Let's break down the problem and solve it step by step: --- ### **Step 1: Write Down What's Given** #### **Process Stream (Hot Fluid)** - **Flow rate:** 600 kmol/hr - **Composition:** 40% Benzene, 30% Toluene, 30% Water (mole basis) - **Inlet Temp:** 180°C - **Outlet Temp:** 120°C - **Allowable ΔP:** 0.8 bar #### **Cooling Medium (Brine, Cold Fluid)** - **Flow rate:** 35,000 kg/hr - **Inlet Temp:** 15°C - **Allowable ΔP:** 0.8 bar #### **Fouling Factor:** - \( R_{f} = 0.00025 \) (on both sides, units: \( (W/m^2 \cdot °C)^{-1} \)) #### **Physical Properties Table** | Component | Density (kg/m³) | Viscosity (cP) | Conductivity (W/m·K) | Cp (J/mol·K) | MW (kg/kmol) | |-----------|-----------------|----------------|----------------------|--------------|--------------| | Benzene | 880 | 0.35 | 0.12 | 172 | 78 | | Toluene | 910 | 0.55 | 0.13 | 190 | 92 | | Water | 998 | 0.25 | 0.61 | 75 | 18 | | Brine | 1030 | 0.9 | 0.6 | 85 | ~58 | --- ## **Step 2: Calculate Heat Duty (\( Q \))** ### **1. Calculate the average Cp for process stream** Mole fractions: - Benzene: 0.4 - Toluene: 0.3 - Water: 0.3 Average \( C_{p,avg} = \sum x_i C_{p,i} \) - \( = 0.4 \times 172 + 0.3 \times 190 + 0.3 \times 75 \) - \( = 68.8 + 57 + 22.5 = 148.3 \) J/mol·K ### **2. Calculate the heat load** - \( n = 600 \) kmol/hr \( = 600,000 \) mol/hr - \( \Delta T = 180 - 120 = 60 \) °C \( Q = n \times C_{p,avg} \times \Delta T \) First, convert to consistent units (J/hr to kW): \[ Q = 600,000 \text{ mol/hr} \times 148.3 \text{ J/(mol·K)} \times 60 \text{ K} \] \[ Q = 600,000 \times 148.3 \times 60 = 5,338,800,000 \text{ J/hr} \] Convert to kW: \[ 1 \text{ kW} = 3,600,000 \text{ J/hr} \] \[ Q = \frac{5,338,800,000}{3,600,000} = 1,483 \text{ kW} \] --- ## **Step 3: Brine Outlet Temperature** Let \( T_{c,out} \) be the outlet temperature of brine. \[ Q = m_{brine} \cdot C_{p,brine} \cdot (T_{c,out} - T_{c,in}) \] - \( m_{brine} = 35,000 \) kg/hr - \( C_{p,brine} = 85 \) J/mol·K First, convert \( m_{brine} \) to kmol/hr: \[ MW_{brine} \approx 58 \text{ kg/kmol} \] \[ n_{brine} = \frac{35,000}{58} = 603.45 \text{ kmol/hr} \] So, \[ Q = 603,450 \text{ mol/hr} \times 85 \text{ J/(mol·K)} \times (T_{c,out} - 15) \] \[ 1,483,000 \text{ W} \times 3600 = 5,338,800,000 \text{ J/hr} \] \[ 5,338,800,000 = 603,450 \times 85 \times (T_{c,out} - 15) \] \[ 603,450 \times 85 = 51,293,250 \] \[ T_{c,out} - 15 = \frac{5,338,800,000}{51,293,250} = 104.17 \] \[ T_{c,out} = 15 + 104.17 = 119.2 \degree C \] But this is nearly the same as the process outlet temperature. In reality, with equal heat capacity rates, the temperature approach can be very close, but let's check the heat capacity rates: ### **Hot stream:** \[ C_{ph} = n_{hot} \cdot Cp_{hot} = 600,000 \text{ mol/hr} \times 148.3 \text{ J/mol·K} \] \[ = 88,980,000 \text{ J/hr·K} = 24,717 \text{ W/K} \] ### **Cold stream:** \[ C_{pc} = 603,450 \text{ mol/hr} \times 85 \text{ J/mol·K} = 51,293,250 \text{ J/hr·K} = 14,248 \text{ W/K} \] Since \( C_{pc} < C_{ph} \), brine is the limiting stream. So, the outlet temperature of brine: \[ Q = C_{pc} \cdot (T_{c,out} - T_{c,in}) \] \[ 1,483,000 = 14,248 \cdot (T_{c,out} - 15) \] \[ T_{c,out} - 15 = \frac{1,483,000}{14,248} = 104.1 \] \[ T_{c,out} = 119.1 \degree C \] **So, brine will leave at 119.1°C.** --- ## **Step 4: Calculate LMTD (\( \Delta T_{lm} \)), Counterflow** \[ \Delta T_1 = T_{h,in} - T_{c,out} = 180 - 119.1 = 60.9^\circ C \] \[ \Delta T_2 = T_{h,out} - T_{c,in} = 120 - 15 = 105^\circ C \] \[ \Delta T_{lm} = \frac{ \Delta T_2 - \Delta T_1 }{ \ln{\left( \frac{\Delta T_2}{\Delta T_1} \right) } } \] \[ = \frac{ 105 - 60.9 }{ \ln{ \left( \frac{105}{60.9} \right) } } = \frac{44.1}{ \ln{(1.724)} } = \frac{44.1}{0.546} = 80.8^\circ C \] --- ## **Step 5: Estimate Overall Heat Transfer Coefficient (\( U \))** ### **First, estimate individual heat transfer coefficients (\( h \))** **Assume shell-and-tube exchanger (most typical).** Let's estimate using Dittus-Boelter for tubes and shells (use typical values for initial sizing). #### **Hot Side (Process; inside tube):** - Take average properties (180°C to 120°C); use average \( T = 150°C \). - Take density \(\sim\) weighted average: - 0.4 × 880 + 0.3 × 910 + 0.3 × 998 ≈ 924 kg/m³ - Viscosity \(\sim\) average: - 0.4 × 0.35 + 0.3 × 0.55 + 0.3 × 0.25 ≈ 0.37 cP = 0.37 mPa·s = 0.00037 Pa·s - Thermal conductivity \(\sim\) 0.4 × 0.12 + 0.3 × 0.13 + 0.3 × 0.61 ≈ 0.26 W/m·K - Cp (mass basis): - 148.3 J/mol·K × (1 mol / avg MW) - avg MW = 0.4×78 + 0.3×92 + 0.3×18 ≈ 61.2 kg/kmol - So, Cp (J/kg·K) = 148.3 / 0.0612 = 2,423 J/kg·K #### **Brine Side:** - 1030 kg/m³, 0.9 cP = 0.0009 Pa·s, 0.6 W/m·K, Cp = 85 J/mol·K (85/58 = 1,465 J/kg·K) ### **Assume tube-side: process stream; shell-side: brine** #### **Assume tube ID = 20 mm = 0.02 m; OD = 25 mm = 0.025 m** #### **1. Calculate mass flow rates per side** - Process: 600 kmol/hr × 61.2 kg/kmol = 36,720 kg/hr = 10.2 kg/s - Brine: 35,000 kg/hr = 9.72 kg/s #### **2. Calculate velocities (initial guess: 1 m/s in tubes)** - Assume 1 m/s in tubes. - \( Q = \rho \cdot A \cdot v \) - \( A = \frac{\dot{m}}{\rho v} \) - So, for process stream (in tubes): \[ A_{tube} = \frac{10.2}{924 \times 1} = 0.011 m^2 \] Each tube area: \[ A_{1 tube} = \frac{\pi}{4} d^2 = 3.14/4 \times (0.02)^2 = 0.000314 m^2 \] Number of tubes: \[ N = \frac{0.011}{0.000314} = 35 \] #### **3. Reynolds Number for tubes** \[ Re = \frac{\rho v D}{\mu} = \frac{924 \times 1 \times 0.02}{0.00037} = \frac{18.48}{0.00037} = 49,946 \] **Turbulent flow** (Dittus-Boelter applies). #### **4. Prandtl Number** \[ Pr = \frac{C_p \cdot \mu}{k} = \frac{2,423 \text{ J/kg·K} \times 0.00037}{0.26} = \frac{0.8965}{0.26} = 3.45 \] #### **5. Nusselt Number (Dittus-Boelter, heating/cooling):** \[ Nu = 0.023 Re^{0.8} Pr^{0.4} \] \[ Nu = 0.023 \times (49,946)^{0.8} \times (3.45)^{0.4} \] Calculate: - \( (49,946)^{0.8} \approx 5,100 \) - \( (3.45)^{0.4} \approx 1.64 \) So, \[ Nu = 0.023 \times 5,100 \times 1.64 = 192 \] #### **6. Tube-side heat transfer coefficient** \[ h_{tube} = \frac{Nu \cdot k}{D} = \frac{192 \times 0.26}{0.02} = 2,496 \text{ W/m}^2\cdot K \] #### **Repeat for shell-side (brine):** Assume similar velocity in the shell. For estimation, use a typical shell-side \( h \) for brine, say 1,000 W/m²·K (brine is more viscous, less favorable than process stream). #### **7. Add fouling factors and tube wall resistance** - Fouling each side: \( 0.00025 \) (W/m²·K)\(^{-1} \) - Tube wall: Stainless steel, \( k = 16 \) W/m·K, thickness = 0.0025 m \[ R_{wall} = \frac{\Delta r}{k} = \frac{0.0025}{16} = 0.000156 \text{ m}^2\cdot K/W \] #### **8. Overall U calculation** \[ \frac{1}{U} = \frac{1}{h_{tube}} + R_{f,hot} + R_{wall} + R_{f,cold} + \frac{1}{h_{shell}} \] \[ \frac{1}{U} = \frac{1}{2,496} + 0.00025 + 0.000156 + 0.00025 + \frac{1}{1,000} \] \[ = 0.000400 + 0.00025 +