Question

A counterflow heat exchanger is used to cool oil $\left(c_{p}=2.20 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( from \)110^{\circ} \mathrm{C}\( to \)85^{\circ} \mathrm{C}\( at a rate of \)0.75\( \)\mathrm{kg} / \mathrm{s}\( with cold water \)\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( that enters the heat exchanger at \)20^{\circ} \mathrm{C}$ at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)

Short answer

(a) 0.750 m^2 (b) 0.760 m^2 (c) 0.780 m^2 (d) 0.790 m^2 Given data: Mass flow rate of oil = 0.75 kg/s Mass flow rate of water = 0.6 kg/s Specific heat capacity of oil = 2.20 kJ/kgK Specific heat capacity of water = 4.18 kJ/kgK Initial temperature of oil = 110°C Final temperature of oil = 85°C Initial temperature of water = 20°C Overall heat transfer coefficient = 800 W/m²K Solution: After calculating the heat transfer area (Step 4), we find that the required heat transfer area is 0.7871 m^2. Comparing this value to the given options, the closest value is 0.790 m^2. Answer: (d) 0.790 m^2

Step-by-step solution

Calculate the heat transfer rate (heat load)

We'll first calculate the heat transfer rate, which is the rate at which heat is exchanged between the oil and water. We can find this using the mass flow rate, specific heat capacity, and temperature difference for one of the fluids (we'll choose oil). The formula we'll use is: $$ q = \dot{m} \times c_p \times \Delta T $$ Where \(q\) is the heat transfer rate, \(\dot{m}\) is the mass flow rate, \(c_p\) is the specific heat capacity, and \(\Delta T\) is the temperature difference. For the oil, we have: $$ q = 0.75 \ \text{kg/s} \times 2.20 \ \text{kJ/kgK} \times (110^{\circ} \mathrm{C} - 85^{\circ} \mathrm{C}) $$ Calculating this gives us: $$ q = 41.25 \ \text{kJ/s} $$

Calculate the temperature difference for the cold water

Next, we'll calculate the final temperature of the cold water once it has absorbed the heat from the oil. Using the heat transfer rate calculated earlier and applying the same equation as in Step 1, we can solve for the temperature difference of the water: $$ \Delta T_{\text{water}} = \frac{q}{\dot{m}_{\text{water}} \times c_{p_{\text{water}}}} $$ Plugging in the values, we get: $$ \Delta T_{\text{water}} = \frac{41.25 \ \text{kJ/s}}{0.6 \ \text{kg/s} \times 4.18 \ \text{kJ/kgK}} $$ Calculating the temperature difference gives us: $$ \Delta T_{\text{water}} = 17.27^\circ \mathrm{C} $$ Now we can find the final temperature of the water: $$ T_{\text{water}}^{\text{final}} = T_{\text{water}}^{\text{initial}} + \Delta T_{\text{water}} = 20^\circ \mathrm{C} + 17.27^\circ \mathrm{C} = 37.27^\circ \mathrm{C} $$

Calculate the Logarithmic Mean Temperature Difference (LMTD)

We'll now use the initial and final temperatures of the oil and water to calculate the logarithmic mean temperature difference (LMTD). The formula for LMTD is: $$ \text{LMTD} = \frac{(\Delta T_1 - \Delta T_2)}{\ln (\frac{\Delta T_1}{\Delta T_2})} $$ Where \(\Delta T_1\) is the temperature difference at one end of the heat exchanger, and \(\Delta T_2\) is the temperature difference at the other end. In this case, $$ \Delta T_1 = 110^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 90^{\circ} \mathrm{C} $$ $$ \Delta T_2 = 85^{\circ} \mathrm{C} - 37.27^{\circ} \mathrm{C} = 47.73^{\circ} \mathrm{C} $$ Substituting the respective values in the LMTD formula: $$ \text{LMTD} = \frac{(90^{\circ} \mathrm{C} - 47.73^{\circ} \mathrm{C})}{\ln (\frac{90^{\circ} \mathrm{C}}{47.73^{\circ} \mathrm{C}})} = 65.68 \ \mathrm{K} $$

Calculate the heat transfer area

Finally, we can calculate the heat transfer area of the heat exchanger using the overall heat transfer coefficient, heat transfer rate, and LMTD. The formula for the heat transfer area is: $$ A = \frac{q}{U \times \text{LMTD}} $$ Where \(A\) is the heat transfer area, \(U\) is the overall heat transfer coefficient, and LMTD is calculated from Step 3. Given \(U=800 \ \mathrm{W/m^2 K}\), \(q=41.25 \ \mathrm{kJ/s}\) (which we'll convert to \(\mathrm{W}\)), and \(\text{LMTD}=65.68 \ \mathrm{K}\): $$ A = \frac{41,250 \ \mathrm{W}}{800 \ \mathrm{W/m^2 K} \times 65.68 \ \mathrm{K}} $$ Calculating the heat transfer area, we get: $$ A = 0.7871 \ \mathrm{m^2} $$ Comparing the calculated area to the options given in the problem, it's closest to option (d) \(0.790 \ \mathrm{m^2}\). So the correct answer is: (d) \(0.790 \ \mathrm{m^2}\)