flow rate: \(\dot{V}_{\text{c}} = 5000.0\ \text{L h}^{-1}\)
- Hot‑side volumetric flow rate: \(\dot{V}_{\text{h}} = 5323.0\ \text{L h}^{-1}\)
- Milk density: \(\rho = 1.03\ \text{kg L}^{-1}\)
- Specific heat of milk: \(C_{p} = 3930.0\ \text{J kg}^{-1}\text{ K}^{-1}\)
- Overall heat transfer coefficient (design): \(U_{\text{des}} = 2000.0\ \text{W m}^{-2}\text{ K}^{-1}\)
- Fouling safety factor: \(SF = 1.2\)
- Minimum approach temperature: \(\Delta T_{\text{min}} = 2.0\ \text{K}\)
- Calculate mass flow rates.
Cold side:
\[
\dot{m}_{\text{c}} = \frac{\dot{V}_{\text{c}} \cdot \rho}{3600} = \frac{5000.0 \cdot 1.03}{3600} = 1.431\ \text{kg s}^{-1}
\]
Hot side:
\[
\dot{m}_{\text{h}} = \frac{\dot{V}_{\text{h}} \cdot \rho}{3600} = \frac{5323.0 \cdot 1.03}{3600} = 1.523\ \text{kg s}^{-1}
\]
- Compute the heat duty and verify energy balance.
Cold stream gain:
\[
\dot{Q}_{\text{c}} = \dot{m}_{\text{c}} \cdot C_{p} \cdot (T_{c,o} - T_{c,i}) = 1.431 \cdot 3930.0 \cdot (70.0 - 4.0) = 371\,058\ \text{W}
\]
Hot stream loss:
\[
\dot{Q}_{\text{h}} = \dot{m}_{\text{h}} \cdot C_{p} \cdot (T_{h,i} - T_{h,o}) = 1.523 \cdot 3930.0 \cdot (72.0 - 10.0) = 371\,058\ \text{W}
\]
The duties match; the energy balance is closed.
- Determine the temperature differences at each end.
\[
\Delta T_{1} = T_{h,i} - T_{c,o} = 72.0 - 70.0 = 2.0\ \text{K}
\]
\[
\Delta T_{2} = T_{h,o} - T_{c,i} = 10.0 - 4.0 = 6.0\ \text{K}
\]
Both \(\Delta T_{1}\) and \(\Delta T_{2}\) meet the minimum approach criterion (\(\ge 2.0\ \text{K}\)).
- Calculate the log mean temperature difference (LMTD) for counter‑current flow.
\[
\mathrm{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln\!\left(\Delta T_{1} / \Delta T_{2}\right)} = \frac{2.0 - 6.0}{\ln(2.0 / 6.0)} = 3.641\ \text{K}
\]
- Compute the required heat transfer area (pure counter‑flow, \(F = 1\)).
\[
A_{\text{req}} = \frac{\dot{Q}_{\text{c}}}{U_{\text{des}} \cdot \mathrm{LMTD}} = \frac{371\,058}{2000.0 \cdot 3.641} = 50.956\ \text{m}^{2}
\]
- Apply the safety factor to obtain the design area.
\[
A_{\text{des}} = A_{\text{req}} \cdot SF = 50.956 \cdot 1.2 = 61.147\ \text{m}^{2}
\]
Final Answer: The regeneration section requires a minimum heat transfer area of 50.956 m²; with a 20 % fouling margin the design area is 61.147 m².
