π Regenerative Heat Exchanger
Thermal analysis of rotary (Ljungstrom) and fixed-matrix regenerators with effectiveness-NTU and Cr* corrections.
β‘ Fortran 90 Engine
Double Precision (IEEE 754)
β ISO / ASME Validated
Heat Transfer
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Calculation Results
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Calculation Methodology
Mathematical Model & Theory
Regenerators operate by cyclically exposing a porous storage solid (matrix) to hot and cold gas streams. The thermal effectiveness is modeled using the effectiveness-NTU method, incorporating the solid matrix heat capacity ratio ($C_r^*$):
$$C_r^* = \frac{m_{matrix} \cdot c_{p,matrix} \cdot N}{\min(C_h, C_c)}$$
Where $N$ is rotation speed [Hz] (or cycle frequency). For an equivalent ideal counterflow exchanger, the effectiveness ($\epsilon_{cf}$) is:
$$\epsilon_{cf} = \frac{1 - \exp[-NTU(1 - C_r)]}{1 - C_r \exp[-NTU(1 - C_r)]}$$
The actual effectiveness ($\epsilon$) corrected for matrix capacity (under Kays & London rotary model) is:
$$\epsilon = \epsilon_{cf} \cdot \left[ 1 - \frac{1}{9 \cdot (C_r^*)^{1.93}} \right]$$
Valid for $C_r^* \ge 2.0$. For stationary matrices, valves switch at period $P \implies N=1/(2P)$.
Academic References:
- Kays, W. M., & London, A. L. (1984). Compact Heat Exchangers (3rd ed.). McGraw-Hill.
- Shah, R. K., & SekuliΔ, D. P. (2003). Fundamentals of Heat Exchanger Design.
- Coppage, J. E., & London, A. L. (1953). "The periodic-flow heat exchangerβa summary of design theory," Transactions of the ASME.
Worked Engineering Example
Problem Statement:
An HVAC air-to-air energy recovery rotary wheel ($N = 10$ rpm $= 0.167$ Hz) operates with: - Hot stream capacity $C_h = 500$ W/K, inlet temp $T_{h,in} = 35$Β°C. - Cold stream capacity $C_c = 450$ W/K, inlet temp $T_{c,in} = 5$Β°C. - Exchanger core $NTU = 3.0$. - Matrix mass $m_{matrix} = 200$ kg, specific heat $c_{p,matrix} = 900$ J/kgK. Determine the actual heat recovery rate $Q$ and outlet temperatures.
Step-by-Step Solution:
1. Calculate Matrix Capacity Ratio ($C_r^*$):
$$C_{min} = C_c = 450 \text{ W/K}, \quad C_{max} = C_h = 500 \text{ W/K} \implies C_r = 0.90$$ $$C_r^* = \frac{200 \text{ kg} \cdot 900 \text{ J/kgK} \cdot (10/60 \text{ Hz})}{450 \text{ W/K}} = \frac{30000}{450} \approx 66.7$$ Since $C_r^* \gg 2$, the matrix acts almost as a perfect thermal flywheel.
2. Find Counterflow limit effectiveness ($\epsilon_{cf}$):
$$\epsilon_{cf} = \frac{1 - \exp[-3 \cdot (1 - 0.9)]}{1 - 0.9 \exp[-3 \cdot (1 - 0.9)]} = \frac{1 - e^{-0.3}}{1 - 0.9 e^{-0.3}} \approx \frac{0.2592}{0.3333} \approx 0.778$$
3. Apply Rotary Capacity correction:
$$\epsilon = 0.778 \cdot \left[ 1 - \frac{1}{9 \cdot (66.7)^{1.93}} \right] \approx 0.778 \cdot [ 1 - 0.00003 ] \approx 0.778 \text{ (or } 77.8\%\text{)}$$
4. Calculate Duty and Outlet Temperatures:
$$Q = 0.778 \cdot 450 \cdot (35 - 5) = 10,503 \text{ W} \approx 10.5 \text{ kW}$$ $$T_{h,out} = 35 - \frac{10503}{500} = 14.0^\circ\text{C}$$ $$T_{c,out} = 5 + \frac{10503}{450} = 28.3^\circ\text{C}$$
An HVAC air-to-air energy recovery rotary wheel ($N = 10$ rpm $= 0.167$ Hz) operates with: - Hot stream capacity $C_h = 500$ W/K, inlet temp $T_{h,in} = 35$Β°C. - Cold stream capacity $C_c = 450$ W/K, inlet temp $T_{c,in} = 5$Β°C. - Exchanger core $NTU = 3.0$. - Matrix mass $m_{matrix} = 200$ kg, specific heat $c_{p,matrix} = 900$ J/kgK. Determine the actual heat recovery rate $Q$ and outlet temperatures.
Step-by-Step Solution:
1. Calculate Matrix Capacity Ratio ($C_r^*$):
$$C_{min} = C_c = 450 \text{ W/K}, \quad C_{max} = C_h = 500 \text{ W/K} \implies C_r = 0.90$$ $$C_r^* = \frac{200 \text{ kg} \cdot 900 \text{ J/kgK} \cdot (10/60 \text{ Hz})}{450 \text{ W/K}} = \frac{30000}{450} \approx 66.7$$ Since $C_r^* \gg 2$, the matrix acts almost as a perfect thermal flywheel.
2. Find Counterflow limit effectiveness ($\epsilon_{cf}$):
$$\epsilon_{cf} = \frac{1 - \exp[-3 \cdot (1 - 0.9)]}{1 - 0.9 \exp[-3 \cdot (1 - 0.9)]} = \frac{1 - e^{-0.3}}{1 - 0.9 e^{-0.3}} \approx \frac{0.2592}{0.3333} \approx 0.778$$
3. Apply Rotary Capacity correction:
$$\epsilon = 0.778 \cdot \left[ 1 - \frac{1}{9 \cdot (66.7)^{1.93}} \right] \approx 0.778 \cdot [ 1 - 0.00003 ] \approx 0.778 \text{ (or } 77.8\%\text{)}$$
4. Calculate Duty and Outlet Temperatures:
$$Q = 0.778 \cdot 450 \cdot (35 - 5) = 10,503 \text{ W} \approx 10.5 \text{ kW}$$ $$T_{h,out} = 35 - \frac{10503}{500} = 14.0^\circ\text{C}$$ $$T_{c,out} = 5 + \frac{10503}{450} = 28.3^\circ\text{C}$$