To solve this problem, we'll use the formula for heat transfer in a counter-flow or parallel-flow heat exchanger.

Given Data:

• Internal diameter of the pipe: $d = 2.5 \, \text{cm} = 0.025 \, \text{m}$
• Mass flow rate of milk: $\dot{m} = 0.4 \, \text{kg/s}$
• Initial temperature of milk: $T_{m, \text{in}} = 49^\circ \text{C}$
• Final temperature of milk: $T_{m, \text{out}} = 18^\circ \text{C}$
• Temperature of the cooling bath: $T_w = 10^\circ \text{C}$ (constant)
• Overall heat transfer coefficient: $U = 900 \, \text{J/m}^2 \cdot \text{s} \cdot \text{°C}$
• Specific heat of milk: $c_p = 3890 \, \text{J/kg} \cdot \text{°C}$

Step-by-Step Solution

1. Calculate the Heat Duty (Q):
   The amount of heat that must be removed from the milk to lower its temperature can be calculated as: $Q = \dot{m} \cdot c_p \cdot (T_{m, \text{in}} - T_{m, \text{out}})$ Substituting the values: $Q = 0.4 \, \text{kg/s} \times 3890 \, \text{J/kg} \cdot \text{°C} \times (49 - 18) \, \text{°C}$ $Q = 0.4 \times 3890 \times 31 = 48268 \, \text{J/s} = 48268 \, \text{W}$

2. Calculate the Logarithmic Mean Temperature Difference (LMTD): The LMTD for a single pipe heat exchanger can be calculated using the formula: $\Delta T_{\text{lm}} = \frac{(T_{m, \text{in}} - T_w) - (T_{m, \text{out}} - T_w)}{\ln \left( \frac{T_{m, \text{in}} - T_w}{T_{m, \text{out}} - T_w} \right)}$ where:

   • $T_{m, \text{in}} - T_w = 49 - 10 = 39^\circ \text{C}$
   • $T_{m, \text{out}} - T_w = 18 - 10 = 8^\circ \text{C}$

   Substituting these values: $\Delta T_{\text{lm}} = \frac{39 - 8}{\ln \left( \frac{39}{8} \right)}$ Calculating the logarithmic mean temperature difference: $\Delta T_{\text{lm}} = \frac{31}{\ln(4.875)} \approx \frac{31}{1.584} \approx 19.57^\circ \text{C}$

3. Determine the Required Heat Transfer Area (A):
   Using the heat transfer equation: $Q = U \cdot A \cdot \Delta T_{\text{lm}}$ Rearranging for $A$: $A = \frac{Q}{U \cdot \Delta T_{\text{lm}}}$ Substituting the values: $A = \frac{48268}{900 \cdot 19.57} \approx \frac{48268}{17613} \approx 2.74 \, \text{m}^2$

4. Calculate the Required Length of the Pipe (L):
   The area $A$ for a cylindrical pipe is given by: $A = \pi \cdot d \cdot L$ Solving for $L$: $L = \frac{A}{\pi \cdot d}$ Substituting the values: $L = \frac{2.74}{\pi \cdot 0.025} \approx \frac{2.74}{0.07854} \approx 34.9 \, \text{m}$

Conclusion:

The required length of the pipe is approximately 34.9 meters.

Would you like further details or have any questions?

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Related Questions

1. How would the required pipe length change if the flow rate of milk was reduced?
2. What if the cooling bath temperature was increased to 15°C; how would that affect the LMTD?
3. How would increasing the pipe diameter impact the heat transfer area and length of the pipe?
4. What factors could impact the overall heat transfer coefficient in real applications?
5. How would the calculation change if the flow was turbulent rather than ideal?

Tip:

When working with heat exchangers, be mindful of assumptions like constant bath temperature; real-world conditions can vary, requiring adjustments in calculations.