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First, we need to calculate the mass flow rate and heat capacity rate of milk using its properties and flow rate: 1. Convert the milk flow rate from L/h to m³/h: \( 250 L/h = \frac{250}{1000} m^3/h = 0.25 m^3/h \) 2. Calculate the mass flow rate of milk: \( \dot{m}_{milk} = \rho_{milk} \times \dot{V}_{milk} = 1030 kg/m^3 \times 0.25 m^3/h = 257.5 kg/h \) 3. Calculate the heat capacity rate of milk: \( C_{milk} = \dot{m}_{milk} \times c_{p,milk} = 257.5 kg/h \times 3860 J/kgK = 994025 J/hK \) Now let's calculate the mass flow rate and heat capacity rate of the ground water: 1. Calculate the mass flow rate of water: \( \dot{m}_{water} = \rho_{water} \times \dot{V}_{water} = 1000 kg/m^3 \times 0.72 m^3/h = 720 kg/h (assume water density as 1000 kg/m^3) \) 2. Calculate the heat capacity rate of water: \( C_{water} = \dot{m}_{water} \times c_{p,water} = 720 kg/h \times 4180 J/kgK = 3011040 J/hK (assume specific heat capacity of water as 4180 J/kgK) \) Using the heat transfer equation: \( q = U A \Delta T_{lm} \) We have all variables except UA and \(\Delta T_{lm}\). So, we can use the heat capacity rate to find the required heat transfer, then use the LMTD method to determine the temperature difference, and finally calculate UA. Calculate the required heat transfer: \( q = \min(C_{milk},C_{water})\times(\Delta T_1-\Delta T_2) \). Since \(C_{milk}

To find the outlet temperature of the water, we can use the following energy balance equation for water: \( \dot{m}_{water}c_{p,water}(T_{out,water} - T_{in,water}) = \dot{m}_{milk}c_{p,milk}(T_{in,milk} - T_{out,milk})\), where \(T_{in,water} = 10^{\circ}C\), \(T_{in,milk} = 38.6^{\circ}C\), and \(T_{out,milk} = 13^{\circ}C\). Now solve for \(T_{out,water}\): \(T_{out,water} = T_{in,water} + \frac{\dot{m}_{milk}c_{p,milk}(T_{in,milk} - T_{out,milk})}{\dot{m}_{water}c_{p,water}} \) \(T_{out,water} = 10 + \frac{257.5 \times 3860 (38.6 - 13)}{720 \times 4180} = 10 + 16.5 = 26.5^{\circ}C \) The outlet temperature of the water is 26.5°C.

To find the milk outlet temperature for different water flow rates, we can use the effectiveness-NTU method of heat exchanger analysis. The effectiveness (\(\epsilon\)) is defined as the actual heat transfer divided by the maximum possible heat transfer. First, we need to calculate the effectiveness of the original system when water flow rate is not changed: \(\epsilon = \frac{q}{\min(C_{milk}, C_{water})(T_{h, in} - T_{c, in})} = \frac{2982075}{994025 \times 28.6} = 0.979\) Now, let's examine the two scenarios: 1. If the water flow rate is doubled: \(\dot{m'}_{water} = 2 \times \dot{m}_{water} = 1440 kg/h\) Calculate the new heat capacity rate of water: \( C'_{water} = \dot{m'}_{water} \times c_{p, water} = 1440 kg/h \times 4180 J/kgK = 6022080 J/hK \) Since \(C_{milk}C''_{water}\), the heat transfer is determined by the capacity of water. Using the same effectiveness, we can calculate the new outlet temperature of milk: \(T''_{out,milk} = T_{in,milk} - \frac{\epsilon \times C''_{water}}{C_{milk}}(T_{in,milk}-T_{in,water}) = 38.6 - \frac{0.979 \times 1505520}{994025}(38.6 - 10) = 16.96^{\circ}C\) In conclusion, when the water flow rate is doubled, the milk outlet temperature is 13.97°C, and when the water flow rate is halved, the milk outlet temperature is 16.96°C.