Let's break down the problem and I'll show you how to **size a heat exchanger** using **Excel** (can be adapted to MATLAB too): --- ## **Given Data** - **Hot fluid (ethylene glycol):** - Inlet temperature, \( T_{hi} = 165^\circ F \) - **Cold fluid (octane):** - Inlet temperature, \( T_{ci} = 70^\circ F \) - Outlet temperature, \( T_{co} = 110^\circ F \) - **Both flow rates:** \( \dot{m} = 110,000 \) lbm/hr - **Fouling factors:** Assume typical values, unless specified: - Ethylene glycol: \( R_{f,hot} = 0.001 \) hr·ft²·°F/Btu - Octane: \( R_{f,cold} = 0.002 \) hr·ft²·°F/Btu --- ## **Step 1: Find Heat Duty (\( Q \))** Heat gained by octane: \[ Q = \dot{m}_c \cdot C_{p,c} \cdot (T_{co} - T_{ci}) \] Assume specific heat values (from typical tables): - **Octane:** \( C_{p,c} \approx 0.543 \) Btu/lbm·°F - **Ethylene Glycol:** \( C_{p,h} \approx 0.57 \) Btu/lbm·°F \[ Q = 110,000 \times 0.543 \times (110 - 70) \] \[ Q = 110,000 \times 0.543 \times 40 = 2,389,200 \text{ Btu/hr} \] --- ## **Step 2: Find Outlet Temp of Glycol** Glycol gives up same heat: \[ Q = \dot{m}_h \cdot C_{p,h} \cdot (T_{hi} - T_{ho}) \] \[ 2,389,200 = 110,000 \times 0.57 \times (165 - T_{ho}) \] \[ 2,389,200 = 62,700 \times (165 - T_{ho}) \] \[ 165 - T_{ho} = \frac{2,389,200}{62,700} = 38.11 \] \[ T_{ho} = 165 - 38.11 = 126.89^\circ F \] --- ## **Step 3: Find LMTD** Counterflow heat exchanger: \[ \text{LMTD} = \frac{(T_{hi} - T_{co}) - (T_{ho} - T_{ci})}{\ln\left(\frac{T_{hi} - T_{co}}{T_{ho} - T_{ci}}\right)} \] Plug in values: - \( T_{hi} = 165 \) - \( T_{ho} = 126.89 \) - \( T_{ci} = 70 \) - \( T_{co} = 110 \) \[ \Delta T_1 = T_{hi} - T_{co} = 165 - 110 = 55 \] \[ \Delta T_2 = T_{ho} - T_{ci} = 126.89 - 70 = 56.89 \] \[ \text{LMTD} = \frac{55 - 56.89}{\ln(55/56.89)} \] \[ \text{LMTD} = \frac{-1.89}{\ln(0.9668)} = \frac{-1.89}{-0.0338} = 55.9^\circ F \] --- ## **Step 4: Calculate Required Area** General equation: \[ Q = U \cdot A \cdot \text{LMTD} \] \[ A = \frac{Q}{U \cdot \text{LMTD}} \] **Assume clean overall heat transfer coefficient (\( U_{clean} \)):** - For oil/glycol, typical: 60 Btu/hr·ft²·°F **Account for fouling:** \[ \frac{1}{U_{overall}} = \frac{1}{U_{clean}} + R_{f,hot} + R_{f,cold} \] \[ \frac{1}{U_{overall}} = \frac{1}{60} + 0.001 + 0.002 = 0.0167 + 0.003 = 0.0197 \] \[ U_{overall} = \frac{1}{0.0197} = 50.8 \text{ Btu/hr·ft}^2 \cdot ^\circ F \] \[ A = \frac{2,389,200}{50.8 \times 55.9} = \frac{2,389,200}{2,842.72} = 840.6 \text{ ft}^2 \] --- ## **Summary Table for Excel** | Parameter | Value | Unit | |----------------------------|------------|-------------------| | Octane flow rate | 110,000 | lbm/hr | | Octane Cp | 0.543 | Btu/lbm·°F | | Octane inlet temp | 70 | °F | | Octane outlet temp | 110 | °F | | Glycol flow rate | 110,000 | lbm/hr | | Glycol Cp | 0.57 | Btu/lbm·°F | | Glycol inlet temp | 165 | °F | | Glycol outlet temp | 126.89 | °F | | Q (heat duty) | 2,389,200 | Btu/hr | | LMTD | 55.9 | °F | | U (clean) | 60 | Btu/hr·ft²·°F | | Fouling factor (hot) | 0.001 | hr·ft²·°F/Btu | | Fouling factor (cold) | 0.002 | hr·ft²·°F/Btu | | U (overall, fouled) | 50.8 | Btu/hr·ft²·°F | | **Required Area** | **840.6** | **ft²** | --- ## **Excel Formulas** You can set up your Excel sheet as follows: - **B2:** `=110000` - **B3:** `=0.543` - **B4:** `=70` - **B5:** `=110` - **B6:** `=0.57` - **B7:** `=165` - **B8:** `=(B9)` (will be calculated) - **B9:** `=B2*B3*(B5-B4)` - **B10:** `=B2*B6` - **B11:** `=B9/B10` - **B12:** `=B7-B11` - **B13:** `=B7-B5` - **B14:** `=B12-B4` - **B15:** `=(B13-B14)/LN(B13/B14)` - **B16:** `=60` - **B17:** `=0.001` - **B18:** `=0.002` - **B19:** `=1/(1/B16+B17+B18)` - **B20:** `=B9/(B19*B15)` --- ## **MATLAB Code** ```matlab % Given m_dot = 110000; % lbm/hr Cp_octane = 0.543; % Btu/lbm-F Cp_glycol = 0.57; % Btu/lbm-F T_ci = 70; % F T_co = 110; % F T_hi = 165; % F % Q Q = m_dot * Cp_octane * (T_co - T_ci); % Glycol outlet temp T_ho = T_hi - Q/(m_dot * Cp_glycol); % LMTD deltaT1 = T_hi - T_co; deltaT2 = T_ho - T_ci; LMTD = (deltaT1 - deltaT2)/log(deltaT1/deltaT2); % Fouling Rf_hot = 0.001; % hr*ft^2*F/Btu Rf_cold = 0.002; U_clean = 60; U_overall = 1/(1/U_clean + Rf_hot + Rf_cold); % Area A = Q/(U_overall * LMTD); % Display results fprintf('Heat duty Q: %.2f Btu/hr\n', Q); fprintf('Glycol outlet temp: %.2f F\n', T_ho); fprintf('LMTD: %.2f F\n', LMTD); fprintf('U_overall: %.2f Btu/hr*ft^2*F\n', U_overall); fprintf('Required Area: %.2f ft^2\n', A); ``` --- ## **Final Answer** **The required heat exchanger area is approximately 841 ft².** You can use the equations above in Excel or MATLAB to check or refine your answer as needed. **Remember to update Cp and fouling factors if you have more precise data for your application!**