Question

Calculate the temperature of milk coming out from a homogenizer if temperature at inlet is 64\(^\circ\)C and inlet and outlet pressures are 205 bar and 5 bar respectively

• A. 66\(^\circ\)C
• B. 69\(^\circ\)C
• C. 70\(^\circ\)C
• D. 72\(^\circ\)C

Hint:

For rapid calculations under exam conditions, use this dairy engineering rule of thumb:
Every 40 bar pressure drop in a homogenizer increases the milk temperature by 1\(^\circ\)C.
For \(\Delta P = 200\text{ bar}\):
\[ \Delta T = \frac{200}{40} = 5\ ^\circ\text{C} \]
This gives the answer immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Homogenization is an irreversible thermodynamic process where mechanical energy is converted into heat.
When milk is forced through the small homogenization valve under extremely high pressure, the mechanical energy supplied by the pump is dissipated as heat due to high shear forces and friction.
Key Formula or Approach:
The temperature rise (\(\Delta T\)) of milk can be calculated by equating the mechanical work per unit mass to the thermal energy absorbed:
\[ w = \frac{\Delta P}{\rho} \]
\[ q = C_{\text{p}} \Delta T \]
Assuming all mechanical work is converted into heat (\(w = q\)):
\[ \Delta T = \frac{\Delta P}{\rho C_{\text{p}}} \]
Where:
- \(\Delta P\) is the pressure drop across the homogenizer (in \(\text{Pa}\) or \(\text{N m}^{-2}\)).
- \(\rho\) is the density of milk (\(\approx 1030\text{ kg m}^{-3}\)).
- \(C_{\text{p}}\) is the specific heat capacity of milk (\(\approx 3.9\text{ kJ kg}^{-1}\text{ K}^{-1} = 3900\text{ J kg}^{-1}\text{ K}^{-1}\)).
As a rule of thumb, the temperature of milk increases by approximately \(1\ ^\circ\text{C}\) for every \(40\text{ bar}\) drop in pressure.

Step 2: Detailed Explanation:
Let us calculate the parameters:
- Pressure drop, \(\Delta P = P_{\text{in}} - P_{\text{out}} = 205\text{ bar} - 5\text{ bar} = 200\text{ bar}\).
Convert this pressure drop into Pascals (SI units):
\[ \Delta P = 200 \times 10^5\text{ Pa} \]
Now, calculate the temperature rise using the thermodynamic relation:
\[ \Delta T = \frac{200 \times 10^5\text{ Pa}}{1030\text{ kg m}^{-3} \times 3900\text{ J kg}^{-1}\text{ K}^{-1}} \]
\[ \Delta T = \frac{20,000,000}{4,017,000}\text{ K} \approx 4.98\ ^\circ\text{C} \approx 5\ ^\circ\text{C} \]
Using the rule of thumb:
\[ \Delta T = \frac{200\text{ bar}}{40\text{ bar/}^\circ\text{C}} = 5\ ^\circ\text{C} \]
Now, calculate the outlet temperature of the milk:
\[ T_{\text{out}} = T_{\text{in}} + \Delta T = 64\ ^\circ\text{C} + 5\ ^\circ\text{C} = 69\ ^\circ\text{C} \]

Step 3: Final Answer:
The temperature of the milk coming out from the homogenizer is 69\(^\circ\)C. Hence, the correct option is (B).