Chapter 11, [[incropera.pdf]] ## Heat Exchanger Variants A heat exchanger transfers heat between two fluids without mixing them Parallel-flow vs counterflow ![[Pasted image 20250414112154.png|300]] ![[Pasted image 20250414112225.png|325]] Crossflow: ![[Pasted image 20250414112442.png|625]] ## 1st Law $\dot{Q}_{H} = (\dot{m}c_{p})_{H}\Delta T_{H}$ $\dot{Q}_{C}=(\dot{m}c_{p})_{C}\Delta T_{C}$ $\dot{Q}_{H}=\dot{Q}_{C}$ ## LMTD Method - all four inlet and outlet temps are known - you want to find required heat exchanger area $\dot{Q} = UA \Delta T_{\text{LM}}$ $U =$ overall heat transfer coeff $A =$ heat transfer surface area ${} \Delta T_{LM} =$ log mean temperature difference $\Delta T_{LM} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln(\Delta T_{1}/\Delta T_{2})}$ | Type | $\Delta T_{1}$ | $\Delta T_{2}$ | | ------------- | ----------------------- | ----------------------- | | Parallel Flow | ${} T_{h,i}-T_{c,i} {}$ | ${} T_{h,o}-T_{c,o} {}$ | | Counterflow | ${} T_{h,i}-T_{c,o} {}$ | $T_{h,o}-T_{c,i}$ | For crossflow and shell-and-tube, use a correction factor $F$ from charts: $\Delta T_{LM,eff} = F \cdot \Delta T_{LM}$ ## Effectiveness-NTU Method - Heat exchanger effectiveness is known - Missing some fluid temperatures - You want to solve for required area or UA 1. Find heat capacity rates $C~\mathrm{[W/K]}$ $C = \dot{m}c_{p}$, $C_\text{min}=\text{min}(C_{c},C_{h})$, $C_\text{max}=\text{max}(C_{c},C_{h})$, $C_{r}=\frac{C_\text{min}}{C_\text{max}}$ 2. Determine $q_\text{max}$ $q_{\text{max}} = C_\text{min}(T_{h,i}-T_{c,i})$ 3. Determine effectiveness $\epsilon$ from charts $\varepsilon = \frac{\dot{Q}_\text{actual}}{\dot{Q}_{\text{max}}} = f\left( \mathrm{NTU}, \frac{C_\text{min}}{C_\text{max}} \right)$ 4. Determine ${} Q_{\text{actual}}$ from effectiveness $Q_{\text{actual}}=\varepsilon Q_{\text{max}}$ 5. Use $\varepsilon$-NTU relations to determine Number of Transfer Units $\mathrm{NTU} = \frac{UA}{C_\text{min}}$ 6. Solve for UA or required area High NTU = high heat transfer for a given minimum heat capacity rate ![[Pasted image 20250414121113.png]] | Heat Exchanger Type | $\varepsilon$ Formula | NTU as a function of $\varepsilon$ | | ------------------------------------------------------------ | ------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------- | | Counterflow | $\varepsilon = \frac{1 - e^{-\text{NTU}(1 - C_r)}}{1 - C_r e^{-\text{NTU}(1 - C_r)}}$ | ${} \mathrm{NTU} = \frac{1}{C_{r}-1}\ln\left( \frac{\varepsilon-1}{\varepsilon C_{r}-1} \right) {}$ | | Parallel flow | $\varepsilon = \frac{1 - e^{-\text{NTU}(1 + C_r)}}{1 + C_r}$ | $\text{NTU} = -\frac{\ln(1 - \varepsilon (1 + C_r))}{1 + C_r}$ | | Crossflow:<br>$C_\text{max}$ mixed<br>$C_\text{min}$ unmixed | ${} \varepsilon = \frac{1}{C_{r}}(1-e^{-C_{r}(1-e^{-C_{r}(\mathrm{NTU})})}) {}$ | ${} \mathrm{NTU} = -\ln\left( 1 + \frac{1}{C_{r}}\ln(1-\varepsilon C_{r}) \right) {}$ | | Crossflow:<br>$C_\text{max}$ unmixed<br>$C_\text{min}$ mixed | ${} \varepsilon = 1-e^{-\mathrm{NTU}} {}$ | $Q\mathrm{NTU}=-\frac{1}{C_{r}}\ln(C_{r}\ln(1-\varepsilon)+1)$ | ![[Pasted image 20250421183147.png]]