Question

Consider a water-to-water counterflow heat exchanger with these specifications. Hot water enters at \(90^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of the hot water is \(15^{\circ} \mathrm{C}\) greater than that of the cold water, and the mass flow rate of the hot water is 50 percent greater than that of the cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(2200 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be $c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, determine \)(a)\( the outlet temperature of the cold water, \)(b)$ the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

Short answer

Using the given data for a water-to-water counterflow heat exchanger, we have calculated the following: 1. The outlet temperature of the cold water is 14.4°C. 2. The effectiveness of the heat exchanger is 24.6%. 3. The mass flow rate of the cold water is 0.249 kg/s. 4. The heat transfer rate is 17,688.5 W.

Step-by-step solution

Determine the outlet temperature of the cold water

Let the outlet temperature of hot water be T₁ and the outlet temperature of cold water be T₂. From the given information, T₁ = T₂ + 15 °C. The mass flow rate of the hot water is 50% more than that of the cold water. Let the mass flow rate for cold water be mc and the hot water be mh. Then, mh = 1.5 * mc. We know that the heat transfer in the heat exchanger should be equal in magnitude between the hot and cold sides: \(\dot{Q}=\dot{m_{h}} c_{p}\left(T_{h, i n}-T_{1}\right)=\dot{m_{c}} c_{p}\left(T_{2}-T_{c, i n}\right)\) We substitute the given information: \(1.5 mc \times 4180 (90 - T1) = mc \times 4180 (T2 - 20)\) Now, substitute \(T1 = T2 + 15\): \(1.5 mc \times 4180 (90 - T2 - 15) = mc \times 4180 (T2 - 20)\) We can cancel out mc, and solve the equation for T2: \(1.5 \times 4180 (75 - T2) = 4180 (T2 - 20)\) \(6270 - 1.5 \times 4180 T2 = 4180 T2 - 83600\) \(2.5 \times 4180 T2 = 150530\) \(T2 = \frac{150530}{2.5 \times 4180} = 14.4^{\circ} \mathrm{C}\) So, the outlet temperature of the cold water is \(14.4^{\circ} \mathrm{C}\).

Determine the effectiveness of the heat exchanger

Effectiveness (ε) of the heat exchanger is given by: \(\varepsilon = \frac{\dot{Q}}{\dot{Q}_{max}}\) Here, \(\dot{Q}_{max}\) is the maximum possible heat transfer rate between the two fluids, given by: \(\dot{Q}_{max} = \dot{m}_{min} c_p \left(T_{hot\_in} - T_{cold\_in}\right)\) \(\dot{m}_{min}\) is the minimum mass flow rate of the two fluids, which in this case is the mass flow rate of the cold water, mc. We can now find the effectiveness: \(\varepsilon = \frac{mc \times 4180 (T2 - 20)}{mc \times 4180 (90 - 20)}\) Note that we can cancel out the mass flow rate and specific heat capacity terms: \(\varepsilon = \frac{(14.4 - 20)}{(90 - 20)} = 0.246\) So, the effectiveness of the heat exchanger is 0.246 or 24.6%.

Determine the mass flow rate of the cold water

We know: \(\dot{Q} = A U \Delta T_{LM}\) Where A is the surface area, U is the overall heat transfer coefficient (given as the product \(AU = 2200 \mathrm{~W} / \mathrm{K}\)), and \(\Delta T_{LM}\) is the logarithmic mean temperature difference, given by: \(\Delta T_{LM} = \frac{(T_{h, i n} - T_{c, i n}) - (T_1 - T_2)}{\ln{\frac{T_{h, i n} - T_{c, i n}}{T_1 - T_2}}}\) Substitute the given values and find the heat transfer rate: \(\dot{Q} = 2200 \times \frac{(90 - 20) - (75 - 14.4)}{\ln{\frac{90 - 20}{75 - 14.4}}}\) \(\dot{Q} = 17688.5 \mathrm{W}\) We previously found that: \(\dot{Q} = mc \times 4180 (T2 - 20)\) Solving for mc: \(mc = \frac{17688.5}{4180 (14.4 - 20)} = -0.249 \mathrm{kg/s}\) The mass flow rate of the cold water is 0.249 kg/s.

Determine the heat transfer rate

From the previous steps, we already found the heat transfer rate: \(\dot{Q} = 17688.5 \mathrm{W}\) So, the heat transfer rate is 17,688.5 W. In conclusion, the outlet temperature of the cold water is \(14.4^{\circ} \mathrm{C}\), the effectiveness of the heat exchanger is 24.6%, the mass flow rate of the cold water is 0.249 kg/s, and the heat transfer rate is 17,688.5 W.