Let's break down the solution step by step: --- # **Given Data** - **Cold water (tube side):** - Flow rate, \( \dot{m}_c = 15,000 \) kg/h = 4.167 kg/s - Inlet temperature, \( T_{ci} = 30^\circ \)C - Outlet temperature, \( T_{co} = 60^\circ \)C - **Hot water (shell side):** - Flow rate, \( \dot{m}_h = 10,000 \) kg/h = 2.778 kg/s - Inlet temperature, \( T_{hi} = 105^\circ \)C - **Tube velocity:** \( v = .4 \) m/s - **Tube size:** 3/4 in. (.75 in.) 18-gage - Outer diameter, \( d_o = .75 \) in = 19.05 mm = .01905 m - 18-gage: wall thickness = .049 in = 1.245 mm → Inner diameter, \( d_i = d_o - 2 \times \text{thickness} = .75 - 2 \times .049 = .652 \) in = 16.56 mm = .01656 m - **Overall heat transfer coefficient, \( U \) = 175 W/m²·°C** - **Tube material:** Water (both shell and tube side) --- ## **Step 1: Calculate Heat Duty (\( Q \))** ### \( Q = \dot{m}_c c_p \Delta T_c \) Assume \( c_p = 418 \) J/kg·K (water). \[ Q = (4.167\, \text{kg/s}) \times (418\, \text{J/kg·K}) \times (60 - 30)\, \text{K} \] \[ Q = 4.167 \times 418 \times 30 = 522,906\, \text{W} \approx 523\, \text{kW} \] --- ## **Step 2: Number of Tubes per Pass** ### **Tube Flow Capacity** \[ \dot{m}_{\text{tube}} = \rho \cdot A_{\text{tube}} \cdot v \] Where \( \rho = 100 \) kg/m³, \( A_{\text{tube}} = \frac{\pi}{4} d_i^2 \), \( d_i = .01656 \) m \[ A_{\text{tube}} = \frac{\pi}{4} (.01656)^2 = 2.154 \times 10^{-4} \,\text{m}^2 \] \[ \dot{m}_{\text{tube}} = 100 \times 2.154 \times 10^{-4} \times .4 = .0862\, \text{kg/s (per tube)} \] ### **Number of Tubes (per pass):** \[ N = \frac{\dot{m}_c}{\dot{m}_{\text{tube}}} = \frac{4.167}{.0862} \approx 48.34 \] **So, round up:** \[ \boxed{49~\text{tubes per pass}} \] --- ## **Step 3: Required Length of Tubes** ### **Heat transfer area needed:** \[ Q = U \cdot A \cdot \Delta T_m \] Where \( \Delta T_m \) is the log mean temperature difference (LMTD). ### **For 1-shell, 1-pass (Counterflow):** \[ \Delta T_1 = T_{hi} - T_{co} = 105 - 60 = 45^\circ \text{C} \] \[ \Delta T_2 = T_{ho} - T_{ci} \] First, find \( T_{ho} \) (hot outlet temperature): \[ Q = \dot{m}_h c_p (T_{hi} - T_{ho}) \] \[ T_{ho} = T_{hi} - \frac{Q}{\dot{m}_h c_p} \] \[ T_{ho} = 105 - \frac{523000}{2.778 \times 418} = 105 - \frac{523000}{11601} = 105 - 45.1 = 59.9^\circ \text{C} \] So, \[ \Delta T_2 = 59.9 - 30 = 29.9^\circ \text{C} \] \[ \Delta T_{m} = \frac{45 - 29.9}{\ln(45/29.9)} = \frac{15.1}{\ln(1.506)} = \frac{15.1}{.410} = 36.8^\circ \text{C} \] ### **Surface Area:** \[ A = \frac{Q}{U \Delta T_m} = \frac{523000}{175 \times 36.8} = \frac{523000}{64400} = 8.12\, \text{m}^2 \] ### **Tube Length:** \[ A = N \times \pi d_o L \] \[ L = \frac{A}{N \pi d_o} = \frac{8.12}{49 \times \pi \times .01905} = \frac{8.12}{2.932} = 2.77\, \text{m} \] --- ## **Step 4: Tube-Side Convective Heat Transfer Coefficient (\( h_i \))** **Assume water properties at average temperature (45°C):** - \( \mu = .6 \) mPa·s - \( k = .64 \) W/m·K - \( Pr \approx 4.5 \) ### **Reynolds Number:** \[ Re = \frac{\rho v d_i}{\mu} \] \[ \mu = .6 \text{ mPa·s } = .0006 \text{ kg/m·s} \] \[ Re = \frac{100 \times .4 \times .01656}{.0006} = \frac{6.624}{.0006} = 11,040 \] ### **Nusselt Number (Dittus-Boelter, turbulent):** \[ Nu = .023 Re^{.8} Pr^{.4} \] \[ Nu = .023 \times (11040)^{.8} \times (4.5)^{.4} \] \[ (11040)^{.8} \approx 1654,\quad (4.5)^{.4} \approx 1.78 \] \[ Nu = .023 \times 1654 \times 1.78 = 67.7 \] ### **Tube-side heat transfer coefficient:** \[ h_i = \frac{Nu \cdot k}{d_i} = \frac{67.7 \times .64}{.01656} = \frac{43.33}{.01656} = 2,617 \text{ W/m}^2\cdot\text{K} \] --- ## **Step 5: Shell-Side Heat Transfer Coefficient (\( h_o \))** Shell-side calculations are more involved; use the Kern method for estimation. ### **Equivalent diameter (\( d_{eq} \)) (for square pitch):** Assume 1 in. pitch (\( p \)): \[ d_{eq} = \frac{4 \times (\text{flow area})}{\text{wetted perimeter}} \] \[ d_{eq} \approx 1.27 \text{ cm} = .0127 \text{ m } \quad (\text{typical for 3/4" tubes, 1" pitch}) \] ### **Shell-side velocity:** \[ \text{Total flow area} = (\text{pitch} - d_o) \times \text{length} \times N_{\text{rows}} \] For a rough estimate, assume average velocity = .2 m/s. ### **Reynolds number (shell):** \[ Re_s = \frac{\rho v_s d_{eq}}{\mu} \] \[ Re_s = \frac{100 \times .2 \times .0127}{.0006} = 4,233 \] ### **Nusselt number (shell):** \[ Nu = .33 Re_s^{.6} Pr^{.33} \] \[ Nu = .33 \times (4233)^{.6} \times (4.5)^{.33} \] \[ (4233)^{.6} \approx 188,\quad (4.5)^{.33} \approx 1.6 \] \[ Nu = .33 \times 188 \times 1.6 = 99.3 \] \[ h_o = \frac{Nu \cdot k}{d_{eq}} = \frac{99.3 \times .64}{.0127} = 5,005 \text{ W/m}^2\cdot\text{K} \] --- ## **Step 6: For 1-shell, 2-tube Passes** - Number of tubes per pass is halved, so \( N = 49/2 \approx 25 \) tubes per pass, but 2 passes. - Tube-side velocity doubles since flow per tube increases for the same total mass flow. \( v = .8 \) m/s. - Repeat above calculations with adjusted velocity and tube count. --- # **Summary Table** | Case | Tubes per pass | Tube length (m) | \( h_i \) (W/m²·K) | \( h_o \) (W/m²·K) | |------------------------------|:--------------:|:---------------:|:------------------:|:------------------:| | 1-shell, 1-pass (counterflow)| 49 | 2.8 | 2,617 | 5,005 | | 1-shell, 2-pass | 25 | 5.6 | 4,600 | 5,000 | --- ## **Conclusions and Best Design Parameters** - **Counterflow arrangement** is more efficient due to higher LMTD. - **2-tube pass** increases tube-side velocity and heat transfer coefficient but requires longer tubes. - **Best compromise:** Use counterflow, 1-shell, 1-pass if physical constraints allow tube lengths ~2.8 m. If shorter exchangers are needed, use 2-pass design. --- ### **Final Answers** **a) 1-shell, 1-tube pass (counterflow):** - **Number of tubes:** 49 - **Tube length:** 2.8 m - **Tube-side \( h_i \):** ~2,617 W/m²·K - **Shell-side \( h_o \):** ~5,000 W/m²·K **b) 1-shell, 2-tube pass:** - **Number of tubes:** 25 per pass (2 passes) - **Tube length:** 5.6 m - **Tube-side \( h_i \):** ~4,600 W/m²·K - **Shell-side \( h_o \):** ~5,000 W/m²·K --- **Design Recommendation:** Prefer 1-shell, 1-tube pass (counterflow) for compactness and efficiency, unless tube length is a limiting factor. --- **If you need detailed step-by-step for the 2-pass case or more detail for any step, let me know!**