In a textile manufacturing plant, the waste dyeing water \(\left(c_{p}=4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to preheat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(15^{\circ} \mathrm{C}\) at the same flow rate in a double-pipe counter-flow heat exchanger. The heat transfer surface area of the heat exchanger is \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient is \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the rate of heat transfer in the heat exchanger is \(35 \mathrm{~kW}\), determine the outlet temperature and the mass flow rate of each fluid stream.

Let's denote the waste dyeing water's mass flow rate as \(\dot{m}_h\) and the fresh water's mass flow rate as \(\dot{m}_c\). According to the energy conservation principle, the energy gained by the cold fluid (fresh water) must equal the energy lost by the hot fluid (waste dyeing water): \(\dot{m}_h c_{p,h} (T_{h,i} - T_{h,o}) = \dot{m}_c c_{p,c} (T_{c,o} - T_{c,i})\) Given that both fluid streams have the same flow rate, we can write: \(\dot{m} c_{p,h} (T_{h,i} - T_{h,o}) = \dot{m} c_{p,c} (T_{c,o} - T_{c,i})\) where \(\dot{m}\) is the mass flow rate, \(c_{p,h}\) and \(c_{p,c}\) are specific heat capacities for hot and cold fluids, and \(T_{h,i}\) , \(T_{h,o}\) , \(T_{c,i}\), and \(T_{c,o}\) are the inlet and outlet temperatures for hot and cold fluids, respectively.

Given that the heat transfer rate is \(35 \mathrm{~kW}\), we can write: \(\dot{Q} = 35,000 \mathrm{~W} = \dot{m} c_{p,h} (T_{h,i} - T_{h,o})\) We are given the initial temperatures \(T_{h,i} = 75^{\circ} \mathrm{C}\) and \(T_{c,i} = 15^{\circ} \mathrm{C}\), and the specific heat capacities \(c_{p,h} = 4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p,c} = 4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Using the energy balance equation, we can find the mass flow rate \(\dot{m}\) and the outlet temperatures \(T_{h,o}\) and \(T_{c,o}\). Rearange to get the mass flow rate: \(\dot{m} = \dfrac{\dot{Q}}{c_{p,h} (T_{h,i} - T_{h,o})}\)

In order to find the outlet temperatures, we need to first determine the LMTD. For a counter-flow double-pipe heat exchanger, the LMTD is given by: \(LMTD = \dfrac{\Delta T_{1} - \Delta T_{2}}{\ln{\dfrac{\Delta T_{1}}{\Delta T_{2}}}}\) where \(\Delta T_{1} = T_{h,i} - T_{c,i}\) and \(\Delta T_{2} = T_{h,o} - T_{c,o}\).

We are given the heat transfer surface area of the heat exchanger as \(1.65 \mathrm{~m}^{2}\) and the overall heat transfer coefficient as \(625 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). We can now write the heat transfer rate equation as: \(\dot{Q} = U A LMTD\) Substituting the known values and LMTD expression, we get: \(35,000 \mathrm{~W} = 625 \mathrm{~W} /\mathrm{m}^{2} \cdot \mathrm{K} \times 1.65 \mathrm{~m}^{2} \times \dfrac{T_{h,i} - T_{c,o} - (T_{h,o} - T_{c,i})}{\ln{\dfrac{T_{h,i} - T_{c,o}}{T_{h,o} - T_{c,i}}}}\) Solve this equation simultaneously with the energy balance equation to find the outlet temperatures, \(T_{h,o}\) and \(T_{c,o}\). Calculating the outlet temperatures, we get: \(T_{h,o} \approx 61.326^{\circ} \mathrm{C}\) \(T_{c,o} \approx 56.326^{\circ} \mathrm{C}\) Now, we can plug the outlet temperature back into the mass flow rate expression to find the mass flow rate:

Now that we have the outlet temperatures, we can calculate the mass flow rate using the expression obtained in Step 2: \(\dot{m} = \dfrac{35,000 \mathrm{~W}}{4295 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}(75 - 61.326)}\) Calculating the mass flow rate, we obtain: \(\dot{m} \approx 0.2825 \mathrm{kg/s}\) So the mass flow rate of both fluid streams is approximately \(0.2825 \mathrm{kg/s}\), and their outlet temperatures are approximately \(61.326^{\circ} \mathrm{C}\) for the waste dyeing water and \(56.326^{\circ} \mathrm{C}\) for the fresh water.