Let's solve the **first problem** step by step: A double-pipe heat exchanger is used to **cool 200 kg/h of oil** (specific heat 2.1 kJ/kg·K) from **120°C to 70°C** using cooling water (specific heat 4.18 kJ/kg·K) at **25°C**. The **water flow rate is 350 kg/h** and the overall heat transfer coefficient U is **450 W/m²·K**. We need to calculate: 1. Required heat transfer area (A) for counter-current flow. 2. Effectiveness of the exchanger. 3. Number of Transfer Units (NTU). 4. Outlet temperature of the water. ### Step 1: Calculate Rate of Heat Transfer (Q) The rate of heat removed from the oil: \[ Q = \dot{m}_{\text{oil}} C_{p,\text{oil}} (T_{\text{in,oil}} - T_{\text{out,oil}}) \] \[ = 200\, \text{kg/h} \times 2.1\, \text{kJ/kg}\cdot\text{K} \times (120-70)\,^\circ\text{C} \] Convert flow rate to kg/s: \[ 200\, \text{kg/h} \div 360\, \text{s/h} = .5556\, \text{kg/s} \] Now calculate Q: \[ Q = .5556\, \text{kg/s} \times 2.1\, \text{kJ/kg}\cdot\text{K} \times 50\, \text{K} \] \[ = .5556 \times 2.1 \times 50 = 58.33\, \text{kJ/s} = 58,330\, \text{W} \] ### Step 2: Water Outlet Temperature Water heat absorbed: \[ Q = \dot{m}_{w} C_{p,w} (T_{\text{out,w}} - T_{\text{in,w}}) \] \[ \dot{m}_{w} = 350\, \text{kg/h} \div 360 = .9722\, \text{kg/s} \] \[ 58,330 = .9722 \times 4.18 \times 100 \times (T_{\text{out,w}} - 25) \] \[ 58,330 = 4064.8 \times (T_{\text{out,w}} - 25) \] \[ T_{\text{out,w}} - 25 = \frac{58,330}{4064.8} = 14.35 \] \[ T_{\text{out,w}} = 39.35\,^\circ\text{C} \] ### Step 3: Heat Capacity Rates (C) Oil: \( C_{\text{oil}} = .5556 \times 2,100 = 1,166.76\, \text{W/K} \) Water: \( C_{\text{water}} = .9722 \times 4,180 = 4,065.8\, \text{W/K} \) Minimum: \( C_{\text{min}} = 1,166.76\, \text{W/K} \) (oil) Maximum: \( C_{\text{max}} = 4,065.8\, \text{W/K} \) (water) \( C_r = C_{\text{min}}/C_{\text{max}} = 1,166.76/4,065.8 = .287 \) ### Step 4: NTU Calculation \[ NTU = \frac{UA}{C_{\text{min}}} \] First, we need to find the area. Rearranging the effectiveness–NTU relationship for a counter-current exchanger: \[ \varepsilon = \frac{1 - \exp[-NTU(1-C_r)]}{1 - C_r \exp[-NTU(1-C_r)]} \] The effectiveness: \[ \varepsilon = \frac{Q}{C_{\text{min}} (T_{\text{hot,in}} - T_{\text{cold,in}})} \] \[ = \frac{58,330}{1,166.76 \times (120-25)} = \frac{58,330}{1,166.76 \times 95} = \frac{58,330}{110,842.2} = .526 \] Now, solve the effectiveness equation numerically for NTU (set up and solve): \[ .526 = \frac{1 - \exp[-NTU(1-.287)]}{1 - .287 \exp[-NTU(1-.287)]} \] Let \( x = NTU(1-.287) = NTU \times .713 \) So: \[ .526 = \frac{1 - e^{-x}}{1 - .287 e^{-x}} \] Solve for \( x \) numerically, for \( x \approx 1 \), we get \( NTU = x/.713 \approx 1.4 \) ### Step 5: Find Required Area (A) \[ NTU = \frac{UA}{C_{\text{min}}} \] \[ A = \frac{NTU \times C_{\text{min}}}{U} \] \[ A = \frac{1.4 \times 1,166.76}{450} = \frac{1,633.46}{450} = 3.63\, \text{m}^2 \] ### Final Answers Required heat transfer area \( A \) is **3.63 m²** Effectiveness is **.526** Number of Transfer Units (NTU) is **1.4** Outlet temperature of the water is **39.4°C**